The Mathematical visitor

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THE

MATHEMATICAL
VISITOR.

EDITED AND PUBLISHED BY

ARTEMAS MARTIN, M. A.,

Member of the London Mathematical Society.

VOLUME I.

1877—1881.

1881.

// ' / ■ /

LIST OF CONTRIBUTORS.

Albbbt, Frahk, Profeaeor of Mathematics Fenneylyania State Nonnal School, MiUeisville, Lancaster Co., Pa.
Adams, John H., Cohnmton, Crawford County, Fa.
Abnold, J. M., Boston, Haasachnsetts.

BowsEB, Bdward a., Profesfor of Mathematics ard Bngtr.eerlng, Ratgers CoUeee, New Bronswick, N. J.

BnooKs, Edward, M. A., Ph. D., Piincipal Pennsylvania Slate Normal School, Millersville, Lancaster Co., Pa.

Bbmam, Woobter W.. Member of the London Mathematical Society, AssisUi.t Professor of Mathematics at the Univer-

feity of Michigan, Ann Arbor, Michigan.
Babbour, L. O., Professor of Mathimatics, CentrsI University, Richmond, Kentucky.
Banks, Prof. Hugh S., lustmctor in Bngl.sh and Classical Literature, Newborg, N. T.
Bakbr, Marcus, U. S. Coast and Geodetic Survey Office, Wa-hingtt n, B. C.

Bbli^ Capt. W. T. R., Principal King's Mountara High School, Kmg*8 Mountain, Cleaveland Co., North Carolina.
Bcix, Br. John. Manchester, Hillsborough County, New Hampshire.
BuBLXsoN, B. F., Oneida Castle, Oneida Co., N. Y.
Batbs, W. B., Earlville, La Salle Co., 111.
Bachxldbb, Dr. Samuxl F., South Boston, Massachusetts.
Bbiohav, Jambs Q., Walton, Harvey County, Kansas.
Bbacb, S. C, Philadelphia, Pa.

Baoot, Thomas, Connjy Superintendent of Riplev Co., New Marion, Indiana.
Bbrbt, John W., C. £., Piitston, Luzerne Co.. Pa.
BuLXJS, Abram R, Ithaca, Tompkins Cj., N. Y.
Brown, Lucius, Hudson, Middlesex Co., Mass.
BuNH, J. P., Judge of Probate Court, Tiffin, Seneca County, Ohio.

Cabby, Prof. W. P., O. E., San Francisco. California.

Christie, Albz. S., U. S. Coast and Geodetic Sarvey Office, Washington, D. C.

Clark, John I., Moran, Clinton County, Indiana.

CRAine, Miss Buza D., Brooklyn, N. Y.

Drummond, Hon. Josiah H., LL. D., Portland, Maine.

DoouTTLB, M. H., U. S. Cotiitt and Geodetic Survey Office, Washington, B. C.

DbLand, Theodore L., U. S. Treasury Department, Washington, D. C.

Davis, Reuben, Bradford, Stark County, Illinois.

Day, Geobqb M., Lockport, Niagara Co., N. Y.

Donovan, J. W., Acsonia, Darke Co., Ohio.

Edmunds, B. J., B. S., Professor of Mathematics, Southern University, New Orleans, Louisiana.
Eastwood, George, Saxonville, MiddlcHCX Co., Mass.

FioKi.nr, Joseph, M. A., Ph. D., Professor of Mathematics and Astronomy, University of the State of Missouri, Columbia,

Missouri.
Frbbland, Frank T., B. S., Instnictor in Mechanics, University of Pennsylvania, Philadelphia, Pa.
FLbMiNo, Robins, Readington, Hunterdon Co., N. J.
Pagan, J. R., Erie, Brie Co., Pa.

Gould, S. C, Manchester, Hillsborough Co., N. H.
Gilpin, Chaklks, Jr., Piiiladelphia, Pa.
Glashan, J. C, Ottawa, Ontario, Canada.

Hendricks, Dr. Joel E., M. A., Editor and Publisher of the Analyst, Des Moines, Iowa.

Hart, Dr. David S., Stonington, New London Co., Conn.

HoRNUNo, Christian, M. A., Professor of Mathematics, Heidlebnrg College, Tiffin, Seneca Co., O.

Hough, H. H., Principal Linden Female Semiiiarv, Doylestown, Bucks Co., Pa.

Haynb^, ARruuft Edw n, M. Ph , Profes or of Mathematics and Physics, Hillsdale College, Hillsdale. Michigan.

Heaton, Henry, Perry, Dalla'* Co., Iowa.

Hoover, Wiiliam, Superintendent of Schools, Wapakoneta, Auglalae Co., Ohio.

Hbadlbv, Benjamin. Dill»boro, Dearborn Co., Inaiania.

Harvev, W. L., Maxtield. Penobsc'.t Co., Maine.

Heath, V. Webster, Rodman. Jefferson Co., N. Y.

Heal, W. E., Marion, Grant Co.. Indiana.

Haldbman, Ctbi's B., Ross, Butler Co., Ohio.

Hinkle, W. S., Leacock. I^ncaster Co., Pa.

Hawlet, George, San FYancit^co, Califomia.

Harvill, George H. Colfax, Grant Co., La.

Johnson, William Woolsey, Member of the London Mathematical Society, Professor of Mathematics St. John's CoUefS,

Annapolis, Maryland.
Jepprie!), Winpield v.. Instructor in Mathematics, Vermillion Institute, HayesviUe, Ashland Co., Ohio.
JopLiN, Gbobok a.. Danville. B >yle Co., Kentucky.
JoNE9, D. C, St Charley Butler Co., Ohio.

Ktrkwood, Daniel, LL. D , Professor of Mathematics, Indiana State University, Bloomineton, Indiana.

Kersbner, Josfph H., late Pr>feH««-ir of Mathematir>s, Mercrsburg College, Mercer<tburg, Franklin Co., Pa.

Knisely, Rev. U. Jes!(b, Ph. D.. Newcomerstown, Tu^cirawas Co., Ohio.

Kite, W. A., M. A., Profesnor of Matht matics, GreencviUe and Tusculam College. Tusculum, Tennesee.

Kummbll, Charle<* H., U. S. Coanr and Geodetic Survey Office, Wofehington, D. C.

Knecht, RsrBEv, Pbotographer, Easton, Nrithampton Co., Pa.

Kinney, T. A., St. Albans, Fr.Anklin, Co., Vermont.

Ladd, Miss Christive, B. A., Fellow of Johns Hopkins University, Baltimore. Maryland.

LuDWio, H. T. J., Professor of Mathematics, North Carolina College, Mount Pleasant, Cabarrus Co., N. C.

Lantz, F. W., Washington, D. C.

LiLLEY, George, M. A.. Coming, Adam** Co., Iowa.

Leland, George H., Windsor, Windsor Co., Vermont.

Leahy, F. F., Covington, Kenton Co., Kei tucky.

Lakin, B. B., Streator, La Sa!le Co., lUinois.

iv

Hartxn, ARTBXAg, M. A., Member of the Loodon Mathematical Society, Brie, Brie Coonty, Pennsylvania.

Mabtih, D. W. K., Webeter, Darke Ck>ant7, Ohio.

Macfarlanb, Alkxandbb, M. a.. D. Sc., Eoinbarg, Scotland.

M::Ada]i, Duni^ap J.. M. A., Profeesor of Mathematics, Washington and Jeffereon College. Washington, PennsyWania.

Matz, F. p., M. a., late Professor of Mathematics, Military and Scientiiic School, King^ Mountain, North Carolina.

Manoe, p., F., Alamos, Sonora, MeJdco.

MfcRBiLL, OsoAB H., MannsvUJe. Jefferson Coan!y, N. T.

McLAUGHUif, Jaju!*, Mantorvllle, Dodxe Conntv, MInne oCa.

Maoruobr, W. T., Stevens Institute of Technologv, Hoboken, New JerMy.

Mb AD, O. F., Uniontown, Fayette Coauty, Peunsyivani.i.

NtonoLS, Waltbb S., Editor Insurance Monitor, New York, N. Y.
Nichols, Ubmbt, Hampton, Kock Island County, Illinois.
NuBTON, B. P., Allen, Hillsdale County, Michigan.

Oathout, Oblanim) D., Lnana, Clavton County, Iowa.
O'CoNNBR, D. B., Union City, Bandolph County, Indiana.

Pbibcb, Bkhjamin, LL. D., F. R. S., Professor of Mathematics, Harvard University, and Consulting Geometer to the U. B.

Coast Survey, Cambridge. Massachusetts.
PuTHAM, K. S., Rome, Oneida County, N. Y.
PoLLARO, Julian A., Govhen, Orange County, N. Y.
PuRDiB, Dr. A. J., Otselic, Chenango County, N. Y.
Place, Edwim, Cincinnatus, Cortland County, N. Y.
PiiiBST, FRAxas M., Bryan, Williams County, Ohio. '

QuiBOZ, J. M., Alamos, Sonora, Mexico.

BoBBBTS, Savurl, M. A., F. R. S., President of the London Mathematical Society, London, BngUnd.

KortBLL, C. A. O., B. A., Teacher of Mathematics at the Carroll Insutute, Reading, Berks County, Pennsylvania.

KuTBB, John S., Editor and Publisher of the School Vititor, Ansonia, Darke County, Ohio.

Robins, Stlvbstbb, North Branch Depot, Somerbet Coun^, New Jersey.

Rubins, C. C. Princeton, New Jersey.

ttBA, John, Hiirit Fork. Adams County, Ohio.

RiCB, Mil TON, Moran, Clinton County, Indiana.

Rowan, B. J., Shawnee, Perry County, Ohio.

S)rLV£sTiiK, J J., LL. II , F. IL i^,, Com^ponding Member of the Institute of France, Professor of Mathematics at Johns

liopkin* Uulv entity, UflltimdPO, Maryland.
BEirf.. E. B.. M(?ml)«r of the U>n(irin Maihe^nmtical Society, Professor of Mathematics, North Missouri State Nonnal School,

KtrksviHc, MU«en)ari.
ScttErfEit, J. ¥". W., Jat« ProFetiHor of Mar hematics and German, Mercers burg College, Mercersburg, Franklin Co., Pa.
STosfB, Profe'^ior Orhpjjd, M. A., Af<tra,icinrK>j at the Cincinnati Observatory, Mt. Lookout, Hamilton Co., Ohio.
bHUfT, L. E^. IL a. VoAdi H]id l>eud^ti^ i^utvay Office, Washington, D. C.
SrvRHLT. WAi,tsi^ on CUy. Venau^j L'minly, Pennsylvania.
Sbaw, Gavin, 3r^ Kemblu. Ontario, Canada.
fiAHDEfis, J. B., Btoti[Dlii|£tim. IndiitUL
tiAXiiBaa, W. F. t., ?!*w AltMny, Floyd Coonty, Indiana.
Sitrwsu^ TaoxA* P , Rothtater, NY.
S^rrsEic, Mn. Ai«)sa T., Chicago, iHluote.
SAKrKi:^,, FRVDJEKtcK 9^ . CecTu (j<irdo, Inyo (Tounty, Calf om la.
tiTMWABT, J. V.t Mud cits IX-lawart? Couniy, Indiana.

S^tf^KIt*, KUJAH A , T... c;nirl-| \t,,r.kpt|l I'-n-fPtty, lowB.

Tbowbbidgb, Professor D vvm, M. A.. Waterbarg, Tompkins County, N. Y.
TUBBBLL, Isaac H.. Principal 4th DUtrict School, Cincinnati, Ohio.
TuTfON, Chablbs H., Wilkes Barre, Luzerne County, Pennsylvania.
Thompmn, B. p., M. E., BUzabeth, Union County, New Jersey.
Tatlob, J. M., MUton, Umatilla County, Oregon.
Tubnbctll, Josbph, Bast Liverpool, Columbiana County, Ohio.

Wood, DbYolson, M. A., C. B., Profeswr of Mathematics and Mechanic4, Stevens Institute of Technology, Hoboken, N. J.

Wood, Hchmon A., M. A^ Principal Coxsackie Academy, Coxsuckie, Greene County, N. Y.

Wbioht, Rev. W. J., M. A., Ph. D., Burlington, Chi.tunlen County, Vermon .

Wbiobt, Dr. S. H., M. A., Pb. D., late Mathematical Editor of the YaUs Vounty ChronieU, Penii Yan, N. Y.

Wbioht, D. L., Mallet Creek/Medina County, Ohio.

Wilt, John M., M. A., Fort Wayne, Allen County, Indiana.

Wicxebbhaji, David, County Surveyor, Wilmington, Clinton County, Ohio.

WiLBY, WiujAM. Detroit, Michigan.

WiLUAMS, J. D., Superintendent of Schools, Sturgis, Michigan.

Waulcb, L. C, New Madison, Darke County, Ouo.

Washbubn, Obobob G., North Bast, Brie County, Pennsylvania.

Walton, J. G., Covington, Kenton County, Kentucky.

The/oUowing name$ were omitted hy miUake on the preceding page:
Caowin, Samukl O., New London, Oneida County. N. Y.
Evans, Professor AsSbb B., M. A., Principal Lockport Union School, Lockport, Niagara Co., N. Y.

Oontributore deceaeed since the publication qf No. I.
Bachbldbb, Dr. Savubl F., South Boston, Mas^achusetto.
Psuio<£, Professor Bbnjamin, LL. D., F. R. 8., Cambridge, MaasachuMtto.

CONTENTS OF VOLUME I.

NTTMBEBl.

Introduction.

Junior Problems.

Senior Problems.

Unsolved Problems.

Solution of a Problem.

Solution of a Problem in Probabilitiee.

The Intrinsic Equation of a Curve.

Notices of Periodicals.

Page
1
2
4
6
8
9
10
12

NUMBEB 2.

Solutions of Junior Problems Proposed in No. 1.

Solutions of Senior Problems Proposed in No. 1.

Solutions of < 'Unsolved Problems'* Published in No. 1.

List of Contributors.

Junior Problems.

Senior Problems.

Unsolved Problems.

Editorial Notes.

Notices of Books and Periodicals.

Corrigenda.

13
23
39
42
42
44
48
49
49

NUMBEB 3.

Junior Department— Solutions of Problems Proposed in No. 2.

List of Contributors to the Junior Department.

Problems.

Senior Department— Solutions of Problems Proposed in No. 2.

List of Contributors to the Senior Department.

Problems.

Editorial Notes.

Notices of Books and Periodicals.

Corrigenda.

51
60
61
62
82
83
85

NUMBEB 4.

Junior Department — Solutions of Problems Proposed in No. 3.

List of Contributors to the Junior Department.

Problems.

Senior Department — Solutions of Problems Proposed in No. 3.

List of Contributors to the Senior Department

Problems.

EditorUl Notes.

Notices of Books and Periodicals.

Corrigenda.

87

96

97

99

114

114

118

119

120

NXTMBEB 5.

Junior Department— Solutions of Problems Proposed in Kos. 1, 2 and 3.

Note on the Solution of Problem 110.

Problems.

Senior Department— Solutions of Problems Proposed in Ko.

Solution of "Unsolved" Problem 40.

On Evolutes and Involutes.

Computation of Sturm's Functions.

Expansion of Polynomials.

The Day of the Week for a Given Date.

Demonstration of two Diophantine Theorems.

Solution of the "Three-Dice" Problem.

Problems.

Editorial Notes.

Notices of Books and Periodicals.

Corrigenda.

Page
121
124
125
126
147
148
151
151
155
156
157
158
159
159
160

NUMBEB e.

Junior Department— Solutions of Problems Proposed in No. 4.

List of Contributors to the Junior Department.

Problems.

Senior Department— Solutions of Problems Proposed in No.

List of Contributors to the Senior Department.

Problems.

Editorial Notes.

Notices of Books and Periodicals.

Corrigenda.

161
175
175
177
192
193
195
196
198

4

MiiiniBTg

Vol. I.

OCTOBER, 1878.

No I.

THE

/

MATHEMATICAL

VISITOR.

EDITED AND PUBLISHED BY

ARTEMAS MARTIN, M. A.,

y'

MEMBER OF THE LONDON MATHEMATICAL SOCIETY.

c

SECOND EDITION.

PRICE, FIFTY CENTS.

ERIE, PA.;

JMO. M. QLAZItR, PfUNTCfl, OOR. ftlXTCCNTH AND PtACM BTRIfTS.
1.873.

THE

MATHEMATICAL VISITOR.

[CNTKMO AOOOROINO TO ACT OF OONQREBS. A. O. 1878, BV AIITCMA8 MARTIN, M. A.. IN THE OFFIOE OF TH» LIBRARIAN OF CONQRCSB, AT WABMIMQTOM.]

Vol. 1. OCTOBER, 1878 No. 1.

INTRODUCTION.

In England and Europe periodical publications have contributed much to the
diffusion of mathematical learning, and some of the greatest scientific characters
of those countries commenced their mathematical career by solving the problems
proposed in such works.

It was stated nearly three-quarters of a century ago that the learned Dr. Hutton
declared that the Ladies^ Diary had produced more mathematicians in England
than all the mathematical authors in that kingdom.

Similar publications have produced like results in this country. Not a few of
our ablest teachers and mathematicians were first inspired with a love of mathe-
matical science by the problems and solutions published in the mathematical de-
partment of some unpretending periodical.

Turning over the leaves of the Maihemaiical Miscellany, published about forty
years ago, we find there the names of Professors Peirce, Strong, Kirkwood, Root,
Docharty, and many othei^s, who have distinguished themselves in various depart-
ments of mathematics.

Some years ago we asked a young mathematician of great promise, who dis-
played great ingenuity and ability in solving difficult probability problems, what
works he had treating on that subject; imagine our surprise when he replied that
he had no works on probabilities, but had learned what he knew about that diffi-
cult branch of mathematical science by studying the problems and solutions in
the Schoolday Magazine.

Believing that there is room for one more mathematical periodical, we have ven-
tured to issue The Mathematical Visitor, which will be devoted chiefly to prob-
lems and solutions ; and we invite professors, teachers, students, and all lovers of the
"bewitching science," to contribute their best problems and solutions for its pages.

Two lists of problems will be published in each number; one, headed "Junior
Problems," for students and persons who have not advanced very far beyond the
elementary branches ; the other, headed "Senior Problems," will contain questions
of a more difficult nature. For convenience of reference the problems will be
numbered continuously.

A prize problem will be published in each number, for the best complete, cor-
rect, solution of which ten copies of The Visitor will be given, and for the second
best solution eight copies will be given.

2

All problems and solutions intended for publication in number two should be
received by September 1, 1877.

Number two will be issued about the first of January, 1878 ; it will contain
about 32 pages, and the price will be 50 cents.

Only a small edition will be printed, hence persons desiring to secure copies
should send in their names at an early date.

Artemas Martin, Lock Box 11, Erie, Pa.

Note. — The first edition being exhausted, a second edition is issued to supply the demand for thiF
No. to complete sets. Some new matter has been inserted to fill out the last leaf. A. >i.

JUNIOR PROBLEMS.

1. — Proposed by Prof. Edward Bbooks, M. A., Principal of Pennsylvania State Normal School ^ Mil-
lersville, Lancaster Co., Pa.

Two trains, one a and the other b feet long, move with uniform velocities on
parallel rails ; when they move in opposite directions they pass each other in m
seconds, but when they move in the same direction the faster train passes the other
in n seconds Find the rate at which each train moves.

2. — Proposed by Mrs. Anna T. Snyder, Alquina, Fayette Co., Ind.

Two tiees of equal hight stood on a level plane, 50 feet apart; one being cut off
at the ground lodged against the other 20 feet from the top. What is the night of
the trees?

3.— Proposed by S. C. (ioiLD, Manchester, Hillsborough Co., \. H.

It is required to enclose the exact area of a given square by means of circular
arcs only.

4. — Proposed by Dr. I)Avn> S. Hart, M. A., Stoninglon, New I^ondon Co., Conn.

The straight line drawn, bisecting any angle of a triangle, to the opposite side,
is less than half the sum of the sides containing that angle.

5.— I'roposed by Artemas MAKTrx, Erie, Erie Co., Pa.

If 22 oxen and 28 cows eat 24 acres of grass in 18 weeks, and 20 oxen and 38
cows eat 30 acres of grass in 27 weeks, and 41 oxen and 26 cows eat 50 acres of
grass in 60 weeks, how long will 40 acres of the same grass last 35 oxen and 14
(•ows, the grass in all cases growing uniformly?

e.— Proposed by Thp:o. L. DeLaxd, t'. S. Treasury Department, Washington, D. C.

Four men. A, B, C and D, agreed to reap a circular field of grain in 28 days for
$249 36, and divide the money among them in proportion to the actual service each
rendered The first, second, third and fourth quadrants contained, lespectively,
wheat, oats, barley and rye. To the end of the 7tli day A cut wheat, B oats, C bar-
ley and D rye; they each then advanced one quadrant, and to the end of the 13th
day A cut oats, B barley, C rye and D wheat ; they each again advanced one quad-
rant, and cut to the end of the 18th day ; again they advanced one quadrant each
and so worked to the end of the 22d day, and finished the wheat; again they ad-
vanced one quadrant each and finished the oats at the end of 8 days more; they
again advanced one quadrant each and finished the barley at the end of the 27th
day ; and then advanced one quadrant each and completed the work according to
contract.

Required — the time for each to reap the field, and an equitable division of the
money, provided one quadrant is as difficult as another.

7.— Proposed by Prof. David Trowbridge, Waterburg, Tompkiiis Co., N. Y.
If A, B, C be the angles of a triangle, prove that

cotA+cotB cotA+cotC cotB+cotC _^

cot^A +"cotiB cotiA -TcotJC cotjB + coUC"^ '

8.— Proposed by Marcus Baker, U. S. Coast Survey Office, Washington, I). C.

In an equilateral triangle ABC lines drawn from each vertex to the opposite
side, dividing that side so that the ratio of the side to one of the segments is n,
form by their intersections another equilateral triangle ; determine the ratio of the
areas of these triangles.

9.— Proposed by Mias Christine Ladd, Union Springs, Cayuga Co., X. Y.
Show that in any plane triangle

. asinA+6sinB+csinC

3 p /a2+62+e2x

a cos A + b cosB + c cosC

10.— Proposed by H. A. Wood, M. A., Professor of Higher Mathematics, Keystone State Normal
School, Kntztown, Berks Co., Pa.

A hollow paraboloid, depth 4 feet, stands on its vertex with axis vertical. After
a shower the depth of wat^r in the paraboloid, measured on the axis, was 2 inches.
What was the uniform depth of the rainfall, the radius of the top of the parabo-.
loid being three feet?

11.— Proposc^d by B. F. Burlksox, Oneida Castle, Oneida Co., N. Y.

The sides of a (quadrilateral field are, respectively, and in order, 17, 35, 40 and
45 rods in length.

Determine the length of the straight line which, in passing through a point
equally distant from all its angles, shall divide the field into two equal parts.

12. — Proposed by Jamks McLatghlin, Mantorville, Do<1ge Co., Minn.

A gentleman has a rectangular garden in latitude 41° 40' north, the diagonal of
which is a meridian line. At the south corner of the garden stands a perpendicu-
lar pine tree, the shade of which at noon, when the days and nights are equal, ex-
tends to a small rivulet which comes in at the east corner of the garden and runs
due west through it, crossing the said meridian line 240 feet from the north corner
of the garden, and going out of it 135 feet from the pine tree.

Required the area of the garden and the hight of the tree.

13. — Proposed by K. P. Noktox, Allen, Hillsdale Co., Mich.

In a triangle ABD the base AB is 40 rods. A line AC drawn from the angle A,
and perpendicular to AB. intersecting BD in C, is rods. If a point F be taken
in AB, 10 rods from B, and a line drawn from this point through C, and produced,
it will cut AD produced in a point E, 5 rods from D. Required the sides of the
triangle.

14.— Proposed by Prof. Horatjo M. Bloomfield, Reading, Berks Co., Pa.
Given, w''^+y'^=Cj z^-\-y^z=b, z^'{-x^=a, x+y=w+z;

to find the value of x by quadratics.

15.— Proposed by Henry Heatox, B. S., DesMoines, Polk Co., Iowa.

If n persons meet by chance, what is the probability that they all have the same
birth-day, supposing every fourth year a leap year ?

Bolutloni of these problemi fthonld be reoelved by Beptember 1, 1877.

SENIOR PROBLEMS.

16,— Fro|««ed bjr Prof, iMviii TBowBBiiir.E, Waterbarg, TompldiiA Co^ X. Y.

Kii'luirfA the exf>an($ion of ff^ in sl series of cosines of maltiples of 9, thus:

^=Ai;-^AjC08/?-*-A2C082iJ'-r . . - -fAnCOsn/^-r - - .

n.^Vwpontd hy MIm Chbiktise Laiiii, Union Springs, Gayoga Co^ X. Y.

The [K/Iar with respect to a triangle of its orthogonal center is the radical axis
of the circumscribing circle, the nine-point circle and the circle wilh respect to
which the triangle is self-conjugate.

IS^—PropOMcd hy L H« Tubbell, ComminsiiUe, Hamilton Co., Ohio.

Bupfiose n fixed lines, radiating from a fi)ced point, are touched two and two by
n circles of given radii ; that is, that each line touches two of the circles, and each
circle touches two of the lines. The order of arrangement being known, it is re-
quired to construct geometricallv a polygon of n sides, whose vertices shall lie on
tne n lines, and whose sides shall touch the given circles.

19.— PropoMd br Dr. B. H. Wbioht, M. A., Ph. D., Mathematical Editor TaUs QmtU^ Chnmide^ Pens
Yaa, YaUi Co., X. Y.

In latitude i=42®N., when the sun's declination north ==^=20°, the sun shone
on the top of a high peak in a mountain /9=4 minutes before it shone on the plane
of the horizon. IIow high was the peak?

20. Pro|)OM5d by William Kckiveu, Kuperintendent of Public Schools, Bellefontaine, Logan Co., O.

The weights Wj and W2 are suspended, d degrees apart, from the circumference
of u vcjrtical circle free to move about its center. Find the angle made by the
nulii drawn from the points of suspension with the horizontal through the center
of the <!ircl(; when the system is in equilibrium.

21." PropoHcd by Dr. David H. Hakt, M. A., StoniDgton, New London Co., Conn.

Hitrjuired five numbers whose sum shall be a biquadrate, and the sum of anj^
four of them a square.

22. Propoftcd by Waltku Hivkulv, Oil City, Venango Co., Pa.

An ellipsoid, axes 2a, 2/>, 2^, is intersected by a cylinder, radius A, the axis of the
cylinder coinciding with the axis 2c of the ellipsoid. Find the volume common to
both.

28. I'ropofiC'd by Aktkmas Maktin, Erie, Erie Co., Pa.

Kin<l the least integral values of x and y that will satisfy the equation

y2— 9781y2=l.
24.- IMopoMcd by Jamkh Mc Lacohlin, Mantorville, Dodge Co., Minn.

An arnjy in the form of a circle, 2a miles in diameter, marches due north at the
uniform rale of n miles an hour. An officer starts from the rear of the army and
rides nround it at the uniform rate of m miles an hour, keeping close to the army
all tlie time. Kecjuired the equation of the curve the officer describes, and the dis-
tance he rides while going once around the army.

26. ProiMwod by DkVolson Wood, M. A., C. E., Professor of Mathematics and Mechanics, Stevens
luHtitutr of T(Thnolo)ty, Ilobokeu, Hudson Co., N. J.

A perf(»ctly homogeneous sphere has an angular velocity co about its diameter.
If the bpliero gradually contract, remaining constantly homogeneous, required the
angular velocity when it has half the original diameter.

— —

26.— Proposed by E. B. Skitz, Greenville, Darke Co., Ohio.

Two equal spheres,, radii r, are described within a sphere, radius 2r ; find the av-
erage of the volume common to the two spheres.

27,— Proposed by Abtkmas Maktin, Erie, Erie Co., Pa.

The first of two casks contains a gallons of wine, and the second b gallons of
water. Part of the water is poured into the first cask, and then part of the mix-
ture is poured back into the second. Required the probability that not more than
i of the contents of the second cask is wine.

28.— Proposed by F. P. Matz, B. E., B. S., Reading, Berks Co., Pa.

The center of a variable circular disk moves along the major axis, between the
left-hand extremity and center,of a variable elliptic surface; to determine the aver-
age area of all variable ellipses whose major axes are coincident with that of the
first elliptic surface and whose right-hand vertex is limited by the right-hand ex-
tremity of said elliptic surface while the left is tangent to the periphery of the disk.

29.— Pn.poeed by £. B. Seitz, Greenville, Darke Co., Ohio.

Two points are taken at random in the surface of a given circle, but on opposite
sides of a given diameter. Find (1) the chance that the distance between the points
is less than the radius of the circle, and (2) the average distance between them.

80.— Proposed by Abtemas Mabth^, Erie, Erie Co., Pa.

A straight tree growing vertically on the side of a mountain was broken by the
wind, but not severed ; find the chance that the top reaches to the ground.

31.— Proposed by Aktkmas Martin, Erie, Erie Co., Pa.

A sphere, radius r, and a candle are placed at random on a round table, radius
-B, the bight of the candle being equal to the radius of the sphere. Find the aver-
age surface of the sphere illuminated by the candle.

32.-— Proposed by E. B. Seitz, Greenville, Darke Co., Ohio.

A radius is drawn dividing a given semicircle into two quadrants, and a point
taken at random in each quadrant ; find the average distance between them.

33.— Proposed by E. B. Seitz, Greenville, Darke Co., Ohio.

Three points beinff taken at random within a sphere, find the average area of
the triangle formed by joining them.

34,— Prize Problem. Proposed by Abtemas Martin, Erie, Erie Co., Pa.

A boy stepped upon a horizontal turn-table, while it was in motion, and walked
across it, keeping all the time in the same vertical plane. The boy*s velocity is
supposed to be uniform in his track on the table,- and the motion of the table to-
wards him. The velocity of a point in the circumference of the turn-table is n
times the velocity of the boy along the curve he describes.

Required — the nature of the curve the boy describes on the table, and the dis-
tance he walks while crossing it (1) when n is less than 1, (2) when n=l and (3)
when n is greater than 1.

Bolutiong of these problems should be received by September 1, 1877.

UNSOLVED PROBLEMS.

By Abtemas Mabtin, Erie, Pa.

I am not aware that the following problems have ever been completely solved.

Perhaps some of those in the Diophantine Analysis may be impossible.

If correct solutions to any of these problems are received in time, they will be
published in the next number of the Visitor.

1. To find three biquadrate numbers whose sum is a biquadrate number.

2. To find four biquadrate numbers whose sum is a biquadrate number.

3. To find four positive whole numbers, the sum of every two of which is a
biquadrate.

4. To find three square numbers whose sum is a square, such that the sum of
every two is also a square.

5. To find four square numbers such that the sum of every two of them shall
be a square.

6. To find four positive cube numbers, the sum of any three of which is a cube.

7. To find three positive integral numbers the sum of whose squares is a
square, and the sum of their cubes a cube.

8. To find three positive whole numbers whose sum, sum of squares, and sum
of biquadrates are all rational squares.

9. To find three positive whole numbers whose sum, sum of squares, and sum
of cubes shall all be rational cubes.

10. To find three positive integral numbers whose sum, sum of squares, and
sum of biquadrates shall all be i ational cubes.

11. To find five integral numbers, the sum of the squares of every four of
which shall be a square.

12. A coin, radius r, is placed at the bottom of an empty hemispherical bowl,
radius jB, and the bowl is filled with water, when the whole coin is just visible to
an eye in a given position looking over the edge. How much must the bowl be
tipped from the eye to hide the coin ?

13. Two horses, coupled together with a rope b feet long, run at uniform rates
of speed on a circular race-course whose radius is a feet, the fleeter horse being
outside of the track and the other on it. They start even, with the rope stretched.
Required the curve described by the fleeter horse, and the distance each has run
when he comes to the track.

14. A cylindrical cask, radius R inches and depth a inches, is full of wine.
Through a pipe in the top, water can be let in at the rate of b gallons per minute,
and through a pipe, radius r inches, in the center of the bottom the mixture can
escape at a faster rate when the cask is full. If both pipes be opened at the same
instant, how much wine will remain in the cask at the end of t minutes, supposing
the two fluids to mingle perfectly ?

15. From an urn containing 20 balls, a number of balls are taken at random
and put into a bag, and then a number are taken at random from the bag and put
into a box ; if, now, a number of balls be taken at random from the box, what is
the probability that the number is odd ?

10. If n bricks be piled upon one another at random, (like they would be in a
wall one brick thick,) what is the probability that the pile will not fall down?

17. Three radii are drawn at random in a circle and a point is taken at random
in each sector ; find the chance that the triangle formed by joining them is acute.

18. A round flat pie is cut into three equal pieces ; if they be piled up at ran-
dom on a level table, what is the probability that the pile will not fall down?

19. Three pennies are thrown horizontally at random, one at a time, into a cir-
cular box whose diameter is twice the diameter of a penny ; find the chance that
only one of the pennies rests on the bottom of the box.

20. Three points are taken at random, one in each of the sides of a given tri-
anprle ; find the chance that the triangle formed by joining them is acute.

21. If four pennies be piled up at random on a horizontal plane, what is the
probability that the pile will not mil down?

22. Three points are taken at random in the surface of a given triangle ; find
the chance that the triangle formed by joining them is acute.

23. If four dice be piled up at random on a horizontal plane, what is the prob-
ability that the pile will not fall down ?

24. A plank, length 2a, is cut at random into three rectangular pieces ; if the
pieces be piled up at random on a horizontal plane, (like three bricks in a wall,)
what is the probability that the pile will not fall down ?

25. Three arrows are sticking in a circular target; what is the probability that
the distance between any two of them does not exceed the radius of the target?

26. If five pennies be piled up at random on a horizontal plane, what is the
chance that the pile will not fall down ?

27. The bottom of a circular box is covered with an adhesive substance. Two
rods equal in length to the radius of the box are dropped horizontally into it at
random, one at a time. What is the chance that the rods are crossed in the box?

28. Find the average distance between two points taken at random in the sur-
face of a given rectangle, one on each side of a diagonal.

29. A straight rod, length 2a and mass M, is suspended by a string attached to
its middle point. A mouse, mass m, crawls down the string and along the rod to
the end. Required (1) the equation to the curve the mouse describes in space, (2)
the equation to the curve the rod always touches and (3) the inclination of the rod
to the vertical when the mouse is at the end.

30. A straight rod, length 2a and mass M, is balanced on a cylinder, radius r.
A large caterpillar, mass m, crawls from the middle of the rod towards one end.
Required (1) the equation to the curve the caterpillar describes in space, and (2)
the inclination of the rod to the horizon when the caterpillar is at the end, the
surfaces of the rod and» cylinder being rough enough to prevent sliding.

8

SOLUTION OF A PROBLEM.

By E. B. Seitz, Greenville, Ohio.

Find the equation to the locus of thp centers of all the circles that can be in-
scribed in a given semi-ellipse

fThlB is problem 87 proposed In Owr SOiooldtty Visitor for June, 1871, p. IW. No eolation was published.— Ed. J

Put CM=a?, OM=y; then representing the
co-ordinates of the point of tangency of the
circle and ellipse by m, n. we have for the
equation to the circle

(»— y)2+(wi— x)e=/ (1),

and for the equation to the ellipse

a2n2+6W=a262 ^2),

Differentiating (1) and (2), we have

dm ^ n — y , dm aki

dn m—x^ dn lAn-

Bixt for the poinl of tangency of the circle and ellipse, these two values of -^ are

equal ;

n-^y_aH .

• • ^r=i~62^ W-

From (3) n= — g- — . .^^ .^ Substituting this value of n in (2), clearing of

fractions and reducing, we have

(a2— 52)2^4— 2(a2— ft^^cmH [oV+iV — (a^— bJ^(M^+ 2(a2— b^)a''xm - aV= (4).

From (3) and (2) (.--|) = ^ = -i^.

and from (1) (iLzJL) ^ = ( _J^) L i ,.

^ \m—xj \m—xj

. aKal'-m^) ^/ y y ^

Wv^ \m — x) *

or (a2— 6>i*— 2(o2— 62)a:m3+[(rt2_62)a^+iy_a4]m2+2a<arm— a<x2=0 . . (5).
Subtracting (4) from (a^ — ft'^) times (5), we have

2(o2— 6V»'— [(2o*— iV+*V+aV*— *^)K+a^ar2^0 (6).

Subtracting (4) from c? times (5), we have

(a2— 62)m»— (ar2+2a2— 6>2m+2a<a:=0 (7).

Subtracting (6) from 2a; times (7), we iiave

[(2o2-62)a;2^.iy^.a2(„2_52)]^2_2(aJ2+2o2-62)oi^tm+3a<a:2==0 . . . (8).
Subtracting x times (7) from twice (6), we have

3(a2— 62)xm2— 2[(2a2— 62)«2+6y+a2(o2— 6^]'»+aMa^+2a2— 62)=0 . . (9).
Subtracting [(2o2—6V+*V+«V—*^)] times (9) from 3x(o2— 62) times (8), we have

2[(a<-a2i24-6<)x<+2(2a2-62)6!^2+6y_(aM»2)(2a2-62)a%c2+2(a2-62)a26y-faV-*'')^'»
-a^[(2o2-62)a;4+6!%ry-(4o*-4o262-i^)a:2^.(2o2-62)iy +o2(o2_fc2)(2o2-62)] =0 . (10).

9

Subtracting 3a?a? times (9) from {a^+2a^—b^) times (8), we have

— 2ahia^—{2a'^b^x^—3bY+a'^a^b^+b^^=:0 (11).

Eliminating m from (10) and (11), we have

6V+2(2a2-6V3r'+i^V~2(2a2--62)5ii^__4(3^4_3^2^^26^yy2

-10{2a^-b^)bh^--Uy+{6a^-6a%^+b^)bh^+2{2a^---b^^^^^
—{9KX^'-Sa^l^¥)bY-2{a^b^){2^^b^)a^^
which is the required equation.

SOLUTION OF A PRO BLEM IN PROBABILITIES.

By Abtemas Mabtix.

If three pennies be piled up at random on a horizontal plane, what is the proba-
bility that the pile will not fall down ?

The pile will stand if the common center of
gravity of the second and third coins falls on
the surface of the first or bottom coin.

Let r be the radius of a penny ; then the cen-
ter of the second coin may fall anywhere in a
circle whose radius is 2r and center the center
of the surface of the first or bottom coin, and
the center of the third coin may fall anywhere
in a circle whose radius is 2r and the center
the center of the surface of the second coin.

The number of positions of the center of the
second coin is therefore proportional to 4;rr^,
and for every one of these positions the center
of the third coin can have 4;rr2; hence the total number of positions of the second
and third coins is proportional to 16;rV^.

We must now determine in how many of these 16;rV positions the pile will stand.

Let A be the center of the first or bottom coin, and B the center of the second
coin. Take AD=AB, and with center D and radius 2r describe the arc CHJ. If
the center of the third coin is on the surface CFJH, the second and third coins will
remain on the first, since BN=NH, BT=TC, and the pile will not fall down.

When AB is not greater than Jr, the circle CHJ will not cut the surface of the
second coin, and the pile will stand if the center of the third coin is anywhere on
the second.

Let AB=AD=x, S=surface CFJH, andp=the probability required;

then

Ax

DG = -

Ax

CG=~-ri6r4-(5r2-4a:2^2j^, arcCI=rcos-i(^^^^^ arc CH=2rco8-i(^^^^

r^cos^^

(

and

3r2— .

4ra;

/.

/{S—7rr^)xdx,
Mr

Mr 16

8?IT*t

16H— (5r2— 4ry ,

/r T^ /5r2-43r2\ 1 r 1^

il 16H-(5r2-4x2)2 a:da;=:Jr*cos-»f ^^^-j^^J — r|<5r2-4x2) 16r<- (5ra-4x2)2 .

THE INTRINSIC EQUATION OF A CURVE.

By Artemas Mabtin, M. A., Member of the London Mathematical Society.

I am not aware that this subject has been treated in an American work.

If s denote the length of an arc of a curve measured from a fixed point, <p the
inclination of the tangent at the variable extremity of the arc to the tangent at the
fixed point, then 8=f(f), the relation between s and f , is called the intrinsic equa-
tion of the curve.

Let y=J{x) be the rectangular equation of a curve, the origin being a point on

the curve, and the axis of y a tangent at that point; then -,— =/'(.r) = cot f, from

dx
which X is known in the trigonometrical functions of ^, say = F{(p) ; then ^ == ^{<p) \

also -, = cosec (p ; therefore -, = cosec (pF\<p) ; from this equation s may be found

in terms of if by integration. A similar result will be obtained when the axis of
.a: is a tangent at the origin. If the given equation of the curve be x=f{y), we
may proceed in a similar manner.

Examples. — 1. Required the intrinsic equation of the circle.

SOLUTION.

The required equation is obviously 8=a<py but we will deduce it from the ordi-
nary equation by the method explained above.

dv a X
The rectangular equation beingy= i/{2ax — x^)^ we have J^ = — -^ -^ =cotf ;

ox 1/ {jbtOX^X^j

whence x=a{l±cosf), . * . -p= diasinf. Taking the lower sign, T-=a,
and 8= Of.

— 11 —
2. — Required the intrinsic equation of the tractrix.

SOLUTION.

The retangular equation of the tractrix is
x = alog( — — ~ — — J -^ \/{2ay — y^^ the axis of y being a tangent at the origin.

. • . -^ - = a tanf . Integrating, «= a log secy + C. When « = 0, f = and C= ;
. • . « = a log secf is the intrinsic equation of the tractrix.
3. — Required the intrinsic equation of the parabola.

SOLUTION.

From the equation y2 = ^ax we have -f- = ^- = cotf , ;ir = a tan2y,

dx v^

whence by integration «= -^r log [^ r-^ ) + :—-.

•^ 2 ® \1 — sm^/ 1 — smY

The intrinsic equation of a curve being given, to deduce the ordinary equation, we

have, when the axis of y is a tangent at the origin, —,- =sinf , -J = cosf ;

. ' . .T= I simpdg] y= r cosfcfe.

When the axis of ^ is a tangent at the origin, ^ = | cosf cfe, y= i siufpds.

Now 8 being known in terms of <f from the intrinsic equation, by integration we
can find x and y in terms of f , and then by eliminating f we obtain the ordinary
equation of the curve in terms of x and y.

Example. — Required the rectangular equation of the curve whose intrinsic equa-
tion is« = atanf.

SOLUTION.

d8 = a sec^fdf, y=j y-^ = asecf + C,

= C^^=^a\og(p:^) + a.

J cosf \1 — smf /

0^ = 0, and

a; = ialog(j^;-^^) . . . (1), y+a = asecf . . . (2).

But from (1) secf = J log"M — )+log"^r j , where the notation log"^ «

has the same signification as e«, e being the base of the Naperian system of loga-
rithms; . • . y-^-a^ ja[^log-i(-^^ ■^^^^A~"f') J'
which is the well-known equation of the catenary.

X

cosf " " \l — smf ,
Whenf =0, a;=0, y = 0; .-. C^a, 0^ = 0, and

12

NOTICES OF PERIODICALS.

The Educational Times, and Journal op the College of Peeceptors.

Mathematical Questions, with their Solutions, Reprinted from the Educational Times.

London, England : C. F. Hodgeon & Son.

The Educational Times is published monthly, and contains, besides other valuable matter of a scholastic
nature, an important Mathematical Department devoted to problems and solutions and short mathematical
papers, edited by W. J. C. Miller, B. A., Registrar of the General Medical Council, to which many of the
leading mathematicians of England, Europe and this country contribute.

Each number contains three or four large double column pages of mathematics. From 20 to 25 prob-
lems are proposed each month, and about half as many solutions are given.

Every mathematician who wishes to keep '^posted" should take the EducatUmal Times.

The Reprint is issued in half-yearly volumes of 112 pp, 8vo., boards, and contains, in addition to the
mathematics published in the TVmes, about as much more original matter. Twenty-five volames have
been published. Vol. xxvi will be ready soon. Vol. xxv contains solutions of 102 problems. These
Reprints are rich in '^Average," '^Probability" and ''Diophantine*' solutions. In fact, nowhere else can be
found so many elegant solutions of difficult problems relating to these fascinating branches of mathemat-
ical science. They are a mine of mathematical treasures ; no mathematical library is complete jivlthout
them.

The editor of the Visitor will be happy to procure for American subscribers the Tim/es at $2 a year,
and the Beprint at $1.85 per vol.

The Analyst. A Journal of Pure and Applied Mathematics. Dee Moines, Iowa: Edited and

published by J. E. Hendricks, M. A.

The Analyst has entered upon its fourth volume, a feat never before accomplished by an American peri-
odical devoted exclusively to mathematics. It is very efficiently and satisfactorily conducted, and de-
serves a large circulation.

The AnaJyHt is issued in bi-monthly numbers of 32 pages at $2 a year. Each number contains several
excellent papers and a well selected collection of problems and solutions.

Educational Notes and Queries: A Medium of Intercommunication for Teachers. Issued
monthly, except in the vacation months, July and August. 8vo., 16 pp. Prof. W. D. Henkle, Editor,
Salem, Ohio. Price, $1 per year.
Edneational Notes and Queries contains, besides a vast fund of other instructive matter, a Mathematical

Department devoted to problems, solutions and mathematical notes.

The Yates CJounty Chronicle, a weekly newspaper published at Penn Yan, N. Y., by the Chronicle
Publishing Co., has a valuable Mathematical Department edited by Dr. S. H. Wright, M. A., Ph. D.,
occupying about a column, devoted to problems and solutions. $2 a year.

The Wittenberger, a monthly college paper published at Springfield, Ohio, contains an excellent Math-
ematical Department admirably conducted by William Hoover, of Bellefontaine, Ohio. $1 a year.

The National Educator, published semi-monthly at Kutztown, Pa., by Uhrich & Gehring, has an in-
teresting department of Science and Mathematics ably conducted by H. A. Wood, M. A., Professor of
Higher Mathematics in the Keystone State Normal School. $1.25 a year.

The Maine Farmers' Almanac, Masters <& Livermore, Hallowell, Me., contains, besides the calendar
and some interesting miscellaneous matter, a Puzzle Department and a Mathematical Department.
The number for 1877 contains solutions to the 6 questions proposed last year and 7 new ones to be an-
swered in the next number. Price 10 cents.

Vol. 1. JANUARY. 1878.

1
No. 2.

THE

t

i

1

1

MATHEMATICAL 1

i,

1 visri'OR.

EDITED AND PUBLISHED BY

1

i

ARTEMAS MARTIN, M. A.

i

PRICE, FIFTY CENTS.

I

* - ERIK. PA.:

JNO. M. QLAZi£H, PRINTER. COR. SiXTMNTM AHO P«AOM STRBETt.

1

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i

THE

MATHEMATICAL VISITOR.

[CNTCRKD AOOOROINO TO AOT OF OONOflCtS. A. D. 1877, BY ARTCMAS MAimN. M. A., AT THE OrPIOC OF THE LlSRARIAN OF OONOmU. AT WAtHINOTON.]

Vol. 1. JANUARY, 1878. No. 2.

SOLUTIONS OF JUNIOR PROBLEMS,

Proposed in No. 1. ^

1.— Proposed by Prol Edwabd Brookb, M. A., Ph. D., Principal of Penngylvania State Normal School,
MillersTille, Lancaster County, Pennsylvania.

Two trains, one a and the other b feet long, move with uniform velocities on
parallel rails ; when they move in opposite directions they pass each other in m sec-
onds, but when they move in the same direction the faster train passes the other in
n seconds. Find the rate at which the trains are moving.

I.— Solution bvDr. S. F. Baohelder, South Boston, Massachusetts; K 8. Putnam, Rome, Oneida County,
New York; and W. L. Habvet, Maxfleld, Peaobscot Co., Maine.

Let a;=rate per second, in feet, of the faster train, and y=rate of slower train.
When they move in opposite directions the rear ends of the trains approach
each other at the rate of x-^-y feet per second,

. . a + b_ ,-v

. — ; — =w (1)

z+y

and when they move in the same direction the faster train gains on the slower at
the rate of x—y feet per second,

••.' ^*=n (2).

Solving (1) and (2) we find

^_ (q+fc) (m+n) {a+b){n—m)

2mn ' ^ 2mn

Rlmilar solutions furnished by B. F. Burleson^ WilHam Hoover and O. D. Oathout.
II.— Solution by Maecus Baker, U. 8. Coast Survey Offloa, Washlagbon, D. C.

Lett? =the velocity at which the train a moves,
and t?'= " " " " " b moves;

and suppose v greater than v'.

First; when the trains move in opposite directions.

Let x^the distance the train a moves, and y=the distance the train b moves
in order to pass each other ;

then x=^mv and y=mv\
whence a:-i-y=:a+6=m(t?+v') (1).

Second; when the trains move in the same direction.

Let a;'=the distance the train a moves, and y'=the distance the train b moves
in order to pass each other ;

then
whence
Dividing (1) by (2),

whence

— 14 —
x'^=nv and y^=inv\

V —

(2).

2t?

m'

n+m , 2t?' n — m
: — ' — and ;= .

Substituting for r+u' and v — v* their values from (1) and (2) we have

2.— Proposed by Mrs. Anna T. Snyder, Orange, Fayette Ck)unty, Indiana.

Two trees of equal hight stood on a level plane, 50 feet apart ; one being cut
off at the ground lodged against the other 20 feet from the top. What is the hight
of the trees.

I.— Solution by F. P. Matz, B. K, B. S., Reading, Berks (bounty, Pennsylvania; and O. D. Oathout,
Read, Clayton County, Iowa. •

Put 50=m, 20=n, and the required hight A; then 'rv?'-\-{h—nf=^l?.
Whence h=^^'=11\ feet.

Solved In a similar manner by B, F, Burleson and James McLaughlin.

II.— Solution by D. W. K. Martin, Webstef, Darke County, Ohio; and the Proposer.

Let a?= hight of the trees. Then after one of the treas is cut off and lodged we
have a right-angled triangle, the base of which is 50 feet, or the distance between
the trees; the perpendicular x — 20 feet, or the distance from the plane to the (op
of the lodged tree, and the hypothenuse x, or the length of the tree cut off.

Consequently " (50)2+(a:— 20)^= x^

whence we easily find x=12\ feet.

Dr. Bachelder^ William Hoover, K. f. Putnam and W. L. Harvey solved by the same method.

3.— Proposed by S. C. Gk)ULD, Manchester, Hillsborough County, New Hampshire.

It is required to enclose the exact area of a given square- by means of circular
arcs only.

I.— Solution by the Proposer; and F. P. Matz, B. E., B. S., Reading, Berks County, Pennsylvania.

Let ABCD be the ffiven square ; draw the circum-
scribing circle and the circumscribing square, abed,
the corners of the given square being at the middle
points of the sides of the circumscribed square.
With centers c and d and radii equal to that of the
circumscribing circle, draw the arcs AED, CFD, and
the area of the pclicoid DEABCF is equal to the
area of the square ABCD.

Proof. — Let S be the area of the square, s the area
of one of the equal segments cut off from the circle
by the square, and P the area of the pelicoid.

Then P=8—2s+'2s=S. -

18

n.— Solution by O. D. Oathout. Read, Clayton County, Iowa; BIarcus Bakbr, U. S. Coast Survey Office,
Washington, D. C. ; and William Hoovsr, Mathematical Editor of |
the WittenberffeTj Belief ontaine, Logan County, Ohio.

Let ABCD be the given square, of which AC is a
diagonal. With AC as a radius, describe the circle
AFEI, and on AF as a diameter, describe the semi-
circle AGF. By the property of the lune of Hip-
pocrates, the lune

AGFH=iAAFE=i square AFEI = square ABCD.

This problem was also solved In a very IngenlouN manner by B. F,
Burlewn and K, & Putnam.

4.— Propoeed by Dr. David S. Hart, M. A., Stonington, New London County, Connecticut.

The straight line drawn, bisecting any an^le of a triangle, to the opposite side,
is less than naif the sum of the sides containing that angle.

L— Solution by Whjjam Hoovsr, Mathematical Editor of the
WittenberffeTf Belief ontaine, Logan County, Ohio.

Let ABC be the triangle, AE the bisector of the
base and AF the bisector of the vertical angle.
Produce AE to D, making ED=AE, and draw CD.
Then AD<AC+CD, or 2AE<AC+ AB,

or AE<i(AC+AB).
But AF is less than or equal to AE ; hence
AF<i(AC+AB).

Good solatlons given by Dr, Bachelder and K. S. Puincmt.

IL— Solution by Miss Christinb Ladd, B. A., Professor of Natural Scieoces and Chemistry, Rowland
School, Union Springs, Cayuga County, New York.

Let the line in question be x, then we have

c sinB ae sinB 2i/ bcsis — a) _ j^ bm^

'^ (h+ef

0?=-

ac sinB 2i/bc8{8-

(b + c) sin J A 6 + c

= ^j^■

co8i(B— C)
But this quantity is less than ^{b+e}, since

because H(*+c)^— «^]<i(i+c>

or ' bc{a+b+o){b+c-^a)<i{h+ef(b+cf

as bc<l{b+cY and (a+6+c)(6+c— a)<(6+c)2.

Mr. Burle9on*$ solution was similar to this.

6.— Proposed by Artkmab Martin, M. A., Erie, Erie County, Pennsylvania.

If 22 oxen and 28 cows eat 24 acres of grass in 18 weeks, and 20 oxen and 38
cows eat 30 aces of grass in 27 weeks, and 41 oxen and 26 cows eat 50 acres of grass
in 60 weeks, how long will 40 acres of the same grass last 35 oxen and 14 cows, the
grass in all cases growing uniformly ?

16

Bolution by Thbo. L. DeLand, U. S. Treasury Department, Washington, D. C; K S. Pctnaji; Rome,
Oneida CJounty, New York; and .

Let a?=the grass and its growth eaten by one ox from one acre in one week ;
y=the quantity eaten by one cow; ^^the weekly growth on one acre^ and ^=the
required time.

Then 18(22a;+28v)=what 22 oxen and 28 cows can eat in 18 weeks; and
24(1 +18^)= quantity of grass eaten from 24 acres in 18 weeks. By the problem
the^e two expressions are equal ; hence if we equate them, and extend the reason-
ing to the other conditions we will have :

18(22x+28y)=24(l+18^) . . (1); 27(20j-+38y)= 30(1+27^) . . (2);

60(41a;+26y)=50(l+60(7) . . (3); t{S5x+Uy) = A0{l+tg) . . (4).

From (1), (2) and (3) we get

3ar+42y— 36<7=2 (5), 90.r+171y~ 135(7=5 .... (6),

246a;+156y— 3005r=5 (7).

Solving these equations, we find

— A —A. _J^

^— 54> y— 108' ^"12'

From (4),

^^S5x-tUy^40g^^^ ^^^^'

The solntlon by W, Ij. Harvey was similar. An IngenloaB solution was furnished by R, F. Burleson.

6.— Propoeed by Thio. L. DeLahd, U. S. Treasury Department, Washington, D. C.

Four men, A, B, C and D, agreed to reap a circular field of grain in 28 days for
1249.36, and divide the money among them in proportion to the actual service each
rendered. The first, second, third and fourth quadrants contained, respectively,
wheat, oats, barley and rye. To the end of the 7th day A cut wheat, B oats, C barley
and D rye. They each then advanced one quadrant, and to the end of the 13th
day A cut oats, B barley, C rye and D wheat ; they again advanced one quadrant,
and cut to the end of the 18th day ; again they advanced one quadrant each and
so worked to the end of the 22d day, and finished the wheat ; again they ad-
vanced one quadrant each and finished the oats at the end of three days more ;
they again advanced one quadrant each and finished the barley at the end of the
27th' day ; and then advanced one quadrant each and completed the work accord-
ing to contract. Required — ^the time for each to reap the field, and an equitable di-
vision of the money, provided one quadrant is as difficult as another.

Solution by Mrs. Anna T. Sntder, Orange, Fayette County. Indiana; Jambs McLaughlin, Mantorville,
Dodg^ Ck>unty, Minnesota; and W. L. Haryby, Stozfield, Penobscot Ck)unty, Maine.

Let i(?=part of the whole work A does in a day,

x= " " " B " "

y= " " " C " "

2= " " ** D " "

By the conditions of the problem,

7w+ 4x+ 5y+ 62 =i (1),

(k;+10a:+ 4y+ 5z=i (2),.

5w+ 8x+10y+ 4z=i (3),

4k? + 6x+ Sy+10z=i (4).

From these equations we easily find

^""8312' '^'""8812' ^""8312' ^""8312-

17

1 8312 67

-^^ = -T-yx= "^^ST^' *^^ number of days in which A can cut all the grain;

_L_8312_-^q31 ,, ,^ ^^ ^, -^ ^^ ^^ j^

J_=^'^-'^^=:148— " " " « C " " "

y 56 7 '

A worked 22 days and his share of the money is $249.36X^^X22 =$114.84;

8312

B « 28 « " « " , « f249.3GX5|?:rX28=$41.ie>;

C " 27 " " " " " $249 36xJ?^X27=$45.36;

D " 25 " " " " " $249.36X-3§^*^X25=|48.00.

ooUJ
Solved also In an elegant manner by the Propoter and B, F, Burleson,

7.— Proposed by Prof. David TbowbRidge, M. A., Waterbui^g, Tompkins County, New York.

If A, B, C be the angles of a triangle, prove that

cotA+cotB , cotA+cotC cotB+cotC ^

cotJA+cotJB cotJA-j-cotJC cotiB+cotJC"

I.— Solution by Bfiss CHBismfB Ladd, B. A., Professor of Natural Sciences and Chemistry, Howland
School, Union Springs, Cayuga County, New York.

We have cotA+cotB= ^ , cotiA+cotiB= ^

hence

2K' ^ ' • ^ K

cotA + cotB c

cotiA+cotJB 2«"

Q. .| , cotA+cotC h , cotB+cotC a

J^imilarly, ^^^^__^_____ -^ and ^^^p.^^-^^^- g^ •

cotA+cotB , cotA+cotC , cot B+cotC c+b+a ^ . ^ ...

• • "ii A~i ~i.i D I Ti A~r — iTV^H TTW, rrri= — s >=l» since z«=a+6+c.

cotiA+cotJB cotJA+cotJC cotJB+cotJC 2«

II.— Solution by ; and the Pboposxr.

Let pif Pi, p^ be the perpendiculars let fall from A, B and C respectively upon a,
h and c ; and r be the radius of the inscribed circle.

Then ;>,(cotB+cotC)=a=r(cotJB+cotiC) ;

cotB+cotC r

cotJB+cotJC pi'

q. ., , cotA+cotC r cotA+cotB r

oimiiariy, cot JA+"cotJC~ p^ cotiA+^tfB" j^ '

- cotA+cotB cotA+ cotC cotB+cotC _ / 1 , 1 , 1 \ _ / 1 \_i

• * cot|A+cotiB'^cot|A+cotiC cotJB+cotiC~''Vi>i P2 PU ~^\r)

Solved also by B. F, Burlemn, Jamea McLaughlin and David Wiekeraham,

18

8.— Proposed by Marcus BakkBi U. S. Coast Survey Office, Washingtoa, D. C.

In an equilateral triangle ABC, lines drawn from each vertex to the opposite
side, dividing that side so that the ratio of the side to one of the segments is n,
form by their intersections another equilateral triangle ; determine the ratio of the
areas of these triangles.

Solution by Dr. S. F. Bachsldbr, South Boston, Massachusetts; and B. F. Burlxson, Oneida Castle,
Oneida County, New York.

Draw AE, BG and CE, making

AD=BE=CG=^^.
n

Put .r=the required ratio, then

AFIH_(FH)^
/rABC~(AC)2 • • • • W-

Put a=AB, then " =AD.
n

By trigonometry, C09CAD=(^^'^'t^|?^-i^'^^' (2).

But cosC AD = \ . Substituting in (2), " = | ;

whence C'D= ' -.which put =6.

The triangle ADF is similar to CDA, consequently

CI) : CA :: AD : AF or CH (3),

and CD:AD::AD:FD (4).

Substituting known values in (3) and (4),

Ctt= f, FD= f,, HF=CI)-(CH I FI))=''^''-=^'^"+-^->.
fin hn^ ' /«r

Substituting in (1),

_ A FIH _ /6'V-«2(n + 1)\ *_ (n— 2)2

"^"/vABC^V abn^ )—n^—n+V

Good Holutionn given by the Propter, K, B, 8eUz and K. S, Putnam,

0.— Proposed by Miss Chribitnk Ladd, B. A., Professor of Natural Sciences and Chemistry, Howland
School, Union Spniigs, Cayuga County, New York.

oT_ xu * • 1 X • 1 «sinA+6sinB+<^sinO n /ct^+b^+c\

Show that m an}' plane triangle, r——. ;k-^ >. = -B i , i •

^^ ^ 'acosA+6cosB4 ccosC V obe )

Solution bv David Wickbrsham, Wilmington, Clinton County, Ohio; Marcus Bakbr, U. 8. Coast Sur-
vey Office, Washin^n, D. C. ; and William Hoovkr, Mathematical Editor WiiieiibeTgeT^ Bellefontaine,
Logan County, Ohio,

Let ABC be the triangle; and let ABDC be a circle circumscribing the triangle,
and its center ; and join OA, OB and OC, and draw OE, OF and OG respect-
ively perpendicular to BC, AC and AB.

Let jB=AO or BO; then BE=/2 sinA, and BC=a=2/J sinA, and sinA=/„,

and OE=iJ cosA, and cosA =

OE
R'

-19-

a sinA= ,and

2xv

in the same way A sin'B=^r-^and c sinC=z^ ■

a sinA+6sinB+csinC=

2R

and

OE OF

AndacosA = aX ^; but aX ^ is double the area

of /\COB-r-/?; in the same way b cosB = double the
area of i^AOC-^R; and c cosC = double the area of

AAOB-i-iJ. But double the area of A ABC is = a6 sinC=''*'^; double the area of

AABC-=-i? is=^^; and acosA+6 cosB^ c cosC =

(2);

and dividing (1) by (2),

a sin A +6 sinB+r sinC

^.(

a cosA+6 cosB h^- cos(

.Solved also in an ele^anl manner by the Proposer and B. F, Burleson.

abc J'

10.— Proposed by Prof. H. A. Wood, M. A., Principal Coxsackie Academy, Cozsackie, Oreene Co., N. Y.

A hollow paraboloid, depth 4 feet, stands on its vertex with a*is vertical. After
a shower the depth of water in the paraboloid, measured on the axis, was 2 inches.
What was the uniform depth of the rainfall, the radius of the top of the parabo-
loid being 3 feet ?

L— SolutioQ by B. P. Burleson, Oneida Castle, Oneida County, New York.

Put D:=entire depth of paraboloid, =48 inches,
d= depth of rain caught, =2 inches,
and i2= radius of the top of the paraboloid, =36 inches.

Let ?'=radius of the top of the water in the paraboloid, and F= volume of the
same. Then, as the squares of ordinates to the axis in all parabolas are to each
other as their corresponding abscissas, {Loomis^ Conic Sections, Prop. VIII, Cor. 1,)

R^d
R} :r^ \: D '. d\ . • . r^= ^-. The volume of the paraboloid of rain caught being

D

one-half its circumscribing cylinder {Loomis^ Ccdcidm, Art. 328,) we have F=

2D '

7:R?d^ d^ . . 1

Obviously, therefore, ^jj^ — ^^^^"^on* ^^^^^ ^Y restoring numerical values, . of

an inch,=the uniform depth of the rainfall required.

As the general expression giving the uniform depth of the rainfall does not con-
tain the factor R we infer that the condition giving the radius of the top of the
paraboloid is superfluous — the answer being easily ootainable without its use, being
constant for every value that may be given to i?.

II.— Solution by Marcus Baker, U. S. Coast Survey Office, Washing^ton, D. C; D. J. McAdam, M. A.,
Professor of Mathematics, Washington and Jeflferson College, Washington, Washington County, Pennsylva-
nia; and Orlando D. Oathout, Head, Clayton County, Iowa.

The general equation of the parabola, y^=2pxy gives for the parabola which gen-

1/^9 9

erates the paraboloid of our problem 2j)= - = \^ ; whence y^^= jx.

20

For the volume of a paraboloid of revolution, we have F=i;ry%, and hence
when a;=2 inches,=i foot, y^=f and therefore V= ' ^, which is the amount of
rainfall upon the surface Qtt ; hence the uniform depth of the rainfeU is, in inches,

32^^-24-

This problem was also solved by Messrs. Baehelder, Hoover, MaU and Pulitam,

11.— Proposed by B. F. Burleson, Oneida Castle, Oneida County, New York.

The sides of a quadrilateral field are, respectively, and in order, 17, 35, 40 and
45 rods in length. Determine the length of the straight line which in passing
through a point equally distant from all its angles, shall divide the field into two
equal parts.

Solution by the Pboposer; and David Wickbrsham, Wilming^n, Clinton County, Ohio.

Let ABCD represent the quadrilateral, and the point
O the center of its circumscribed circle. Let EOt rej)-
resent the straight division line required, the two ends
E and F of which must fall, it may be easily deter-
mined, on the lines AD and BC of the quadrilateral,
respectively. Draw the diagonals AC, BD; also join
AO and BO, producing the latter until it meets the
side AD in M.

Put AB = 40 = a, BC = 35 = 6, CD = 17 = cand

DA = 45 = d. Let x =:BD, y = AC, r = AO and A= area of the quadrilateral ABCD.

By Geometry, a?y = ac+6d . . (1). Also 6cH-ad : cd+a6 :: a; : y . . . \2).

and M=h/[(«-a)(«-6)(«-c)(«-d)]=537.467578 square rods, where 2«=a+6+c-rd.

By Trigonometry we now easily determine the following angles: ABC =83° 13'
21", BAD=63° 51' 3" and ABM=37° 19' 7". We also determine further, by
Trigonometry, that BM = 36,59924 rods, OM=11.45062 rods, AM =24.71801 rods,
the angle EMO = 78° 49' 50" and the angle OBF=45° 54' 14". We now find
the area of the triangle ABM =443.762137 square rods, which lacks just 93.705441
square rods of fulfilling the conditions of the problem. Supposing EF to represent
the true division line, it will be evident that the difierence in the areas of the two
triangles BOF and EOM must be 93.705441 square rods

Put m=OM = 11.45062, ^(7=93.705441 square rods, a=^OBF = 45° 54' 14"
and /9=Z_EMO=78° 49' 50". Let 2=0F, w=OE and ^=/_EOM or BOF. In

the triangle BOF, sin«: 2:: sin(a+^):r; .-. z=4-^^.. In the triangle EOM,

sin(a+^) ^

sin^ : w :: sinO^-hf) :m; .-. w=-!^^^ Now area B0F = ir2 sincp, and area
EOM = Jmir sinf, hence r^ sinf — Witt? sinf=:<7 (3).

21

By substituting in (3) the values found for z and ?r, and reducing, we have

rf_ _ irf _ _^ .^

cotfj+cottt cotf-f-cot/? ^ ''

In equation (4) we can easily find the value of cotf , and hence the angle if or
EOM, which is 36° 20' 17". Having found the angle EOM or BOF we can find
the remaining parts of the two triangles EOM and BOF. The parts EO in the
one and OF in the other, when added together, give EF= 30.63992 rods.

Solved also by />r. Bachelder and James McLaughlin,

12.— Proposed by James McLaughlin, Mantorville, Dodge County, Minnesota.

A gentleman has a rectangular garden in latitude 40° 41' north, the diagonal of
which is a meridian line. At the south corner of the garden stands a perpendicu-
lar pine tree, the shade of which at noon, when the days and nights are equal, extends
to a small rivulet which comes in at the east corner of the garden and runs due
west through it, crossing the said meridian line 240 feet from the north corner of
the garden, and going out of it 135 feet from the pine tree. Required the area of
the garden and the hight of the tree.

I.— Solution by E. B. Seitz, Greenville, Darke County, Ohio; B. F. Burleson, Oneida Castle, Oneida Co.,
New York; and the Proposer.

Let NESW be the garden, NS the meridian line, E the
east corner, and EHK the rivulet.

Put NH=a=240 feet, SK=6=135 feet, SH=.r,
A=hight of the tree, and /=41° 40', the latitude. Then
from the right angled triangles we have

SH : NH :: SK : NE and SN : SK :: NE : SH.
Taking the products of the extremes and the means, we
have (SH)2.SN=(SK)^NII, or x'+(u;'=ab\

Restoring the numbers, we have ar'^+240.r^= 4374000,
whence a:=111.544738 feet.

.-. area NESW=NS . EH=(a-( a:)|/(a.r)=57ol8.926
square feet=1.3204o3 acres, and A =.r cot ^=125.342 feet.

Solved also by Dr. Baehelder and David Wickeraham,

II.— Solution by Dr. Samuel Hart Wright, M. A., Ph. D., Mathematical Editor Yates County Chronicle,
Penn Yan, Tates County, New York.

Let x=aside and y an end of the garden, a=135, 6^=240. Then
shadow=(ic24-y2)^— 6, and y : x :: a : y; ,\ y'=ax, b : x :: y : (j^'^+y*)^ ;

.-. y^=-|:iT2=«^, and .r^— Wr=a62, whence ir= 290.467 and y=198.013.

Area = 57519.1 square feet = 1.32046 acres. Shadow = 111.545 == 6i . Declination of
sun's upper limb= +16'= r.

Hight of tree = 6i cot(/— r)=6i cot(41° 10'— 16')=128.765 feet.

Id.— Proposed by E. P. Norton, Allen, Hillsdale County, Michigan.

In a triangle ABD the base AB is 40 rods. A line AC drawn from the angle A,
and perpendicular to AB, intersecting BI) in C, is 9 rods. If a point F be taken
in AB, 10 rods from B, and a line drawn from this point through C, and produced,

22

it will cut AD produced in a point E, 5 rods from D. Required the sides of the
triangle

Solution by E. B. Seitz, Greenville, Darke County, Ohio.

Draw DHN perpendicular to AC produced,
and draw GEM parallel to DN, meeting BD
produced in G.

Let AB = a = 40 rods, BF=6=10 rods,
AC = c=9 rods, DE = d=5 rods, EG = a-, and
HD = y. Then from similar right angled tri

angles we find CM = % CN= ^, EM =^^lZ^^
bo

DN="|and AM : AN :: EM : DN,

or b+x : b-\-y :: (o — b)x .ay (!)•

We also have (CM-CN)2+(EM-DN)2= (DE)2, or cV-y)H[(a-6)a;-ayJ' = Wf . (2).

From(l)y= ^"~~ "^ . Substituting this value of y in (2), and reducing, we have

[(a-ft)2+c2]x*+26cV+(c2— <P)6V-2rtWRc-rt262d2=0 . . . (3).
Restoring the numbers in (3), we have

98U-<+1620rH5600jr2— 200000^—4000000 = (4).

Solving (4), we find x = 8.172543366 rods. . • . y= 5.089544455 rods,

BD=(a2+c2)X+ 1 (aHc'^'^^ei. 86713226 rods,and AD='^^-*±^=24.47218584rods.

Good solutions were given by Mettsrs.Bachelder, Burleson^ McLaughlin and the Proposer,

14.— Proposed by Prof. Horatio M. Bloomfdelj), Reading, Berks County, Pennsylvania.

Given M?2+y^=c, z^+y^=bj 2^-\-x^=ay x+y=w+z; to find the value of .r by
quadratics.

a.

Solution bv B. F. Burlbson, Oneida Castle, Oneida County, New York; and P. P. Matz, B. E., B.
Reading, Berks County. Pennsylvania.

Let 8=x+y=w-\-z and we evidently have the three equations
^+{^-yr=a . . (5), z^+f=b . . (6), f+(s-zf=e . . (7).

Eliminating y and z from (5), (6) and (7), we obtain

28^—(2a+2cy=2be+2ab — {a^+2b^+(^) (8).

Resolving (8) «, or x+y,=± {i{a+c)±^[4ab+4bc+2ac — a^—G^—4hVi^y^ . . (9).
Subtracting (2) from (3), x'^—f=a—b (10).

Solving (9) and (10), we find x= ^rj^. — -^ — kr i l\ al \ o 2— 0I2 — Jnin^-

^ ^ ^ ^ ^ ±2{2{a+c)±2l4ab+4bc+2ac-a^-2b^-c^>^}^

Unflnlshed solutions given by Marcits Baker and William Hoover.

16.— Proposed by Henry Hbaton, B. S., Superintendent of Schools, Sabula, Jackson County, Iowa.

If n persons meet by chance, what is the probability that they all have the same
birthday, supposing every fourth year a leap year?

23

SoluUoQ by the Pboposxr; and K. 8. Putnam, Rome. Oneida County, New York.

Since the twenty-ninth of February occurs but once in every 1461 days, the prob-
ability that it is the birthday of any given person is ^ , and the probability that

it is the birthday of n persons is .^ .^^v . Since any other given date occurs 4 times

(4 \ «
TiftT ) ' ^^^
since there are 365 such dates, the required probability is

1 1365^ ^ V_l +365(4)"
(1461)».^'''^Vl46iy ~ (1461)" ■

-®:»-=;^j^s>->=^-

SOLUTIONS OF SENIOR PROBLEMS,

Proposed in No. 1.
16.— Propoeed by Prof. David Trowbridok, M. A.» Waterburg, Tompkins County, New York.

Required the expansion of ff^ in a series of cosines of multiples of 0, thus :
^=Ao+Ai cos6^+A2COs2/?+ . . . +AnCosntf+ . . .

Solution by the Profosbr; and Prof. H. T. J. Ludwick, Mount Pleasant, Cabarrus Co., North Carolina.

We have
C^J^0^dd=i7:^==f^JiA(^^ . . ]=2;rAo, .'. Ao=i;r2.

Also, \0^cosm0dO=\ -sin mO+ .. cos mO— , sinm^ =+ o(— 1)*",

= j '' d/?cosm/?[Ao+AiCOS^+ . . +AniCOsmfl+ . . ]=A„,.t;

... A„=+^^^. We hence have <?2=|;r2-4[cos «?-^|f+ '-°|^- .. ].

cw.-lf (?=o,4=i-i, + ^-^+ . . ;

Solved also by O. M. Day,

1 7 •— Proposed by Miss Christine Ladd, B. A., Professor of Natural Sciences and Chemistry, Howland
School, Union Springs, Cayuga County, New York.

The polar with respect to a triangle of its orthogonal center is the radical axis of
the circumscribing circle, the nine-point circle and the circle with respect to which
the triangle is self-conjugate.

Solution by the Proposer.
The equation to the nine point circle (Salmon, p. 127,) may be thrown into the form
(acosA+/9 cosB+r cosC)(a sinA+/J sinB-f /- sinC)-2(/3y'8inA+/'a 8inB+a^8inC)==0;

—^24

that of the circle with respect to which the triange is self-conjugate {Salmon, p. 243,)
may be written

(a sinA+z? sinB+;- sinC)(« cosA+/9 cosB+t'CosC)— ()97'sinA+7'asinB+a/9sinC)=0.
Hence it appears that their common radical axis with the circle

^Y sinA+T'tf sinB+«/5 sinC=0 is a cosA+^9 cosB+;' cosC=0.

Rolved alfio In an elegant manner by O, Jf. Day.

18.— Proposed by I. H. Turbbll, Camminsville, Hamilton County, Ohio.

Suppose n fixed lines, radiating from a fixed point, are touched two and two by
n circles of given radii; that is, that each line touches two of the circles, and each
circle touches two of the lines. The order of arrangement being known, it is re-
quired to construct geometrically a polygon of n sides, whose vertices shall lie on
ine n lines, and whose sides shall touch the given circles.

Solution by the Profossr

Let Xy Xu X2y . . Xn-i be the n lines given in position, and Fi, Fi, . . . Vn the
given circles. Place Fi touching the lines X, Xi; Y^ touching the lines- Xi, X2\
.... Yn touching the lines Xn-\ , X,

From any point a in X draw a tangent to Fi, meeting X^ in some point a\ ; from
ai draw a tangent to Fj, meeting X^ in a^^ and so on, until a point On-i in Xn i is
determined. From On-i draw a tangent to Y^ meeting X in On. If a, On are coin-
cident the polygon is constructed. In like manner take two other points 6, o in X,
and obtain the corresponding points bnjCn-

Suppose p to be the required point in X\ then jo, pn are coincident. Now the
four tangents to F,, viz : aaj, 66^, cc,, pp^, are cut by the tangents JT, Xi ; hence
{Muloahy^s Modem Geometry, p. 39,) the anharmonic ratio «, 6, c, ^=ai, 6|, Ci,/>,.
Similarly ai, 6,, C], pi =02,621 ^2^ P2^ and so on ; hence abcp^s^aJbrfinPn^ Therefore in
X we have six known points, a, i, c, a„, 6„, tv, to find a seventh point p in the same
line such that abcp^=aJhfinPn-

This is an elementary problem in the Modern Geometry (See Mulcahy, p. 23,)
and gives two positions for y. Hence the original problem admits of two solutions,
when the lines oaj, 661. Ac, are direct tangents to the given circles. If some of the
tangents are travjiverse, the number of solutions is vasUy increased, and it is evident
that numerous impossible cases may arise.

10.— Proposed by Dr. S. H. Wright, M. A., Ph. D., Mathematical Editor Yates County ChronieU, Penn
Yan, Yates County, New York.

In latitude /=42°N., when the sun's declination north=5=20*^, the sun shone on
the top of a high peak in a mountain ^=i minutes before it shone on the plane
of the horizon. How high was the peak?

L— Solution by E. B. Sxrrz, Greenville, Darke County, Ohio; and Gborob Eastwood, Saxonville, Middle-
sex County, Massachusetts.

Let Z be the zenith of the mountain peak, S the place of the sun's center when
it first shone on the top of the peak, S' the place when it first shone on the plane
of the horizon, N the north pole, and let ZH be perpendicular to SN produced.
PutNZ = J;r-;i, SN = S'N = i-~o^ S'Z=i;r+/+r = ^, /= 34', the refraction of

28

light at the horizon, r = 16', the sun's radius, ZH = a,
HN=6, SH = c, SZ=d, Z_S'NZ = #, /_SNZ = y?,
,^SNS'= J)9=l°, jB=3956 miles, the radius of the earth-,
A=the hight of the peak. Then in the spherical trian
gle S'NZ we find

.-. y?=<?+i/9=lll°23'54". In the triangle ZHN we
have a = sin-' (cos;i sin^) = 43° 46' 55", and h = cos-' (sin / sec a) = 22° 3' 29" ;
.-. c=Jff— 5+6=92° 3' 29".
In the triangle SZH we have d=cos-»(cosa cos <;) = 91° 29' 8 7".

.-. A=i?8ec(rf— i«')—J8= 2/2 8in2j((/—^)sec(d—f'') = 1354 215 feet.

Holved In a similar manner by D, J, Me Adam,

II.— Solution by the Propobbr.

Let the zenith distance of the sun's center when its upper limb risas on the
plane = (90°+ refraction 34'+sun's radius 16') = 90° 50'=5;; hour angle then = P;
Eenith distance when rising on the peak = 2', and hour angle then=P';
72=3956X5280 feet. Then

P=2sin-V[sinK>l+«— ^)sini(^— ^+2)secAsec/J]=110° 2H' 54'.

P'=P+ Jj9=lll° 23' 54", and 2'=cos-i[sin(d"+f)sin;isecy?]=91° 29' 8.7",
where tanf=cot^cosP'. Dip of horizon=2'— z=39' 8.7"=^. Then hight of
peak = i2(sec^ — 1)=2R sin4A sec A = 1354.215 feet.

20.— Proposed by William Hoovsr, Mathematical Editor Wittenbergery Belle/ontaine, Logan Co., Ohio.

The weights Wi and W2 are suspended, 9 degrees apart, from the circumference
of a vertical circle free to move about its center. Find the angle made by the radii
drawn from the points of suspension with the horizontal through the center of the
circle when the system is in equilibrium.

Solution byDuNLAP J. MoAdajl M. A.. Professor of Mathematics, Washington and Jefferson College,
Washington, Washington Comity, Pennsylvania; and E. B. Ssm, Ureenville, Darke County, Ohio.

Let /= angle which the radius drawn from Wi makes with the horizontal
when in equilibrium, then ^r— ;f — ^=f , the other required angle, and taking mo-
ments about the center of the circle we have Wi cos /= W2 cos(;r — ;f — fl), which gives

X l/Wi+W2COS0\ Q. .1 1 ^. . 1/1^2+ »»^lCOSfl\

^=^^ (— iT^-sinT )• ^'"^'^^'^^ ^' ^""^ f =tan-(- TTTsintf O'

Solved in a similar manner by MeMgn, Hoover and Jld/s, and geometrically by G, M, Day.

21.— Proposed by Dr. David S. Hart, M. A., Stonington, New London County, Connecticut.

Required five numbers whose sum shall be a biquadrate, and the sum of any
four of them a square.

Solution by the Pbofoskr.

Let r, IT, Xy y, z represent the numbers. Then t?+w7+a;+y +2=biquadrate=m* (1),

t7+ic+ic+y=D=n^, v+w+x+z=\Z]=p^, v-\-w+y+ziB=z[Jr=zf^

v+x+y+z=\J=r^y w+x+y+z={J—»^.

26

Adding the last four equations, and dividing by 4, we have

r+i«+a?+y+5;=KnHl>2+<r'+r2+«2) (2),

From this, subtracting the last four equations, we have v=^{n^+p^+^+f^—StP)f
w=J(n24-j>2+52+^_3r2), x=i{n^+p'+f^+f^-'Sq'), y=^n?+^+f^+ffl-3p^.
«=:J(p2^g2_|_^^^_3^2j Equating the right-hand members of (1) and (2) and
then multiplying by 4, n^+p^+<f+r^+tl^=Am'^.

Let n=^am, p=lmf q=<m, r=^dm, «=eiii; then, substituting and dividing by »i^,
we have a^+h^-\-<?+d^+e^=Am\

Let a=^m — A, b=m-—B, e=m—Q d^^m—D; then, substituting, canceling,

transposing, Ac, we have m= — 2M-kg4-C4-^ — ' ^^ which expression, in order

that the values of r, w, x, y, z may be positive. A, J?, C, 2) must be taken consecutive
numbers in the natural series, or nearly so, and e must be=2(-4+J?+ C+D), or
nearly so.

Let ^=1,5=2, 0=3, i)=4; then e =20, m=^^, a=-^ , b=-^-, e=~^

or

d= , or multiplying all these quantities by 2, m=43, a =41, 6=39, c=37,

rf=35, €=40; whence «=4lX43,p=39x43, 9=37X43, r=35x43 and «=40x4a
^ Then by substitution we have t;=249X(43)^460401, t(7=624X (43^^1153776,
a:=480x{43)2=887520, y=328X (43)^=606472, ^168 X (43)^=310632

From this solution we see how n numbers can be found whose sum is a biquad*
rate^ and the sum of any n— 1 of them a square.

22.— Proposed by Walteb Sivsrlt, Oil City, Venango Coonty, Pennsylvania.

An ellipsoid, axes 2a, 26, 2c, is intersected by a cylinder, radius 6, the axis of the
cylinder coinciding with the axis 2c of the ellipsoid. Find the volume common

to both._Proin the Normal Manihly, vol. lit, p. 98.

Solution by E. B. Seitz, OreenvUle, Darke County, Ohio; and the Pboposkb.

The equation to the ellipsoid is— j+ ^j + ~^~^ • • • (^)» ^^^ ^^® equation to

tlie cylinder is ar^+y^=62 . . . (2). The volume common to the ellipsoid and cyl-
inder is

V=8fffd^dydz=.8ffzd^y=8cff(l-^,^^yda^y.

The limits of z are and y^(b^—y^) ; of y, and ft ;

28.— Proposed by Artkmas Hartik, M. A., Erie, Erie County, Pennsylvania.

Find the least integral values of x and y that will satisfy the equation

a?_9781y2=l.

27

Solution by the PBoPOgnL

Put ^=9781, then y'A=y'{9781)=r+l_

ui+1

^h+1

ws+etc.,
where r is the integral part of y^A. The last quotient of every complete period is

2r. Let m be the number of quotients in a complete period, and - the last con-

9m

vergent in the first period; then, when m is even, x=pm9 y=qm9 and when m is
odd, x=p2m, y=q2m- Let tin, Wn-ui be any two consecutive partial quotients; then

^=tt„ + etc., Ipn±l^Un+,+ete.;
aoF=0, 6oF=l; ai=r, 6i=^— r^; ti<y=r, ui=--j — -«; anfi=t^6n— an,6nfi= — r-^^-
IfP^^ Pl^ be any two consecutive convergents and Un+i the quotient correspond-
mg'to^-^Sthen£'- = ^, ^ =-!2^L±l, . . . . gn±2^ t^M;>n^.+J>»

The partial quotients are easily found to be 98 ; 1, 8, 1. 8, 1, 1, 12, 1, 1, 1, 15, 1, 4>
1, 2, 2, 6, 1, 1, 1, 3, 2, 1, 1, 2, 5, 9, 4, 3, 2, 21, 1, 1. 5, 7, 6, 1, 12, 3, 16, 6, 3, 7, 1, 1, 2,
9, 2, 48, 1, 38, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 65, 3, 2, 1, 27, 1, 1, 3, 1, 7, 1, 4. 1, 1, 1, 1, 4, 4,
1, 1, 1, 1, 4, 1, 7, 1, 3, 1, 1, 27, 1, 2, 3, 65, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 38, 1, 48, 2, 9, 2, 1,
1, 7, 3, 6, 16, 3, 12, 1, 6, 7, 5, 1, 1, 21, 2, 3, 4, 9, 5, 2, 1, 1, 2, 3, 1, 1, 1, 6, 2, 2, 1, 4, 1,
15, 1, 1, 1, 12, 1, 1, 8, 1, 8, 1, 196,=M,67=2r.

As 157, the number of quotients in a period, is odd, therefore x=piiff, y=qva
satisfy the equation x*— 9781y*= — 1; and x^pm, y=qsu satisfy the equation
af*— 9781y*=+l. It is not necessary to compute the numeratora of the convergent
fractions as p^rqm-\-qm-i . Computing the values of gi, qi, qs, qt, etc., we find

g„„=493415265416374912814471103134560111323246466844777158418926677585
7369909961,

p,5,=487982458985070727523051082184114842770783358477341808754197715936
903166164370.

x=P3u== 2Dfs7+ 1 =47625376075423266962291555142064377580641746864784592
0709133116505163927786611046291325633404816631400075031779842394788655329
052356895448229542978234993801,

y=53u=2p,57gi67=48155598903730791575885817698096793W7125906711321806
07164384581211216970331509974781382264086340917459934751261746227.674,74943
Q227294^525618O3^3(;330)579a4O.

These prodigious numbers are believed to be the largest of the kind ever computed.

24.— Proposed by Jamcb HoLauohun, Mantonrille, Dodge Coonty, Minnesota.

An army in the form of a circle, 2a miles in diameter, marches due north at the
uniform rate of n miles an hour. An officer starts from the rear of tlie army and
rides around it at the uniform rate of m miles an hour, keeping close to the army
all the time. Required the equation to the carve the officer describes, and the dis-
tance he rides while going once around the army.

28

Solution by E. B. SsiTZ, Greenville, Darke Co., Ohio; and Artemas Martin, M. A., Brie, Erie Co., Pa.

Let BPC represent the army, B being the rear, the
center, A the initial position of B, P the place of the
officer at any time. Draw PM perpendicular to AO,
and join and P.

Put OP=a, — =c, AM=a:, PM=y, AB=t(?, arc
m

AP=z.

Then OM=v/(a2—y2), and tD=^—a±i/(a2—y2) , (i)^

the double sign being taken + when P is in the first

or fourth quadrant, and — when P is in the second or third quadrant.

We also have z : w :: m: n, whence nz=mw . . (2), and d:?=^dx^+dy^ . . (3).

From (1) and (2) we have z=~~[x—a±i/{a^—f)'] (4).

Differeptiating (4), we have &= — (dc=h -y^^Z^pj • • (5).

Substituting the value of dz from (5) in (3), reducing, and solving for dx, we find

the double sign being taken as in (1).

Integrating (6), and observing that when a;=0, y=0, we have for the required
equation,

where E(cy ^y^i—cA represents an elliptic arc, semi-major axis unity, eccentric-
ity c and ordinate — y/l — c^.

When the oflicer has gone one-fourth of the way around the army, we havey=a,
and from (5) and (7) we find that the distance he rides is «i = — ^- — 5-{l+-^c)],

where E{o) represents the length of an elliptic quadrant, semi-major axis 1 and ec-
centricity c. When he has gone half way around the army, y=0, and the distance

he rides is 22=— y— -^1+J9(c)]. When he has gone three-fourths of the way

around, y= — rt,and the distance he rides is z^ =—^ 2[l+3-EI(c)]. When he has

gone once around the army, y = 0, and the distance he rides is z^ =-^ — £^^)1'

Good solution given by Henry Healon,

26.— Proposed by DbVolson Wood, M. A., C. E., Professor of Mathematics and Mechanlcfs Stevens In-
stitute of Technology, Hoboken, Hudson County, New Jersey.

A perfectly homogeneous sphere has an angular velocity (o about its diameter.
If the sphere gradually contract, remaining constantly homogeneous, required the
angular velocity when it has half the original diameter.

29

L— Solution by Waltbb Sivbbly, Oil City, Venango County, Pennsylvania.

Let r=radius of the sphere before, and ri=the radius after, contraction and
w' = the required angular velocity.

The moment of the momentum about the fixed diameter before contraction was
miPwy where m is the mass of the sphere and k is the radius of gyration, and after
contraction was mk'^'y and as no forces acted 6n the sphere during the interval of
contraction, mi^ = mk'V. {RoiUh^a Rigid Dynamics^ zd edition, Art. 129, p. 195.)

.•, tb'=z^-—-w. ButA2=^r2, k''^=—r^\ .-. ro'= — s^w = 4ro when n =Jr.
m«'^ 5 5 r^

This solution supposes the mass of the sphere to remain constant during con-
traction. .

Solved also by Prof, David Trowbridge^ the Proposer and F, P. McUz,

n.— Solution by Abtbmas Mabtii?, M. A., Erie, Erie County, Pennsylvania.

Let R be the radius of the sphere before contraction, r the radius after contrac-.
tion and wi the angular velocity after contraction.

The area described by the radius R in the unit of time is JiJ^f**, and the area de-
scribed by the radius r in the same length of time is ^Aoi . From the principle of

the coTiservcUion of areas we hsLVQir^iOi =iR!^(o] whence o>i =— „ {o,=4i(o when r:=^R.

Solved in a Bimilar manner by Henry Heaton,

26.— Pmopoeed by K B. Sbitz, Greenville, Darke County, Ohio.

Two equal spheres, radii r, are described within a sphere, radius 2r ; find the av-
erage of the volume common to the two spheres.

Solution by Hbnby Hbaton, B. S., Superintendent of Schools, Sabula, Jackson County, Iowa.

Let be the center of the sphere whose radius is 2r,
BED a section of a sphere whose radius is r and whose I
center is 0, and A the center of one of the small spheres.

Put a;=OA and y=distance between the centers of the
small spheres. Let BEF be a section of a sphere whose
center is A and whose radius is y.

If y<(r — x) the sphere BEF will lie wholly within
BED, and the center of the second small sphere may be I

on any point of the surface. If y>(r — x) and <(r+a:), I

the sphere BEF will intersect the surface of BED as in the figure, and the center
of the second small sphere must lie on that part of the surface of BEF which
lies within the surface BED. Draw BC perpendicular to OC, and join BA and

BO. Then since (BO)2=(AB)2+(AO)2+2CAX AO we have CA= '^~^^~-^ . Hence

y+- sz — )* A maybe on any point of the sphere

whose radius is x and whose center is O, x being less than r. The volume common
to the two small spheres when the distance between their centers is y is

t?=;rf -H-r^— r^+^ oy* )• The whole number of different positions of the two

spheres equals the square of the number of points in the sphere BED Hence the
average volume required

-ao —

'fi — a? — y2>

136
'315'

Solved also In an elegant manner by the Proposer.

27.— Proposed by Abtbmas Martin, M. A., Erie, Erie County, Pennsylvania.

The first of two casks contains a gallons of wine, and the second 6 gallons of wa-
ter. Part of the water is poured into the first cask, and then part of the mixture
is poured back into the second. Required the probability that not more than ^ of
the contents of the second cask is wine.

Solution by Hi>:nby Heaton, B. S., Superintendent of Schools, Sabula, Jackson Ck)unty, lovra.

Let z=number of gallons of water poured into the first cask, and y=number of
gallons of the mixturejpoured back into the second. Then the number of gallons

in the second cask=6 — x-\-y, of which — ^is wine. But — ^— must not exceed

^ a-\-x a-\-x

~~^~^y hence y must not exceed- ^\ \ Since the cask from which y gal-

Ions are taken contains a-\-x gallons, the probability that y does not exceed

, A — is 7 rr , and that x should have any particular value,-7— .

(n — V)a—x (n— l)a— 0?' . j i > f^

Hence the required probability is

^-Ir" (fc-^y^ . (n-l)a-ft / (n-l)a \

28.— Proposed bjr F. P. Matz, B. E., B. 8., Mathematical Editor National Educator^ Reading, Berks
County, Pennsylvania.

From a point taken at random in the left-hand half of the major axis of an el-
lipse whose minor axis is unknown, a circle is drawn at random, but so as to lie
wholly in the surface of the ellipse. Find the average area of the ellipse whose
major axis is that portion of the given major axis between its right-hand extremity
and the circumference of the circle.

Solution by E. B. Sbitz, Greenville, Darke County, Ohio.

Let ACBD be the ellipse whose major axis AB
is known, and minor axis CD unknown, M the
center of the random circle, NRBS the ellipse
whose average area is to be determined, RS its
minor axis, and MK the normal line to the ellipse.

PutOA=a, 0C=W7, AM=a?, MN=y, PR=2,

UDT

NP=r, MK=yi , xi= — = the radius of curva-

ture at A. Then we have v=^(2a — x — y),

yi = w?^ f 1 — ^2_y^ )> ^^d area NRBS=/T17s. The limits of z are and v; those

of y are and x when x is less than xi , and and yi when x is greater than xi ;

-31-

those of ;)• arc and a ; and of w, and a. Hence the required average is

/III ^^2d.r(lydz+ 1 1 I ~vzdxdydz \dw

J=-

1 r r ' r r <i"%«fe+ r fc dxdydzidw

But rTr'^ r C'ij:dy<h+C'' ^ f dxdydzidw

= ij"[P f{1ii-x~y)dxdy+j^^ ^\^-x-y)dxdy'Yw,
= \f\p {Anx— ^xyx+C [(2o— a;)2-(2a— ;c — y, )2]rfa;ldw,

= ^T' r3aw(a-2-w2)«cos-i ( '" \ +'2ahv-\-aic^—1w^+^~\dw=^^^a%57:+n).

J=

-2.^i0mflC />— ^)^'^-'^+/; £ i2a-.-yfd.dy]du,,

+ J^ [(2o— ar)* — (2a— a;— yi )*]d:e1dw,

== leaMtr+n/Il^^"^^"'^-'^'^)''^^^-^ I ) +136a.«,-129aV-128a%.^

_ ^<^^ / 2205.T+2012 \
""672 V 10-+17 A

92«/7« . lfiK?«-

29.— Proposed by E. B. Sbitz, Greenville, Darke County, Ohio.

Two points are taken at random in the surface of a given circle, but on opposite
sides of a given diameter. Find (1) the chance that the distance between the
points is less than the radius of the circle, and (2) the average distance between them.

Solution by the Proposer.

1. — Let AB be the given diameter, OD the radius |
perpendicular to AB. With A as a center describe
the arc OC, and from any point M in the surface
AOC and with a radius equal to AO describe the arc
PN. If M is one of the random points, and the
second point Q be taken anywhere in the surface
APSN, the distance between them will be less than I
the radius of the circle. Put 0A=1, OM=a?,
/_AOM=tf, zLONM=^, /.MOP=s^,areaAPSN=u.
Then a:=cosec sin ^=2cos^,

rf.7!=cosec d cos <pd<p:= — 2sin <pd(/fj and '
a=POR+PMS-POM+NMS-AOR— MON,

=^— sin ^ cos ^ — J^ — Jsin ^ cos ^+ Jcot (? sinV-

-32-

When M is anywhere in the surface ODC, the number of favorable positions of
Q is ui = J;r — ^ — sin ^ cos f . Hence the required chance is

I uxdxd0+ I I tixdx+ I Ui xdx dO

*>^ ^ l^irL>^ ^ 2C08« J

I I ^Kxdxdd
= — 5^ I I uxdxdff-\- ^ I I ttj:cb+ I ui xdx \d0.

^ J {i J Q '•- •/ i^»L*^ *^ 2oos« J

But when a; =0, f =0 and ^=J/t; when ar=l, ^=^ and ^=J;r; and when
a:=2cos/», f=T. — 2d and ^=1?.

8 /*^'r /*^'

. • . p= ^ j I ^^[} — sin ip cos 4") sin ^'^ cos (/'di/f

— I J( ^ + sin ^ cos ^ — cot sinV) cosec^^ sin <p cos ^cif dtf
8 /^M^r /^M^

— I Kf +^i^ ^ cos f — cot (? sin^ f) cosec^ ^ sin (p cos ^d^?
+ I (i^ — f — sin if cos f ) cosec^ d sin ^ cos ipdf dO^

= o^¥r (4;:— 4/?+<?cosec2<?— cot» — 3i/3)d»

+ ir^V^— 16 <?--;: cosec2<?+4flcosec2<? — 4cotff—8sintf COS tf)dtf, = § — ^-^

2. — Let M be any point in the semicircle ADB, and Q any point in the
semicircle APB. Put OA=r, OM=a:, OQ=y, MQ=z, Z_AOM=(?, ^AOQ=p,
Then 2=[r^4-y^ — 2rycos(/?+^)]^, and the average distance between the points is

I I I I zdOdifxdxydy

f f C Cdddifxdxydy '^^ -^ 0^ 0^ 0^

= -3|pr/'/'/'[8sinK<'+f )+40sinK<'+f ) cos2j((?+f )-48sinJ((?+^)cos*{<?+f )
+3cos2((?+f )-l+6sin2(<?+f )cos((?+f) log (^ JnTf^^f^f ^)1<^'^<^?'^^-«»

= ^r'[328inJ<?+32cosJ<?+16sinVcosJ^+16sinJ/?cos2j(?

— 24sin<J<? C08i<?— 24sini<? cos<i^+3sin«(? log tanj»+3sin»(? log tani(.T— ^)]cW,

1472r_

~ 135^^2 •

33 —

30.— Proposed by Abtbkas Martin, M. A., Erie, Erie County, Pennsylvania.

A Straight tree growing vertically on the side of a mountain was broken by
the wind, but not severed ; find the chance that the top reaches to the ground.

I.— Solution by E. B. Sbitz, Qreenville^ Darke County, Ohio.

Let OA represent the tree, AOM a vertical plane per-
pendicular to the mountain side, and intersecting it in
OM, and AON any vertical plane intersecting the mountain
side in ON. Take as the center of a sphere, and the
intersection of its surface with the planes AOM, AON
and MON will form a right-angled spherical triangle
BCD. When the wind blows m the direction of the
plane AON, let H be the point at which the tree must
break, that the top A may just reach the ground at K.

Put OH=a?, OA=a, /_AOM=J;r— ^, the complement

of the elevation of the mountain, diedral MOAN=f , /_AON=<f'. Then in the

cot u sec (D
spherical triangle BCD we have tan ^=cot ^ sec^p, whence sin ^= — TW2~'

In the right-angled triangle HKO we have HK=OHsin^, or a — x=a:sin^,

whence a;=--— ^ — j. Substituting the value of sin (f% we have

x=a[l+ cot^tf sec V — cot sec f |/(1 + cot^^ sec^f )].

X

When f <i?r, the required chance for a given direction of the wind, is — ; but when

a

!?>i^, the required chance is J. Hence, integrating first with respect to #, we find
or the chance required

"= — rr^ — ■ " * ^^•^« ^' ^'^"'■''"^

— cot sec f |/(1 +cot2tf 9Q(i^(p)'\d<pdd^
= .+ 21 \^ — ^ sec^^ — cot/? sec^f +sec f v (1+cot^^ sec^f )

dip,

1 2 X^'
= -7- + ^1 [ J7r(l — sec -f ) + if tan f sec f + sec f]dipj

1 2 r^ / ^ , , -1^' 3 2

^ 4 "^ ;r2"L*''-^~ ^^^^"^^^^^^Jo ^ 4 ^ ;r^ •

II.— Solution by Hbnrt Hbaton, B. S., Superintendent of Schools, Sabula, Jackson Ck>nnty, Iowa.

In the figure AOD is the side of the mountain. AB the stump of
the tree, ABC the plane passing through the tree perpendicular to
ACD, BAD the plane in which the tree is supposed to fall, BCD the
horizontal plane through B, and BE the perpendicular from B
upon AD. Puta=hight of tree, BE=a:, ^BAC=^ and /_CBD=/?.
If the length of the top=x, AB=a — x,

BD= ,f"-^^ „ BC=(«-x)tan,9and BD=fc^/^
l/{a^ — 2ax) \ / I cos # '

-84-

If tf<Jff and the top shorter than x, or if <?>|;r and <.ir and the top shorter
than ia, it can not reach the ground. Hence the required probability is
3 tan,9 fii'r ^, „, • mv , .1 <^

3 tanySr flf , o. -MX * ail^ t an ^ se c ^/'^ cos'^ sin^tfrftf
= 4 ^^L**n<?(,W^-sm¥)-tan^}J^ ^-J^ ^ (l-cos^^sin'^) '

= |~*^^-^ [j^(co8Ai:r)-l^(cosAi.].

If ^ is supposed to be unknown, the required probability is

3 2 r^ r^'ain^sinWddd^ 3 2 z*^^ • ^^^ 3 2
^= 4 - ;r2 J J v'(l-C08*^8in2<?)=T-l^J „ ^ "'^ ^'^^ 4 " t^ '

Cfta«. ^. JTummtfa gave an excellent Rolution on the supposition that the elevation of the mountain is Icnown.

81.— Proposed bj Abtcmab Mabtin, M. A., Erie, Erie Coonty, PennsylTaoia.

A sphere, radius r, and a candle are placed at random on a round table, radius
J2, the hight of the candle being equal to the radius of the sphere Find the ave-
rage surface of the sphere illuminated by the candle.

Solution by K B. Sbitz, GreenYille, Darke County, Ofaia

Let BCED be the table, A the point at which the
sphere touches the table, M the place of the candle, the
the center of the table. With A as a center describe the
arc DKE through M.

Put AM=x, OA=y, OD=JS, ^OAD=tf, ^AOD=f,
/_ADO=<&, w=the area of the surface of the sphere il-

luminated. Then cos(?= — -^ (1),

y8ind=R3\n(fi .... (4), y cos <p=I{—x cos <p .... (5), w=2;rr2f 1— V

Now if we regard x a constant, and less than JB, A may move over the surface
of the circle, center and radius JB — x, and for each position of A, M may move
over the circumference of the circle, center A and radius x, and the amount of the
surface illuminated will remain the same. But when x<R and y>R—x, or when
^>JB and y^x—R, for each position of A, M may move over the arc DKE, and
the amount of surface illuminated will remain the same. Hence the required av-
erage is

8=

I u.27CQsdx.27:ydy+ I I u.2dxdx.27n/dy+ I I u,20xdx.27rydy

r •^ J r^ R—x • ^ R^ x~R

f f 27ra:dx .27rydy+ C C 2dxdx.27n/dy+ ) ) 20xdx.27rydy

•/ r •^ J r*^ R—z *^ R^ x—R

But C2dydy=df— Cfdd. From (1) and (2\ xy sm ddO^—R cos fdy (6).

From(4)and(6),a;sin^d^=— cosfrfy . . . (7). From(3),JRa?sin^d^=ydy . . (8).

SB

From (7) and (8), yddz=—Rcoa<i>dii> . . . (9), and from (5) and (9),

fdd=:—B{R—zcos<p)d<l>. .-. C2dydy=ef+R/>—Itx sin </>.
Hence, observing that when y=B — x, O^tt and yJ=0; when y=x — B, (?=^=:0,
and when y^R, <?==^=cos-* (-sp)> ^e have

j^hmoa-^ (^—xi/(4IP-x^^xdx

r24iP'(i22+r2) cos-i f^\—r{2QIP+i^)i/{4:Ri—r^

=2;rr^ I

[^ 2iIP{R^—r^ cos-i ( 2**^ ) +3r(2 ft2+r2)|/(4i?2-r2)

-g=

=2^2-

]

32.— Propofled by K B. Sbitz, Greenville, Darke County, Ohio.

A radius is drawn dividing a given semicircle into two quadrants, and a point
taken at random in each quadrant ; find the average distance between them.

Solution by the Pboposbr.

Let ACB be the given semicircle, its center,
AOC and BOC the two Quadrants, M any point in
the quadrant AOC, A'C'B' the concentric semicircle
through M, and N any point in the quadrant B'OC
It is evidently necessary to consider only those posi-
tions of the two points in which N is confined to
the quadrant B'OC.

Put OM=a;, ON=y, OA=r, LCOM=0,
^CON=f. Then U^=[x^+y^—2xyco8{0+ip)y^; the limits of y are andx;
of X, and r: of f', and |;r ; and of d, and ^;r. Hence the required average is

I I I U^+jr^ — 2ay cos(»+f) M^^i^iwicydy
*^ *^ */ L A

I I I I dddfxdxydy
= ^J J f fl:^+f-2:i^yQ0s{d+<p)y^dd<fxdxydy,

= _!__ I^'^' J^' ^ [8sinJ(«+f)+40sinK#+f )co82J(#+^p)-48sinK<>+^)cosX<'+^)
4 3cos2(/?+sp)-l+6sin2(#+f)co8(tf+iP)log(-t^^^^^

-86-

16r /•5*' /•'^T

16r rJiT

lor /»•- .
= ^^J I 24cosi(?+32cos»J<?-24cos5i(?-3sin2tf-24cosJ{iT+«)-32cos«J(iT-|-^)

32r

135;r2

.[9V2-95+61og(-i^)].

33.— Proposed by R B. Skitz, Greenville, Darke County, Ohio.

Three points being taken at random within a sphere, find the average area of the
triangle formed by joining them.

Solution by the Proposer.

Let J be the required average, and Ji the
average area of the triangle when one of the
points is fixed on the surface of the sphere.
Then if x is the variable distance of one of the
points from the center of the sphere, the other
two points being confined to the concentric
sphere whose radius is x, we have

r^Ji (^:TcA\A7:T'dx ^
, Jor2 ^ V3 / 9 .

X(3-)

ATTOT^dx

'"-yC^

W^ y

Let O be the center of the sphere, A the fixed
point on the surface, B, C the two random
points within the sphere, ANEF the great circle
through B, AKNH the small circle through B, C, S its center, M the middle point
of AN. Put AB=a;, AC=y, AS=2, AO=r, ^BAO=^, ^OMS=f, ^CAB=v'',
/_SAM=(o. Then AM=rcos^=2cosro, SM=rsin /^ cosf =2sin w,

z=r (cos2 ^+sin2 tf cos*-^ (fY^, area ABC=|.ry sin ^'.

An element of the sphere at C is y^sin (/> dy(l(f^d<f . 'Izji? sin dxdtt ; the limits of y
are and 2zcos{</f''(o)=zi ; of ^, f«i-^;r=c«ii and (o-\-^:t=(02 ; of x, and 2rcosW=.Ti :
of d, and J;r ; and of ^, and tt. Hence, since the whole number of ways the

1 f*
two points can be taken within the sphere is -^^''^, we have

9r2

= j—^ I I 1 I ^cos^(f'' — w)Qmy'd(f&in0d0.7:^dxd(/\
= 7^1-9 I I I (cos^/^+6sin^tfcos''^^cosV+5sin^^cos*f)rf^sin#(Z(/r'^c/ar,

y» /• ^

^^To \ i (5cos*f + Ocos^f eos^^ — lOcos^f cos^/!^ — Gcos2^cos*/?
lb»/ 0*^

+ 5cos%cos^tf + cos*d)d4fsin0cos*ffdff^

37

1^ f^ 1 9 9 9

~ T12J (16+10cos2^+cos4f )d^r=^ ;rr2. .-. J=— Ji= -;rr2=-^ of the area of

a great circle.

34.— PuzB Problem. Proposed by Abtbmas Martin, M. A., Erie, Eri^ County, Pennsylvania.

A boy stepped upon a liorizontal turn-table, while it was in motion, and walked
across it, keeping all the time in the same vertical plane. The boy's velocity is sup-
posed to be uniform in his track on the table, and the motion of the table towards
nim. The velocity of a point in the circumference of the turn-table is n times the
velocity of the boy along the curve he describes Required — ^the nature of the
curve the boy describes on the table, and the distance he walks while crossing it (1)
when n is less than 1, (2) when n=l and (3) when n is greater than 1.

Solution by K B. Seitz. Greenville, Darke County, Ohio.

Let AMN represent the turn-table, and let MN be the
intersection of the fixed vertical plane with the table,
the center of the table, A the point in the circumference
at which the boy stepped upon the table, P his position
at any time while crossing it, AP the corresponding arc
described by the boy, OB the perpendicular on MN, the
motion of the table being from N towards M.

Put OM=a, OP=r, 0B=6, BM=c, arc AP=s,
/_AOP=», ^AOM=f, ^OPN=/^, ^OMN=/?.
Then arc AM=n . arc AP, or a<f=nz . . . (1),

ip=d—[x+^ .... (2), r=6cosec/>e . . . (3), dz^=dr^-\'T¥^^ . , . (4). Ditibreii-

tiating(l),(2)and(3),wefinddf=^cfe (5), dif=dd—d/i (6),

dr=—b cosficosec^fidfji (7). From (5) and (6), dz= — {dO—d/i) . . . (8).

Substituting the values of r, dr and dz from (3), (7) and (8) in (4), and solving for
dd, we find

• I.— When n<l, by integrating (9), taking the double sign +, we have

'^ 2(a2-n%=«)« **^ V(o2-n262)'^-n6cot/J + J sinix{ahin^n-n^b') ^ ''
Putacos/'=(a'^+n262)>^os// .... (11). Then we have

J sin/i(a%in2//-n2ft2) " '-^ ^ > ^J (ah\n'4'^n'b^){a*sivi\>-n*b*)

Substituting in (10), then substituting the values of the functions of ft and ^,
determined from (13) and (11), and observing that ^=0 when r=^a and /i=,'9, we
have for the polar equation to the curve

<?=8in-> ( * )-sin-» ( * ) +cos-» (± ^ (r2-/>2)>^) - cos-' ( "^^ )

_ nb _ / [(a*-n^6')>^(a'-nV )H-H-^6c] [( «^-n''6» )H ±w(r^-6^)^ \

(a''-«%2)H^**S V[(a2-n2A2)^+«c][(a2-n*)H(a2+n26-'^-M2r2)5<+n26(r*-6«)H]^ ' ^ ''

38

the sign ±: being taken + when /^<j7r, and — when //>!?:; and the sign h= being
taken — when /i<i;r, and + when //> J;r.
When r^=a and /i=;r — ^, from (1), (2) and (12) we we find

_ 2a . _, / ne \ 2ab . ( a{a'^—nW)'>^-\-iMW, \

^1— „ sin \^)+ (a2--n252)>^^^^ V(a^~n%2)^(ai-nV)>i--nW ' ' ^^^'
the distance the boy walks while crossing the table.

lb J — cos-^(n), which is the equation to a
circle, radius ^ , passing through O and A ; hence if the boy walks across the
center of the table, he will describe the arc of a circle, whose length is sin~^ (n).

II. — ^When n=l, the logarithmic part of (12) becomes infinite, which shows that
it is impossible for the boy to cross the table in a plane that cuts the table on the
side from which it is moving. But if the vertical plane cuts the table on the side
toward which it is moving, we must substitute —6 for 6, in (12), and then making
n=l we find

fl=eos-^(±^— ^^--)-sin-^(-)+ ,log(-^^.+j.^;)^j-(^ (14)

Substituting —b for 6, and making n=l, in (23), we have

zi =2asin-^ ( — j + log ( j, the distance the boy walks.

If 6=0, (14) becomes (?=cos~^ ( ± — V which is the equation to a circle on OA

as diameter; hence if the boy walks across the center of the table, he will describe
the circumference of a circle, and will leave the table at the same point at which
he stepped upon it

III — When n>l, the vertical plane must cut the table on the side toward which
it is moving, in order that the boy may cross the table Hence, if we substitute
— 6 for 6 and —fjt for /i in (9), and take the double sign — , we have

c/#=. °:^f /% - ^-j^^—^^ v^[a'-(a«+n*6«)cosV] . . (15).

1. When /9 is not greater than 45°, or when /9 is not less than 45° and n<-7- .

we have by integrating (15) for the equation to the curve

<?=9in-i (A^-sin-' ( *-) +co8-» (± ^{r^-b^A-co8-' (^ )

tA / [n»6c+(o'-n»6'0?^a'-nV)>f|[w(r»-fe')X±(o«-n»6^)H] \

"^(a«— n*6'')« '^^ \[nc+(a='-n«6=')H][n='6(r*-6«)«±(a«-n«6«)J<(a«+n%*-nV)H]j ' ^ ^'

If ^<4o°, n can not be greater than- . When ri=a and ft^n — /9, from (1), (2) and

na\ a A 2a. ./o\, 2ah . /n*6c+(o«-n*6*)H(o«-nV)}<\ ,-„

(16) we find ., = - sm- (J + ^^,_^,^,^ log (--A_._!^__^L^ (17),

the distance the boy walks.

2. When ^ is not less than 46° and n> -j-, we find by integrating (15) for the

equation to the curve.

»=-- (v)-™- (1) +--'(* ^c--'^")—- (v)

+

(n'b^^a?)M ^^^ \- n(n^6'^-a2)H(r«;~62)>^[n6--(aHn%'.r-nV2)K] j

in which n can not be greater than — . From (1), (2) and (18), when r^=a and

/i=;r— /?, wefind

2a . , /wc\

zi = sin~^ ( - I

n \ a J

the distance the boy walks.

3. When ^ is not less than 45® and n=~, (15) becomes

dd=:i8in^fidfjt — (1 — 2cos^/^)^ecV cosec/id// (20).

Integrating (20), we find for the equation to the curve

-«"*"( 26^) +-*"(^!2^=;^) (21).

From (1), (2) and (21) we find

..=2atan-(^^_^)_f-+4(263-a^)«. . . (22).

the distance the boy walks.

In II and II the curve is convex toward the center of the table, and the motion
of the table is not toward the boy for all positions in the curve.

Solved also by Henri/ HeaUm,

SOLUTIONS OF ''UNSOLVED PROBLEMS,"

Published in No. 1.

14. A cylindrical cask, radius r inches and depth a inches, is full of wine.
Through a pipe in the top water can be let in at the rate of h gallons per minute,
and through a pipe, radius r inches, in the center of the bottom the mixture can
escape at a faster rate when the cask is full. If both pipes be opened at the same
instant, how much wine will remain in the cask at the end of t minutes, sup-
posing the two fluids to mingle perfectly ?

Solution by nxVoLsoN Wood, M. A., G. E., Profesaor of Mathematics and Mechanics, Stevens Institute
of Technology, Hoboken, Hudson County, New Jersey.

Let V=;rajB2, the volume of the cask; a=its altitude; g=:^=number of gal-
lons of inflow per second; A=the contracted section of the issuing vein: K=^7zll!^,

the cross-section of the cask ; §=the quantity of liquid, both water and wine, at
time t; w?=the quantity of wine at that time; :r=the variable head or depth of

liquid in the cask, and A=k^( ^ ). Then ^=-^r, and the quantity of out-
flow in an element of time is ki/{2gx)dt^=ky/{2g)^ ( "^ ) ^' "^ AQ^t, and the
quantity of inflow being qdt in the same time, we have — dQ=qdt — -4§^ . . (a).

w

The quantity of wine which flows out in an element of time will be ^ of the

liquid which flows out in that time. Hence dw=— ^-^ dt,=^— y^dt .... (6).
To integrate (a) let q — AQ^=y, and we have

Substitute dt from (a) in (6) and we have

= ^ . i^t <2^=r, then integrating and

observing that for «?= F, ^= V=^r^, we have

w _( q-AQ^ \ « . ^_[{AV ^-q)w+qr^f
V~\q-AV^J ' ' • ^~ A^V^

and this value in the equation (c) gives a direct relation between t and w.

To find the limit of Q, make w=0, and we find Q=~3^^.^ and this value in
equation (o) gives <=oc as it should.

20 Three points are taken at random, one in each of the sides of a given tri
angle ; find the chance that the triangle formed by joining them is acute.

• Solatlon by K B. Sbite, Greenville, Darke Ck>unty, Ohio.

Let ABC be the given triangle, M, N, P the random
points Draw AD perpendicular' to BC, MS and MT
perpendicular to AM and BC, and MR to MN. When I
the random point in BC falls on BD, as at M' drawl
M'S', M'T'and M'R' perpendicular to AM', BC and
M'R If N falls on CS, the angle M of the triangle
MNP will be obtuse if P falls anywhere on AB ; and I
if N falls on ST the angle M will be obtuse if P falls |
anywhere on BR.

Put CM=2;, CN=y, BR=«, CS=u, CT=v, BM'=;r', BP=y', GR'=z\
BS'=u', BT'=< CD=m, BD=n, cotAMD=<, cotAM'D=<', cotC=<i , cotB=fe .

tx i^x'

Then we have u = - — .^ , . ■^ , v=x secC, x = 6(cosC — t sinC), w' =-, — p , . p ,
^cotC+sinC ^ rcosB+smB'

r' =a;'secB, a;'=c(cos B— <'sinB), da?= — fc sin Cdt, dx^= — c sinBcft',

ysinC : x — ycosC :: a — x — «cosB : zsinB, whence

(a — x)z sinB sinC (a — x)cosC

z=

eosA(a;co8B+ycosA) cosA

-41-

^. .• ■ /» ■; , (a — rr'VsinBsmC (a — xOcosB

Similarly we find r= . . , ^ , — ; rr ^^ 1 •

•^ cosA(a;' cosC+y' cos A) coaA

Hence the chance that the angle M is obtuse is

5sinC /^ncosBsinC cosBcosC , sin^Bsin^C,^ , . .,. ^., /^+taiiC\"| .^

— r-J,L-^SA- ssj— + -5SA-<'+''*'-'"<«(.-rHr)J'*

+ J(3+2cotA tanC) tanK) log sinC— Jtan^A log sinA.
The expressions for the chances. No and P© , that the angles N and P will be ob-
tuse, are similar to that for Mo . Hence, since a triangle can have but one obtuse
angle, the chance that the triangle is obtuse is Mo +No +Po , and the chance that
it is acute is

;>=l-(Mo+No+Po), ■

= i (^+ ^sinBstnct ) ^^^^^ ^"^ cosecA+ 1 (5+ ^|^|^) ^n^B log cosecB
. 1 /^ . 2sinCtanC\^ „^, ^ /cosAcosB . cosAcosC . cosBcosCx

1/1 1 J; \^^ 1 / COSC CQSB COSA \ to

2 \ cosA cosB cosC / 3 VcosAcosB cosAcosC cosBcosC/
Or.— When a=b=e, ;>=y (27 log 3 -?).

28. Find the average distance between two points taken at random in the sur-
face of a given rectangle, one on each side of a diagonal.

Solution by Abtemas MARTiif, M. A., Erie, Erie Ck>unt7, Pennsylyania.

Let M be the mean distance required, Mi the mean distance between two points
taken indiscriminately in the rectangle and M2 the mean distance between two
points taken both on the same side of a diagonal.

Then ( WilUamsan'a Integral CalcubiSy p. 305,)

M=2Mi-M2 (1).

But ( Williamson^a IrUegral Calculus^ p. 348,)

and {Edueadonal Times Reprkd, vol. xxiii, p. 92,)

M2 = j^n-j+ Vco3B+sin^logcotJB)+f -^+ y VcosA+sin'AlogcotiA) ,

Substituting in (1),

-42-

Cbr.-If a=b, M=^6+(16-|/2)log(l+»/2)].

This problem was solved in a rery elegant manner by JB. B, SeUz,

LIST OF CONTRIBUTORS.

Solutions of the Problems published in No. 1 have been received as follows :

¥hrcmiE.B.8BiTz, Greenville, Ohio, 8, 12, 13, 191, 93,28, 24. 26, as, 29,80,81,82,88 and 84, and 14. 20 and 28 of **Uo-
solved Problems ; " B. F. Burleson, Oneida Castle, New York, 1, 2, 3, 4, 5, 6, 7, 8. 0, 10, 11, 12, 18 and 14; Jakbs
McLAUOHiiiK, Mantorville, Minnesota, 2, 4, 6, 7, 8, 9, 10, 11, 12 and 18; WiiiLiAM Hoovbb, Bellefontaine, Ohio,
1, 2, 8, 4, 8, 9, 10, 14 and 20 ; K. 8. Putnam, Rome, New York, 1, 2, 8, 4, 5, 8, 8 and 15 ; Dr. S. F. Bachbldkb
South Boston, Massachusetts, 1, 2, 4, 8, 10, 11, 12 and 18; Henry Heaton, B. S., Sabula, Iowa, 15, 24, 25» 26, 80
and 84; F. P. Matz, B. E., B. 8., Reading, Pennsylvania, 2, 8, 10, 14, 20 and 25; Marcus Baker, U. 8. Coast
Survey Office, Washington, D. C^ 1, 8, 8, 9, 10 and 14; Miss Christine Ladd, B. A., Union Springs, New York,
4, 7, 9 and 17; O. D. Oathout, Read, Iowa, 1, 2, 8 and 10 ; W. L. Harvey, Maxfleld, Maine, 1, 2, 5 and 6 1
David Wiokebsham, Wilmington, Ohio, 7, 9, 11 and 12; Prof. David Trowbridge, M. A., Waterburg, New
York, 7, 16 and 25; G. M. Day, Lockport, New York, 16, 17 and 20; D. J. McAdam, Washington, Pennsyl-
vania, 10, 19 and 20; Walter Sivekly, Oil City, Pennsylvania, 22 and 25; Dr. 8. H. Wright, M. A., Ph.
D., Penn Yan, New York, 12 and 19; Prof. DeVolson Wood, Hoboken, New Jersey, 25, and 14 of "Unsolved
Problems ;" Mrs. Anna T. Snyder, Orange^ Indiana, 2 and 6 ; Theo. L. DeLand, U. 8. Treasury Depart-
ment, Washington, D. C, 5 and 6; 8. C. GK>ULD, Manchester, New Hampshire, 8; Prof. W. W. Johnson,
Annapolis, Maryland, 24; Dr. David 8. Hart, Stonington, Connecticut, 21; D. W. K. Martin, Webster,
Ohio, 2 ; £. P. Norton, Allen« Michigan, 18 ; I. H. Turrell, CumminsviUe, Ohio, 18 ; H. T. J. Ludwick,
Mt Pleasant, North Carolina, 16 ; Qborge Eastwood, Saxonville, Massachusetts, 19 : Chas. H. Kummbll,
Detroit, Michigan, 80.

^

JUNIOR PROBLEMS.

36.— Proposed by Oblando D. Oathout, Read, Clayton County, Iowa.

A and B propose to trade horses. A asks $20 "to boot," and B asks $10. They
finally agree to "split the difference " How much "boot money" should A receive?

36.— Proposed by J. B. Sandebs, Bloomington, Monroe County, Indiana.

A merchant bought 400 yards of cloth at $4 per yard, payable in 3 months, and af-
ter holding it for 15 days sold it at $4.25 per yard, receiving therefor a note paya-
ble in 4 months. When the purchase money became due, he had this note dis-
counted at the bank to meet it. What did he gain by the transaction ? What
would he have gained if he had borrowed at 6 per cent, interest, until the matur-
ity of the note he had received, sufficient to pay for the cloth, apd why should
there be any difference in the results ?

87.— Proposed by B. F. Bublbson, Oneida Castle, Oneida County, New York.

A parsimonious farmer has a board 6| feet long and 2 feet wide which he wishes
to cut and form .into a square without any waste of lumber. How must he cut the
board and arrange the pieces?

38.— Proposed by Prof. Edward Bbookb, M. A., Ph. D., Principal Pennsylvania State Normal School,
MillersTille, Lancaster County. Pennsylvania.

A man gave me his note for $500 payable in 8 years, interest at 6 per cent.; if he
pays the interest annually, to what rate is this equivalent if the same amount of
interest had been paid at the end of the time?

43

30.— PropoBed by W. B. Batk, EarlviUe, LaSaUe County, lUinoia.

If Dr. A kills 3 patients out of 7 ; Dr. B, 4 out of 13 ; and Dr. C, 5 out of 19,
what chance has a sick man for his life who employs all three of these doctors at
the same time ?

40.— Proposed by J. O. Walton, Coylngton, Kenton CJounty, Kentucky.

Two circles have the same center. If a chord, length 2a, be drawn across the
larger circle, it will just touch the circumference of the smaller. Required the area
of the ring.

4 1 •— Proposed by Abtemas Martin, M. A., Erie, Erie County, Pennsylvania.

Find X and y from the equations

42.— Proposed by Benjamin Headlet, DiUsborough, Dearborn Ck>unty, Indiana.

Cut a board 3 feet wide and 7 feet long into three pieces, so that they will make
a square.

48.— Proposed by Abteicas Mabtin, M. A., Erie, Erie Ck>unty, Pennsylvania.

Two men, A and B, plow a rectangular field a rods long and b rods wide, A go-
ing ahead all the time. What part of the field does each plow, estimating n fur-
rows to the rod ?

__ . sedbyV

Ashland County, Ohio.

From a given point without a circle, to draw a secant that shall be bisected by
the circumference.

46.— Proposed by Mrs. Anna T. Snyder, Orange, Fayette County, Indiana.

A father having 640 acres of land in a circle gives to each of his six sons a cir-
cular farm touching the circumference of the circle and also the circumferences of
the farms of two of his brothers, the six farms being equal in size. Required the
number of acres in the farm of each of the sons, the number remaining to the
father, the number remaining in the center and the number in each of the other
six equal pieces.

46.— Proposed by Pn)f . J. F. W. Scheffbr, CoDege of St James, Washington County, Maryland.

The sides of a plane triangle are proportional to the roots of the cubic equation

a?*— oo^+fta?— c=0.
Find the sum of the cosines of the angles of the triangle.

47.— Proposed by Daniel Kirkwood, LL. D., Professor of Mathematics, Indiana State Uniyersity,
Hloomington, Monroe County, Indiana.

The number of diagonals that can be drawn in a certain polygon is equal to 7
times the number of sides. How many sides has the polygon?*

48.— Proposed by Joseph Ficktjn, M. A., Ph. D., Professor of Mathematics and Astronomy, University
of the State of Missouri, Columbia, Boone County, Missouri.

Find the radius of a sphere that shall circumscribe four equal spheres which
touch each other.

49.— Proposed by Dr. 8. F. Bachbldeb, South Boston, Massachusetts.

P and Q owned a triangular piece of land, the sides of which were 20, 18 and 4
chains. The longest side bordered on a road which ran east €tnd west. They
agreed to divide it by a line parallel to the shortest side so that Q, who took the re-
sulting trapezoid, should have f of the land. A surveyor located the line ; then
P, setting a stake at the middle of it, proposed that a new line should be run north
and south through that point, so as to give both a square corner on the road. Q
assented ; did he grain or lose, in area, by the change of line, and how much?

44

50 — Proposed by Mabcus Baker, U. 8. Coast Survey Office, Washington, D. C.

In any plane triangle ABC the angles B and C are bisected, the bisectors meeting
the sides in B' and C respectively. Join B' and C and from any point P in B'C
let perpendiculars, pa , Pb and pc , fall upon the sides a, b and c respectively ; prove
that p^=pf,-\-pc,

51.— Proposed by Artemas Martin, M. A., Erie, Erie Ckiuniiy, Pennsylvania.

There is a series of right-angled triangles whose legs differ by unity only. Call-
ing the one whose sides are 3, 4, 5 the fird triangle, it is required to find general
expressions for the sides of the nth triangle, and compute the sides of the 100th
triangle.

52.— Proposed by James McLaughlin, Biantorville, Dodge Ckiunty, Minnesota.

Three circles, radii a, 6, c, are drawn in a triangle, each circle touching the other
two and two sides of the triangle. Find the sides of the triangle.

53.— Proposed by Artemas Martin, M. A., Erie, Erie County, Pennsylvania. '

The towns C and D are 40 miles apart. A set out from C to travel to D at the
same time that B started from D to go to C. A overtook a drove of sheep moving
forward at the rate of 2 miles an hour just 10 minutes after he crossed a creek
known to be 10 miles from C ; B arrived at C 3 hours and 55 minutes after he met
the same drove. B was overtaken by an express proceeding at the rate of 5 miles
an hour just 5 minutes before he came to an inn 12 miles from D ; A arrived at D
8 hours and 20 minutes after he met the express. Required the hourly speed of A
and B.

54.— Proposed by E. J. Rowan, Shawnee, Perry County. Ohio.

What length of line, fastened to a point in the circumference of a circle whose
area is one acre, will allow an animal to graze upon just one acre outside of the
circle ?

55.— Prize Problem. Proposed by Artemas Martin, Erie, Erie County, Pennsylvania.
Gi\eny=x+ax^+br^+C2!*+dafi+€a:f^-^fx'+ga^+h^+ix^^+jx^^

+kx^^+lx^^+mx^*+nx^^+px^^+&c. . . (1),
and x=y+Ay^+Bf+Oy'+I)f+Ef+Ff+Gh/'+Hf+Iy''+Jy''+Ky'^

+Ly'^+My'^+Ny'^+Py'^+&c. . . (2);
to find A, B, C, D, E, F, (?, JZ, J, /, K, L, M, iV, P in terms of a, 6, c, d, ej, g, A, i,
J, k, /, m, n, p.

Solutions of these problems should be received by September 1, 1878. Ten copies of the Visitor will be
given for the best, complete, correct solution of the prize problem, and eight for the second best solution.

'-^^^''^^^^^s^^s^-

SENIOR PROBLEMS.

56 —Proposed by George Eastwood, Sazonville, Middlesex County, Massachusetts.

In a planie. the equation of a straight line in terms of the perpendicular (p) from
the origin and the angle iff) which it makes with the axis being ysiiiO-\-xQO^O-=p\
prove tLat the same form holds for the equation of a great circle of the sphere,
when Xj y and p are put for tan a?, tan y and tanp.

57.— Proposed by Oscar H. Merrill, MannsviUe, Jefferson County, New York.

Prove that the cube of any given number is greater than the product of any
other three numbers whose sum is throe times the given number.

4B

58.— PropoBed by Mias Christins Lapd, B. A., Pk-ofessor of Natural Sciences and Chemistry, Rowland
School, Union Springs, Cayuga County, New York.

An ellipse and a parabola have a common focus and the other focus of the el-
lipse moves on the directrix of the parabola. Show that the points of contact of a
common tangent subtends a right angle at the common focus.

59.— Proposed by Dr. S. H. Wright, M. A., Ph. D., Mathematical Editor Yates County Chnmide, Penn
Yan, Yates Coonty, New York.

When the sun's declination was 3 north he rose ji^ farther south than when his
declination was S' north. Required my latitude.

60.— Propofljd by axoROE Lillet, Kewanee, Henry County, niinois.

Find the minimum eccentricity of an ellipse capable of resting in equilibrium
on a perfectly rough inclined plane, inclination /9.

61.— Proposed by Mies Christine Ladd, B. A., Professor of Natural Sciences and Chemistry, Howland
Sdiool, Union Springs, Cayuga County, New York.

If ABC be a triangle inscribed in a conic, P'P"P'" the points in which its sides
\neet tlie directrix of the conic, and Q'Q"Q'" the poles of focal chords through
FT"?'", then will AQ', BQ", CQ'" meet in a point.

62.— Proposed by F. P. Matz, B. E., B. S., Mathematical Editor National Educator, Reading, Berks
County, Pennsylvania.

In an equilateral triangle ABC lines are drawn as in Problem 8 ; find the ave-
rage area of the equilateral triangle formed by the intersections of the lines.

63.— Proix)sed by Isaac H. Turrell, Cumminsville, Hamilton County, Ohio.

Within the space enclosed by three given circles which touch each other external-
ly it is required to inscribe, geometrically, three circles each of which shall touch
the other two and also two of the given circles. Analytical solutions also desired.

64 — ^Proi^osed by J. J. Stlybster, LL. D., F. R. S., Corresponding Member of the Institute of France,
Professor of Mathematics, Johns Hopkins University, Baltimore, Maryland.

If there be two equations in x, (which for greater simplicity may be supposed to
be of the same decree n) find the most general form of M=a, rational integral func-
tion of the coeflScients of these equations such that ifo, Mx^, ]l&^-^ shall

each of them also be rational functions of the same.

65.— Proposed by Dr. David S. Hart, M. A., Stonington, New London County, Connecticut.

Find six square numbers whose sum is a square, and the sum of their roots plus
the root of their sum a square.

66.— proposed by Benjamin Pbircb, LL. D., F. R. B., Professor of Mathematics, Harvard University, and
Consulting Geometer to the U. S. Coast Survey, Cambridge, Middlesex County, Massachusetts.

What are the probabilities at a game of a given number of points, but at which
there is only one person who is the actual player? When the player is successful
he counts a point, but when he is unsuccessful he loses all the points which he has
made and adds one point to the score of his opponent.

67^— Proposed by Dr. Josl E. Hbndbicks, M. A., Editor of the Analyst^ 0es Moines, Polk County, Iowa.

Suppose the earth to be an indefinitely thin spherical shell, equatorial radius,
rotary velocity and gravitating force the same as at present except that gravity is
assumed to be all concentrated at the center, and none in the shell ; and suppose a
particle at the equator, and on the inside of the shell, to be separated from the
shell, and to move henceforth from its centrifugal force, resulting from its motion
while attached to the shell, and from the gravitating force at the earth's center.
Re(^uired the axes of the ellipse the particle will describe, and the time required
for its return to the same point in space at which it was detached from the shell.

68.^Propo8ed by Prot David Tbowbridob, M. A., Waterburg, Tompkins County, New York.

A sphere is divided at random by a plane, and then two points are taken at ran-
dom within the sphere ; find the chance that both points are on the same side of
the plane.

46

69.— Proix)sed by DandelEIirkwood, LL. D., Professor of Mathematics, Indiana State UniverBity, Bloom-
ington, Monroe County, Indiana.

Given the radius of Mars, 2250 miles, and the radius of the orbit of its inner
satellite, 5800 miles ; to determine whether the latter can have an elastic atmos-
phere, supposing its diameter to be 45 miles, and its density equal to that of the
primary.

70.— Proposed by John H. Adamb, Cochranton, CJrawford CJounty, Pennsylvania,

In digging a well 6 feet in diameter a log 3 feet in diameter was found lying di-
rectly across the center of the well. How many cubic feet of the log must be re-
moved from the well ?

71.— Proposed by Benjamin Peibcb, LL. D., F. R H., Professor of Mathematics, Harvard University, and
Ck>nsulting Geometer to the XJ. S. Ckutst Survey, Cambridge, Middlesex County, Massachusetts.

Given the skill of two billiard players at the three-ball game, to find the chance
of the better player gaining the victory if he gives the other a grwnd discount.

— [From Our Skhoolday VUitor^ vol. xv, p. 220.
72.— Proposed bv D. J. McAdam. M. A.. Professor of Mathematics, Washington and Jefferson College,
Washington, Washington County, Pennsylvania.

The center of an epicycle whose radius is ^^ » revolves on a deferent, radius a,
with uniform angular velocity v ; a particle revolves in the epicycle so that the ra-
dius drawn in the small circle to the particle is always parallel to itself. Supposing
the particle to be moving under the influence of a force at the center of the large
circle, find the law of the force.

73.— Proposed by Artemas Mabtin, M. A., Erie, Erie County, Pennsylvania.

Find the average distance between two points taken at random in the surface of
a given semicircle.

74.— Proposed by Rev. W. J. Wright, 31 A., Ph. D., Member of the London Mathematical^Society, Car-
lisle, Cumberland Uounty, Pennsylvania.

Transform the Eulerian integral r(n)= j e-^x^-^dx to the Legendreian,

J

75.— Proposed by Prof. H. A. Wood, M. A., Principal Coxsackie Academy, Coxsackie, Greene Co., N. Y.

A sphere 4 inches in diameter, specific gravity 0.2, is placed 10 feet under water.
If left free to move, what will be its velocity at the surface of the water, and what
will be the maximum hight it will attain ?

76.— Proix)6ed by J. J. Stlvbstsr, LL. D., F. It S., Corresponding Member of the Institute of France,
Professor of Mathematics, Johns Hopkins University, Baltimore, BCaryland.

Prove that the eo nation to the sums and the equation to the products of the roots
of an equation of the nth degree taken two and two together may each be put un-
der the form of a determinant of the order i(n—l) or of the order Jn according as
n is odd or even, and wjrite down the determinant of the third order which equated
to zero is the equation (of the degree 21) to the binary products of an equation of
the 7th degree clear of any irrelevant factor.

77.— Proposed by E? B. Ssrrz, Greenville, Darke County, Ohio.

Two points are taken at random in the surface of the quadrant of a circle, and
a line drawn through them. Find the chance (1) that the line intersects the arc in
two points, (2) that it intersects the arc in one point, and (3) that it does not inter-
sect the arc.

78.— Proposed by DbVolson Wood, M. A., C. E., Professor of Mathematics and Mechanics, Stevens In-
stitute of Technology, Hoboken, Hudson County, New Jersey.

One end of a fine, inextensible string is attached to a fixed point, and the other
end to a point in the surface of a homogenous sphere, and the ends brought to-
gether, the center of the sphere being in a horizontal through the ends of the
string, and the slack string hanging vertically. The sphere is let fall and an an-

47

gular velocity imparted to it at the same instant, the sphere winding up the string
on the circumference of a great circle until it winds up all the slack when it sud-
denly begins to ascend, winding up the string, the sphere returning just to the
starting point. Required the initial angular velocity, the tension of the string
during the ascent of the sphere, the initial upward velocity of the center of the
sphere, and the time of the movement.

79.~-Propo8ed by Artemas Martin, M. A., Brie, Erie County, Pennsylvania.

Find the average distance of the center of an ellipse, axes 2a and 26, from its
circumference.

80«->Fropo8ed by E. B. Seitz, Greenville, Darke Ckiunty, Ohio.

A chord is drawn through two points taken at random in the surface of a circle.
If a second chord be drawn through two other points taken at random in the sur-
face, find the chance that it will intersect the first chord.

81.— Proposed by Artemas Martin, M. A., Erie, Erie County, Pennsylvania.

A fixed hemispherical bowl, radius R, is full of water and contains a heavy cyl-
indric rod, length 2a, radius r and specific gravity /?, having one end against its
concave surface, and resting on its rim. Determine the inclination of the rod.

82.— Proposed by John M. Wilt, M. A., Fort Wayne, Allen County, Indiana.

An unknown cone is cut at random by a plane; find the chance that the section

is an ellipse.— From the Normal MoniMy, vol. ill, p. 108.

83.— Proposed by Artemas Martin, M. A., Erie, Erie County, Pennsylvania.

Find the average distance of one corner of a rectangular solid, edges a, 6, c, from
all points within it.

84.— Proposed by E. P. Norton, Allen, Hillsdale County, Michigan.

A fox starts from a point 100 rods due north from a hound, and runs due west
at the rate of 10 miles an hour; at the same instant the hound starts in pursuit, at
the rate of 15 miles an hour, always keeping in a direct line between his starting
point and the fox. Required the equation of the curve described by the hound,

and the distance he runs to catch the fox.— From the Normal Monthly, vol. lU, p. 85.

85 — ^Proposed by J. J. Sylvester. LL. D., F. It S., Corresponding Member of the Institute of France,
Professor of Mathematics, Johns Hopkins University, JBaltimore, Maryland.

Let X be the coefficient of any power of a in the expansion of

1

{l—xa){\—a?a){l—a?a) .... (1— a:%)
where i is any integer, and let X be arranged according to the powers of x ; prove
that its coeflBcient will form a series reading the same from left to right as from
right to left, possessing this property that as we pass from either end toward the
central term or terms the coefficients may increase or remain unaltered but can
never decrease.

86.^Propo8ed by E. B. Seitz, Greenville, Darke County, Ohio.

Find the average distance between two points taken at random Vithin a rectan-
gular solid, edges a, 6, c.

87.— PrijRe Problem. Proposed by Artemas Martin, M. A., Erie, Erie Couty, Pennsylvania.

A cask containing a gallons of wine is placed upon another containing h gallons of
brandy. Water runs in at the top of the wine cask at the rate of m gallons per
minute, the mixture escapes into the brandy cask at the same rate, the mixture
in the brandy cask escapes at a like rate into a tub containing c gallons of water
and the tub overflows. Find the quantity of each fluid in the tub at the end of
t minutes supposing them to mingle perfectly.

SolatloDS of these problems should be received by September 1, 1878.

48

UNSOLVED PROBLEMS,

By Abtebcab Mabtin, M. A^ Erie, Pa.

31. Solve the equations

cy'{b+e+x){a+c+y) — cy^{a+b+c){c+x+y)=2i /al)xy,

bi/{b+c+x){a+b+z) — bi/{a+b+c){b+x+z)= 2i/acaa,

ai/{a+c+y){a+b+z)—ai/(a+b+e){a+y+z)=2}/b(yz,
by quadratics.

32. Two rods, lengths 2a and 26, have their middle points connected by a string,
length c. If they be thrown on a level floor, what is the chance of the rods being
crossed ?

33. One end of a fine string, length I, is attached to a wooden sphere, radius r
and specific gravity />, and the other end is fastened to a point in the bottom of a
river, depth a and velocity v. Determine the position of the sphere.

34. A cubic equation is written down at random ; what is the chance that its
roots are the sides of a real triangle?

35. A cask in the form of a frustum of a cone, radius of top r inches, radius of
bottom R inches and depth a inches is full of water. Required tne time of emptying
through a pipe in the bottom, radius m inches, the quantity of water in the cask at
the end of any time <, and the velocity of discharge. A rigorous solution from
first principles is desired.

36. What is the chance that the roots of a biquadratic equation written down
at random are all real ?

37. The first of two casks contained a gallons of wine and the second contained
b gallons of water. Part of the water was poured into the first cask, then part of
the mixture was poured into the second, and part of the mixture in the second
poured back into the first. Required the probability that not more than -^ of the
(Contents of the first cask is wine,

38. A sportsman saw a duck in a circular pond, which dodged behind a tree

f rowing in it. The sportsman ran along the edge of the pond, but the duck kept
ehind the tree all the time and swam towards the shore. Required the equation
of the curve the duck described, and the distance it swam to reach the shore.

39. Two men start from the same side of a square field and walk across it in
random directions. Find (1) the chance that botn men will cross the opposite side
of the field and (2) the chance that their paths will intersect within the field.

. 40. The first of two casks contains a gallons of wine, and the second b gallons
of water. From the first is poured into the second as many gallons as it already
contains, and then as much is poured from the second into the first as was left in
the first. How much wine remains in the second cask after n such operations?

41. Three circles, radii a, 6, c, touch each other externally. A point is taken at
random in each circle. Find the chance that the triangle formed by joining the
points is acute.

42, Suppose a hawk a yards north, and an eagle b yards south, of a pigeon.
The pigeon flies due east at the rate of m miles an hour ; the hawk flies continually
towards the pigeon ; the eagle flies continually towards the hawk, and the eagle
catches the hawk at the same instant that the hawk catches the pigeon. Required
the equation to the curve the eagle describes, and the distance each has flown when
the hawk and pigeon are caught.

49

43. The first of two casks contained a gallons of wine and the second contained
6 gallons of water. Part of the water was poured out of the second cask and then
it was filled up out of the first cask, and the deficiency in the first supplied with
water. After n such operations, what is the probability that less than -^ of the con-
tents of the second cask is wine?

44. The velocity of a river a yards wide is v miles per hour. A deer that can
swim n miles an hour in still water starts to swim across the river, all the time
aiming for a point in the bank directly opposite the point he started from. At the
same instant a dog, b yards up stream from the deer, that can swim m miles an
hour in still water, starts in pursuit, and swims continually towards the deer. Re-
quired the equation to the curve described by the deer ; the equation to the curve de-
scribed by the dog, and the distance each swims before the deer is caught.

45. A heav^ flexible cable, length a, slides endwise down a smooth inclined
plane, inclination ^, the lower end of the plane being at a distance above the ground
greater than the length of the cable. Required the velocity of the cable at any
time t during the motion, and the form of the cable at the moment the upper end
leaves the lower end of the plane.

EDITORIAL NOTES.

We are grateful for the very liberal patronage extended to the first No. of the Visitob, and hope that the
present No. will meet with alike favoraUe reception at the hands of the mathematical public. It will be
our constant aim to make each Na more desirable than tte one preceding it.

Mr. R B. Sbitz and Prof. Benjamin Psisck have our sincere thanks for valuable asedstaooe.

Na 8 will be published about the 1st of January, 1879; it will contain about 82 pageiL and the price will
be 50 cents. Persons desiring to secure copies shoiild send their orders at an early date, as only a limited
number will be printed.

Copies of No. 1 cannot be supplied, as the whole edition is exhausted.

The Visitor wiU be pubUshed semi annually, and perhaps quarterly, as soon as our subscription list is
large enough to meet the expensa

We have left a few copies of Our Sehodday Visitor Mathematieal Annual for 1871. Pi ice, 25 cents.

The first prize is awarded to E. B. Setts, Greenville, O., and the second to Hknby Heaton, Sabula, Iowa.

Send an orders to ARTEMAS MARTIN, Lock Box 11, Brie, Pa.

NOTICES OF BOOKS AND PERIODICALS,

Lectures on the Elements of Applied Mechanics. By Morgan W. Crofton, F. R 8., Professor of Mathematics

and Mechanics at the Royal Military Academy. 8 vo. pp. 107. London, England: C. F. Hodgson & Son.

"A svnopsis of a course of Lectures on the elements oi the Theory of Structures and the Strength of
BCaterials, forming the first part of the Course of Applied Mechanics, at present studied by the Gentteman
Cadets of the Royal Military Academy." The subject is admirably treated in an elementeu^ manner, with-
out the Calculus, and illustrated by numerous examples.
The Principles of Elementary Mechanics, By DeVolson Wood, Professor of Mathematics and Mechanics

in the Stevens Institute of Technology. 12mo., pp. 851, Price 12.00. New York: John Wiley & Sons.

A hurried examination of this work convinces us that it is one of the best treatises on Elementuy Me-
chanics we have yet seen. It is written In Prof. Wood's ludd style, and the different subjects are amply
elucidated by a great number of interesting and well-selected problems and solutions.
Tracts Rdating to the Modern Higher Mathematics, Trtust No, 2. Trilinear Co-ordinates, By Rev. W. J*

Wright, Ph. D., Member of the London Matnematical Society. 8vo., paper, pp. 77. London, England:

C. FT Hodgson & Son.

A full and clear presentation of the subject of Trilinear Co-ordinates, with application to eTamplea
A list of writings relating to the method of Least Scares, with Historical and Critical Notes, By Mansfield

Merriman, Ph. D., Instructor in the Sheffield Scientific School of Yale College. From the Transactions

of the Connecticut Academy. Vol IV, 1877. 8vo.. pamphlet, pp. 82.

This is a work of great value to students. It contaiiu the titles of 406 papers, books and parts of books,
written in eight languages, ranging in date from 1722 to 1870.
On the Transcendental Curves whose emuUion is sin |/jdn my » a sinap sin tuv +5. By H. A. Newton, LL. D.,

Professor of Mathematics in Yale College, and A. W. Phillips, Ph. D., Tutor in Matehmatics in Yale Col-
lege. From the Transactions of the Connecticut Academy. 8vo., pamphlet, 11 pp. of text and 24 plates.

An interesting discussion of some of the curves represented by the equation given in the title.

60

On a New Method of Obtainfna the Differentials of Functions with espeeial reference to the Newtonian Con-
ception of Rates or Velocities. By J. Minot Rice« Profeasor of Mathematics in the United States Nayy
and W. Woolsey Johnson, Professor of Matiiematica in Saint John's GoUeee, Annapolis, Maryland. Re-
vised Edition. 16mo., pamphlet, pp. 32. Price 50 cents. New York: D. Van Noetrand.

On the Part of the Motion of the Lunar Perigee which is a Function of the Mean Motions of the Sun and
Moon. By O. W. Hill, Ph. D., Assistant in the Office of the American Ephemeris and Nautical Almanac.
Quarto, pp. 28.
An able mvestigation of an important problem in Astronomy.

The Method of Leaat Squares applied to a Hydraulic Problem, By Mansfield Merrimao, Ph. D., Instructor in
the Sheffield Scientific School. 8vo., pamphlet, pp. 9. Reprinted from the Journal of the Franklin In-
stitute for October, 1877.

The problem considered is that of determining the yelooity of a river at different depths below the sur-
face of the water.

The History of MalfattVs Problem. By Marcus -Baker, IT. S. Coast Survey . 8vo., pamphlet, pp. 10. From
the Bulletin of the Philosophical Society of Washington.
In this interestlDg paper Mr. Baker has £^ven an account of several solutions of this famous problem, and

a list of the authors who have considered it.

The Educational TVmes, and Journal of the College of Preceptors. Lond^ England: C. F. Hodgson & son.
The mathematical department continues under the able editorship of w. J. C. Miller, B. A. fCach num-
ber contains three or four double-column pages of problems and solutions. Many of the leading mathema-
ticians of this country and Europe are numbered among its contributors.

Reprint of the Mathematics from the Educational Times. Same publishers. Issued in half-yearly volumes
of 112 pp., 8vo., boards. Contains besides the mathematics published in the Times about as much more
original matter. VoL xzvi contains solutions of 86 problems and 2 papers: voL zxvii contains solutions
of o5 problems and 9 papers. The Reprints are rich in *Trobabillty" and * 'Average" solutions.
The editor of the Visitor can furnish the Times at <2 a year, and the Rqorint at $1.85 per voL

The American Journal of Pure and Applied Mathematics. Prof. J. J. Sylvester, LL. D., F. R SL Editor in
Chief; W. R Story, Pk D^^, (Leipeic) Associate Editor: with the co-operation of Profs. Benj. Peiroe, LL.
D., F. R. S., of Harvard University, Simon Newcomb, LL. D., Ph. D., of the U. a Naval Observatory,
and H. A. Rowland, C. E. Published under the auspices of the Johns Hopkins University, Baltimore,
Maryland. To be issued in quarterly numbers of 96 quarto pages. Price $5 a year. The flnit number
will appear in January, 1878.
The illustrious names connected with the Journal are a sufficient guaranty that the matter will be of the

highest order of excellence.

The Analyst, A Journal of Pure and Applied Mathematics. DesMoines, Iowa: Edited and published by J.
E. Hendricks, M. A. Bi-monthly; each No. contains 82 pp. Price $2 a year. No. 1, vol. 5, contains sev-
eral valuable papers, and the usual number of problems and solutions.
The Analyst is conducted with marked ability, and should have a wide circulation.

The Wittenberger^ published at Sprinfield, Ohio, contains a mathematical department ably conducted by
William Hoover.
The WittenJberger is a neat monthly magazine, devoted to the interest of Wittenberg College. Price $1.10

per year.

Educational Nolea and Queriea, published monthly, except in the vacation months of July and AugustjCon-
tains a mathematical department devoted to problems, solutions and mathematical notes. Prof. W. D.
Henkle, Salem, Qhio, Editor and Publisher. Price $1.

The NatUmal JSidtico/or, published monthly at Kutztown, Pennsylvania, by A. B. Urick, contains a depart-
ment of Science ana Practical Mathematics, conducted by F. P. Mats, B. E., B. S.

The Yates County Chronicle^ published weekly at Penn Yan, New York, by the Chronicle Publishing Co., con-
tains a Mathematical Department conducted by Dr. S. H. Wright, M. A., Ph. D.
The Dr. is publishing a valuable coUecUon of formulas in Trigonometry, Astronomy, &c. $2 a year.

The j€^erwon OmuUy Journal, published at Adams, New York, by Hatch ft Allen, contains a Mathematical
department conducted by O. H. Merrill.

The Maine Farmers^ Ahrumac tor 1878, Masters & livermore, HaUowell, Maine, contains the usual Puzzle and
Mathematical Departments. Solutions of the 7 questions proposed last year are given and 5 new ones
proposed.

CORRIGENDA.
Page 18, solution of problem 8, line 1, for '"CE" read CD.
Page 28, solution of problem 16, line 5, for "^^""^j"'* read *^""^^'

Page 24, solution of problem 18, line 4 from the end of the solution, for *y read p.
Page 27, solution of problem 2S, line 12 from the end of the solution, for "p" readpm .
Page 99, second line from the end of the prize solution, for *<£[ and IP' read II and III.
Page 40, line 4 from the bottom, in the value of u, for "cotC" read cosC.

> %^

•

Vol. 1. JANUARY. 1879. No. 3.

THE ^^7

1

MATHEMATICAL

VISITOR.

1 ' 1

! !
1

EDITED AND PUBLISHED BY

ARTEMAS MARTIN, M. A.,

MEMBKR OF THE LONDON MATHBMATKWL SOCIETY.

V

iPRIOHJ, B'lir'TTr OH2Sra7S.

Q^ ERIt; PA.: '

THE MORNING DISPATCH STEAM BOOK AND JOB PRIHTINO HOUSE.
X879.

THE

MATHEMATICAL VISITOR.

[Ottered aooortUng to Act of Oongreat, A, D, 1878, by Abtrmas Martin, M. A., in the Qfie^ of the ZMnxarian of Oottgrem, at Wcuhmglon,]

Vol. I. JANUARY, 1879. No. 3.

JUNIOR DEPARTMENT.

Solutions of Problems proposed in No. 2.

S5.— PropoKd by Orlando D. Oathout, Read, Gl«yton County, Iowa.

A and B propose to trade horses. A asks $20 *'to boot," and B asks $10. They finally agree to
*' split the difference." How much *' boot money " should A receive ?

I. — Solution by Frank Ai.brrt, Professor of MAtbemattcs, PennsylTanU State Kormal School, MIUersTille, Lancaster Co., Pa.

B*s horse plus $20 is A's yalnation of his own horse, and B*s horse minus $10 Is B*s valuation of A's
horse ; hence if they " split the difference," the value of A's horse is, by the agreement, one-half the sum
of A's and B's valuations, or B*8 horse plus $5 ; therefore B should pay A $5.

II.~8olntion by Oscar H. Merrill, South Butland, Jefferson Co., K. T.; K. S. Putnam, Borne, Oneida Co., N. T.; Tiiso. L. Dk-
Land, U. 8. Treasury Department, Washington, D. C; William Wiley, Detroit, Wayne Co., Mioh.; and D. W. K. Martim, Webster,
Darke Co., Ohio.

Since they are $80 apart at first, each will have to throw off $15 ; hence A should receive $5.

Answered in a similar manner by the Propoterf B. F, BuHeton, J, B. JPagan, V. WebUar Heath, E, P, Norton, John I. Clark, Prof. F.
P. Mats and Walter 8. NichoU.

III.— Solution by Gavin Shaw, Kemble, Ontario, Canada.

The difference between A and B is $20-|-$10===$80. If they split the difference, A must come down
$15; consequently the *'boot" that A should receive is $5.

36. — ^Proposed by J. B. Sanderb, Bloomington, Monroe County, Indiana.

A merchant bought 400 yards of cloth at $4 per yard, payable in 3 months, and after holding it for 15
days sold it at $4.25 per yard, receiving therefor a note payable in 4 months. When the purchase money
became due, he had this note discounted at the bank to meet it. What did he gain bjr the transaction ?
What would he have gained if he had borrowed at 6 per cent, interest, until the maturity of the note he
had received, sufficient to pay for the cloth, and why should there be any difference in the results ?

Solution by K. 8. Putnam, Bome, Oneida County, N. Y.; Prof. Frank Albert, MlllersTille, Lancaster Co., Pa.; and E. B. Seitz,
Oreenville, Darke Co., Ohio.

His gain was $100, less the interest on $1700 for 1 month and 18 days, or $86.40.
' In the second instance his gain would be $100, less interest on $1600 for 1 month and 18 days, or $87.20.

In the first instance he pays for the goods and receives his profit at the end of the 3 months. In the
second instance the transaction is not completed and profit realized until the maturity of the note.

fi. F. Bntieaon estimates true discount and does not count the days of grace. The Propoter neglects the dajrs of grace.

37.— Proposed by B. F. Burlbbo.v, Oneida Castte, Oneida County, New York.

A parsimonious farmer has a board 6^ feet long and 2 feet wide, which he wishes to cut and form into
a square without any waste of lumber. How must he cut the board and arrange the pieces ?

Solution by Benjamin Hradlky, Dillsborough, Dearborn Co., Indiana; George H. Lrlano, Windsor, Windsor Co., Vermont ;
y. Webster Heath, Bodman, Jefferson Co., N. Y.; and John I. Clark, Koran, Clinton Co., Indiana.

— 52 —

Let ABDC represeDt the board. Produce AB till AL=AC,
and on LB as a diameter describe a semicircle, and produce AG
to E ; then will AE = the side of the square which is equivalent
to ABDC. Describe a square on A E as a side, and draw EB.
Now, the three pieces ACGHI, HIB and GDB of the board are
respectively equal to the three pieces ACGHI, ECG and EKH
of the square.

Also answered Id an ingenious manner by the Propomrr, Hem^f Heaton^ Prof.
Mak and A'. S, PMtnatn,

88.— Proposed by Prof. Edwaud Bro<»k.s, M. A., Ph. D., Principal Pennsylvania State Normal School, Millervrille, Lancaster
Co., Pa.

A man gave me his note for $500, payable in 8 years, interest at (i per cent.; if he pays the interest
annually, to what rate is this equivalent if the same amount of interest bad been paid at the end of the
time?

Solation by the Pkup4>8Kr; William Wilkt, Detroit, Wayne Co., Mich.; and F. P. Hatz, M. E., H. S., Professor of Higher
Mathematics and Astronomy, King's Mountain High School, King's Mountain, Cleaveland Co., Korth Carolina.

Interest for 8 years -=$30 x 8 =$240 ; interest on interest =$30x06 = $1.80 ; sum of periods =28 ;
$1.80X28 = $50.4 » ; amount of interest =$240 -h $50.40 = $290.40 ; one year's mterest = $290.40 -i- 8=
$86.30; rate = $36.80 — 500= 7iJ per cent.

Solved in a similar manner by E. B. Seitz and A'. S, PutwiTH, B. F, Burlemtu and Johu f. Clark compute cotHjKmml interest.

f a9.~Proposed by W. B. Batch, Earlvllle, LaSalle County, Illinois.

If Dr. A kills 8 patients out of 7 ; Dr. B, 4 out of 18 ; and Dr. G, 5 out of 19, what chance has a sick
man for his life who employs all three of these doctors at the same time ?

SolnUon by Thbo. L. DkLand, U. 8. Treasury Department, Washington, D. C; Svlvcstkb Bobims, Korth Branch Depot, 8omeriet
Co., N. J.; and Walter 8. Nichols, New York, N. Y.

If Dr. A kills 8 out of 7, i remam ; if Dr. B kills 4 out of 13, V>, remam ; if Dr. C kills 5 out of 19,
It remain. Then chance of life after all have *' dosed " him is

♦ xAxiJ=^A

Answered in like manner by Prof. Mats^ Heur^ Heatou and K. S. Pnt»tim,

40.— Proposed by J. G. Walton, Covington, Kenton County, Kentucky.

Two circles have the same center. If a chord, length 2a, be drawn across the larger circle, it will just
touch the circmnference of the smaller. Required, the area of the ring.

Solution by K. J. Edminos, B. S., New Orleans, Orleans Co., La.; Mabcvh Bakkb, U. 8. Coast Surrey Offloe, Washington, D. C;
and 0. D. Oathout, Read, Clayton Co., Iowa.

Let R and r be the radii of the circles. Then R2-'i^ = o», and area of ring = n{I^—f^)== na^.

Substantially the same were the solutions by Messrs. AVbtrl, Bmrlfmm^ Clark^ DeLandj Heath, IfeaJ, Maiim, Sfata. Norton, Fttttttnu,
Jtobint, Sett:, Shaw and WOetf.

41.— Proposed by Abtksi ah Martin, M. A., Erie, Erie County, PennsylTania.

Find X and y from the equations

a^-h(^-+t^) (1 +xy-hx*^y2-hjr»y-h.ry3) = 87, xyia^-^]^) (a^-^+xy+y-) (ic^+y--hu:y-hay»-har»y) = 1190.
I.— Solution by E. P. Nobton, Allen, Hillsdale Co., Mich.

The given equations may be put in the following form :

ixy-\-x^'\-y^)-\~xy(2f^-\'i/^)^xy(a^'{-if')(xy^2^^^f) = S7 (3),

^Xa^-+2^-^)(a^+a:^+»')-+xV(^-+yOX-^^^ W-

Put V = jcy-\'X-'{-y^, and w=xy(pi?-\-y^) ; then we have r-f-t(7+rii; = 87 (6), vic(r-f-t£;) = 1190 (6).

From (5) and (6) we readily find v = 7 and w? = 10, or acy-f-ai^+j^ = 7 (7) and a:y(j^-hy3) = 10... .(8) ;

whence a; = 2 and y = 1 .

II.- Solution by Mabcis Bakek, U. S. Coast Surrey Office, Washington, D. C.

From(l)

a-^+ai^4-y''4-a^(a^'+2^)+^K«^*4-3/')(«'-h-*^-hy') = 87 (3);

hence we have the sum and product of the two expressions

a^'-h«y-h2^-h«y{aJ2-h2^) and xy{x^'-\-y^) (a^'+»y-hy').
whence a;2-|-xy-f-y3-f-a^(.r2-f-y2) = 17, xy(j:*-|-y^) {^'-{-xy^y-) = 70.

— 53 —

Here we have also a 8um and product,

whence .^-H-J^y+y-' = 7, xyiaj^+j^) = 10.

In a similar manner we find

whence x = 2 and y=l.

Good Bolatioiis giveu by Messn. ^HmoH, LeUmd, Jfots, Putttam^ MobUu and Seits.

49.— Proposed by Benjamix Hkaulby, Dillsborough, Daartwrn County, Indiana.

Cut a board 3 feet wide and 7 feet long hito three pieces, so that they will make a square.

f8eeBolntionof37.1

43.— Proposed by Aktrmas Maitix, H. A., Krie, Erie Ck>anty, Pennsylvania.

Two men. A and B, plow a rectangular field a rods long and 6 rods wide, A going ahead all the time.
What part of the field does each plow, estimating n furrows to the rod ?

Solution by Exocii Bbeby Skitz, GreenTille, Darke Oonnty, Ohio.

Conceive the field to be divided into squares, the side of each of which is equal to the width of a
furrow ; there will be af/n- squares. At each comer after the first A gains two squares on B.

1.— -When bn is an even number, and the first two furrows are plowed along the side of the field, a
being greater than 6, there will be \bn rounds, and A will gain 2(6n— 1) squares ;

1 6ii — 1 1.1 1 ., ^ . I bn — l 1 1,1 ^, ^

•• 2-^- abn^ =2-^an-abn^ = ^^^'^^'^^ 2- abn^ = 2 - a» + oAn^ = ^« P^^.
2.— When bn is an even number, and the first two furrows are plowed across the end of the field, there
will be ^6n rounds, and A will plow one square from the end of B's last furrow; hence A will gain ibn
squares :

8.— When 6n is an odd number, and the first two furrows are plowed alon^ the side of the field, there
will be \{bn—\) rounds, and a furrow of (an— 6n-f3) squares, which A wiU plow; hence A will gain
2(6n— 2)-f-aw— 6»H-3=(on-h6ii— 1) squares, which will also be the gain when the first two furrows ara
plowed across the end of the field, as may easily be shown ;

I an-^-bn — l _\ 11 1 _ a' rt

• • 2 ''■ 2abn^ - 2 +2a» "^ 2bn~' 2a6/*^ " ^ ® P*"'

•«^ 1 an-\'bn-l 11 1 . I «. *

"^^ 2- 2alm^ = 2 ^ 2an ~ 26n + 2a^^ = » « P*'*'

Solved in a similar manner by Frank Albert, B. F. Dmiewm, K, 8, Putmm and Waller Sieerlif.

44.— Proposed by Winfibld Y. JKFrRiES, Instructor in Mathematics, Vermillion InsUtute, Hayesville, Ashland Co., 0.

From a given point without a circle, to draw a secant that shall be bisected by the circumference.

Solution by E. B. Skitz, Greenville, Darke Co., Ohio; and William B gov ek, Mathematical Editor WittcHherger, Bellefontaine,
Logan Co., Ohio.

Let A be the given point, O the center of the circle, and B
the middle point of AO. With B as a center and a radius
equal to half the given radius describe an arc cutting the given
circumference in C. Draw OE parallel to BC, meeting AC
produced in E. Now since AO = 2AB, we have by similar tri.
angles OE = 2BC = the given radius, and EC = AC ; hence E is
on the given circumference, and the secant AE is bisected at C.

Good solutions received from Messrs. ^2&er<, Baker, Burleson^ Edmtindu, Heathy
Me.idam, PtUnaw^ SireHy and Wileft,

45.— Proposed by Mrs. Anna T. Snyokb, Orange, Fayette Co., Indiana.

A father having 640 acres of land in a circle gives to each of his six sons a circular farm touching the
circumference of the circle and also the circumferences of the farms of two of his brothers, the six farms
being equal in size. Required the number of acres in the farm of each of the sons, the number remaining
to the father, the number remaining in the center, and the number in each of the other six equal pieces.

— 54 —

Solutfon by K. 8. Putnam, Borne, Oneida Co., N. T.; B. F. Buuebok, Oneida CMtle, Oneida Co., N. Y.;and Marvik Bastk,
U. S. Coast Surrey Office, Washington, D. G.

Let radius CD = B, radius of each smaller circle = r. The lines joinisg
the centers of the inscribed circles will be the ddes of a regular hexagon,
center at C; therefore CB = BA = 2r, CD = 3r and r = }/?.

Similar snifaces are as squares of similar parts, therefore each inscribe d
circle = ( of the father's farm, =7H acres. The part remaining to the
father =i of 640 = 213^ acres.

The area of the sector ECD = ^ttR*. If from this we take the equilateral
triangle ABC = r«i/3 = }i?«i/3, and the two sectors EAF and DBF, each J
of an inscribed circle and both = | tcr* = f^nR^^ there will remain

Whole circle = tcR* = 640 acres ;

... .i?.:^'^^'~«^'^^:640:areaEFD;

whence area EFD = ^i^^^^^) ^20.0586 -h acres.

2\^ -^ 6(area EFD) =93.0114 + acres, the part m the center.

.."X Nearly thns were the solntions by Meesn. Martin^ McAdam, OaOtont^ 8eUs^ WS^ and the I\opo9er.

46.~Propoied by J. T. W. Sciibffkr, Profeisor of HathematiGS and German, Meroersburg College, Mercenbnrg, Franklin Co., I*a.

The sides of a plane triangle are proportional to the roots of the cubic equation

ar» — ax« -h 6x — c = 0.
Find the sum of the cosines of the angles of the triangle.

Solution by W. E. Hkal, Wheeling, Delaware County, Indiana.

Let Pf q, r denote the roots of the equation ; kp, kf/, kr the sides of the triangle ; and A, B, C the
angles respectively opposite.
Then

Zqr 2pr 2pq

.: coeA + coeB + coeC = <P'''+^'+^*" + '^*+^*'J-'^*'->-'^'+'^' + '">-
But from the Theory of Equations

P + ? + r = o, p*i-\'qr-{'pr = h, p<p' = c; r. pq* +pr* -{'p*q-{'qr^ -{'p*r-^q*r = ab—Scy

p^-\-fi^-\~r^ = a^^Sab-\-Sc and cosA -h cosB -h cosC = ^^ ~"^'* ~ ^*^-

Answered in a similar manner by the Propoter and Messrs. Albert, BurUaon, Jfafa, McAdam, Pulnom, SfiL: and ^.

47.~Proposed by Daniel Kibkwood, LL.D., Professor of Mathematics, Indiana State University, Bloomington, Monroe Co., lad.

The number of diagonals that can be drawn in a certain polygon is equal to seven times the number of
sides. How many sides has the polygon ?

Solntion by ; W. B. Heal ; K. 8. Putham ; and William Wilky.

The number of diagonals that can be drawn from any angle of a polygon of n sides is n — 3 ; hence
the number of diagonals that can be drawn from the n angles is n(n—S); but each diagonal of the

polygon will thus have been drawn twice. The number of diagonals is therefore ^^^~ K in the
problem on hand we have, therefore, the equation '*>^ ~ -^ =7n, whence n = 17.

Solved also by Messrs. AlbeH, Baker, CTart, Edmunck, DeLand, Heal, M(U», McAdain^ MervUl, X»rtoH, Oathoni, Putnam, Robin*, S 'Us,
Trotebridge and Mxb. Anna T, Snyder.

48.— Proposed by Joseph Ficklin, M. A., Ph. D., Professor of Mathematics and Astronomy, University of the State of Missonri,
Columbia, Boone Co., Mo.

Find the radius of a sphere that siiall circumscribe four equal spheres which touch each other.

Solution by the Propokkr ; V. Wew;tir Hkath ; Prof. T. P. Matz, M* £., M, S.; Marcv a Baker ; B. F. Burleson ; and
E. B. Skitz.

Suppose the radius of the given sphere to be a. Form a tetraedron by joining the centers of the four
spheres. The edge of this tetraedron will be 2a. Again, e = }£i/6 expresses the relation between the
radius of a sphere and the edge of the inscribed regular tetraedron ; therefore 2a = fRV^ ; whence,

— 55 —

R s= ^1/6. This is the radios of the sphere whose surface passes through the centers of the four given
spheres ; therefore \aV6 + a = a(l + ^yQ) is the radius of the sphere which circumscril)es them.

49.— Propoied by tb« late Dr. S. F. Bacheldrs, 8<rath Boston, MaBnchosetti.

P and Q owned a triangular piece of land, the sides of which were 20, 16 and 4 chahis. The lonsest
side bordered on a road which ran east and west. They agreed to divide it by a line parallel to
the shortest side so that Q, who took the resulting trapezoid, should have } of the land. A surveyor
located the line ; then P, setting a stake at the middle of it, proposed that a new line should be run north
and south through that point, so as to give both a square corner on the road. Q assented ; did be gain or
lose, in area, by the change of line, ana how much ?

Solution by B. F. Burleson, Oneida Gestle, Oneida Co., N. Y.; and E. B. Sbitz, Greenville, Darke Co., ().

hbt ABC represent the triangle, and a = 4,
6 = 18, and c = 20 be the sides. Let ED be
the surveyor's located division line. The area
ot the triangle ABC = 8i/(ll9) square chains
Therefore P's share = Ji/(119) square chains.
The triangles ABC and ADE being similar we
have 4 : 1 : : (4;« : (ED)*; .'. ED=s2chains.
In a similar manner we find AD = 10 chains.
Through the middle point H of Ihe line ED
draw FO perpendicular to AB, the points F
and G being in the lines AC and AB respectively. In the right-angled triangle HQD we have given the
hypothenuse HD = 1 chain, and the angle GDH, = the angle ABC, to find the remaining parts.

By Trigonometry, coeB= °''^^~ ' = ih and GD = }J of a chain. WhencelGH = A^(119), and

AG =AD — GD = W chains From E let fall the perpendicular EK upon AB. Then because the
triangles EKD and HGD are similar, and because ED = 2HD, we have KD = 2GD == |}, EK = 2GH «
Ai/(119), and AK = AD — KD = Jj^ chains.

Therefore the area of the triangle AKE = ^ x AV(1 1?) == tHV(ll^) square chahis. Now because
the triangle AKE is similar to the triangle AGF, we have (AK)< : (AG)< : : A AKE : A AGF ;
that is, (W)* • (W)' ' • f*ii/(119) : ||iH|i/(ll9) square chahis = area of triangle AGF.

Therefore by the change of division Unes Q loses fi/(119) — ^iHi/(119) = Tjy^i/(119) =
0.0005651 -f of a square chain, = 0.48904 + of a square rod.

Good lolationt receired from Mean. Albert, Baker, Ludwig, Mat*, Nichob, OaihotU, Pvitnam, Solmu and Wdeff.

50.~Propoeed by Hakous Bakkb, U. S. Coast Survey OfBce, Waahington, D. C.

In any plane triangle ABC the angles B and C are bisected, the bisectors meeting the sides in B' and C
respectively. Join B' and C and from any point P in B'C let perpendiculars pa, pb and />«, fall upon the
sides a, b and c respectively ; prove that pa —ph +?<-.

SoluUon by Frank Alhkkt, Profeesor of Bfatbematics, Penntylrania State Normal School, IflllenTille, Lancaster Co., Pa.

From B' and C draw parallels Pc and Pj to pc and ph reBpectively ; also, from B' and C draw
parallels Pa and Pa^ to pa'<, and suppose Pa < Pa'; then let fall a perpendicular from B^ to P^^; also let m
and n be the parts into which the line B'C is divided ; then by similar triangles

w-fn:Pc:: A:i?c= ^_^^ . . . (1), and m+n : P^ : : m : p4= ^*^ ^2);

also f» + « : Pa. - Pa : : m : ;,„ - Pa = '''^^-'7 ^"l whence p, = ^^•'+-'»-P«

^ w -f n ^" m + n

.(3).

Pa. = P6 and Pa = Pc, hence pa = ^'^^'t ^^^ =Ph +Pc from (1) and (3).

But

Excellent aolntionn sent by K. li. Seitt, J5f. S. Puitnun and Waiiam Hooter.

51.— Proposed by Ahtkm.i« Martin, M. A., Krie, Erie County, Pennsylvania.

There is a series of right-angjled triangles whose legs differ by unity only. CallUig the one whose
sides are 8, 4, 5 the first triangle, it is requh^d to find general expressions for the sides of the nth triangle,
and compute the sides of the 100th triangle.

SolnUon by the Proposeb.

Let ^x^ — 1) =p„ = perpendicular, i(xn -h 1) = 6„ = base, and yn = K= hypothenuse of the wth
tnangle.

— 56 —

Then i('«»-l)' + i(^ + l)* = y»*; whence a^2_2y^« = -i (1),

or (a:«-y«\/2)(a^+y„i/2)=-l (2),

Weid8ohave(ak,-yoV2)(iCd + yoi/2) = --l, and{xo-jtoi/2)*' + Maso + 2tot/2)«» + i==--l (8),

where n may be 0, 1, 2, 8, &c.

ABSuming iB„ — y« i/ 2 = (xo — ;/o i/ 2)2" + » and a5„ + yn "/ 2 = (a^i -f- 3/o "/ 2) *» + S which we are at liberty
to do, we find

x, = .5[(u^ + yol/2)^' + ^ + («:o-?/ol/2)*' + Q. y«=2^2[(^ + ^^2>'""''~(^'"«'«^'^>'""^']"

But it is easily seen that j:^ = 1, yo = 1 ;

... ^,= J[(^2 + l>'« + ^-(i/2-l)^« + ^]. !fn = ,2^^\(V^2-\-\)^'^^-\-(V2-iy'^ + ^'j^

and the sides of the nth triangle arc

^=^(x,.-l)=J[(i/2+l)^" + '-(l/2-l)^'' + '-2]. 6„=.^(x„ + l) =

^[(V2-hl)'"+^-(\/2-l)^» + »+*2]. K = y.= ^y^[(V2-^iy'^ + ^-\'(V^-iy'^'^']'

The operation of involution is very tedious, except when n is a small Dumber.

We have from (1), Xn = i/(2y»^ — 1). When x„ aud .y» are very large numbers, the 1 under the radical

may be omitted without sensible error, and we have, for the superior limit of their ratio, " = i/2; and
the values of Xn and yn are the numerators and denominators of the odd convergents to the square root of
2 expanded as a continued fraction.

1/2=1 + 1

2-fl
2 + 1

2 + etc.;

the successive convergents are {, f . J, \i, H* H» Uh ^^^^

Writing ^^ ^'» ''^ ^% . . . '^'"'*'' for the odd convergents, we have

hn = Tin + I, hn = t (^i^+i + 1 )i P« = iv '/2«+i " 1 ); • '• ''loo = »'30i, ^loo = i(</aoi + 1 ), /?100 = 4(?30l — 1 ).

When n = 1, Ai = ra = 5, 6i = 4, i>i = S, the sides of the /rst triangle.

The sides of the second triangle are 20, 21, 29 ;

of the third, 119.120. 169;

of the fourth, 696, 697, 985 ;

of the fifth, 4058, 4059, 5741 ;

of the sixth, 23657, 28658, 33457 ; etc., etc.

In finding the sides of very large triangles it will be convenient to employ the formulas

72c _ 2^e= + 1 ... 7*. __ 7r(47c' + 3)

r,r - 2qcrr • ' • ^*^' T.^ - iv(4</c^+ 1.) *^^'

Uu» + I _ 4r...,„ + I + 3</.2„ + 1 ^v tlim + 1 _ ^lim + :» — 4/-a« + a

r,,«+:, - 3r,« + i+27,« + i * " * ^^^' r,« + i 3r,« +3--27,« + , ^'^'

where c may be any odd number, and m may be any number.

Take c L 7. then by (5) ^^J = gSS' ""^ "^ '«»'"'« '" = '» <«> «-« ': = SS

n, , ^o •. 7w 128971066941642015967892393 , » . , . oo /m :

Take c = 23. then ;^ = ^j-jQ^3jg^jjj299234022705885' amlby takmg m = 88 (7) gives

7«7 _ 22127936779729111812858639
rr.7 ■" 15646814150613670132332869'

Now take c = 67 and we have by (5)

q.^y = 43389386297227576661095959458572614828030405587898980692163388984677691985393,

rno, = 30645578943282956180057972969833245887630954508753698529117371084705767728666.

. '. p^m = 21669693148618788880547979720286307 164015202768699465346081691992838845992696,

bun = 21669698148618788880547979729286d07164'}1520276869946534608l691992838845992697,

hm = 306455739432329561800579729698832458876309545087536935291 17871074705767728665.

Mr. PtitHoiu^ by an elpgaot but laborious method, couiputee the sides of the lOOth triangle, but does uot find the general values of
the ddee or the nth triangle. Messrs. Albert, Hoocer, Httrt^ Ki^iun and fieitz give general exprfsslons for the sides of the ath triangle, but
do not compute the sides of the ICiOth triangle.

/

— 57 —

53.— Proposed by Ja.m£8 McLaughlin, BfantorTille, Dodge Count3', MJnnewta.

Three ciFclee, radii a, 6, c, are drawn in a triangle, eacli circle touching the other two and two sidee of
the triangle. Find the sides of the triangle.

Solation by E. B. Seitz, QreenyiUe, Darke Goanty, Ohio.

Let ABC be the triangle, M, N, P the centers of the
three circles. I^t MD = ME = a, NF = NG = 6, PH =
PK = c, a + 6-fc = «, ZNMP = 0, / NMD = 9), / PMB
= tfy, Z DMB = CD. Then DF = 2V{ab)y EK =2 v^{ac),

QH = 2i/(&c), sin <p = 11 » cos 9>=^Tl» sin^ =

-^-> /^ cos t6 =

^Tin"r=VC> tani^=^-,:^^=VC> ^^^^^

'^ — i(® + <P + ^); whence tan J<p = — tan i(0 -\- <p + ^),

_ tan ^ -f- ^*^_i<? -f t*° i ^ -- tan J© tan J<p tan i^

"~ tan i© tan i^ -f tan yj tan iV' + tan J<jy tan JV' — 1

__ a v^(aAc)-f- « i/ {a^«) -f- « V (flc^) --^c\/ a

aA-v/c + ac-v/6-l-ov' (^) —^2 y' g

AD — a tan Afij — «^Ca*c) + ai/(a6«) -h rtl/(ac«) — bn/a
Similarly we have BF = V(a_6<0_-Hv'(a6j^^^^^^^

.-. AB=AD+DF+FB= L __ i I _ J.

(a+c)- r6i/(ac)-<n/(aA)~ai/(6c)+6 ji/(cw)+i/(6«)+i/((»)|- -i/Ca6c«)-h6«l

By symmetry, AC= -»- ^ -^ —J,

|ai/6-|-6a/a+i/(a/>8)-ci/«[ j6v'c+(Ji/6+i/(6c«)~oi/«|.

-j a-v/fe-f-6-v/o-f-v/(a6«)— ci/«|- \a\/c-\^a-\'\/(a^)—b\/8 .-
Unfloished solutions vrere receivetl from Messrs. Bakery Burleaon and Mntz,

and

BC=

58«— Proposed by Artkmas Mabtin, M. A., Erie, Brie County, Pennsylvania.

The towns G and D are 40 miles apart A set ont from G to travel to D at the same time that B
started from D to go to C. A ovei-took a drove of sheep moving forward at the rate of 2 miles an honr
just 10 minutes after he crossed a creek known to be 10 miles from C ; B arrived at G 3 hours and 56
minutes after he rntt the same drove. B was overtaken by an express proceeding at the rate of 6 miles an
hour just 5 minutes before he came to an inn 11 miles from D ; A arrived at D 8 hours after he met the
express. Required the hourly speed of A and B.

rowing to a clerical error, the distance of the inn from B and the time A traveled after be met the express were printed wrong

Solution by Dinlap J. McAdam, M. A., Professor of Mathematics, Washington and Jefferson College, Washington, Washington
County, Pennsylrania.

Let a; = A's hourly speed, and y = B's. Then 10 + 1 = distance A traveled till he overiook the drove.
8j|y = distance B traveled after he met the drove (a).

47 X 47 1/— 120 2j"

j2 y — 10 — ^ = distance drove goes from A overtook till B 7ii«< it. -^ — gl ~ = ^^^"^ ^^™ ^ <^^^-

took till B met the drove.

(47 y— 120 --2a)y
24

= distance B traveled in that time..

.(6).

—"^•^rs^tmc A and B had traveled before A (wertooifc the drove. - — J-^-= distance B traveled in that

X Ox

time (c.)

.'. B's whole distance = j2 "»" 24 "*" ~ 6a; ~ '

Reducing, 47^2^ — 2aJ»i^ — 2aay — 960a; + 240y = (1).

— 58 —

By second condition of the qneation,
11 — ^ = Stance of B from D when express overtook him. Sx = distance A traveled after he met the

express [d). 8x — 11 + ?!> = distance express traveled from it overtook B till it met A.

®^ —Jf^ + ^ = time from express overtook B lUl it met A. ^^' "4^^ "^ ^ = distance A goes from
express overtook B till it met A (e). i9~ = ^^'^^ ^ ^^^ ^ ^^ traveled when express overtook B.

wr~^^ = distance A traveled before express overtook B (/).

... A'sdis1ance = 8. + ^--^^^ + ^ + ^^^^^^

Reducing, iry2 + 96a;^ + 343a^+ 660a:- 2400y = (2).

From (1) and (2), by elimination, we find
533a:y+ 1514x2 + 52463; — 35760 = (3), 471y«-h76scy-f 123^-8940= (4).

Plndtog y in (3) = ^^^ ~ ^|^ ~ ^^^^ and substituting in (4),

3417O08ar* -h 24060478ic» — 132461 181x3 — 585140760a; + 2021 1 55200 = 0.
Solving for one value, we find a; = 3, and then y = 4.

This problem was Bolred in an elegant manner by Meitn. Bur/cwm, I\UmaM and 8eU*,

54.^Piopofled by E. J. Bowax, Shawnee, Perry County, Ohio.

What length of line, fastened to a iK)int in the circumference of a circle whose area is one acre, will
allow an animal to graze upon just one acre outside of the circle?

Solution by Enoch Berry Seitz, Greenville, Darke County, Ohio.

Let C be the point to which the line is fastened, A and B the points
in the given circumference, to which the animal can graze, and O the cen-
ter of the given circle Let 0A = o = 4-^'( )i the radius of the given

circle, and Z -A-CO = 0. Then we have AC = 2a cosO, the length of the
line, and the area common to the given circle and the circle upon which
the animal can graze = a\7t -f- 20 cos 26 — sin 20); therefore the area upon
which the animal can graze outside of the circle is
4;ra2cos20 — a^Ti + 20cos 26 — sm 26; = jra^ whence 26 — tan 26 = 2jr.
Solving this equation bv the method of Double Position, we find
6=51oi6'24". .-. AC = '2a cos 6 = 8.92926 rods.

WUliam Hoortr need Calculus in hie solution. Prof. F. P. Matz furnished a good solution without nnmeiical result.

511.— Pbixk Pkobl£M. Propomd by Aktkm.vh Mautix, M. A., Erie, Erie County, Pennsylvania.

Given -^j^

y = a? + OJF- + te^ 4- rar* + c^aj« + «»•' +/«^ 4- //«'' + /.a! • 4- ;a?»" 4- J^^
and

« = y -f- Ay^+By«+Cy<+Dy^'+Ei^+Fy'+Gy«+er+Iy'"+Jy''+Ky'-+Ly>=»+My»^+N^^ . .(2);

to find A, B, C, D, E, F, G, H, I, J, K, L, M, N, P in terms of a, 6, c, (/, e, /, g, h, /, j, k, /, m, «, p.

Solution by E. B. Seitz, GrepRTillc, Darke County, Ohio.

Substituting the value of ^ from (I) in (2), and collecting the terms having like powers of x, we have
x=x -h (A + a>r- -h (2aA + B + &>''+[(«■'+ 2ft)A+3aB-f C+r>^-f.[(2«6-f 2c) A-|-(3a'^+36)B-HaC+D+rf>*
+[(2ac+6'4-2(/)A4-(a«4-6a64-3c)B+(6a-+4/>)C+oaD+£+^]a^4-[(2a</+26c+20
4-(4a=^+12aA+4c)C4-(l0a'+56)D-f-6aE4-F+/>'+[(2ae4-2/«/+r*H-^/^)A4-(3a^^^
4-12a^64-12ac+66^'+4ti)C+(l()a-^4-20aA+5c)bH-(15a-'+6/>)E4-7aF4.G4-^^

+6aftc-h6a«-|-6-''-|-66c?+3c2-h3/) B -f (4a»64-12rt-V+12ai»'^-f 12ac/-|-12/>r-f 4e)C-f (5(*^+30a26^20a(-+lW-f-5rf)D
-}-(20a=» + 30ah -f- 6c) E -f (21a2-f-76)F-|-8aG+H-|-/i]a?'4- [(2a(/+26/-|-2c<'-f f/^+2/0 A-}-(3airf 6aAf7-h3a(^-|-6a/-
+362c4-6^+6rf/4-3</)B+(4aac4-6a'V/^+12a-^rf4-24a6c+12a^+46-"»+126rf+6c^4-4/)^^

-f 20acZ-h206c-|-5f')D+ (15a^-|-60a*6-f-30ac -h l^^ -f- 6^/) E -f (35a3-H2«6-h7r)F-f-(28a-^+86)G-f 9aH-f-I+»>'"
4-&C , (3).

By the principle of Indeterminate Coefficients we have from (3) A -f- a = 0, 2rtA -f B + 6 =0 ,
(a-»+26)A+3aB-|-C-hc = 0, (2a6-|-2c)A 4-{3a' + 8ft)B + 4rtC-f D-|-'/ = 0, &c.

— 59 —
From these equations we find
A = — a, B = 2a«— 6, C = — 6a» +5o6 — c, D = 14a* — 21a«6 -f 6ac -f86» — d, •
B = — 42a6 + 84a'6 — 28o«c — 28ai« + 7ad + 76c — e,

P = 132a6 — 330a*6 + J20a^c + t80a«6« — 86a2rf — 72a6c — 1263 + 8ae -f 8ftrf -|- 4(J» — /,
G = — 42»aT + 1287a«6 — 495a*c — [^©OOa'fca + ISSa^rf + 165a65 + 495a«6c — 46o«c — 45a(j3 — 456«c — 90aW

+ »o/ + 96« -f 9cd — ^,
H = 1430a8 — 5006a«d + 2002a»c + 5006o<6« — 716o*d — 2860a'6<5 — 1430o2ft3 + 220a«« -h 330a«c« -f- 660a«W

+ 6m<a^c + 566* — 55ay — 566«rf — 55b(^ — llOoftc — UQacd + lOag + 106/ -f lOce -h 5rf» — /«,
I = — 4862a9 + 19448a76 — 8C08o«c — 24024a66* + 3003a'^d -f- 15016a*6c + lC010a36» — I00\a*e — 4004a»6d

— 2002o3c» — 6006a«6«c + 286ay + 28663c + 858aa66 -f- S58a^cd + 858a63rf -h 858a6c« — 66a«<7 — 66ad»

— 6662e — 132a6/ — IdStace — 1326crf — 22(r» -h llaA + 116<7 + ll^/" + llrfe — i,

J = 16796a»o — 75582a«6 + 31824aTc + H1384a«6« — 12376a«rf — 74266a'^6c — 61886a*6» -f 4368a*6
+ 21840a*6d + 10920aV + 10920o26* + 43680a362c — 1365ay — 5460o»6« — 6460a'crf — 6460a63c

— 8190a«62d — 8190a»6c« — 87365 + 364a3gr + 8646»ci + 864oc» -h lC92a26/ -f- 1092a^ce + 1092a6'e
+ 54&a^cP + 6466208 + 2184a6cd — 7Sa% — 786y — 7S<^d — ISbd^ — 156a6^ — 156ac/ — 166ad«

— 1566c« -f 12at + 126A + }2cg + I2d/ + Qt^ — j\

K = — 58786a" + 293980a96 — 125970a«c —603880a^62 + 60388aTci -f 352716o«6c + 352716a»63 — 18664a6e

— 111884a66d — 55692a»c« — 2784«0a*6«c — 92820a36* + 6188ay-f 618806* + 80940a*6€ + 80940a*c«i
+ 61880a'62ci + 61880a36(? + 61880a«63c — 1820aV — )8206V — 7280a»6/ — 7280a3cc — 7280a6»d
-- 8640a3d2 — 8640a«c3 — 10920a26«6 — 10920o62c« -- 21840o»6c(i + 456aVi + 4556-V + 4556^3
+ 1365a«6flr + 1865aV + 1365a^de -f I865a6y + 186568crf + 1365ac2rf + 1366a6d« + 2730a6ce

— 91a«i — 9Wg — 91c«c — 91crf2 _ 9ia<j« -- 182a6A — 182acflr — 182ad/ — 1826c/ — 1826rfg + 18ai
+ 136i -f I8c/i + 18dflr + 18c/ — k,

L = 208012a" — 1144066a^o6 + 497420o9c + 2288390a«63 — 203490a«rf — 1627920a76c — 189924ra66»
+ 77520a7€ -f 542640a«6d + 271820a6c« + 1627920a»62c + 678300a*6* — 27182a«/ — 162792a66c

— 162792a«cd — 406980a*6«rf — 406980a*6c» — 542640a363c — 8l396a266 + 8568a*^ + 42840a*6/
+ 42840a*ce + 42840a6*c + 2l^ma*cP + 85680a362e + 85680a263rf + 171360a36<J(i + 28660a»c3
+ 1286a0a262c« -h 142866 — 2880o*A — 23806*rf — 9520a36flf — 9520aV — 9520a^de — 9520a63€ — 9620a6c3

— 14280a26y — 14280a3c2rf — 14280a26€P — 28560a36cc — 285«0a6W — 47606305 + 560a3i -f- 6606-y
-h 1680a«6A + 1680a3c^-|-1680a«d/+ 168006^^7 -|- 16806^06 + 1680a<^c -f- 16806c«rf+ 1680ocrf2 + 840oV
+ 8406«rf» + 8360a6c/ + 3860o6dc + 140c* — lOSa^ — 10663A -- 105cy — 1056«« — 210a6i — 210acA
-- 210ad^ — 210a«/ — 2106cflf — 2106d/ — 210c£fc — 85^3 + 14aife -h 146; 4- I4c» + lidh + \^9

M = — 742900a»3 -#- 4457400a"6 — 1961256aiOc — 9806280a«6« + 817190a9rf -|- 7354710a»6c -h 98062800^63

— 319770a8e — 2558160a76(/ — 1279080aV — 8953560a662c — 4476780a'^6* + 116280oy + 813960a«6c
+ 818960a«crf + 24418800*62^ + 2441880a«6c2 + 4069800a*63c + 813960a36* — 88760a«(7 — 88760a6«

— 2825600*6/ — 283560o6cc — 116280o*rf« — 581400a*68c — 681400a26*c — 1162800a*6cd — 193800o*c3
~ 776800a363«Z — 1162800o362c2 + 11628a*A + 116286*c + 68l40a% -h 68140o*c/ + 58140o*rfc
-h 58140o6*rf + 116280a36y + n6280a3c2rf + 116280o36(? + Il6280a63c« -h 116280a26c3 -f- ll6280a«63c
+ 232560o36cc + 848840a263crf — 8060a*i — 30606*c — 3060oc* — 12240o36A — 12240a3c^ — 13240o3rf/

— 12840a63/— 1224063c(f — 18360o26V - 18360oVc — 18360o«cd2 — \S3l60ab^(fi — 6120a3e3 — 61206«c3

— 86720a«6c/ — 36720o26dc — 36720o62cc — 36720o6c3d -h 680a3^- + 68063(7 -f 680c^d -h 680arf3
+ 2040a36i + 2040a«cA + 204Oa% + 2040oV -f 2040a6«A -|- 20406^ + 20406V« -f 2040acy
+ 20406c«c + 20406crf2 + 2040o6c» -f 4080a6c<jr + 4080 a6d/ + 4080a«fe — 120a«ifc — 1206*1 — 120(^g

— laOePc — 120cc2 — 120a/« — 240o6; ~ 240aci — 240o«i^ — 240a€<7 — 2406cA — 2406d(7 — 2406<?/

— 240cd/ -f 15o/ -h l^bk + 15c; + 15di + 15«A + 15/^ — w,

N = 2674440O"— 17883860oi26 + 7726160a"c -|- 42498880a»o6« — 8268760oiod — 32687600a96c — 49081400o863
-I- 1307504a9g -h 11767536o»6rf + 5883768o«c2 + 47070144oT62c + 274575840*6* — 490314ay

— 3922612o76€ — 3922612a7cd — 18728792a«6»rf — 18728792a66c3 — "874575840*630 — 6864396a*6*
-f- 170544a7(; + 1198808a«6/ -h 1193808a6cc + 596904o«d3 + 596904o«6« + 3581424o*63« + 7162848o*6crf
-f- 1193808o*c3 + 69690400*63^ -h 6969040o36*c + 8953560o*6«c3 — 54264a«A — 825584a*6^ — 325584«*c/

— 826584o«dc — 325584a6»c— 813960a*6»/— 813960a*c«d — 813960o*6rf« — 818960o«6*ci — 16279200*60^-

— 1085280a363€ — 1085280o36o3 — 3255840o36«ccf — 1627920o36»c3 — 77626^ + 15504o*i + I65046*d
+ T7520o*6A + 77520a*c<jr + 77520o*rf/ + 77620a6*e + 88760fl*c« + 387606*o» + 387606a.«c*
-f- 166040a»6V + 155040o3c«e + 155040o»c^3 + 155040a«63/ + 155040o63c» + 232560a«6»rf«
+ 810080a»6c/ + 310080o»6rfc -h 310080a63od + 465120a26«c« -h 465)20a36o»rf — 3876ot; — 38766*/

— 38766c*— 16604o36i— 15504o3cA — 16604o»rf(7 — I5504a»c/— 15601o63fir — 1560463cc — 15504oc»d

— 60 —

— 28256a«6«A — 28256a«cV — 23256a"6e« — 832666«c«rf — 7752o«d« — 77526«d« — 46512a«6cy

— 46512a«W/ — 46612a3crfe — 46512a6«dc — 46512o6«c/— 4«512a6c«c — 46612a6crf» + 8l6a»ifc + HlWh
■+- 8l6c»e -f 816W -|- 2448a»6/ + 2448a«d + 2448a2dA + 2448o«6jy + 2448a6«i + SiiSft'c^ + 24486«4f
+ 2^4Sac^g -f 24486c8/ + 2448o(i«e + 2448ac€« -f- 1224ay« + 12246«c9 + I224c»d» + 4896a^K:A
-f 4896aMj7 + 4896a6<s^ + 4896acrf/ + 48d66«fo — 136a«Z — 1366y — 136<«A - 136^/ — 186d<!«

— 1866/« — 212abk — 272a<:;- 272arfi — 278o«A — 272a/^ — 2726ci — 272WA — 2726e^ — 272ai^

— 272<je/ -f 16am + 166/ + 16c* + I6d; + 16ei + 16/A + 8<7« — n,

P = — 9694845a^« -h 67863915a"6 — 30421765a«c — 18253()530a"6« + 18037895a"d -f- 143416845aW6c
+ 239028075a96» — 5311735ai06 — 5311735ao9W — 26558676oV — 239028075a«6«c — 169352050a»6*
+ 2042975ay + 18386775a»6e + 18386775a8crf + 73547 lOOa^ft^df + 73547100a76<^ + 17l009900a«63c
+ 61482970aS6s _ 786471aV — 6883768a^6/ — 5883768a-c6 — 2941884a7rf« — 20693188a«6««

— 6864396o«c9 — 6864396a»6« — 41 186376a«6cd — 41186876a«ft3rf — 61779564a*6«c« — 51482970a*Mc
-f- 346l57a7A -f 245167a6^ + 1716099o%+ 1716099aV+ 1716099a^6 + 6148297a*6y+ 5148297a6<J^
+ 5148297a*6d» -h 5148297a«66c -^ 10296694a*6c€ + 8580496a*63e + 8580495a56<«/ -f 8580495a*6c3
+ 26741485a*6«cd + 17160990a»63ca _ 74613afii — 746136^6 — 447678a»6A — 447678a*c^ — 447678afioy

— 447678a6»d — 223839a'^€2 — 1119195a^V — in9195aV« — 1119195a*cci» — 1119195o»fr««

— in9I»5a6*c« — 2238390a*6c/ — 2238390a*6(fe — 1492260a36y — 2238390a»6«<^ — 2238390a«^c3

— 4476780a563ce — 447678aa36c2(Z — 4476780a26-'»cd — 873065aV + 20349ay + 2034966« + 101745a<6i
+ 101745a*cA + 101745a*rfflf + 10)745aV/+ 101745a6y + I01745a6c* + I0\145b*cd + 203490a»6«A
+ 203490a')cy + 203490a'>6<!> + 203490a'6»j7 + 20d490aW + 203490a63(;> +406980a9% + 406980a»6<//
+ 406980a3crf« + 406980ot>cc + 610470a26«c/ + 6\0410a^me + 6l0470a«6c2e -|- 610470a96crf»
-h 610470a62cad -|. 67830o3d« -f 6783063<r» — ^8^a*k — 4845^>V — I9980a^bj — 19380o3ci — 19380a»dA

— 19880a»c^ — 19380afr'»A — 193806»c/ — 1938063*? — 19380oc3e — 193806c«rf — 19880aW3 — 29070a«6»t

— 2907(ki«c2flr — 29070o2cf^e — 29070a«c^ — 29070a6V — 2907063c«« — 2907a6«c<? — 29070a<Ai8

— 9690ay3 — 58.4aa«6cA — 5Sl40a^bdg — 58140a«6e/ — 5Sl4(kficdf — 58140a62c^ — 58140a6«((r

— 58140a6cy — 116280a6crfe — 969c* + 969a»/ + 96963i + 969cy -h 969crf» + 2907a«6* + 2907oSq;
-f 2907a2rfi -f 29a7o'«A + 2907ayflr + 2907a6y + 29976«c* + 29076»rf/7 + 29076V + 2907ac«A
+ 29076c2flr + 2907c2rf« + 2907a<i«/ + 29076d^« + 2WJad^ + 29076ce9 + 2907a6/8 + 68.14a6ci
+ 6814a6rfA + SHliabeg + 58l4acd^ + 68l4acc/ + 58l46cd/ — 153a«TO — 16363^; — 153c«t~ ISScPg

— 153c/« — l5Sag^ — 306a6/ — 306acife — dOfyidj — 306aci -- 306a/A — 3066c; — 3066d» — 8066<!A

— 8066/i7 - 806crfA — 306cc^ — 806rfc/— 51c3-h 17(Wi+ 176/»+ 17c/+ 17(0:+ He; + 17/1 + 17^*—/?.

The values of the coefficients to I were found by the solution of the equations obtained from (3).
I then discovered a law by means of which the remaining coefficients were determined. In each coeffi-
cient the sum of the numerical coefficients is — 1 or + 1> according as the number of the co^cient is odd
or even. This may be applied as a test of the accuracy of the results.

Mr. 8eUz also compated the ralae ef Q, which we reluctantly omit for want of space.

Prof. TrotBbridge employed the Differential Calcalue in hii solution, and found the ralaes of the flnt eeren letb^ra.

LIST OF CONTRFBUTORS TO THE JUNIOR DEPARTMENT.

The following persons have furnished solutions to the Problems indicated by the numbers :

E. B. Sbitz, OreenTille, 0., 35, 36, 38, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 54 and 55 ; K. S. Putnam, Rome, N. Y., 35, 36,
37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 4^, 49, 50, 51 and 53 ; B. F. Burleson, Oneida Castle, N. T., 35, 36, 37, 38, 40, 41, 43, 44, 45, 46,
47, 48, 49, 52 and 53 ; Prof. F. P. Matz, King's Mountain, N. C, 35, 37, 38, 39, 40,-41, 46, 47, 48, 49, 52 and 54 ; Prof. Fkank
Albisbt, MillprsYille, Pa , SS, 36. 38, 40, 4:1, 44, 46, 47, 48, 49, 60 and 51 ; John I. Clask, Moran, Ind., 35, 36, 37, 38, 40, 42, 43. 46,
47, 48 and 49 ; Mabcus Bakkb, Washington, D. C, 40, 41, 44, 45, 47, 48, 49 and 52 ; William Wilbt, Detroit, Mich., 35. 38, 40.
44, 45, 47, 48 and 49 ; Sylvksteii Robins, North Branch Depot, N. J., 39, 40, 41, 47, 49 and 51 ; O. D. Oathout, Read, Ipwa, %&,
40, 45. 47, 48 and 49 ; V. Webster Heath, Rodman, N. Y., 35, 37, 40, 42, 47 and 48 ; D. W. K. Martin, WelJster, 0., 35, 36, 38,
40 and 45 ; Prof. D. J. McAdam, Washington, Pa., 44, 45, 46, 47 and 53 ; Gavin Shaw, Kemble, Ontario, Ganoda, 35, 33, 40 and
44; WiLLTAM HoovBR, Bellefontaine, 0., 44, 50, 51 and 54 ; George H. Leland, Windsor, Vt., 37, 40, 41 «od 42 ; Walter 8.
NicnoM, New York, N. Y., 35, 36, 39 and 49 ; E. P. Norton, Allen, Mich., 35. 40, 41 and 47 ; J. R. Fagan, Erie, IHu, 3fi, 30. 3S
and 39; Theo. L. DeLanp, Washington, D. C, 35, 40 and 47; £. J. Edmunds. New Orleans, La., 40, 44 and 47; Hbnrt
Heaton, Siibula, Iowa, 37, 39 and 40; W. E Heal, Wheeling, Ind., 40, 46 and 47 ; G. G. Washburn, North East, Pa., 35, S6
and 38; Mrs. Anna T. Snyder, Chicago, III., 45 and 47 ; Prof. David Trowbridge, Waterbnrg, N. Y., 47 and 55; Waltck
SiTERLY, Oil City, Pa.. 43 and 44; 0. H. Merrill, Mannsville, N. Y., 35 and 47 ; Benjamin Headley, DillsboroMgh, Ind., 37 and
42 ; Dr. David 8. Hart, Stonington, Conn., 51 ; Prof. U. T. J. Ludwio, Mt. Pleasant, N. C, 49 ; ProC J. F. W. Sciietfem,
Meroersburg, Pa., 46; F. F. Leart, Covington, Ky., 40; T. P. Stowell, Roch<>eter, N. Y., 37 ; Hilton Rice, Moraa, Ind.,
42 ; J. B. Sanders, Bloomington, Ind., 36; Prof. Joseph Ficklin, Columbia, Mo., 48; Prof. Edward Brooks, MillerevUle,
Pa., 38.

The first prize is awarded to £. B. Sbitz, Greenville, O., and the second prize to Prof. Dayid
Tbowbbidgb, Waterburg, N. Y.

— 61 —

PROBLEMS.

80.— PropoMd by John I. Glabk, Moran, Clinton Oonnty, Indian*.

When Bilver is 4 per cent disconnt and gold is at 5 per cent premium, taking greenbacks as a standard,
what will silver be worth if we take gold as a standard.

M.—Propowd by G. G. Wabbbubn, North East, Erie Gonnty, Pennaylmnla.

Two men. A and B, each desire to sell a horse to ; A asks a certain price and B asks 50 per cent
more. C n- fused to p«y either price ; A then reduced his price 16f per cent and B reduced his S8| per
cent., at which prices G took both horses, pajing $220 for them. What was each man'fa asking price ?

90.— Propowd by K. S. Pvtkam, Bom«, Oneida County, New York.

Having cut a square, area a\ from one comer of a board 2a wide and 8a long, cut the remainder into
three pieces so that they will make a square.

91.— Propoaed by Fkahk Albert, Profcaior of Mathematica, PninqrlTanla State Nonnal School, MlllenTille, Lancaater Co., Pla.
Show that in any iposceles triangle, the pquare of a line drawn from the vertex to any point in the
base, plus the product of the segments of the base, is constant.

•2.~Pit>po9ed by Jamks Q. Bbiguam, Walton, Hanrey County, Kaniaa.

What is the rate per cent, of interest when a sum of money amounts to ten times itself in 21 vears,
compounded annually ? And what would be the rate for the same time if compounded semi-annualfy ?

•S.— Propoaed by E. J. EnMrWDS, B. S., New Orleans, Orleans County, Louisiana.

A point P being given on the bftee of a triangle, draw a line across the triangle parallel to the base
which will subtend a right angle at P.

94.->Propoaed by Hekry Nichols, Hampton, Bock Island Co., 111.; and Mn. Anna T. Sntdbb, Chicago, Cook Co., HI.
It is required to divide a taperinflr board into two rqual partn by sawing it across, parallel to the ends.
Find the width at each end so that the lengths of the pieces will be expressed by rational numbers.

•5.— Proposed by T. A. Kinney, St. Albans, Franklin County, Vermont

Given two circles and a point without them to draw through the point a line cutting the two circles so
that the portion of the line intercepted between the two circles shall be equal to a given line. — [From
Chaumnet's Oeometry.

96.— Proposed by L. C. Walkrr, New Madison, Barke County, Obio.

A cannon ball, radius r, rolls into the comer of a room whose walls are at right angles, and perpen-
dicular to the floor. What is the radius of another ball just touched by the cannon ball ?

97.— Proposed by AvnuAB Martin, M. A., Member of tbe London Mathematical Society, Erie, Srie County, PennqrlTaaia.

Solve, by quadratics, the equation x* — 2aa:' — 2a&B + 6« = 0.

99.— Propoaed bj Mrs. Anna T. Sntmr, Chicago, Cook County, IlUnoia.

On a ciT-cular griddle 12 inches in diameter, 3 equal circular cakes are baked of such size that each
cake touches the edge of the griddle and the edges of the other two cakes. What is the diameter of each
cake?

99.— Proposed by W. Woolset Johnson, Professor of Mathematics, St. Johns College, Annspolis, Anne Arundel 0>., Md.
The extremities of a line of fixed length which slides along a fixed line are joined to two fixed points.
Find the locus of the intersection of the joming Imes.

100 —Proposed by Daniel Kirk wood, Ll. D., Professor of Mathemattca, Indiana State UniTerslty, Bloomington, Monro*
Oonnty, Indiana.

Required to And a number such that when it is added to 15, 27 and 45 there arise three numbers which
are in geometrical progression.

191.- Proposed by 0. H. Mskrill, Mannsrille, Jefferson County, New Tork.

Two equal circles intersect, the center of ^ach being on the circumference of the other. From A, one
of the points of intersection, a diameter AB of one of the circles is drawn. Fmd the radius of the
circle touching AB and the circumferences of both the given circles.

199.— Proposed by J. D. Williams. Superintendent of Public Schools, Sturgis, St. Joseph County, Michigan.

Solye the equations a? + y -f xy = 75, a? — j/« = 815, by quadratics.

19S.— Proposed by Wxluam WooLstT Johnson, Professor of Mathematics, St. John> College, Annapolis, Anne Arundel Co., Md.

BR is an ordinate, from any point B of a circle, to the diameter passing throusch the fixed point A;
and T is the intersection of the tangent at A with the radius produced through B. Find the locus of the
intersection of AB and TR.

194.— Proposed by W. T. R. Bell, M. A., Principal King's Mountain High School, King's Mountain, Cleayeland Co., N. 0.

The railroad debt of a certain company is $56400, interest at 7 per cent, coupons payable seml-
anoufilly. What tax must be levied to pay the interest and create such a sinking fund as wiU absorb
the debt in 15 equal annual payments?

195.— Proposed t^ John IUa, HI1I*b Fork, Adams County, Ohio.

Find the co-ordinates a, 6, a\ b\ of the centers, and the radii r, r' of the two circles
^ + ai«-20x-.40=0, andaJ»+y»+40y + 60 = 0;
■ud in case the two circumferences cut each other, what are the co-ordinates of the pobits of inteiiectiont
— [From Xoooais' Anaiytieal Oeometry,

— 62 —

!•••— Propowd by S. G. Bkacr, Phllwielpbia, PeanflylTanfat.

A circular saw, one foot in diameter, in cattinf off a round loe, is stopped when it reaches the oenter
of the log, when it is found tliat the wood covers } of the side-surface of the saw. What is the diameter
of the log?

lOT.^Propoted by Stlyrstkr Bobxhs, Nortb Braaeh Depot, SomeiMt Ooanty, New Jeraey.

At a Piremen*s Fair a eilver trumpet is offered to the company exhibiting the ladder that can be used
in the greatest number of streets and alleys for the purpose of reaching windows on either side withoat
chanffioff the location of its foot. All bases and perpendicnlars must be rational lengths, and a company
may mcTude in their count dimenrions having as many decimal places as their ladder has, but no more.
The **Hudsons" bring a lodder 65 feet long, the '* Keystones ** offer one 32^ feet in length, and the
"Delawares" show one of 42^ feet. To whom must the trumpet be awarded, and on what count?

10^— Propoiedby Jusrrs F. W. ScHKirtB, ProAnor of Mathematioi and Germao, Meroenbarg College, Meroeirtnug, I'mkUii
OoQnty, PeniuylTaola.

Three army corps. A, B and G, beins engsged In a campaign, have provisions sufficient for 30 weeks.
On' these provisions B and' G would subsist 9 weeks longer ttian A and B; and A and C, 15 weeks longer
than B and G. After 6 weeks the three armv corps encounter the enemy, In consequence of which A
loses i of its troopf), B loses ( and G loses i ; also } of the remaining provisions are lost. How many
weeks will the remainder of the three corps subsist on the provisions lefi?

14>9«~ P wpo>ed by Thomas Baoot, Prindpal Sbelby School, Omumoi, Jeffenon Comnty, Indlaiw.

A field in the form of a right-angled triangle has a base and perpendicular of 40 and 200 feet
respectively. What leneth of rope attached at the vertex of the right angle will peimit a horse to
graze upon half the field?

110.— Pxzn Pboblbii. PropoMd hf Abtsmaa Mabtik. li. A., Meaaber of the London MatheBiBtioBl Sooleiy, Kria, Eiie Oo., FB.
Give an expeditious method of approximating to the square root of a quantity, and find by it the
square root of 2 to at least one hundred and finy places of decimals.

BotQtIoBi of thaw pioblema ehookL ba raoalTed bj SepCambar 1, 1870.

SENIOR DEPARTMENT.

Solutions of Problems proposed in No. 2.

56.— Propotad by Georof. Eastwood, Sazonvllle, HiddlcMZ CoiiDty, Mnwanhneatta.

In a plane, the equation of a straight line in terms of the perpendicular (p) from the origin and the
angle (0) which it makes with the axis being yB\nO-{'ZCosB=pi prove that the same form holds for the
equation of a great circle of the sphere, when x, y^ptae put for tan x, tan y and tan p.
Solotlon by W. E. Hial, Wheeling, Dehiware Coanty, Indiana.

We wHl assume that the required equation is reduced to the form ^ tan x + ^ tan y = tan/> ; we wish to
find the coefficients A and B. Malting y = Owe find for the intercept on the sxis ot z,

il tan zi = tanp, ,\ A = ^^ . Similarly, J? tan yi = tan », and 5 = ?° ^ .
tanzi " ' ' '^ tanyi

But by Spherical Trigonometry we have ^ "^ = cosO, . ^^ = sin ; . •. ^ = cos 9, i? = sin 9 and the

xtLTkxi taoi/i

equation is reduced to cos Otan x + sin Otan y = tan p. Q. E. D.

Solred alao in an elegant manner by E, B, 8ri/s, F. P. Mats and the Propo»er.

57.^PropoMd by Oscab H. Mxkbill, Mannarille, Jeffenon County, New York.

Prove that the cube of any given number is greater than the product of any other three numbers
whose sum is three times the given number.

SolntiOQ by E. B. Seitz, OreenTille, Darke Gonnty, Ohio.

Let m be the given number, m + 2ii one of the other numbers. Then the product of the three numbers
will be greatest when the remaining two numbers are equal ; hence let m — n be each of the two numbers.
Then tkie product of the three numbers = (m + 2nXm — n)< = m' — Smn^ -f 2n\ which is less than m' for
all admissible values of n.

Good eolatlont giren by Me«n. fleal, IRcAob, Trowbridge and the Prcpowr,

58.~Propotad by Miw Chsistxive Ladd, B. A., Baltimore, Maiyland.

An ellipse and a parabola have a common focus and the other focus of the ellipse moves on the direct-
rix of the parabola. Show that the points of contact of a common tangent subtend a right angle at the
common focus.

— 63 —

Solntioii by 1. B. Sbitz, GrcenviUe, Darke Countj, Ohio ; and Dunlap J. McAdam, M.
A., Profenor of MathetnatioB, Waabington and Jeffnaon CSollege, WashlDgton, WaablDgton
Oounty, PennsylTsaia.

Let F be the common f ocus» F' the other focns of the ellipse, &^'
the directrix of the parabola, T' the InterBection of SS' and the ellipse,
TT' the tangent to the parabola, M the point in which TT' produced
meets the axis of the parabola produced, and TN the perpendicular on
8S'.

Now /FTM=/FMT= /NTT', and TF=TN; .'. /TT'F
= Z TT'N, and Z TFT'= Z TNT'= a right angle.

Again, / F'T'M = / TT'N = / FT'T ; . . TT' is a tangent to the
ellipse.

SolTed alao by the Ptopoter.

69.— Propoaed by Dr. 8. H. Wkxght, M. A., Ph. D., Mathematioal Kditor Y<Ue» OmU^ Chroniele, Peno Tan, Tatea Connty, M. T.

When the sun's declination was S north he rose fi^ farther south than when his declination was S'
north. Required my latitude.

SolntioQ by the Pkopmbb.

Let P be the north pole, and P' the north point of the horizon, S the place of the sun when its decli-
nation is d, and S' when 8',

Then 88' = A SP = co-declination of S, and S'P = co-declination of 6' and the required latitude =
PF=A.. In the spherical triangle 8PS' all the sides are given to find the angle P88'=il. We haye,

therefore ^=2a)s-'^(^^^-^-+ ^'~^^^ and A = sln-»Csin ^sin 8). If 8 = l(P, 8' = 2(F,

and /? = 12°, then A = 32° 12' 41".

SolTed alao by £ B. £Mb and Piof. De Vobom Wood,

60.— Propoaed by Okoboi Lillet, Kewanea, Henry Goonty, lUinola.

Find the minimum eccentricit j of an ellipse capable of resting in equilibrium on a perfectly rough
inclined plane, inclination fi.

I.— Solution by E. B, Sxxn, OreenTille, Darke Coanty, Ohio.

Let TP represent the inclined plane, P the point at which the ellipse touches
the plane, C the center of the ellipse, T the point at which the major axis
produced meets the plane, TN a horizontal line, and let CP be vertical. Let
a, 6 = the semi-axes of the ellipse, CM = x, PM = y, angle CTP = 0, and

angle PTN = /?. Then we have tan = ^^ ...(1), and tan (0 -f /?) = - ...(2).

From (1) and (2) we find a«lan = 6«tan (0 -f-/?)..C3). By development

and redaciion (3) becomes o^tan/SJtan^G — (o«— 6«)ian6-f 6*^wjy^=0 (4)

The two positions of an ellipse, in which its center is verticallj above the
point of contact with the plane, are given by the values of tan 6 in (4).
When the ellipse is in equilibrium, the value of tan 6 is between the two
values in (4). Hence, for the ellipse of minimum eccentricity these two
values are equal, and therefore from (4) we must have (a' — 6^}^ = 4a^&^tan^^...(6).

we flndft« = o« —

2a«in/J_
l + sin/?"

,.(6).

Substituting a«(l — c«) for 6^, we find «« = j-^*°jf^-

Solvmg (5) for 6«,

II.— Solution by I>B VoLBoir Wood, M. A., C. E., Profeawr of Mathematica and Mechanica, SteTena Inatitnte of Technology,
Hoboken, Hndaon Gonnty, New Jei^y.

The center of the ellipse will be vertically over the point of support, and since the plane is tangent to
the ellipse, a vertical through the point of support and a parallel to the plane through the center will
moke conjugate diameters. Hence the acute angle of the conjugate diameters is 90^ — >£

At the point bordering on motion the potential energy is a maximum, and the major axis will bisect
the acute angle of the conjugate diameters; hence the positive angles made by the conjugate diameters
with the major axis will be = 45^ - ^ /?, 0'= 135° + ^ /J.

The condition for conjufntte diameters is a^iu Bind' -)- 6>co6 OcosO' = 0, which, by substituthig the
preceding values, gives a«8hi«(45o — i /5) — 6*co6«(45° — J ^ = ; which reduced gives

SoWed alio by Prof! P, P. XatM,

Ve^J-O— V(ifs6)

— 64 —

•1.— Proposed by Mia CaBLsruri Ladd, B. A., Baltimore, lUryland. .

If ABC be a triann^le inscribed in a conic, VP"F'" the points in which its sid^ meet the directiiz of
the conic, and Q'Q"Q' " the poles of focal choi ds ihroogh P'P'F", then will AQ', BQ", CQ'" meet in a point.

Solatloa by th« Pkopcwrm.

Reciprocating, tbis becomes : If from the yertiCes of a trianele ABC, lines be drawn to I, the center
of the inscribed circle, and perpendiculars to them, IM', IM'^ IM''', meet opposite sides in M^ M'^ M'",
then are M^ M'^ M"^ three colHoear poiotf*. And this is true, since the equations to the lines in question
are sin^Bcos JAa -f- sin ^Acos ^By9 — cos kCy = 0, &c, and tbe lines joining to vertices their intersections
yrith opposite sides areaco6iA-}-y^co6|B = 0, «&c., therefore, &c.

•2.— Proposed by F. P. Matz, H. E., M. 8., Profetaor of Higher Mathematics and Astronomy, King's Mountain High School,.
King*s Mountain, CleaTeland County, North Carolina.

In an equilateral triangle ABC Hoes are drawn as in Problem 8 ; find the average area of the equi-
lateral triangle formed by the intersections of the lines.

fiolntion by the Pbopo«kr ; Prof. DbVouok Wood, M. A., C B.; Wiluax Hoover ; Walter 8. Nichols ; and E. B. Ssin.
Put AB = a and AD = jc, then CD = i/(a2 — oa; -f a^),
fiH- ^^ np -*' -FH ^^ "" ^*

and the required average area is

1/3 ,ri«Ca-2a:)V/ic

ri«^ " '■" 2 Jo a« — ax-hJc«'

^/P\

--m.

6S.— Proposed by I8aac H. Tvrrell, Cnmminsrille, Hamilton County, Ohio.

Within the space enclosed bj three given circles which touch each other externally it is required to in-
scribe, geometrically, tbree circles each of which shall touch the other two sod also two of the given
circles. Analytical solutions also desired.

I.— Solution by the Proposer.

Find the external centers of similitude
PP,,P,. and draw the radical axes OR, OS, OT.
of the pairs of given circles AB, BC, CA, and
designate by X Y, Z« (which are not shown in
the diagram,) tbe circles required to lie respec-
tively opposite A, B and C.

If a pair of circles be touched in the same
way by another pair, the center of similitude
of either pair lies on the radical axis of the
other pair ; hence x, y and z, tbe external
centers of similitude of the pairs AX, BY, CZ.
lie respectively on (>8, OT, OR ; und since AX
is. touched in the same way by the pairs BY,
CZ. BC, its radical axis contains the points ^,
2, P,,. Similarly y. x. P, and «, x, V, are each
three points in directum. Hence xyz may be
regarded as a variable trianele who^e vertices
move along the fixed lines OS. OT, OR, which
meet in a point, and whose sides pass respec-
tively through PP„P,. three given points in a
right line : in other woids, if the sides xy^ yz,
for all positions of the variable triangle, pass
through the points P, P^^, the third side zx will
always pass through P,. (Mukahp's Modem Geometry, p. 18.)

Since the radical axis of a pair of circles is perpendicular to the line joining their centers, the problem
is reduced to finding that position of the variable triangle in which the lines hx, yz, shall intersect at rig^t
angles in some point V. On AP,, as a diameter, df^scribe a circle on which the point V must lie. From
A draw any three Ihies meetbg the circle AP,, m V'V" V", and the Une OS in xit,fc,„,

« J^5^ ^^ ^^"^. P/c,,^ meeting OR fa z^ z,^ z,„ ; also P,;r„ P^,^ P,^,,„ meeting the circle AP,, In
^i^ji^nr (To avoid aimplication these Imes, as well as the cirole AP^ are omitted m the figure ; they can^
however, be easily reproduced.)

— 65 —

The anbarmonic ratio A . VV'V"V'" = P, . 9!Xpt,fc,„-=V„ . zr/s,^,,^ P„ . VV,V„V,„ ; hence we have, on
the circle AP.^, Bix given points, V'V"V"Tjy»V,,, to find a seventh V, such that the anharmouic ratio of
WXV,,, shall be equal to that of VW'T"'.

This is an elementanr problem, (Mulcahy, p. d2,) and g:ives two popitions for the required point V,
which shows that the onginal problem admits of ttpo solutions ; one position gives the group enclosed by
the circles ABC, the other the group enclosing them.

To complete the solution, draw the line AV meetinji: OS in x, T/b meeting OR in 2, and F,/s meeting
OT in y. To determine the circle X, draw a tangent xM. to the circle A, intt rsecting P,^ in it ; from ]N
lay oft towards x a distance N9 = NM. Since P,;z is the radical axis of the circles A and X, the pohit q
(not shown in the figure) will be the point of taugency on xM of the circle X whose center will evidently
be the intersection of Ax with a line through q perpendicular to arM. Similarly the circles Y, Z can be
determined.

NoTB. — A similar construction will apply if A, B and C are three small circles drawn on the surface
of a sphere.

II.— SolatioB by X. B. Skitz.

Let A, B, C be the centers of the three given circles, radii (ij b, c; A', B', C the centers of the three
Inscribed circles, radii a;, y, z, the circle A' touching B and C, B' touching A and C, and C touching
A and B. PutAB = o-f6^ AC = a + c, BC = 6 + c, A'b = b+x, A'C = c + a;, AA'=«, /ABC = A
^ ABA' = Q, Z. OBA' = <pf and let t = the common tangent of the circles A and A'.

Now, shice each of the six circles A, B, C, A^ B', C touches four of the others externally, the square
of the common tangent of any two of the circles, which do not touch each other, is equal to twice tiie
product of their diameters {AnalyH, vol. ii, p. 24) ;

«« = 8aa?, and««=(a — x)«-h<«=a«-f.6aa; + a;« (l)w

By Trigonometry we have

Similarly we find

«»ni9=V((T+-ck6-R)' «^'^*^Wt+6f(6Tc"))' '^^'''^^^=^(^^$h)'

But shi H^ + <P) = Bin i/i, whence shi'ie + shi*^9— sin*^/?-f2co6 i/Sfshi^e sin i<p = (2).

Substituting the values of sln^d, &c., In (2), clearing and reducing, we get an equation of the first

degree, from which we readily find x = ^ , , r — , , , „ rririrj — , . . xt
^* •' 2ab + 2ac-f- 6c + 2v'[2a6c(a + o -f- c)]

By symmetry we have y = gofc + S^ 2b^:\^W^^ab^+bT^y ^ "^ a6 -h 2(^26^'2i/[2^6<«+'* + 01

•4.— Propotad by J. J. Sylvkstbr, LL. D.. F. JL* 8., Gorresponding Member of the iDititnte of Fnnce, ProfeMor of Mathomatie
Johnft Hopklna Unirenity, Baltimore, Maryland.

If there be two equations In x, (which for greater simplicity may be supposed to be of the same
degree n), find the most general form of iV = a rational Integral function of the coefllclents of these
equations such that Mx, Afx\.„JI£3f*^^ shall each of them also be rational functions of the same.

Solution by Ber. W. J. Wsioht, M. A., Fh. D., Member of the Loodon Mathematical Society, Gape May Point, Gape May Go., N. J.

It is assumed that the equations are simultaneous, then JH Is the elimlnant of the given equations, and
It follows as a corollary that Mx, &c., are rational Uitegral functions of the coefficients.

Letx» + a,x^i-Hii'^ + &c. =0 (1), and «" H-6ia*-» -h M"-^ + &c. =0 (2)

be a pair of shnultaneous equations, cr, /3,y, &c., the roots of (1), or, fiu yu &c., those of (2), in which a is
the common root. If the roots of (1) be substituted m (2) It is plain that (2) will vanish for the value a
and therefore the entire product of all these successive substitutions will vanish. But by the Theory of
Equations this product is a symmetric function of the roots of (1) and is expressible In terms of It s cof ffidents

Now 2 a = — ai, ^ a< = oi« — 203, JS o^ = oj, &c.; consequently by Introducing the values of these
symmetric functions we evidently obtain a homogeneous function of the coefficients of the two equations
which is the elimhiant. Mx, Mafi, i&c., must alse be functions of these coefficients since in the course of
elimhiation by Bezout*s or any other method we must arrive at an equation of the first df gree from which
the value of x is found in terms of the coefficients of one or both of the equations, and this multiplied by
M gives us still a rational function of the coefficients.
a b

= Is the elimlnant of ox -f 6 = 0, aia;-f-6i = 0. Mx:=

iau9lnUi(m.—M(tr-^ =:M^

ax 61
Bee my Trad m InvariatUs, Arts. 1—7.

(ii bi

x->

•II.— Propoeed hj Dr. David S. Habt, M. A., Stonlngton, New London Gonnty, Gonnectievt.

Find six square numbers whose sum Is a square, and the sum of their roots plus the root of thdr sum a

eqnare.

— 66 —

fiolutioii by the PaopoesB.

Let 2WW, 2/M, 2p«, 2q8, 2r« and m«-f-n«-h|>« + g« + r« — ««bethe rooto of the squares ; then the sum of
their squares is Cwi« + n« -|- /?» + <?« -f- r« -f- ««)«, whose root is w»» -f- n« + ;?« -H" + »* + *•. which added to the
sum of Iheir roots gives (2wi + 2n + 2p + ag + 2r> + 2(m«-f n« + p«-h5* + »*) = a =4/«;

whence * = ^~^"', ^^, "^^ "^^ "^ - , in which expression m, n, o, q, r, may be consecntiye numbers hi

the natural series, and t such that s may be nearly equal to one of these numbers.

Let m = 2, n = 3. /) = 4, ^ = 5, r = 6 and ^ = 10; then « = V* whence the roots of the squares are
22, 33, 44, 55, 66 and ^*, and multiplying by 4. we have 88, 132, 176. 220, 264 and 239, the sum of whose
squares is (481 )% and their sum is 11 19, . -. 11 19 + 481 = 1600 = (40)S as required.

From the solution above we readily see how any n numbers may be found havingthe same conditions.

66.— Propoeed by Bexjamix Pkircb, LL. D.. F. B. 8.. Profcivor of Mathematloi, H&rrard Univeralty, and OontnltiDg Ofometn'
to the United State* CoMt Surrey, Cambridge, Bflddleiex County, MiMachusetto.

What are the probabilities at a game of a given number of points, but at which there is only one person
who is the actual player ? When the player is successful he counts a point, but when he is unsuccessful
he loses all the points which he has made and adds one point to the score of his opponent.

Solution by the Proposkb.

Jjet <p (t, j) denote the probability that the player will gain the game, when he has t points to make,
and when the opponent hasy points to make. Let A be the whole number of points of the game, and a
the player's chance of gaining a point. It is obvious that (p (ifj) = a^ (i — 1 , j) -f- (I — a)<p (A, j — 1),
which, when j = 1, gives ^ (•. 1) = o ^ (i — 1, 1) = a'; and in general ^ (si) = a* + (1 ~ o') <P (^i ■— IX
because <p (i,jy- <p {hj — 1) = a(^ (i — I J) — g) (h,j — 1)].

This gives ^(A,» - 1 = (1 - a*)[9(A,i- 1)-1], whence <p(^/) = 1 -(1 -a*y,

<P(}J) = 1 — (1 — oO (1 — a*y~S wid at the beginning of the game ^(A, A) = 1 — (1 — a*)*.

67.~Propo«ed by Dr. Joel S. Hritdbicks, M. A., Editor of the Analyit, DeiMoinet, Polk County, Iowa.

Suppope the earth to be an indefinitely thin spherical shell, equatorial radium, rotary velocity and
gravitating force the same as at present except that gravity is assumed to be all concentrated at the
center, and none in the shell ; and suppose a particle at the equator, and on the inside of the shell, to he
separated from the shell, and to move henceforth from its centrifugal force, resulting from its motion
while attached to the shell, and from the gravitating force at the earth's center. Required the axes of
the ellipse the particle will describe, and the thne required for Its return to the same point in space at
which it was detached from the shell.

Solution hy the Propocu,
Let r denote the equatorial radius of the earth
= 3962^ miles, say ; let v = equatorial velocity per

second = , where 2/> = interval, in seconds, of

one rotation of the earth = 86164 seconds. Then Is

t^= J ...(1). Let i/ = the velocity per second of

a body, supposed to move in a circular orbit about
the center of the earth, under the influence of a
centripetal force a, and at the distance r from the
earth's center. Then, from the known laws of
orbital motion, we readily find t/ ='\/(gr);

.-. v'^=gr (2).

Let a, b represent the semi-axes of the required
ellipse AB, F, the center of the earth, and conse-
quently one focus of the ellipse, and let l=LS =
the latus rectum of the ellipse. Then, by Cor's 1 and

2rv^ 26^
2, Prop. XVI, Book I, Principia, /= ^^^ = ^ (by property of the empse)...(3). Bubstituthig for v« and

i/« in (8) their values from (1) and (2), 1= =: ^.^ = 27.374 miles (4). We have PO =r- a=

UP a ^ '

l/(a«~6«), by property of the ellipse; whence 62 = 2ar — r« (5). Substitutmg this value of 6« hi (4)

and reducing, we get a =?^^^= 1984.677 miles (6).

And this value of a hi (5) gives 6« = ^ T^ ^ .(^Tb = 7CrJ(^ / ,) = 164.817 miles.
" 2gp^ — ;rV > \2gp^ — irh-/

Haying determined the Yftlaes of a and 6» we hare, bj Kepler's second law,
^ptfidt I di -.1 nah : <, or ^fWt : 1 : : nah : <,
p being the radioB-rector and vi the Telocity in the ellipse. But for eqaal portions of time ovi is constant
liy Eepler*s second law, therefore, if the unit of time be one secono, pvi = nr, and oonsequentfy

t= -J^ = 29^ 55''. Therefore the particle will retam to the pobit hi space at which it was detached,
in 29 minutes and 55 seconds.

8olT«d in a liiiiilAr manner by £L fi. iSnCz.

•8.~PropoMd by Prof. David Tbowbridok, M. A., Watoiburs, Tompkins Gonntj, K«w York.

A sphere is divided at random bj a plane, and then two points are taken at random within the sphere;
find the chance that both points are on the same side of the plane.

I.— Solntion by Abtxm ab Mabtin, M. A., Member of the London Mathematioal Society, Erie, Xile Oonnty, Penn^ltanta.

Let X be the distance of the dividing plane from the center of the sphere ; then the sphere is divided
into two segments having a conunon base, radius i/(f< — a^}, and altituoes r — x and r -f 0.
Let r= volume of the smaller segment ; then
- K = ;r(r« — aJ»)x K*" — »)+ *<*•— «)*, —\n{^ — 3r«a; -f aj»), and kni^ — T = volume of the other

segment. The chance that both points are in the sesment F is tt .x.» and that both are in the other

segment, f^^^\^ ' ^or a particular value of x.

Hence the chance required is

Solved in a similar mairaer t^ Woilar 8, NicMg,

II.— Solntlon by 1. B. Sbxts, OreenTlUe, Darke County, Oliio.

Let x=the altitude of one of the segments into which the sphere is divided ; then the volume of the
segment is inr2>(3r— x), and the required chance is

26

If n points be taken at random within the sphere, the chance that they will all be on the same side of
the plane is

'= rir -« - , = 2*.,*.+l Jo (^'^^

^Hr^n+i L 2/1 -f 1 "^ (2n + 1)(2« + 2) "^ (2» + 1)(2/» + 2)(2n-|- 3) '^'"'^ (2/i4-l)(2n-h2)..(3/i+l)Jo

(2n + 1)(2« + 2) ^ (2» + 1 )(2/» + 2)(2n -|- 3)

"*'(2»-hl)(2
If» = 2, /j = |f; ifn = 3,p = «, Ac.

2 2^i» 23ii(n— 1) 2»+Ml 2.3...ii)

2/1 + 1 "*" (2» + 1)(2ji -f 2) "*" (2» -I- l)(2ii -f 2)(2n + 3) "*" "" "*" (2fi -f 1) (2n -f 2)...(3»-|-l)

6*.— Proposed by Dattibl Kibkwood, LL. D., Professor of Mathematics, Indiana State UnlTeiBlty, Bloomlngton, Monroe
Oonnty, Indiana. ' *

Given the radius of Mars, 2250 miles,' and the radius of the orbit of its inner satellite, 5800 miles ; to
determine whether the latter can have an elastic atmosphere, supposing its diameter to be 45 miles, and
its density equal to that of the primary.

Bemariu by the Pboposbb.

Soon after the discovery of Mars* satellites my attention was turned to the following passage in La-
place's System of the World : —

''If other bodies circulate round that which has been considered, or if it circulates itself round another
body, the limit of its atmosphere is that point where its centrifugad force, plua the attraction of the ex-
traneous body, exactly balances its gravity. Thus the limit of the moon's atmosphere is the point where
the centrifugal force due to its rotary motion, plus the attractive force of the earth, is in equilibrium with
the attraction of this satellite. The mass of the moon being 7^ of that of the earth, this point is therefore
distant from the center of the moon about the ninth part of the distance of the moon from the earth. If.
at this distance, the primitive atmosphere of the moon bad not been deprived of its elasticity, it would
have been carried towards the earth which might have retained it. This is perhaps the cause why this
atmosphere is so littie perceptible." Book IV, Chap. X,

— 68 —

This statement gives us tiie equation . ^m + »^ = ^' where if = the mass of the central body, m=

that of a satellite, a = the distance between their centers, a; = the distance from the satellite to the limit
of stability, v = the satellite's velocity of rotation. Solvhig this equation for the earth and moon we
obtain the result given in the foregoing extract. When we solve it, however, for Mars and its inner
satellite we find the limit of stability wUhin the mrface of the satelUte.

Hence, accordmg to Laplace, the latter can have no atmosphere, nor can the body itself ever have
existed in a fluid form. Can this result be accepted ? K not, the statement of Laplace is erroneous. Let
us inquire, then, what is the diBtnrbing or divellent force of the earth on a particle of matter between
itself and the moon, but in the vicinity of the latter? It is well known that the moon's attraction pro-
duces terrestrial tides by the difference between its force at the center and at the surface of the earth. The
case would evidently be the same wilh the earth's action on the moon, or with Mars' action on its satellites.
But this will £^ve us a result very different from that obtained by the formula of Laplace. We conclude,
therefore, that for anything we know to the contrary, either satellite of Mars may be surrounded by an
atmosphere.

Note by Profissok Pkircb.

I have satisfied myself that the limits of atmosphere given by Laplace are incorrect ; and that there ia
in reality no Jimitation which can be given to the night of the atmosphere of a satellite.

70.— Proposed by John H. A daub, Ck)chrftatoD, Crawford Ck>iiDty, Pennaylvania

In digring a well 6 feet hi diameter a los 3 feet in diameter was found lying directly across the center
of the w^l. How many cubic feet of the log must be removed from the well ?

I.— Solution by Artexas Martzh, M. A., Member of tiie London Mathematloal Sodely, Brie, Erie Connty, PennsylTsnia.

Put R = d feet = radius of the well and r = IJ feet = radius of the log.

Let the part of the log that will be removed be intersected by a plane parallel to the axes of the log
and well, and at a distance x from them, and 7= the required volume ; then 4(Ifi — ^Xr^ — »^)i = area

of the section, and V= ^f^C^" — ^')K^ - ^)^dx.
But

(/f* ar'}*-/f 27? 8iP 10/?* 128/?^ 256*' 10247?>i mSR'^ "' 3.4.6 8... 2/* ^^^-^ ®^"
Multipljiog this by ^f^—x*)hdx and integratmg each term separately by the formula

j->(^--^)*^= (2^^^'- we have finally

y—^r^Hxy^ 16/?^ \2dR* im8/2« 32768/2« 262l44/?'o 2097152/?"

_ 14l57r" _[l.8.5.7...(2»--3)K2rt- l)r-^>'+'__ \
67108864/?" [2.4 6.8. 10...2»K2» + 2)/f--+» ^^' r
_ /1_ 1 _ 1 _ 5 _ 35 _ 147 _ 693 _ 14157 _ \

~^'"^\2 64 2048 181072 8388608 268435456 8589934592 109951 163777t5 /

= 41.041 + cubic feet.
II.— SolntloD by Abtimas Martin, M. A., Member of the London Matfaematfoal Society, Erto, Brie Ooonty, Penn^ylTanta.

Put r = radius of the log, and R =■ radius of the well. Take the origin of rectangular co-ordinates at
the hitersection of the axes of the log and well, the axis of the log being the axis of p and the axis of the
well the axis of z ; then the equations to the surfaces of the log and well are

«a+2» = ,« (1), afi+t/^=R^ (2>

Let F= the volume removed; then V=^S J jj d3Ddydz = S Jj yditdz :, (3>.

The limite of y are and (/?*— a^; of «, and (*«— jr«>; of x, and r.

.-. r = 8fj(r«— flr0*c^»/y = 8 fJ(f^-d')h(R'-x^)kdx (4).

Let n = f , and z = rv; then

= f JP [d + «=)B(0 - (1 - ^)FW] .

-69 —

71.— Proposed by Bkmjamin Pkxscb, LL. D., F. B. 8., ProfeMor of Mathematioe, Harvard UaiTenity, aod Consolting Ctoometer to
tho Unitod States Goast Sarrey, Oambridge, Middlesex Goanty, Maasachnsetts.

Given the skill of two billiard players at the three-ball game, to find the chance of the better player
gaining the victory if he givea the other a gmnd cKMouti^.— [From Our Schoolday Visitor, vol. xv, p 220.

Solatlon by the PmoposKiu

Let the two players be (A) and (B), whose probabilities of making a sin^ shot are respectively a and 6,
and let (A) give the grand discount. (A) looses all he has made when (B) succeeds in making a shot.

Let A = the number of points to be counted per game, A = ^ ° . » ^= — , ,_-7 so that
l-A=a-a)B, l-5 = (l-6)^, ^-a = ^(l-a)(l-6), 5 — 6 = 5(l-a)(l-6).

When (A) has « shots more to make and (B) has^' shots more, let F (i, j) = (A*8) probability of winning
when he has the play, and f (t,^) = (A's) probability of winning when ( B) has the play. The fundamental
equations are, obviously, F(t,i) = aF(t-l,/) + (l-o) f (»,/), f(t,i) = ftf (A. j-l)+(l-6)F(i »,
which give by substitution, transposition and division F (t, j) = A.F(i— 1, j) + (1 —A) f(h,j— 1).
In special cases, we see that F (i, 0)= f (t, 0) = 0, F (1,1) = ^, F(0,j) = l, F(i, l) = ^.F(t — 1, 1)=^';
whenceF(»,/)~f(A,i~l) = ^[P(i-l,i)~f(A,i-l)] = ^'[l-f(/^i-l)].=^' + (l-^Of(M~l^^
F(t — l,j) = -4*-»4-(l — ^<-»)f (A,i — 1), which, substituted in the first fundamvDtal equation,
giveB^'(l-6)[l-f(^i-l)] + f(A,i-l)==:f(i,i);andifc = ^*(l-Hf(/^i) = a-c)f(A,i---l) + c,
l-f(A,i) = (l-c)[l-f(A,i-l)] = (l-c)i;F(si) = ^«-f(l-^0[l-(l-cy-»],
= 1 _(l _c)/-i(l _^<); p(/^ A) = 1 - (1- c)*-»(l - ^*).

79.— Propoeed by Dcnlap J. McAdam, M. A., Professor of Mathematics, Wasbinston and Jefferson College, Washington,
Washington County, Pennsylvania.

The center of an epicycle whose radius is ^ revolves on a deferent, radius a, with uniform angular
velocity v; a particle revolves in the epicycle so that the radius drawn in the Fmall circle to the particle is
always parallel to Itself. Supposing the pvticle to be moving uuder the influence of a force at the center
of the liu^ circle, find the law of the force.

Solution by Waltbb Siverly, Oil City, Venango County, Pennsylrania.

Let 8 be the center of the deferent, and a its radius, O the
center of the epicycle and P the position of the particle at any
time, ABB the diameter of the deferent to which the radius
drawn in the epicycle to the particle is always parallel. On
SB take 8C = OP, join SO, OP, PC and 8P; SOPC is a paral-
lelogram, CP = 80, hence the path of the particle is a circle,
center C and radius o. Let O', P' be the next consecutive
positions of O and P; 80', CP' being parallel, angle 080' =
angle PCP'; PP = 00', and the particle moves with the uni-
form velocity that the center of the epicycle moves on the
deferent.

Supposing the particle to move under the influence of a
center of force at 8, draw a tangent to its path at P, produce
SO to meet the tangent in H; 8H is the perpendicular from
the center of force on the tangent, which p\xi=p; and put

8P = r,OP = 6. Then cos 08P = ^^*'"t^» 8H=rcos08P;
' 2ar

For the velocity we have the well-known formula, v= » h being a constant.

a3 — 62-f r3

. *. » = , _: jLr4r's' ^^* ^^ being constant, the velocity varies inversely as o* — 6» -f- r«, and the given

velocity being constant, the supposition that the particle moves under the influence of a center of force
at 8 with the given velocity is impossible. Supposing it possible for the particle to move with the
velocity found for a center of force at 8, to find the law of the force we have the well-known formula,
« h* dp „ ^ ... ..__.^ _ , ^ _ ^_^ „ SaT-hh-

V dr

Substituting the value of p we find F =

varies inversely as

(a^ — 62 -f- r^y

(a^—b^-^r^f

,. or, 8a^A« being constant, the force

Note by the Proposer. —Problem 72 gives the conditions involved in the Hipparchian method of
representing the moon's motion. Hlpparcnus imagined the moon to move with uniform velocity in a
circle of which the earth occupied a point at a distance of ^ radius from the center. Another method of
considering the motion was by means of an epicycle such as given in the problem. He constructed his
system so as to *'save the observations." If he had constructed a svstem whicii would save the laws of
motion of a body under a central force, he would have left out the condition of ** imiform velocity. "
His entire theoiy would then have been abandoned.

I

— 70 —

73.— Proposed by Abtkmab Mabtiw, M. A., Xrie, Erie Ooaa^, P«nnqrltania.

Find the average distance between two points taken at random in tlie surface of a given semicircle.

I.— Solatlon by E. B. Sutx, OreenTlUe, Dark* County, Ohio.

Let ACB be the given semicircle, O its center, M, N two
points such that the Ime through them intersects the arc of the
semicircle in two points C, D, and M', N' two points such that
the line through them is parallel to CD, and intersects the arc of
the semicircle in C and the base in jy. Draw OH perpendicu-
lar to CD.

Let OA = r. CM or C'M'=aj, MN or M'N'=y. CD = tt,
CD' = r, / AOH = B, Z COH or C'OH = tp.

Then tt = 2r8io^, t; = r(sin cp + tan dcos ^); an element of
the semicircle at M or MMs r sin (pd<pdx, at N or N' it is dOjfdy,
The limits of 9 are and ^tt, and doubled; those of x are
and tt, when ^ < 0, and and v, when ^ > 6; those of j/ are
and X, and doubled; and those of ^ are and B, and 6 and n — B,

Hence, since the whole number of ways the two points can be taken is {ith*, the required average is

= ^-J'*'( r* 16sin»<pd<?)+J''^(8in<p + tanOcos<p)*smipd^)rfe,
= ^r p. (8-7co9e-2sin^cose)de= 256r _ 147?r.

II.— ^lutioD by the Psoposxb.

liCt P and Q be the random points, P being nearer than Q to AO, 0Q = 2;, OP = y, /A0P = 9^

Z AOQ = 0, PQ = z and J = average distance required. Then 2 = [aj« -f- .v« — 2a;^cos (B — <p)]k, and

Jo Jo Jo Jo ^^y^^

^ di?SoSofo [l^^'^'KO -<?>)+ 128in3K0 - <p}co8 (B - cp)-2 + 8co8*(e - <p)

+ 8sin«(0 — 9))cos(0 — <p)log-( 1 +co8ecK0— <?>) '^'^dxdBdg),

"" l£^/o'/o* [Itein'JCO - ^) + 126in3J(e - ^)co8(e - <p) - 2 + 8coe«(0 - cp)

+ 38in«(e — <p)co8(e-<?))log^ l + coscci(0 — <?>))• ]rf«W^

= 4^^/o' r^Ssin^JOcos iO+ 128hi»J0cos J8 — 328in«4©cos J© — Gshi JOcos J©

— 64COS i© + 64 + 9am 6co9 B + Gsin^eiogCl + cosec iB)}dB,
^ 256r _l472r a t- ^ r 6v ^ , /j ,

45«r 1.35;r«'
This problem wm alio solved in an elegant manner by Henry Heatou.

74.— Proposed by Rer. W. J. Wxioht, M. A., Ph. D., Member of the London Mathematical Society, Oarlisle, Cumberland County,
PonnsylTanla.

Transform the Eulerian integral F (n) = J er-'oif^-^dx to the Legendreian F (w) = J (^log ) dx.

Solution by ProC D. J. McAdam ; ProU H. T. J. Lvdwio ; Prof. F. P. Mats ; and E. B. Skits.

Let€-' = y, then — a; =logy, ora; = log(-^ andda;= ^. Porx=oo, y=0; fora;=0, y=l; there-
fore the limits must now be and 1. Substituting, F(n) = — J ^ {log j dy = ( (log-^ <^.

Shice the value of the last result is independent of the name of the variable we may write x for y,
which gives the result required.

75.— Propeeed by Prof. H. A. Wood, M. A., Principal Gozsackie Academy, Gozsackie, Qrsene Oonnty, New York.

A Sphere 4 inches in diameter, specific gravity 0.2, is placed 10 feet under water. If left free to move,
what will be its velocitv at the surraoe of the water, and what will be the maximum hight it wiU attain?

— 71 —

SoIntioB by Waltkb SirsBLY, OU City, Venango Oonnty, Pennqrhrmnia.

Let r = J = radkiB of the Bphere, p = J = its specific gravity, A = 10 f eiet, r = the velocity acqaired in
ascending a distance x, F = the velocity at the snif ace of the water, i; = the resistance.

For the motion of the sphere, vj = g\- — ^}— ^' Putting gy 1}= fir', da; = zt^Tjj-

The required hight = J- = ^ "^"(l -^)^ ^'^ nearly.

The value of k deduced from Newton's Princlpia, Book 2, Prop. 38, and used by Atwood, Hutton,
Simpson, Emerson, and all English authors until recently is t^— • This value of k gives r= [V^^— P)}^

= W(fig) nearly, and A = £' = ^^^ = ^ = 4^ faiches nearly.

Poisson rejects Newton's Theory of Resistances, and in Vol. 2, Art. 367 of his Mechanics gives an

9
approximate value of 1;= ^ . This value of k gives A = 31 inches. Whewell, Eamshaw and Tait do

not give any method for finding the numerical value of k^ and Rankine in his Applied Mechamcs, p. 599,

gives for the resistance in a fluid, ^=~q^~* where e is the wei^t of a unit of the fluid, A the area of the

greatest cross-section, and k for a sphere = 0.51.

SoItmI alio by Prot De VoImoh Wood^ who, by uatng the coefficient of raditanoe glren by Bankine, flnda A— 0.338 + fMty«4.056

7€.— Propoeed by J. J. Stlvkstbb, LL. D., F. B. 8., Goireepondlng Member of the Inatitnte of France, ProtaMr of Mathemati ce
Johns Hopkina UniTenlty, Baltimore, Maryland.

Prove that the equation to the sums and the equation to the products of the roots of an equation of
the nth degree taken two and two together may each be put under the form of a determinant of the order
^fi — 1) or of the order |n according as n is oidd or eveu, and write down the determinant of the third
order which equated to zero is the equation (of the degree 21) to the binary products of an equation of the
7th degree clear of any irrelevant factor.

[We hare reoetred no lolntion of thla problem. We hope to reoeiTe one f6r pobUoation In No. 4.]

77.— Proposed by B. B. Sxirz, GieenTllle, Darke Oonnty, Ohio.

Two points are taken at random in the surface of a quadrant of a circle, and a line drawn through
them. Find the chance (1) that the line intersects the arc m two points, (2) that it intersects the arcm
one point, and (8) that it does not intersect the arc.

Solation by Henbt Hxatom, Principal of Public Schooli, Sabnla, Jackson County, Iowa.

Let />!, ps and pz equal the required probabilities in the order named.
Then it is evident that ;h + j>s + i^ = 1- To find pi let M in the figure
be one of the points, then the other must be either on the surface MGB
or on AMF when the line intersects the arc in two points. Put AC = 1,
AM = iv, /MAB = ©, surface MGB = tt and surface AMF = ty. It is
evident that the sum of all values of u equals the sum of all values of
V, Hence we may use 9tt instead of tt + V. u = — sin^ — |'i/2iz;sind.

The probability that the first point falls on the element of surface at

M is — . — - The probability that the second point falls on either u or
V is -7^^- The limits of a; are and 2cos (iir + 0) = a/ ; of 0, and i^r.
Hence Pi = ]^ (^ C'^2uxdx(»

irJQ Jo

28

To flndi>3> let N be one of the points, then if the line does not intersect the arc, the other point lies
either on the surface ENB or on AND.

Put AN=y, /CAN = ^, AENB = t«i, and A AND = wi. Then, as before, 2ui may be used histead
of ui -f- t?i. Ill = I — |tan (p — |v^2ysin (iK--<p). The limits of y are and sec <?> = y' ; of <p, and i^.

Hence ii5=^,j^ Jo ^"^^^^^'^'^ 3^ Jo (»c«^-tan<?)secV)rf<P=g^- Ih=^-Pi-P^ = ^^-

This proMett was also solred rery elegantly by the Propottr,

= l^/o^'Wl - 2sin ©COS e) - sfai'e — 2sin ©coe»0 + iem^cos^yiB = 1 -

— 72 —

7ft.— PropoMd by Dx Vouoir Woo», M. A., O. E^ ProfMwr of HiUhMMtioB ud MMhuici, SleT«M laitftate of T6eliiiologj»
Hobok«D, Hadson Oomtj, New Jenej.

One end of a fine, inextenaible etrinff is attached to a fixed point, and the other end to a pohit in the
Burface of a homogeneouB sphere, and the ends brought together, the center of the sphere bemg in a hori-
zontal throng the ends of tne string, and the slack string hanging yertically. The sphere is let fisll and
an angnlar velocitj imparted to it at the same instant, the sphere winding np the string on the circomfer-
ence of a great circle nntil it winds np all the slack when ft suddenly logins to ascend, winding up the
string, the sphere returning just to the starting point. Required the initial angular yelocitj, the tension
of the string during the ascent of the sphere, the initial upward Telocity of the center of the sphere, and
the time of the movement.

Solution by Waltu Siterlt, Oil City, Yenango Oonnty, PeansylTMiia; and the PROPOSBm.

Let / = the length of the string, r = radius of the sphere, m = its mass, k = \Xa radius of gyration, x =
the length of the unwound portion of string at the end of descent, v = velocitv at the end of aeecent, v' =
velocity t with which the sphere begins to ascend, oo = the initial angular velocity, co* = the angular velocity
with which the sphere begins to ascend, ^i = thne of descent, ta = time of ascent, T = tension ot the
string during the ascent and /=:the impulse communicated by the string to the sphere at the end of
descent.

The length of string wound up at the end of descent = rcoti =r <»'%/( - )> <i being the time of falling

freely through the hight x. .*. x = l—r<M>^yj-f^ whence « = -^^-^/r^J (1).

Ir I Ifi

For the impulsive motion, i» — q>' = -^» r + 1/ = ^* Eliminating /, - (co - a>') = r -f- 1/.

ButA« = ^»-, i;=^(2a«); .'. «' = a>~|;[^(3iA«) + i/] (2).

For the upward motion, the origin being at the point where the center begins to ascend and the axis
of y positive upwards, m -^ = T—mg... (8).

For the angular acceleration, mJ^'^'^^^f^, or 711 '^'i = -^/.. (4). Also y =rOor^ = !^ ...(5).

BUminating^ft ^]^*rom (3), (4), (5), 'T=^mg = } the weight of the sphere. Eliminating T,^^

from (3), (4). (5),^f =:-^. Integrating, observing that when <= 0, ^^ = < ^|^ = t/-fflf< (6)

dv
When^' =K>,t = ti; .-. v* = \gt2..,{1). Integrating (6), observing that when < = 0, y = 0, y = t^« — A^<«.

Whenif = x,«« = ^; .-. x = i»%- A^^«...(8). From(U (S)v'=V(W^) (9), U='yjQ^)..^(m

Prom (5)^ = r^' .'. ij'=ra)',(M)'=^^/(i^gx), (11). Substituting these values of t/, a' in (3),

x=/(6-"/35). 8ubstitutingthisvalueofii:in(J).(9),(»=^^^^^J^ ,/=•-{ i^jfl:6~-i/(85)] \

The whole thne = «i -f <« = ( 1 + ti/86) ^/( -^^ "" ^^^) J).

Tbto problem was also lolTed by Prof. F, P. Matz.

79.~Propoeed by Arvrmas Haktix, M. A., Erie, Erie Ckninty, PennBylTanbt.

Find the average distance of the center of an ellipse, axes 2a and 26, from its circumference.

Solation by Henry Hbaton, Prindpal PnbUo Schools, Sabnla, Jackson County, Towa.

irdz
Let J = average distance required ; then ^='^— , where r is the distance from the center to any

jdz

point P in the circumference of the ellipse, and dz is the differential of the arc estimated from the ex-
tremity of the minor axis.

Let X and y be co-ordinates of the point P; then the equation of the ellipse is a V -h ^'^^ = 0*^1

r = i/(j-'-f-y«)aod«/2=v'(</a^+</yO. Buty=* i/(a2— ar^), therefore r« 1/(62 -fe»a;*), puttmg «^ = *^^,

^/{a^ — e^Jfi)dx

— 78 —

•••— Proposed bj £. B. Seits, GreenTille, Du-ke Goun^, Ohio.

A chord is drawn through two points taken at random in the surface of a circle. If a second chord
be drawn through two other pdmts taken at random in the surface, find the chance that it will intersect
the first chord.

I.—Solntion by Hsnrt Hkaton, Principal of Public Schooli, SabaU, Jackton Goontj, Iowa.

Let EE be the first chord, drawn through the points
D and N. Let G be one of the two other points. Put
BC=:1, CD = x, DN=y. EG = 2, /CDN = 0, /CEG
= <f», Z CEK = i>=: sln-»(a«faie), -surface KGI = u; eur-
twoe EGH = V, required probability =/>, and probability
the two chords idll not intersect =p'»

Then u = ^ — ^^in 2^H-isin 2^>--zco6 ipmn (g>^i>).
If the fourth point falls upon either u or i? the chords wUl
not intersect. If G occupies every point of the segment
EEBHIy the sum of all values of v equals the sum of all
values of u, hence in taking the sum 2u may be used
for « -f r.

Confining G to one side of the chord EE, and N
within the circle through D, will evidently each give
half the result obtained by allowing them to occupy every point of the circle.

The probabilities, that x will have any particular value, that N will fall upon the element of surface
at H, that G will fall upon the element of surface at G, and that the fourth point will faU upon u or v,

are respectively —

%nxdx ydydB zdzdg) , u + v

and

The limits of z are and 2 cos<p, = 2/; of <^, ^ and ^tt; of y, and ^txoo%H=i^\ of Xy and 1;
of 0, — Jjrand + Jjr.

Hencep' = ^J_^^ JJ^ J^ ) ^ udB2:d^dyd<pzdz,

^ d^S-i!^ So fo fl' [8cofl»«p(24?>-2V^-8ind^4^in2<p)-8cos^«wfi»<^

^ sl* Xit' fo /o" ^^^ "~ ^^^* + *° — 1«C08«^ + 4an^i/JCOi^)flBxdxydy,

= ola/tf/^ (8^ — *2?r^ + 12^ — 16cos«^ + 4sin^^osV)co8^rf^^"*«^»

= ut^ ft*' r* (3^' — I27ct + 12^ — 16cos«^ + 4sin2^«oS2^)co8«6co6ec^6sin»^N50s ifxiOdi/.^,

= igL rtr [720«cos^ — 27e^co6'»6cosec^ + 366cos»eoot -j- o4$isoV^ — 73?r6co8^
-f 27«^co8«©co8ec^ — ISTrcos^Ocot 0— 27?rcot»6 — 27cot^ + (ISjt^ — 37)coe^ — 56cos^0 - 1 Scor-O |<I0

2 246 _ ^ , , , 1 , 245

= 8-73r^- B»*i'+/>=1» ••• /' = 3 + 72;r^-

— 74 —

Il.^-Solation by the Piohmks.

Let M, N be the first two random points, AB the chord
through them, P, Q the second two random pomts, CD the
chord throngh them and O the center of the circle. Draw
OH and OK perpendicular to AB and CD.

Let OA = r, AM = tr, MN = x, CP = y, PQ = z,
/ AOH =0, ^ COK = <?),/ AOC = ^, and// = the angle
AB makes with some fixed line.

Then we have AH = rain 0, CE = rsin <pi an element of
the circle at M is rsin BdSdw; at N, d^xdlx; at P, rsin <pd<pdy\
at Q. dindz. The limits of are and \n\ of cp, and G,
and doubled; of ^, and 2^, and doubled ; of //, and 2^;
of tr, and 2r8in 6, = to'; of a;, and tr, and doubled; of
y, and 2rsin<p, =^; and of s, and y, and doubled.
Hence, since the whole number of ways the four points
can be taiien is it*i*^ the reqmred chance is

= ^^ Jj^(89« - aOsin 6cos 9 - 468fai»©co8 + 3 sin«6 + sin^) sfai^d©^

SI.— Proposed by Artbmas BCaktik, M. A., Srl«, Erie Oonnty, PentiiylTMUa.

A fixed hemispherical bowl, radius Ry is full of water and contams a heavy cylindric rod, length 2a,
radius r and specific gravity p, having one end against its concave surface, and resting on Its rim. De-
\ inclination of the rod.

termine the i

Solution by B. B. Sens, CheenTille, Derke Coanty, Ohio.

Let O be the center of the bowl, BDD^' a vertical section
of the rod through its axis, A the point at which the rod
touches the rim, B the point at which it touches the concave
surface, G the center of gravity of the rod. Join E and E',
the middle points of AB and BT ; then the center of sravity
of the immersed portion of the rod is at some point of EE' ;
let H be the point. Draw AC perpendicular to AB, meeting
BO produced in C ; draw the vertical line CS, and johi BS.
Draw HE and GL perpendicular to BS, and HR and GN per-
pendicular to AB.

LetBN = ND=a, AO = i?, BB'=2r, AB = »i, B'F = n,
/OAB=/ AB8 = 0, the inclination of the rod, and tan ^==t.
Then we have m = 2i?coeO, n = 25co6 — 2rcot©, ^BC8 =
900 — 2©, BS = 2i?cos20, BL = ocoeO — rski 0, LS = 2i?cos26
— acosO + rsinO,

HP — !<^+ ^'») _ SiJrcosO — 5f^cote _ 8/?rco8ecot^— br^eol^^

"""■ 4(iii + n) ~ 8i?coflQ-'^XM>te' 16iJcosO-8rootQ

^ _ 16ig«C06^ e — 16/?rco8 ecqte -f St^^cot^© r^__ _ S/grsJnQ — 5r«
" " 16Jf?co8 6 - 8rcot » ^^ - g/jcos Q - 4rcot O'

— 75 —
_ 16Jg %m«ecofl»e — leigrrin B + &t^H^^ + 5i^ xrc, _ 32^rsiD ^ + (16ig» — 5f^) 8in«9 ~ 48i PBm<6 — 5i ^

The forces acting on the rod are a downward force at = 2^09^/9, an upward force at H =
^f*(m-\-n)=7tr^2RcoBB — TCotB\ and the reactions of the bowl at A and B, which forces meet at C.
Taking moments of the forces at G and H about C, we have L8 . 2ieaf«p = KS . itr^(2Eco8B — rcoiB).

Substituting the yalues of LS and ES, and transforming the equation, we find
5f«<w - aOJPi" — (128aErp -|- 64a2rp + G4J^ — 40r«)<«> -\- (612oif»/> — 128af«/5 + 256a«i?p -|-256i?»— 896i?r«>P
+ (\(Xi4aBrp -f 1664i2«r — 128o«r/o + 66»<»>« -r (2660aiP/5 + 384of«/o — 256a«i?/5 -|- 2816JP -f 376fir«)r
+ 2804ari?p<8— (2560aiP/5H-«84af*/5+266a^i2p— 2816iP~876-Br«>64<l(>24ai2r/9^^

4<512aJP/c>-128<w-/5— 256a»5p— 2MiP+896i2T«)<»— (128a5r/»~6^ — 64iPr-f-40r«)<»+20-Br3< — 5r» = 0,
an equation of the 12th degree for the determination of 9.

9a.— Propoted by John M. Wilt, M. A., Fort Wayne, Allen Oonntj, Indiana.

An unknown cone is cut at random by a phme; find the chance that the section is an ellip6e.-:-[From
the Normal Monthly, vol iii, p. 108.

Solntion by HsimT Hxatoh, Principal of Pnblie Sohooli, Sabola, Jaokaon County, Iowa.

Put a = altitude of the cone, 20 = angle at the vertex, ^ = ang]e the cutting plane makes with the
axis of the cone, i» = the chance that the section is eUiptical and g, that it is a complete ellipse, supposhig
a and known; and P = the chance that the section is elliptical, and Q, that it is a complete ellipse,
supposing all yalues of a and 6 equally probable. The ntunber of planes that can cut the axis at .the
angle g) is proportional to cos ^.

If ^<0 the number of planes parallel to each other that can cut the cone is proportional to 2atan8cos^
the distance between the extreme parallel planes that can cut it.

If ^> B the section is elli^ical and the number of parallel planes that can cut the cone is proportional
toasec6Bin(^+6). In this case if the plane cuts both sides of the cone the section is a oompfefe eOipse.
The number of parallel planes that can do this is proportional to osec 6shi (<p— B).

jr>tanOcotf^+/;'a«cOBin(^+W i + (i* + «rt«'e-l cot'>+6+>

I have not been able to integrate these expressions, but by taking the ordinates 5^ apart and applying

Q !/<*»= jQi;yo+y2 + y4 + y6 + 5(yi-fyii-fy6) + yi,],

.o <»ch 30O. Iget/f SirSS = ••*«^' '-^So'cotT+e + i. = »-^«^- Hence P = 0.8538
and Q = 0.1880.

In a solution of this problem in the Lady's and Oentleman'8 Diary for 1861, given by the Editor, the
values of p and q are the same as found above; but the values of P and Q are found by integrating the

numerator and denominator respectively, thus getting P = *^-^ = 0.2532, and Q= ^~= 0.1202.

This method is evidently incorrect, since the number of planes whose section is an ellipse is divided by
the whole number of cutting planes, all planes not being eqwilty probable. It is the evident intention of
the author of the problem that all cones and also all planes that cut the same cone, shall be equally
probable. Hence it is evident that the probability for any given plane depends on the number of planes
ttiat can cut the cone through which it passes. Consequently all planes are not equally probable, and the
solution referred to above is incorrect.

— 76 —

ftS.— PropoMd by Abtem as Mastw, M. A^ Erie, Bri« Ooontj, PwuiqrlTuia.

Find the average distance of one corner of a rectangular solid, edges a, b, e^ from all points within it.

I.— SolatioD kj B. B. Sens, CkasiiTllla, Dwka GkMin^, Ohio.

Let AF be the rectangular solid, P anr point within it.
From A draw AF, the diagonal of the soud, and AC, AG,
AE, the diagonals of the faces. The solid consists of the
three pyramids, ABOFQ, ACDBF, AEFGH. Produce AP
to K hi the face BCFG, and draw KN perpendicular to BC.
Let AB=a, AH = 6. /J) = c, AF = «, AG = «i, AC = ea,
AE = «3, AP = a;, AK = ir', ^ BAN = 0, / PAN = tp.

^=:tan->(^-), G' = tan-iQ, and let ^ = the required

average, Ji = the average distance of A from all points
within the pyramid ABCFG, Aj = the average diBtauce of A
from all points within the pyramid ACDEF, and J^ = the
average distance of A from all points within the pyramid
AEFGH.

Then e = \/(a^ + 6^+c«), *ii ='/(a»-|-6*), «i=l/(a«+c3),
e^ = -v/C^-l- <J^)» a?' = a sec Qsec (f>; an element of the pyramid

ABCFG at P is a^'^cos^x/Grf^^; the limits of are and ©'; of <p, and <?/; and of «, and y. Hence,
since the whole number of points within the pyramid is ^o^c, we have

^' = Jc/o*^ So* So' ^^ <P'lM<pdx= llf^'^S*' sec^C^rf^.

= g^j;^[6coe V(«^ + ^^) + a»log |*cos9 +>/fa»+ 6"-coe^)J ]^^^

. •. J=^(A+ J, + A) = ^ - jg^tan-

12*le*^-'©

<? . ,/at

r;ar. If a = 6 = c, J = ia[>/3 — ijrH-21og(2H-i/3)].

II«-^Iation Xxj the Proposeb ; and H. T. J. Ludwio, Profenor of MathontaUca, North Oarollna Oollege, Moant PlMMuat^
Oabamu Coantj, North GaroUna. ^ -« » •»

Let X, ^, z be co-orduiates of any point P within the solid, the orighi behig at the comer A. Then AP
= V{^'¥t^-\-^\ and if ^ = the average distance required we have

=Ls:s>^--^^^w^s:s>^'^^^c-^^$^r')'^-
=Ls:^^-^^^'^^'^Ls:^^^-M'''%w'>

^Ls:<^^^^^^^^%Uir')^-^s:'-^<vi^^.^,^

+^))<te.

+^--s"V-+'^^"'^*''^"*^'

6«

= 4 +

-,i--c:)-,L'"-(")-,.r^""-C)-

12a

— 77 —

»C~PioiK)Md bj I. p. HoBTQir, AllM^ HUlidU* Goanty, MkdMcMi.

A fox starts from a point 100 rods due north from a hound, and runs due west at the rate of 10 miles
an hour; at the same instant the hound starts in pursuit, at the rate of 15 miles an hour, always keeping in
a direct line between his starting point and the fox. Required the equation of the ettrve described by the
hound, and the distance he runs to catch the fox.— [From the Normal Monthly^ vol. iii, p. 85.

I.— Solntlcm by Ammmab Mabtim, M. A., Member of the London Mjtthematioal Society, Erie, Srle County, PennqrltanU.

Let A be the stai^g point of the fox, the starting
point of the hound, i^tne point where the fox isxsaughC
and the curve CB the path of the hound. Take any point
H in the curve, and from C draw a line through H meeting
AB in F; then when the hound is at H the fox will be at F.
Draw HE parallel, and HD perpendicular to AB. Take the
origbi of rectangular co-ordinates at C, and put AC = a»
« = fj = ratio of speed, EH = AD = a;, CE = y, AF = w
and curve CH = z. From the similar triangles HCE, FCA

we have xiy. : w : a, whence to = - (O-

Also, z = nw= - ^ (2), and z = fv(cb^ + rfy») (3).

y ^

f\/ida^ -f df^) = ^°*...(4), whence, differentiating,

•«fa.+rf».) = '«-(??^^>..! (5). Pnttingg=p.

(6) becomes yV(l+p') = 'x>py—naai (6).

Differentiating (6) we get y.('[^'^ + ^»^/(^ +p')=nadp. (7).

Dividing by 2(1+P»)*. rfy(l+p.)»+-yg^=g^J^...(8). Integrating(8).KH-P')i=~'/^jg,^j ..(9).

Letp=2«n/(»-l). thendp=?^^^>;'?.«d KH-p')* = ««/^'^g^/> :..(10).

^1 + p')^= naH(>/2, to) + C, where the notation H(>/2, w) denotes an arc of an equilateral
hyperbola, semi-axes 1 and abscissa to. When y=0, a;=rO, p = and to= }; therefore C = 0, and

y(l+j,«)i=naH(i/2,u;) (11).

But. = ,/[i + i,/0>. + l)] and from(6)p = '^!^4_^^;^^^^^^

is the equation to the curve described by the hound.

When the hound overtakes the fox, y = a and (6) gives x = ap — ^CL±^*^.

II

We can integrate (9) by series, and then by reversion find p in terms of y; and then when y = a we

have p in terms of n; but the resultin^c series converges slowly when n is less than 2.

II.— Solution by AKTXMAfi Martin, M. A., Member of the London Mathematioa) Society, Erie, Eiie Co., Pa.
Let CE = y, HE = x, AF = to, CH = 2, CA = a, n = |f = f a : to : : p : x. And w : z : . I : n;
.-. aai = ioy...{l) and z = nw,..{2). Differentiating (1) and (2), adx = wdy'hydw..,(3u and rfs = wrfir...(4).
Assume cix = sin <pdzy then dy = cos gxiz and by substitution we get from (8), y = na sin <^ — iitr cos q>..,{5).
Differentiathig (6), dy=7ia cos q)d<p -\- nw sin <pdtp — n cos <pdto. But dy = cos (fdz = n cob (pdto; substituting
weget2cos<^io — u;8iu^^=^acoe<^<p (0). Dividing by i/(co8 <p),

2diri/(cos <p) — ^^^^^^ = adq)v^{cm<p) (7). Integrating, 2wV{co8<p)=:af d(p\/ {cob (p) (8).

Let cos ^ = ^, then ^ = cos-1.^ d^ = - -^j, and u.. = - a f -^^-^0^^-^^^^ (9).

Let l-.« = ^^ then. = >/(l-^). rf, = -^^-^^. and u.s = af^^}0^^.

— TO —
When y = 0,«=0, 1*^=0, ^=0, « = ! aiid<«0; .-. 0»(%i0f s««/2[JKil/^ 0--*F(h/2, 0]>

= ai/2[B(|i/2, •l-C08^)-iP(h/2, l/l-oos^)] (10).

Flrom(l), ti7=— ; substituting in (5X^ = nay rin 9 — ncu; cob ^. Solving for cob ^ we find

wliich is the rectangular equation to the curve described by the hound.

I am indebted to LeyhownCt MathenuUieal Bepomtary^ New Series, Vol. 3, pp. 905—6, for the methods
of integration employed in this and the preceding solution.

Ml»— PiopoMd by J. J. STLTXffntB, LL. D., F. R. 8., Gorretpondliig Member of the Inetitnte of Franoe, Proftwor of Mathwnetlci,.
Jbime HopkiM UniTenltj, BeltUnore, MatylMid.

Let JT be the coefficient of any power of a in the expansion of j-. :rjz — -rrTi — 3r-c — 7i — zt-v

^ *^ *^ (1— a»)(l— ai«a)(l— a^)....(l— a!*a>

where t is any integer, and let X be arranged according to the powers of x; prove that its coefficient will

form a series reading the same from left to right as from right to left, possessing this property that as we

pass from either end toward the central term or terms the coefficients may increase or remain unaltered

but can never decrease.

BolQtioii t^ Ber. W. J. Wbioht, M. A., Ph. D., Member of the London Mathematicel Society, O^pe May Pdnt, Gape M«y Co., N. J.

Let /(aEto-|-xi<X'Hcia'+")=-^o+-^ia+-^sa*+*-(l) in wliich is represented the expanded denominator of
the given fraction.

By the Theorem of Arbogast any term of this expansion, as X» = l>^^xi . fiXo + />"^-«aji' . /ax^ +

+ /)aji*-» ./•i-i«o + «i"VWBo; in which D, 2>«, 2>»^i are derivatives obtained by differentiating (1) as a

function of two variables, and which are shown to follow in an invariable order with coeffiqients which
are obtained by dMded differentiation, in which also by a similar process of differentiation the derivativee
otfxo are expanded hito a series, as D^/Xo = i)*-'aJi . /lOJb + D"*-^* . /saso + + aJi"» . /^aib.

In this case we have to expand (l — xia-^x^— )-* in which we know the value of xi to be

a;-f a^-f + a^andzo=l; hence Jro = l and Xi=xH-»2-f- -fa^.

The equation above for the values of X^, Xi may readily be found when the derivatives have been

written out as Dxi = xg, D^i = xs; 2>x,« = 2xiX2, Dh:^i = ^xiXa + aji«. This indeed is a little different from
the derivation of Arbogast, but it immediately follows from a divided differentiation of (1). The

derivatives of fxo =/l are seen to be D^f\ = — D"^^^ + D^^i^ — ± ^i" and ii>e expansion ia

(1 — xia + xao2 — )-» = l+XiH-(xi«— 2>aji)a« + (Xi» — Z>Xi'^ + 2>«Xi)a'+

Now it will be observed that in this expansion X| is the leading letter in each term, and since it is the

sum of powers its second and other powers as (x + x^+ +a;*)" will give coefficients which increase to-

the central term or terms and can not decrease, a relation which manifestly can not be changed by any of
the succeeding terms in Xn.

Illustraium.'-Tjeii^S, then Xs = (x + a;« + a^y ~ 2«iX2 -f Xa = a? -f ar« + 2x» -f 2a« + Sx^ -f a;^ + a^»
where aXjXs = 2(a: + a? 4- x*) (a? -I- X* + x») and Xs = a:*'.

Profemor Stltsster eent the following note with the problem :

*' The questioner proposes to call such series graded series. It may be proved that the multiplier of

every combination of powers of a, 6, c, in the expansion of ^r-ri .rj rr— where a, «, i

(a, X, t)(6, x,j){c, X, a;...

represents (1 — o) (1 -- xa) (1 ~x3a)...(l — x*a) will also form a graded series. It seems also to be the <
tliat the expansion of (a x, i) (6, x. j) (c, x, k).., possesses the same property of generating graded series,
and that a very great further generalization may be made of the theorem by introducing powers, and also-
by changing the form of the simple factors, the function (a, x, i) always of course retaining the general
form <pl . tpx . q)3fi . cpx^,.,q>x^\ but beyond the theorem first stated in this note the proposer of the^
question has not proceeded in the way of proof, nor does his method of proof for tliis case appear capable
of being extended beyond it.''

— 79 —
Find the ayerage cUatanoe between two points taken at random wiihin n rccfangular Fr>1id, cdgf b a» 6, e.

I.— Elation by the Pmoposciu

Let AH be the rectangular solid. Draw AH,
the diagonal of the solid, and AC, AG, AE, the
diagonals of the faces. The solid consists of the
three pyramids, ABCHG, ACDEH, AEFGH.

From A draw AP to any pohit P within the
pyramid ABCHG, produce it to E in the face
BCHG, and draw EN perpendicular to BG.
Form the rectangular solid PRUSVHTQ.

Now if AP is the distance and the direction
of the second random point from the first, the
Volume of the solid Pu represents the number
of ways the two points can be taken.

Let AB = a, AD = 6, AF = c, AH = 6 =
(a« + 6« + c«)i, AC = «i = (a» + ft»)*, AG = ea = .
Ca«-f c»)*, AE = ei, = (6»-f <J«)*, AP =«, AE = «', II

Z NAB = ©, Z PAN = ^ ©' = tan"^(0. ^ = tan-'(- '^ -); and let J = the required average,
^1, 2^2* ^3 = the average distances between the two pointH when P is within ihe pyramids ABCHG,
ACDEH, AEFGH, respectively.

Then 2' = a8ec08ec^, volume PR = (a—xcoBBco6(p)(b — zAn(p)(c — xB\nOco&(p); an element of
the solid at P is sficoa <pdBdq)dx\ for 2^1, the lunits of are and O'; of q>, and q>'; and of x, and a/.

r r P(a — «C08©C08^X*~«8^^XC— ««n0CO8<?)>r»CO8<prfOd<^

. A Jo Jo Jo .^

r r* r'(a — srcos© cos^)(6— a;sin ^X« — *BU^ ®cos tp)x'C08 tpdOdgxix

= 01.^-^ r r* r'(a — ajcosOcos^X* — *Bin<pX« — «8i'»®<»»<?*>'*<»8<?^®<'W'!c.
aWtrJo Jo Jo

= ii^ o f^ r* (216c — I4ab tan © — I4ae sec ©tan <p + lOa^sin 0sec«©tan <p)Bec*Qeec^ipd»d(p,
Wj^<?Jo Jo

= :i7rf i73 C fsSo^csec^ — 20a*ghk ©sec^^ + eSb^fWt^(a^ + ft^coe^©)* - 42a6«sin ©Bec*©(a« + 6«cos«©)*

— 28c8ec*©(a« + 6«C0tf«©)l + aOown ©8ec«©(a« + 6acoS«©)l + ft3a«6csec^l. g^^_ ^«>8 Q + f a^ -f l'COgg)*^

^« ,r « r ^. /6 COS + (a« -I- 6«cos8©)i\"l ,^
— 42a368inG8ec*©log(^ ?-^^ — — -) UB^

■" 5 "^ 10562c* "^ 6<;« "^ 66« "^ 15c« "^ ISfr* 1562^2 lOft^c* 56c \ae/

The values of J2 and -^3 are similar to that of Ji, and may be written out wiihout performing the
integrations.

[,w("f)+'HCt")]+i('"?^-";rM^?)

+t(^-ia)'««(^-)+»(''r-''J;iD--Ct'>

+ —

— 80-~

MMtki.NOfftl

ORrolimi ; and Artemas Mamtin, M . A., Member of tb« Loodon MatkMnMkml Bodetjr, liK >rt» OoubIj, tmujlrmaim.

II .-Solution by H. T. J. Lvdwio, Profeaor of MathMMtfa. Nocth QuoHm OoUmi. Moaat PtauMil, GUbttrv OooatJ, Sorth

Taking one corner of the solid as oiigin of rectangalar co-ordinates, denote the points by (x, y, t) and
(w, V, «) t«'«pective1y. and th«* required average distance bj J; then the distance between the points is

[(X - wy H- (y ~ r)* -f (2 - «)«]*, aud

r p p n n p[fx-«T)« + Cy-r)« + (z~*)«]Wirfl«W«cl«^
^ __.>'o J(\ Jo Jo Jo Jo _ ,

Jo Jo Jo Jo Jo Jo

=3„.kJ;/oXpn^-^^-^-^>'<'«rs^r*)^<-^^^^^^

-|»«'^'C-(5H:7+.')i)]'*«^'-

-9^?- Jo Jo L 30 " - —20 ''"——20 +««y(*»+J^)log<^ (aH^SjT-/

8c»aK2«»+«»)-at(a«+«»)« ,«„/y+{*'+»'4^\ ■ 8^2»M:«»)-3jf(j|H:c')» /x-Ha!«+»»-l-<')*\

^ 20 ^V X / ^ 20 ^ V y / 2 V«(«»+y«-K)*/ 2 *™ Vy(a«4y+cs)}/

10*" lc(x'+»H-e')iJ J""^''~8o*6«<^Jo Ll5 iS "^ 15 ~ 8 "

_ x(3?+f)t 4«'a<a«-f-«»)l _ <:<x(ar=+e^ j( a:'+y+<!')i _ 4«(ftM-«') (ar'+^+<^l
16 " '' 15 8 "*■ 15 " " 15 '

''■ 8 ' ^ 5 " ■•" " 20" '**l (a)»+6*)4 " J

8A<!MaB»+c')-36a<i«-H!')« f ft+(j»+y+<»)> "J _ (fty-^) (ft^-g t^f+c*) f x+(a!«+6«+e«)n
f 30 - log 1^ - (^,.j^)j J 40 - log (^ (6«+c«)l J

cjr*. r be 1 **<•* ',r ex 1 *<^* . r 6-« n,

90 L a'^b"' "^ a2c2 "^ b^c' J "^ 12 L c Sa-^^-cJ ^ U— c J "'" 12 L 6 Sa^ftc^J ^ Le— 6 J

30a€

, ^ , r^>+(''"'+<-=)n 2a3 ^ , ffec^ 26» , .foci 2c» ^ , fa^l

— 81 —

87.— Pbibs Pboblxk. PiopoMd by Astexas Martin, M. A., Brie, Erie Gonnty, PennsylTAnia.

A cask containine a gallons of wine is placed upon another containing b gallons of brandy. Water
rans in at the top of the wine cask at the rate of m gallons per minute, the mixture escapes into the
brandy cask at the same rate, the mixture in the brandy cask escapes at a like rate into a tuo contidning
e ^lons of water and the tub overflows. Find the quantity of each fluid in the tub at the end of i
muiutes, supposing them to mingle perfectly.

Solution by Wiluam Wooukt JoHiraox, Prorenor of Mathenwtlce, St John^e College, AnnapoliB, Anne Anindel Co., Vd.

1. Let X denote the number of gallons of wine in the first cask at the end of t minutes, y the number
of gallons in the second cask, and z the number of gallons in the tub.

The rate per minute at which wine is passing from the first into the second cask is ^ -; this fraction

also represents the rate at which wine is entering the second cask, while ^ - is the rate at which wine

is leaving it. Thus we have the differential equations :

dx mx /,x dy mx my /^x dt my mt /o\
di^^'a ^^^•ctt=-a"-"T ^^^'-rft= 6~"" c ^^^'

and to determine the constants of integration we have, when 4 = 0, x = a, y = 0, s = 0. From (1),

dx mdt — "- -^

— = — — ; .-. x = e •+ constant, and, since t = gives x^a, x = ae ' (4). Substituting this

value of X, (2) becomes ^ = me" • — ™ (6).

Since the solution would be of the form y=^Be * ...(6) were the first term of (5) zero, we may assume
the solution to be of the form (6), B denoting a function of t.

dv ^/dR\ m ■* dH ^ ■'

Differentiating (6), ^f = c" * {^^J — ^Bf \ which agrees with (5) it-^ = me » "" • (7) ;

from which ■^=_r«* •+*; but, since < = gives y = 0, and therefore, by (6), 5 = 0, we have

ab ab / —*** _!!*\
k = — __-., and substituting the value of 5 and i& in (6) y= __ji(^ •— < * j (8).

Substituting this value of y, (3) becomes^ = -^;:t( «~ " — « " * ) — ~ *W* "whence, as above,

we assume z=Be~ • (10); differentiating, — = e " * — — — 5e ~ ', which agrees with (9) if

dB

dB ma / ?:*-;• "*-¥\

Integrating, 5 = 7 — rw -— v« • "^ — > — tTv^ -x « ' ' +*; but since t = gives 2=0, and
(a — o){a — c) (o — o){o — c)

therefore i? = 0, we have * = "

(a-c)(6~cy
Substituting the value of B and k in (10),

« = ac(7- ... e •— 7 Tz-rz — -e * +> — vv, — c« •) (11).

\((i^)(a-^c, (a— 6)(6— c) ^((i— c)(6— c) / ^

This gives the wine in the tub; the brandy in the tub is found by the proper change of constants in

be / — ** — — \
(7) to be rzri * * "^ * j'> whence the water in the tub is

^ (or^b) ia^c) ^ " "^ (a— 6) (6— c) * (a— c) (6— c) *

2. The solution takes a different form when two or more of the constants a, 6, c are equal. Thus if
6r=a, we have in place of (7), -^ = m; whence B=^mt -i-k^ in which 4 = gives k = 0; therefore

y:=mte" ' (12). We have now, in place of equation (9,) -,. = — « ~ "• — ^^; and assuming as be-

fore2=5«~ •,^ = ^e • " • ...(18), whence, integrating, J?=^« ^~ • -x^£^,« •" •+ ifein

Qjf^ ^^ _ mt Q^ mt fj^ «t

which t = gives k = ;; ri; hence substituting we have z = t • — ; rr e" ' -{- . rr « " • ...(14).

— 80-^

l1*-Solntloii by H. T. J. Lupwio, Prof«aM>r of MathMMtfa. North OwoUm OoUm. Ummi nmmoL GUbttrv Ooimtj, llofth
Oiroliiw ; and Artkmas Hastin, M . A., Member of the London M«tb«OMiioal Sodotj, Kno, Irto OoaB^, PmnsylTaBla.

Taking one corner of the solid as origin of rectangular c<M>rdinaleB. denote the points by (x, y, t) and
(tr, p, «) i«'^pective1y. and th«* required average distance bj J; then the distance between the points ia

l(x ~ «,)* + (y ~ vy -f (2 - «)«]*. aud

f" p p p p p i(x - wy + (y - 1.)« + (z-^y^dxd^bdwdvdn

•^__»/o ^0 *'o Jo Jo Jo _ _ ,__ ^

p p re p p p j;^^^^^
Jo Jo Jo Jo Jo Jo

-2»«'^*C-(xi+?+.T)i)]''«^'.

- i - f r*r7'^('^y)» 7a!y(a^-+»*-N°)| , 19<fay(a^ +»'-N')* , „^_ . .^,.../e+(a:^-^^yS^)*^
~3a^b'<f-JoJoL ») ' 20 " ''■ 20 ~ -r^x^x-nDfogy (^+^SjJ /

8<;*a(2««+<?)-3*(a«+<!')« /s,+(j,?+j«+c8)i\ 8(<»(2y'-H!')-3y(»'+e'')« /a;+(a*+y«4<.)l\
+ -2y -'*«V— (^»-|-<;«)» /+ " 20 " '***v "(jo-wr /

^ 20 ^V X / ^ 20 "**! V y ' 2 '" Vt(i»+ir»+<»)t/ 2 **" V(«'+»'+«»)*'

~ 10*^ UxH-»H^)U J'^*^'' ~ S.^6^ Jo Lfs" " 16 + is 8 "

16 + 15 " 8 ''" 15 15

a<6»-|-««)«(a«+6»+«»)» , 26«cM*'+6'+tf')* , 86«<!ar(2a<+6«)— 3ct(«»+ft«)» , re+(a?+4»+«»)*1

8A<ftB(aif+<!«)— SM-c'-K")'
80

16 "^ ' 815a'-6-c- "^961 r^ ' "*" >- "*" " a^ J

90 Iw am "^ a%* "*■ "62c- J "^ 12 L c 5a-'6-cJ *"^ le-c J "'" 12 L 6 ^^bd'J ^ Lc— 6 J

— 81 —

87.~P>»B Pboblxm. PivpoMd bj Aktemas Martin, M. A., Erie, Erie County, Penney Wania.

A cask containing a gallonB of wine is placed upon another containing b gallons of brandy. Water
rans in at the top of the wine cask at the rate of m gallons per minute, the mixture escapes into the
brandy cask at the same rate, the mixture in the brandy cask escapes at a like rate into a tuo contdning
c ^lons of water and the tub overflows. Find the quantity of each fluid in the tub at the end of t
rauiutes, supposing them to mingle perfectly.

Solution by William Woolsct JoHNeosr, Profeeeor of Mathematice, St John^s College, AnnapolU, Anne Arundel Co., Vd.

1. Let X denote the number of gallons of wine in the first cask at the end of t minutes, y the number
of gallons in the second cask, and z the number of gallons in the tub.

The rate per mhiute at which wine is passing from the first into the second cask is -; this fraction

also represents the rate at which wme is entering the second cask, while ^ is the rate at which wine

is leaying it Thus we haye the differential equations :

dx mx t\\ ^y ^* ^y rtji\ ^* ^tf *"' /o\
dt^—a ^^^' dl = V-T ^^^' T<= 6" ~ c ^^^^

and to determine the constants of integration we have, when « =: 0, x = a, y = 0, f = 0. From (I),

dx mdt — — - "*

— = ; .'. fl: = e •+ constant, and, since i = gives x == a, a; = oe • (4). Substitutmg this

value of Xt (2) becomes ^ = m« • — ~ (5).

Since the solution would be of the form y = Be * ...(6) were the first term of (5) zero, we may assume
the solution to be of the form (6), B denoting a function of t.

Differentiating (6), ^ = «~^Y^) - 7^<^"""'» ^^ich agrees with (5) if "^ = 7716^""'.... (7) ;

ab "'^"^
from which J? = __r« * • +*; but, since < = gives y = 0. and therefore, by (6), ^ = 0, we have

i; = — _^, and substituting the value of J? and ifc in (6) y= __^fe «— e *j (8).

Substituting this value of y, (3) becomes -r. = ^. ( c~~ « _ g " * j — ?*^ .(9), whence, as above,

— *** dz — *' d iff wi _ ■■•
we assume z= Be • (10); differentiating, j-. = « ' y. Be *, which agrees with (9) if

dB ma / "^-^^ *'-'?\

Integrating, 5 = ;r — ,., — -\e'~ » —7- ., ,, rc • * -j-k; but since 1 = gives 2=0, and

(a — 0) (a — r) (a — 6) (6 — c) *

therefore i? = 0, we have k==

(a-c)(b^c)
Substituting the value of ^ and i; in (10),

(a —— b _"* c — "'X

(a^~b)(a—c^ (a— 6)(6-~c) "^(o— c)(6— c) / ^ ^

This gives the wine in the tub; the brandy In the tub is found by the proper change of constants in

be / —^ — — \
(7) to be rzrl « * — c • ) ; whence the water in the tub is

^_^^___ J i^ _?J c(ac-hbc^ab) -*;*

(a— 6) (or^c) " "^ (a— 6) (6— c) " (a— c) (6— c)

2. The solution takes a different form when two or more of the constants a, 6, c are equal. Thus if
6 = a, we have in place of (7), -^ = m; whence B = mt -i-k, in which t = gives k = 0; therefore

y = nUe'~' (12). We have now, in place of equation (^»)3,= " «~ - _ ~; and assuming as be-

toTez = Be-' '\-^ = ~~e ' ' -....(18), whence, integrating, J? =^e ^ - ^- .-^^^e •-• +*in

a<fl met — "** cuj* — ■** ad' — ^

which t = gives it = 7 xr^\ hence substituting we have 2 = t « — ; rr e ■ + , rr e • . . .(141

® («— c)*' ® a— c {a^-^^ ' (a— c)* ^ ^

dB mH mH^

3. But if o = 6 = c, (13) becomes ^ = — , whence ^=0^ +*» in which < = giyes *=0; hence in

this case z= -^ c • (16).

The expression for the brandy in the tab would now be identical with tliat for y in (12).

4. Again, if while a and 6 are unequal, c=a, after assuming z = Se * , as in (10), we have

'^ = -^- '^f'?"'^; whence^= ^ + 7-^\--c'?"^+ A, and determining A as before,
<U a — b a — b a — b (a — by ' ^

mat -?j a^b -^ M -^ ,,^.

2= — -£C • +; it;« *— X 1^ « • (16).

a— 6 (« — &)* (o— 6)« '

. 5. Finally, a and 6 being unequal, if c = 6 the expression for z is found by interchanging a and b in (16),

and the expression for the brandy is now obviously of the form giyen in (12), that is nUe '^.

Expression (12) might haye been found from (8) by making 6 = a and eyaluatmg by any of the usual
processes for indeterminate forms ; and in like manner (14) might haye been deduced from (11), (15)
from (14) and (16) from (11).

Prot JbAiuoM also gare «n eUbonita lolution for the genoni gm« of n cMks, which m relnctantl j omit Ibr vuit of spaoe.

Thewlotions by Prof. DeVobom Wood, E, B. Seilx, Prof. JYneMdge and (\fnu B. flaUenum are all •zcalleDt, and wa wiah we had
I to publish them.

LIST OF CONTRIBUTORS TO THE SENIOR DEPARTMENT.

The following persons haye furnished Eolutions to the Problems indicated by the numbers :

E B. Sritz, Oreenrille, 0., 56, 67, 58, 69, 60. 62, 63, 67, 68. 73, 74, 77, 80, 81, 83, 86 and 87, and 40 of *«UnK>lTcd ]
Prof. F. P. Hatz, Kln((*s Moaqtain, N. C, 56, 60, 62, 69, 70, 74, 78 and 84; Prof. DByouoK Wood, Hoboken, H. J., 69, 60, 62, 76,
78, f» and 87 ; Waltcb 8. Nichols, New York. N. Y., 57, Gi and 68, and 30 and 40 of ** UnnlTcd Problems '* ; Hkmkt Hkator,
flabula. Iowa. 73, 77, 79, 80 and 82 ; William Hooteb, Bellefontalne, 0., 62, 70, 79 and 83; Pro! H. T. J. Luvwio. Mt. Pleasant,
H. C. 74, 83 and 86 ; Waltm Sitsblt, Oil Qty, Pa., 72, 75 and 78 ; Prof. David Tsowbkidoe, Waterbois, N. T., 67, 60 and 87 ;
Pnr. W. W. Jomraoir, AnnapoUa, Md., 85 and 87 ; SeT. W. J. Wright, Oipe May Point, N. J., 64 and 85 ; Prof. Bbm jawh Pkibcb,
Oambridge, Has., 66 and 71 ; Mte CHKurnrx Ladd, Baltimoro, Md., 68 and 61 ; Prof. D. J. McAdam, Wadiinfton, Fa., 72 and
74 ; W. S. Hkal, Wheeling, Ind., 66 and 57 ; Ocobgr Eastwood, 8axonTill^ Mass., 56 and 81 ; I. H. Tumuix, andnnad, 0.,
68; Prot Dakikl Kibkwood, Bloomington, Ind., 69; Dr. Jobl K. Hxndbicks, DesMoInca. Iowa, 67; T. P. Stowul,
Bodiester, N. Y., 70 ; Datid Wiceushah, Wilmington, O., 84; Crsrs B. Haldemait, Boas, Bntier Go., 0., 87 ; Dr. David 8.
Habt, Stonington, Conn., 66 ; Dr. 8. H. Wright, Penn Yan, N. Y., 59 ; 0. H. Mbrsill, Mannsrille, N. Y., 57.

The first prize is awarded to Prof. W. W. Johhbok, Annapolis, Md., and the seoond prize to Prof.
DbYolson Wood, Hoboken, N J.

PROBLEMS.

111.-— Proposed by Oavix Shaw, KvBobl; Ontario, Oanada.

It is required to divide a giyen number a so that the contmned product of all its parts shall be the
greatest possible.

IIS.— Proposed hj BoBnrs rLsxiKo, Beadlagton, Hnnterdon Oonntj, New J«n«y.

Fmd 34 right-angled triangles haying the same hypothenose.

llS.~Propoaed bj Trkdbrick S. SAxrRLS, Cerro Oordo, Inyo Ooanty, CUIfomia.

What IS the yolnme of a chip cut at an angle of 45 degrees to the center of a round log, radius r?

114.— Proposed bj Dr. David S. Habt, M. A., Stonington, New London Goonty, Oonneeticat.

To find two numbers snch that their sum shall be a square, the sum of their squares a square, and if
the cube of each be added to the square of the other the sums shall be equaL

115.— Proposed by Stltebtkr Bonxs, North Branch Depot, Somexset Ooanty, New Jeisey.

There is a series of parallelopipeds whose dimensions differ from a perfect cube by one unit in only
one of the edges. In eyery case the solid diagonal is an integer. Calling the one whose edges are 1, 2, 2
tile Jlrst parallelopiped, it is required to find general expressions for the dimensions of the nth solid, and
compute the lengUi, breadth and thickness of the 30th one.

11«.— Proposed by Marcus Bakrr, U. S. Coast Sorvey Office, Washington, D. 0.; and Joskpr B. Mott, Neosho, Newton Co., Mo.
Bolye the equation 2»* = a, and find the yalue of x when a = 800.

— 83 —

117.— Proposed bj Dr. Joxl S. Hkmdkxcks, M. A., Editor of the AndlifBlt DeaMolnfl*, Polk Ooanty, Iowa.

Sappope a ball to be projected, in a horizontal direction, and dae west, from a point yertically aboye
the 40th degree of norm latitude; and suppose the projectile yelocitv to be just sufficient to arrest its
motion in space which resulted from the earth's motion on its axis, so that it will descend to the earth in a
vertical plane corresponding with the meridian through the point of projection. And suppose, further,
that the ball, under the circumstances, shall fall to the earth in 5 seconds. How far will the ball strike
the earth north of the 40th parallel ?

llS.~PropoKd by Winrsld V. JcFraixs, Instructor in Matliematios, YormiUion Institato, HftyssTtlle, Ashland Go., 0.
How many different combinations, each composed of n letters, can be formed from m letters, of which
a are one letter, 6 are another and c are another?

119.— Proposed by W. E. Hxal, Wheeling, Delaware County, Indiana.

Show how to trisect an angle, and how to construct two mean proportionals between two giyen
straight lines, by means of the curves called the conchoid and the cisgoid.

190.— Proposed by J. J. Syltkstkh, LL. D., W, B. S., Corresponding Member of the Institute of France, Professor of Mathematics,
Johns Hopkins Unircrsity, Baltimore, Maiyland.

Prove that for all positive values of k less than unity the equation (x + a) (x + ^) = ^(a; + c)> has two
real roots.

191.— Proposed by E. J. Edmunds, B. S., New Orleans, Orleans Oonnty, Louisiana.

Prove that r(^ • r(^ • r(?)...r(^^) = (2^)"* • n-k, n being a positive faiteger and F denoting
the well known Eulerian integral

199.- Proposed by Prof. David Tbowb&idob, M. A., Wateiburg, Tompkins County, New York.

When the earth is in perihelion, suppose the sun*s mass to be suddenly increased by a half of itself, or
so that m' becomes { m'. Required the change in the elements of the terrestrial orbit.

19S.— Proposed by Dr. Datid 8. Habt» M. A., Stonington, New London County, Gonnecticnt.

To 4nd three whole numbers such that the sum of the squares of any two of them increased by the
product of the same two shall be a rational square.

194.— Proposed by Abtemas Mabtik, M. A., Member of the London Mathematical Society, Erie, Erie County, Pennqrlvania.

The first of two casks contamed a gallons of wine, and the second b gallons of water; c gallons were
drawn from the second cask, and then c gallons were drawn from the first cask and poured into the
second, imd the deficiency in the first supplied with c gallons of water; c gallons were then drawn from
the first cask, and c gallons drawn from the second and poured into the first and the deficiency in the
second cask supplied with c gallons of wine. Required the quantity of wine in each cask after n sudi
operations as that described alK>ve.

195.— Proposed by Obulmdo D. Oathoitt, Bead, Clayton County, Iowa.

What is the average thickness of a slab sawed at random from a round log?
196.— Proposed by Fbamcxs M. Pbibot, Bryan, Williams County, Ohio.

Divide unity into three such positive parts that if unity be added to each part the three sums shall be
rational cubes.

197.— Proposed by Saxubl Bobbbts, M. A., F. B. 8., Member of the London Mathematical Society, London, England.

Two random points being taken within a circle (i) on opposite sides of a given diameter, (2) on the
same side, (8) anywhere; find in each case the average radius of the concentric circle touched by the
chord throu£^ them.

198.— Proposed by E. P. Nobtom, Allen, Hillsdale County, Michigan.

There is a circular fish pond surrounded bv palisades, to the outside of which a horse is tethered. The
length of the tether is equal to the circumference of the pond. Required the diameter of the pond,
sopposbig the horse to have the liberty of grazing an acre of grass.

199.— Proposed by Bbubbm Davis, Bradford, Stark County, Illinois.

It is required to find three positive numbers, such that If each be diminished by the cube of their sum
the three reuuunders will be rational cubes.

ISO*- Proposed by Abtemas Mabtik, M. A., Member of the London Mathematical Society, Erie, Erie County, Pennsylvania.

A circle is inscribed at random in a given semicircle, find (1) the average area of the circle and (2)
the chance that the circle does not exceed } of the semicircle.

ISI.— Proposed by DbVolsor Wood, M. A., C. E., Professor of Mathematics and Mechanics, Sterens Institute of Technology,
Hoboken, Hudson County, New Jersey.

A spherical homogeneous mass m, radius r, contracts by the mutual attraction of its particles to a
radius nr; if the work thus expended be suddenly changed into heat, how many degrees F. will the
temperature of the mass be increased, its specific heat being s and the heat uniformly disseminated ?

199.— PropoeiHS by Tbanklin P. Matz, M. B., M. S., Professor of Mathematics and Astronomy, King*s Mountain High School,
Klng*s Mountain, Cleareland County, North Carolina.

Three persons. A, B and C, are banished to a level circular island, diameter 2r feet. At the center of
the island is a cylhidric fort, diameter 2a feet. During a dark and foggy night B and C stray away from
A, and from each other, and lie down to rest. At the first dawn of dear morning, while B and C are yet
repostog, A looks around for them. Required the probability that A, without moving from his place of
observation, can see both of his companions.

— 84 —

ISS.— PropoMd bj Aitsxas Mabtik, H. A«, K«mb«r of th« London Xalhoafttlcftl 8od«^, Srio, Krio Oouty, PonnsjlTMln.

A dack Bwims acrofis a river a rods wide, always aimiDg for a point in the bank b rods np stream from
a point opposite the place she started from. -The velocity of the current is v miles an honr, and the duck
can swim n miles an hour in still water. Required the equation of the curve the duck describes in space,
and the distance she swims in crossing the river.

184.— Proposed by Wiluax Hootxr, Mathematical IGditor of the WUtenhergtrt BeUefontaine, Logan Gonntj, Ohio.

Two equal rings begiu to move freely from the extremity of the horizontal radius of a quadrant of a
circle, one down the arc, the other down the chord; if the radius be 300 feet, how long after the motion
begins will one ring be vertically above the other?

1S5.— Propoeed by Artem as Hartih, M. A., Member of the London Mathematical Society, Erie, Brie County, PenniylTanla.

Three equal circles touch each other externally; find the average area of all the circles that can be
drawn in the space enclosed by them.

1S6.— Propoeed by Johic W. Bkrry, Pittston, Lnierne Ckmnty, Penn^ylvmnia.

A smooth, straight, thin tube is balancing horizontally about its middle point, and a particle whose
weight is ], that of the tube is shot into it horizontally with such a given velocity that it just arrives at the
middle point of the tube. Find the angular velocity communicated to the tube.

187.— Propoeed by E. B. Skitz, OreenTlUe, Darke Oonnty, Ohio.

Two points are taken at random in the surface of a circle, but on opposite sides of a given diameter;
find (1) the chance that the chord drawn through them does not exceed a line of given length, and (2) the
average length of the chord.

188.— Proposed by Artrmas Martin, M. A., Member of the London Mathematical Society, Erie, Erie Gonnty, Penntylvania.
Two sides of a plane triangle are a and b; find its average area.

180.— Proposed by M. H. Doolittlr, TJ. S. Coast Barrey Office, Washington, D. C.

Suppose infinitesimai aerolites equally distributed through all space, everywhere moving equally in all
directions with a given uniform and constant absolute velocity. The aggregate mass intercepted in a
given time by a given stationary sphere is supposed to be known. Determine the effect upon the
eccentricity of a spherical planet of given mass and volume moving in an eccentric orbit, all or whose
elements are given.

140.— Proposed by Artemas Martin, M. A., Member of the London Mathematical Society, Erie, Erie Cooniy, PennsylTania.
A circle is drawn intersecting a given circle, its center being at a given distance from that of the
given circle. Find the average area common to both circles.

141.— Proposed by J. J. Stlvestbr, LL. D., F. B. S., Corresponding Member of the Institute of Fnnoe, Professor of Mathematics,
Johns Hopkins Unirersity, Baltimore, Maryland.

A row of particles in contact are charged some with positive and some with negative electricity.
Under the effect of their mutuid actions or the influence of some external body, these electricities are
subject to continuous variation so conditioned that when the electricity of any intermediate particle
becomes neutral the electricities of the particles on either side of it are of opposite signs.

Hupposing the electricities of each of the particles to be known for one moment of time and for some
Bul)sequent moment, show that a certain logical inference can be drawn as to the joint changes undergone
by any two extreme or any other two particles in the interval, and state what it is.

149.— Proposed by E. B. Sritz, GroenTille, Darke County, Ohio.

Two points are taken at random in the surface of an ellipse, one on each side of the major axis; find
(I) the average distance between the points, and (2) the average length of the diord drawn through them.

148.— Proposed by Artsm as Martin, M. A., Member of the London Mathemattcal Society, Erie, Erie County, PennsylTania.

A chord is drawn through two points taken at random in the surface of a circle; if a second chord be
drawn through two other points taken at random in the surface, find the average area of the quadrilateral
formed by joining the extremities of the chords.

144.~ r ropossd by E. B. Sritx, OreenTiUe, Darke Conn^, Ohio.

A circle is circumsciibed about a triangle formed by joining three points taken at random in the
surf ace of a drcumscriptible polygon of n sides; (1) find the chance that the circle lies wholly within
the polygon; and (2), a second circle being described in the same manner, find the chance that both
circles lie wholly witliin the polygon and one of the circles is wholly within the other.

145.— Proposed by Bcnjamin Pbirce, LL. D., F. R. 8., Professor of Mathematics, Harrard UniTenity, and Consulting Oeomater
to the U. 8. Coast Surrey, Cambridge, Middlesex County, Mssiachusetts.

Prove that if two bodies revolve about a center, acted upon by a force proportional to the distance
from the center, and independent of the mass of the attracted body, each will appear to the other to
move in a plane, whatever may be then* mutual attraction.

14«.~Proposed by Bot. W. J. Wright, M. A., Ph. D^ Member of the London Mathemaflcal Society, Cape May Point, Qipa Maj
County, Nov Jersey.

If Sa be the coefficient of oa in the determinant i> = 2 Jb<iu<itt***<^n, and J denote the determi-
nant ^±8nBi2.:Sn^ prove that J^D^K

— 85 —

147.— Fbize Phoblem. Propowd by Abtemab Martik, M. A., Member of the London Matbematical Society, Erie, Brie Go., Pa.
A right cone, whose equation is x^-\-y'^ = e-if^.,.(l), and a paraboloid of revolution, whose equation is
j/8-|-j^r=pa;...(2), have their vertices coincident, the axis of the cone being perpendicular to the axis of

the paraboloid. Find the volume common to both by the formula ^= J J i dxdydg,

SolotioDB of these problems should be receired by October 1, 1879.

EDITORIAL NOTES.

We roKret to have to record the death of a valued and talented contributor, viz.: Dr. 8. F. Bacheldeb, of South Boston, Mass.
The pari of the following notice enclosed by quotation marks Is condensed ftx>m a letter received from Mrs. Bachelder, dated Imdi 2,
1878.

'* Dr. Samuel T. Bachelder was born in the town of London, N H., on the 14th of October, 1829. He received an academical
education at Gilmanton, N. II., studied medicine in his native town, and attended lectures at Cambridge, Mass. He settled in Feabody,
Ham., where he had"n good practice; but twelve years ago he moved to Boston, where he has since lived and practiced his profession
Dntil his death. He was a member of the MaaiacluMutU Medical Society, and a Master Mason of St. Paulas Lodge, South Boston. He
was much beloved by all who knew him fur his kindly spirit and gentle bearing. Four years ago he began to have symptoms of spinal
difficulty, which gradually increased and extended to the brain and caused his death January 1st, 1878."

Dr. Bachelder was an acceptable contributor to ttie Visitor, and to the mathematical departments of the NatUmal Educator and the
WUtenberger. He was very skiliful and ingonioui in the solution of algebraic and geometrical problems.

We tender our best thanks to Mr. £. B. Seitz for valuable assistance, and for his successful efforti in extending the circulation of
the Visitor; and also to Mr. Bbi'bex Knkcht, of Easton, Pa., for the subscriptions secured by him.

The present No. has been dela^'ed a month in consequence of the ditHculty of procuring Greek type. We trust a like delay wIU
not occur again.

No. 4 will bo published about the first of January, 1880; it will contain about 32 pages and the price will be 50 cents. Persons
desiring to secure copies bliould send their orders at an early date, as only a limited number will be printed.

Having reprinted No. 1, in fine style, wo are now prepared to supply copies of that No. Ptioe, 50 cents. The new edition
contains an article on the Intrinsic Equation of a Curve, by the Editor, inserted to fill out the last leafl We can also supply copies
of No. 2 at 50 cents each.

If every one of our present subscribers would procure one new one, we could publish the Visitok aeml*aQnnally. Bow many
will secure the extra subscriber?

Send all oiders to ABTSMAS MABTIN, Lock Box 11, Erie, Pa.

NOTICES OF BOOKS AND PERIODICALS.

Th6 BlemenU of Analytical Mechanka, By DeVolfon Wood, M. A., G. £., Professor of Mathematics in Stevens Institnte of
Technology. Second Edition. Bevised, Corrected and Enlarged. New York : John Wiley A Sons. 8vo., cloth, pp. 249. Price |3.00.

The plan of the new edition is the same as the former one. " It is designed especially for students who are beginning the study
of Analytic Mechanics, and is preparatory to the higher works upon the same subject, and to Analytical Physics and Astronomy.
The Oalcnlus is freely used."

The work is written in a remarkably clear and elegant stylo and beautifully printed, and illustrated with numerous diagrams.
The general formulas are applied to the solution of a large number of interesting problems. In short, it is an excellent work upon
the subject of which it treats, and we heartily commend it to all persons in want of an elementary treatise on this branch of
mathematical science.

The PkUoeophy of Arithmetic as Developed from the three Fundamental Processes of Synthesis, Analysis, and Comparison.
Containing also a History of Arithmetic. By Edward Brooks, Ph. D., Principal of Pennsylvania State Normal School, and Author
of a Normal Series of Mathematics. 8vo., cloth, pp. 070.

This work is Dr. Brooks* crowning effort. It sbould be in the hinds of every teacher in the land.

The chapters on the Origin and Development of Arithmetic, Early Writers on Arithmetic, Origin of Arithmetical Proce§S68 and
the Origin of Arithmetical Symbols are deserving of special attention.

The whole book is written in Dr. Brooks' happiest style, and is as fascinating as a novel; it contains a vast fund of historical and
other valuable information that the student can not find elsewhere without poring over many huge and musty volumes.

i%me Triijncmetry and Funciionul Anali/ftM. By Alfred H. Welsh, M. A., late Professor of Mathematics in the Akron Bnchtel
Colleee, and present Instructor of Bhetoric and English Literature in the Columbus High School. 8vo., cloth, pp. 100. Columbus :
Q. J. Brand & Company.

A good elementary work on Logarithms, Plane Trignometry and Trignometrical Analysis. It is wall printed on heavy tinted
paper, but the composition is apparently the work of inexperienced hands.

Trade Bdatiag to tfte Modern Higlter MathemaUca. Trad No, S. Invariants. By Rev. W. J. Wright, M. A., Ph. D., Member of
the London Mathematical Society. In press.

The author has kindly sent us advance sheets of the first 54 pp. A perutol of them convinces tis that the Tract will be a valuable
addition to the literature of the subject.

The Normal lligher AriOimftic, Designed for Common Schools, High Schools, Normal Schools, Academies, etc. By Edward Brooks,
A. M., Ph. D., Principal and Professor of Mathematics in Pennsylvania State Normal School. 12mo., pp. 514. Price $1.25.
Philadelphia : Sower, Potts & Company.

A most excellent work ; we unhesitatingly pronounce it the he^ Higher Arithmetic we have ever seen.

The Normal Union Arithmetic^ Designed for Common Schools, Normal Schools, High Schools, Academies, etc. By Edward Brooks,
A. M., Ph. D. 12mo., pp. 424. Philadelphia : Sower, Potts A Company.

A leading feature of the work is the union of mental and written arithmetic in one book—hence Its name. A happy oombliiation,
aad wo predict for it a wide popularity. Just the thing for common schools.

Sesearchee w Graphical SUUict, By Henry T. Eddy, C. E., Ph. D., Professor of Mathematics and Civil Engineering in the University
of Cincinnati. Iteprinted from Van Nostrand's Engineering Magazine. 8vo., pp. 122.

Consists of several articles on Arches, Stresses and kindred subjects of special interest to the civil engineer.

Oil (he Origin of Crnnttn, By H. A. Newton, LL. D., Professor of Mathematics in Yale College. 8vo., pp. 16. From the American
Jonrnal of Science and Arts. Vul. XYI, Sept., 1878.

An interesting comparison of the theories of Kant and Laplace In regard to the origin of these wanderers in space. Kant believed
them to \x formed from the matter of the condensing solar nebula; Laplace considered that they were made from tiie matter that la
scattered through the stellar spaces.

Prot Newton concludes tliat the moss of obser\-ed facts are in favor of the hypothesis of Laplace.

OiMervatione and Orbittt of the Satdlites of Man. By Asaph Hall, Professor of Mathematics, U. S. Navy.' 4to., pp. 46. Washington.
1878.

An interesting account of the important discovery which has immortalised the name of Professor Hall. Hamilton College, at its
lait oommenoement, conferred upon him the well*merited Honorary Degree of Ph. D.

— 86 —

Tk9 EAtmHmal Tmm md Jimmtd ^ <*• CnXkgt nf JVemptorv I0 pnblialMd aoothly bj 0. P. Hodnon A 800, Loodoo, 1 ^ . .
and oontains % Talnable Mathematical Department of three or four double-oolamn quarto pages, ably edited bj W. J. G. Miller, B. A.,
Begietrar of the General Medical Goancil, which nnmbeiB among Its oontribators many of the leading mathematidanB of Sngtand,
Continental Enrope and this coantry.

UdihBmaliioai, QaeafMrna, icttJb Qvtk BolytioHs, SeprUtlfd from the Edfuntimud Time*, Same pnbUdiexB. Issaed in half-yearly Tolnm«a
of 112 pp., 8fO., boards.

The ReprnU contains, besides the mathematics published In the 2lmet, many additional solutions and papen. Tol. XXVUI
contains 8 papers and solutions of 71 problems; Vol. XXIX contains 10 papers and solutions of 91 problems.

Sereral of the solutions of problems in the Diophantine Analysis are by our able contributor, Dr. David S. Hart, M. A., and quite
a number of those in " Arerage " and '* Probability " are by our able contributor, Mr. E. B. Beits.

The Editor of tiae Visitor can ftamlsh the Timee at 12.00 a year, and the BepruU at $1.75 per Tolume.

The American Jawmat of MathematieB, Pure mtd AppUed, Prof. J. J. Sylvester, LL. D., F. B. S., Editor in Chief; W. E. Story,
Ph. D., (Lelpsic), Associate Editor in Charge; with the co-operation of Benjamin Peirce, LL. D., F. B. S., of Harvard Unirersitj ;
Simon Newcomb, LL. D., F. B. S., of the Naral Observatory ; and H. A. Bowland, C. E. Pnblt8h«>d under the auspices of Uie
Johns Hopkins University. Baltimore, Md. Issued in quarterly numbers of 96 quarto pages. Price ffi.OO a year.

The first volume of this rery able mathematical periodical Is completed, and contains papers of the highest order of ezoeltenoe.

The Journal numbers among its contributors many of the dlstingulshea mathematicians of tiie world.

The Analif$l, A Journal of Pure and Applied Mathematics. DesMoines, Iowa: Edited and published by J. B. Hendricks, M. A.
Bi-monthly; each number contains 82 pp. $2.00 a year.

No. 1, Vol, VI, presents an interesting table of contents. Dr. Hendricks is deserving of great credit fbr the veiy able manner
in which he has so successfully conducted the iitia^, and should have a remunerative subscription list.

The WUImiberger, a neat monthly magadne, devoted to the Interest of Wittenberg College, published at Springfield, Ohio, has aa
ezoellent mathematical department edited by William Hoover. Price $1.10 a year.

BdHMf* EdHcatknua MoiUU^, New York: A. S. Barnes A Co. $1.50 a year.

An ezoellent educational monthly. The mathematical department Is under the management of, and very ably conducted by,
Piof. F. P. Mats, M. E., M. S.

Edueai»!kud Notea and Queriee. ProC, W. D. Henkle, Salem, Ohio, Editor and Publisher. Issned monthly azoept In the vaeatios
mouths of July and August. $1.00 a year.

Contains, besides a vast ftand of other interesting matter, mathematical notes, problems and solutions.

The Mmme Ifarmert' Almamac for 1879 contains, besides Puzzles, Biddies, Ac, Mathematical Questions, and answers to those
proposed last year. Hallowell, Me.: Masteis 4 Livermore. Price 10 cents.

The NevhEHO^fomd Journal of EthuxUion, Weekly. Boston: Published under the auspices of the Ambricak InsTrrurs or
Ikstbvctiox and the Tbachkbs' Assooiatioxs of the New*£ngland States. $2.50 a year.

The mathematical department is very efflclently conducted by Prof. E. T. Qulmby, of Dartmouth College.

The Yatee Oomty Chromele, Weekly. Penn Yan, N. T.: Chronicle Publishing Co. $2.00 a year.

The mathematioal department continues under the able management of Dr. S. H. Wright, M. A., Ph. D,

The TSuMumnoek BepuhUean, Weekly. Tunkhannock, Pa.: Cyrus D. Camp. $2.00 a year.
The BepMiean has an Interesting arithmetical department edited by C. W. Bushnell.

The PuuuybHmia School Journal. Monthly. J. P. Wickeraham, Editor. Lancaster: J. P. Wlckeiaham 4 Co. $1.60 a year.

The AmimlMMia S^ool Journal is one of the best educational peria

but occasionally contains scientific axtioles and ezamination problems.

The Pennmbmnia School Journal is one of the best educational periodicals in the country. It has no mathsaatinsi department,
— sionally oontr' • — '- — -^-« ' »- -• ^^^ —

The Ameriean Jomnal of Edueation. Monthly. St. Louis: J. B. Merwin. $1.60 a year.

A reiy excellent ** Journal of Education,** and it should be liberally patronized by the tsacheis of Missouri and other Statci.

CORRIGENDA.

No. 2.
Page 16, line 3 from bottom, for "aces** read acres. Page 83, line i, for " (tAt)" " ^^^ (t^)** ^Ne* S3, last line of
solution of Problem 16, in part of the edition, for " ^ " read 1. Page 32, line 7 Drom the bottom, ybr ** cos«(0+^r read coe«K#+^).

Page 35, Une 2 fh>m tiie bottom, for •• ^^^ " read ~^^, and «6r " cos*C* + ♦)" **** cm^B + ♦). Page 37, Une 4 flrom bottom,
for « (13) " read (3). Page 38. line 14, for " (23) " read (13). Page 39, first Une of Problem 14, for »r» read B; seoond line fkom
bottom, for « V— voB* '* read V— ^^-t n being the number of cubic inches in a gallon. Page 40, change the signs otdQ. Page 41,
line 12, in part of the edition, for '* ton A log cosec A" read ton* A log cosec A. Page 42, lines 1 and 2, for **d»» read d».

No, 3.

Page 66, solution of 49, last line but on^ for "| r^ll9;— HHM ^O^^^r read Hilll »'(119) — I »'(1W); "olution of 50, line 2, for •^^-
read pa, Fsge 66, the sides of the fifth triangle should be 40q9, 4060, 5741 ; and the sides of the sixth, 23660, 23661, 33461. Page
67, solution of 62, line 8, for ** Un^ " read ton^w ; in the value of BC for **— b /{be)" read — h i^oe). Page 59, Une 3 of the value of
N, In part of the edition for •*— 3728792 " read — 137^8792; expunge h from the last term of line 8 of the value of N. Page 63,
first solution of 60, last line, equation (6), for *MniS'* read sln/3. Page 65, second solution of 63, Une 3, for **A'6** road A'B.
PBge 67, line 3, for ** radius-rector ** read radius-vector. Page 68, second solution of 70, line 6, in part of the edition, for
** (B> — «•) *' read (B* — x')^. Page 72, Une 6 fh>m the end of the solution of 78, for ** <• — f,» read t — f,. Page 73, solution of
80, Une 23,/or ** ^•"read 12^*. Page 76, line 6, for **— 20B*(^« " read — 20Br«<«>. In solution of 82 the value of j> should be

r 'aseoMn(^ +«)cos^fl^ r*'asecMn(^ — 9)cos^<i^

-^^ ^ j^ ^ and tiie value of q should be -^ ^ -_ .

J Satan«cos<^d^+l^ aseoMn(^ + 0)cos^# J 2atan$oos>^<i^ + J * aseoMn(^ +«)coe^4if

y

IQ

k^ '

//' \ n

Vol. I. JANUARY. 1880. Ho. 4.

THE

MATHEMATICAL
VISITOE.

EDITED AND PUBLISHED BY

ARTEMAS MARTm, M. A.,

Member of tke Jjoadon JHhitkentmHcai Society*

f M^%^ ^^^'l^ ^^llf^i

SRIE, PA.:

MINTtS IT THI UITOII.

1880.

' -* ----' ■■ *

THE

MATHEMATICAL YISITOR.

[ENTCnO ACmROiNO TO ACT Of CONOMESS, IN TMB VBAII ItTS, IT AKTEMAS MARTIN, M. A., IN THE OFFICE OF THE UBRARIAN. OF COmMBSS, AT WAENINOTON.]

Vol. I. JANUARY. 1880. No. 4.

JUNIOR DEPARTMENT.

Solutions of Probtetus Proposed in J^, 3.

88*"~Pn>P<MBed by John I. Clakk, Moran, Clinton Coanty, Indiana.

When dilver is 4 per oent. discount and gold is at 5 per cent, premium, taking greenbacks as a standard,
what YfiW silver be worth if we take gold as a standard?

L^Bolation by Hon. JoeiAH H. Druvxond, LL. D., Portland. Me.; Oavin Shaw, Komble, Ont, Canada; Walter S. Nichols,
New York, N. Y.; Robins Fleming. Readington, Hunterdon Co.,N. J.; William Wiley, Detroit, Mich.; Prof. H. S. Banks,
Inetrnctor in English and Classical Literature, Newburg, Orange Co., N. Y.; and Geo. H. Lbland, Windsor, Windsor Co., Vt.

Silver being at 4 per cent, discount Is worth 96 ; gold being at 5 per cent, premium is worth 105 ; hence
silver is worth ^f^ of gold, or 91 J.

II."^oIution by H. T. J. Ludwio, Professor of Mathematics. North Carolina College, Mount Pleasant, Cabarrus Co., N. C. ;
Theo. L. DeLand. Office of the Secretary of the Treasury, Washington. D. C. ; E. P. Norton, Allen, Mich. : J. V. Stewart,
Mnncie. Delaware Co , Ind.; David Wickersbax, County Surveyor, Wilmington, Clinton County, Ohio; and O. G. Washburn,
North East, Erie County, Pennsylvania.

$1.00— $0.04 = $0.96, value of silver dollar in greenbacks ; $1.00 + $0.05 = $1.05, value of gold dollar
in greenbacks ; $f gg = value of greenback dollar in gold, and $0.96 x Jg J = $0.91 f, value of silver dollar
in gold.

III.— Solution by John S. Roter, Ansonia, Darke Co., O.; Prof. W. P. Casey, San Francisco, California; W. T. Maoruder,
Stevens Institute, Hoboken, N. J.; Frank Albert, Professor of Mathematics, Pennsylvania State Normal School, MiUersville,
Lancaster County, Pa.; and V. W. Heath, Rodman, Jefferson Co. N. Y.

Let 100 per cent. = value in greenbacks ; then 96 per cent. = silver and 105 per cent. = gold. When
gold is taken as the standard, silver will be % -i- 1.05, = 91 J.

This problem was also correctly solved by Thomas Bapot, A. R. BvUiif, B. F. Burleson, J. R.Fagan, Dr. HarU Prof. M(U»^
K. S. Putnam. J.A.PoUard, Sylvest^ RolAntt, T. P. Stowell, E. B. Seltz and Walter Siverly.

89«"'Pw>po«?d by O. G. WASHBtTRN, North East, Erie County, Pennsylvania. '

Two men, A and B, each desire to sell a horse to G ; A asks a certain price and B asks 50 per cent,
more. C refused to pay either price ; A then reduced his price 16S por cent, and B reduced his 33J per
cent., at which prices C took both horses, paying $220 for them. What was each man's asking price?

I,— Solution by Prof . W. P. Casey, San Francisco, California; •Gayin SHAW,-£emble, Ontario, Canada; E. J. Edmunds,
New Orleans, La. ; and the Proposer.

A's asking price — 16| per cent, of it = ^ of his asking price, and \ of A's asking price = B*s asking
price ; but B's asking price — 33 J per cent, of it = A's asking price, and therefore, per question, J of A's
asking price + once his asking price = $220, or y of his asking price = $220 ; whence A's askiiu; price
= $120 and B's = $180:

II.— Solution by D. W. K. Martin, Webster, Darke County, Ohio; D. B. O'Conner, Union City, Randolph County, Indiana;
A. R. BULUS, Ithaca, N. Y.; and J. S. Royer, Ansonia, O.

100 per cent. = A's asking price, and 150 per cent = B's. 100 per cent. — 16f per cent =.- 83J per cent.
= A's selling price ; 150 per cent. — 331 per <5«nt. of 150 per cent. = 100 per cent, B*s selling price. Then
83J per cent. -|- 100 per cent. = S220. 100 per cent = $120, A's asking price, and 150 per cent = $180,
B's asking price.

Solved in a eimUar muimer by John I. Clark, Bobins Fleming and Prof. K B. SeUz.

UIw— Solntion by Gsorob A. Jopuk, Centre College, DanvUle, Ky.: I>r. David 8. Hart, M. A., Stoninjgton. New London
Co., Conn.; Waltkb Sitbklt, Oil City, Pa.; K. 8. Putnam, Borne, N. Y.: Thomas Baoot, Principal Shel^ School, Canaan,
Indiana; T. P. Stowbll, Bochester, N. Y.; Prof. P. P. Mats, M. K, Beading, Pa.; O. D. Oathodt, Loana, Iowa; J. B. Faoah,
Erie, Pa.; and B. F. Bubueson, Oneida Castle, N. Y.

Let X = price A asked ; then Lc = what B asked, j^r = A's reduced price and x = B's.

.-. Ja; + a? = $220, or 11a; = $1320 ; whence x = $120, A's asking price, and }a;=$180,B's asking price.

Solutions Bhnilar to the above were given by Mcsmb. Albert, Banks, Drummond, Leland, Norton, NiehaU, PoUard, Stewart,
SMdy, WickersAam and WUey.

rV.— Solution by Thso. L. DeLaitd, Offlce of the Secrctair of the Treasurv, Washington, D. C ; Wiluam Hooter, Snper-
intendent of Schools, and MathemaUcal Editor Wlttenderger, Wapakoneta, Ohio; and Stlvbstbr Bobins, North Branch Depot,
Somerset County, New Jersey.

Let 6x = A's asking price and 9x = B's ; then 5jj = A's selling price and 6x = B's.

.-. 6a; + 5ir = 11a; = $220, and x = $20. ex = $120, A's asking price, and 9a; = $180,B's asking price.

90.— Proposed by K. S. Putnam, Borne, Oneida County, New York.

Having cut a square, area cfl, from one comer of a board 2a wide and 3a long, cut the remainder into
three pieces so that they will make a sqare.

Solution by B. F. Bitblbson, Oneida Ca«tle. Oneida County, N. Y.; Julian A. Poll.ird, Wind'^or, Windsor County, Vermont;
Stlvestbr Bobins, North Branch Depot, Somerset Co., N. J. ; G. II. Lelamd, Windsor, Windsor Co., Vt ; and Bobims Flbmino,
Beadington, Hunterdon County, New Jersey.

Let ABCD represent the board, AB being 3a units
in length and BD 2a units ; also let EFCG represent
the square, whose side is a units in length, which is
cut from the board. The area of the remaining
piece ABEFGD = 5a^ therefore the side of its equiv-
alent square must be a\/5 units in length.

Trisect the side AB at the point P, distant 2a units
from the angle A ; Join DP and £P, and the remain-
der of the board will be divided into the three re-
quired pieces.

Demomiiration. — The angle DP£ is a right angle
and each of the lines DP and £P is a\/5 units in
length ; therefore they must constitute two sides of the required square, and the two extraneous triangles
EAP and DBF when properly arranged must complete it. The arrangement of the pieces is shown in
the right-hand figure in the diagram.

Solved in like manner by Messrs. Albert, BuUis, Clark, Caeey, Heath, Putnam, Seitz, Shidy and SUferly.

91«— Proposed by Fbahx Albbbt, Professor of Mathematics, Pennsylvania State Normal School, MillersvUle, Pa.

Show that in any isosceles triangle, the square of a line drawn from the vertex to any point in the base,
plus the product of the segments of the base, is constant.

I^— Solution by I>r. David 8. ELabt, M. A., Stonington, New London County, Connecticut

In the isosceles triangle ABC, AG = BC. Draw AD perpendicular to and bisecting the bade BO. Also
draw AE dividing BC into two unequal parts BE, EC ; then {Eudid II, 5) BDxDC = BExEC -f (ED)» ;
but (ED)* = (AE)«— (AD)<, whence BDxDC = BE X EC -f(AE)«-( AD K and by transposition we get
(AD)' -f BDxDC = (AE)« -f BExEC ; therefore the truth of the theorem is manifest.

Solved in a similar manner by Prof. Banks, K. S. Putnam, Walter Siveriy and Qacin Shaw.

H.— Solution by Prof. B. J. Edkunds, B. S.. Principal of Academic School No. 8, New Orleans, 1a: J. F. W. Sohxffbb,
Professor of Mathematics and Qerman, Mercersburg College, Mercersburg, Pa.; Chablbs H. Tcttton, Wilkes Barre, Luzerne
County, Pa.; and T. P. Stowbll, Bochester, N. T.

We are to prove that (AE}«-|- BExEC = constant. Let AD be the altitude ; we have BE=BD — ED ;
EC = BD -I- ED, for BD = DC ; multiplying, BExEC= (BD)* — (ED)«. Adding ( AE)* to both members,
(AE)« -f BExEC = (AE)« + (BD)« - (ED)* = (AD)* -f (BD)* = (AB)* = (AC)*.

Solved in a similar manner by Bobins Fleming, E. P. Norton. J. A. Pollard, Sylvester Bobins, E. B. SeUz and L. P. SMdy.

ExceUent solutions received from Messrs. Albert, Baker, BuUis, Casey, Clark, Drummond, DeLand, Miartki, NkhoU,OathMd,
Wlekersham and Wiley.

V

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98^— Proposed by Jaxbb Q. Bbwhah, Walton, Harv^ County, Kansas.

What is the rate per cent, of interest when a sum of money amounts to ten times itself in 21 years,
compounded annually? And what would be the rate for the same time if compounded semi-annually?

I^— Sohitlon hj William Hoovsb, Saperintendimt of Schools, and Mathematical Editor of the WUiatdmver^ Wapakooeta,
Aofl^aiae County, Ohio.

1. — ^Put 10 = M, 81 = <, the principal sp, and let r = the rate per cent, required ; then p(l + ry = iip.
or (1 + ry = !». Taking logarithms, llog(l+r) = logn; whence r=: log- »('^^—l, = 0.1158+.
2. — ^Let n = the rate in this case; thenp(l+|ri)'<=fip, or (l-|-|ri)»=f»; whence

n = 21og-»(^^)-2. =0.1127+.

lYv—fiolptlon by Datid Wioksbsham, Ooonty Snnreyor, Wilmington, Clinton Coonty J[). : Hon. J. H. Dbumvond, Portland,
Maine; K. S. Puthax. Bome, N. T^ Stltbbtbb Robins, North Branch Depot. N. J.; A. R. Bulus, Ithaca, N. Y.; V. Wkbstsr
Hbath, Rodman, N. Y.; Willlam Wilbt, Detroit, Mich.; G. H. Lblaxd, Windsor, Vt; Gatot Shaw, Kemble, Ontario, Canada;
and JoHX L Cuok, Moran, Clinton County, Indiana.

Let X be the rate per cent, and suppose the principal to be one dollar ; then (l+«)^ =s 10, and log(l-f«)
s=jC= 0.047619048. The number answering to the logarithm 0.047619048 is 1.11588; therefore l+a;r=
1.11588, and the rate per cent is 0.11588.

The rate per cent, when compounded semi-annually is found thus : Let y be the rate for six months
and (l+y)«=10 and iog(l+y) = :/f =0.023809524; the number answering to this logarithm is 1.06635,
and the rate per cent for six months is 0.05635, and the rate per cent for a year is double this or 0.1127.

Nearly thus were the sohitiona by Messra. Banki, C|»ey, FletMng^ Hearty NkhoU^ OcOhout^ PoUatd^ Mofer^ Shitfy, Skeetiy and

93^— Proposed by Prof. E. J. Eumuwd s, B. 8., New Orleans, Orieans Coon^, Looisiana.

A point P being given on the base of a triangle, draw a line across the triangle parallel to the base
which will subtend a right angle at P.

Solution by J. F. W. ScHnrsn, Professor of Mathematlca and German, Meccenbnig College, Mercenbnzg, Franklin Co., Pa. ;
and T. P. Stowbll, Bochester, Monroe Co., N. Y.

Cbnsfnicftofi. — ^Let ABC repres^it the given triangle, P the given point in the
base. Draw the medial line CD, connect P with G, describe from D with a ra-
dius = BD an arc which cuts GP produced at £, Join D with £, draw PF par-
allel to D£ and through F draw GK parallel to AB ; then GK is the required
line.

DemoiMerafiofi.— D£ being parallel to FP, we have DE : FP = CD : GF ; but
GK being parallel to AB, BD : FK =^ CD : CF ; hence DE : FP = BD : FK, but
by construction DE = BD, therefore FP = FK = GF, consequently the semi-
circle described upon GK passes through P and the angle GPK is a right angle.

Also constnicted In an ingenious manner by Prof. Frank Albert, Pof . B. B. 8eUx and L. P.
Solved algebraically by Messrs. Casey, Baker, BuUU, Hoover, PoUard, Siveriy, WUey

M«"~Prapo«^ ^ Hnnrr NiOHOLi, Hampton, Rock Island Co.,IU.; and Mrs. Anna T. Sntdbb, Chicago, Oook Cb., Dl.

It is required to divide a tiH;)ering board into two equal parta by sawing it across, parallel to the ends.
Find the width at each end so that the lengths of the pieces will be expressed by rational numbers.

I,«8olntion by J. F. W. Bcbmitmb, Professor of Mathematics and German, Mercersboig College, Meroersborg, Pa.

Denoting AB by x, CD by y, EF by «, we have A ABG : ADCG = a^ : ^,

or ABGD : ADGG = jj«— y» : y^. . . (1), and similarly ADGG : EFGD = w* : ir'— y«. . . (2).

Multiplying (1) by (2), and taking into consideration that ABGD = 2EFGD, we obtahi

a!«-|-2^ = 22f2 (3).

In order to find rational numbers for x, y, z, we put x = «-|-2m, y = «— 2(m-f 1), and
then find 2 = 2m< -4-^^+1* a;=:2m^-4-4m-)-l, 2/ = 2iii'' — l, in which any integers may be
substituted for la. Letm= 1, thenz = 5, x = 7, y=l. If m = 2, 2= 13, a; = 17, y = 7;
Ac, Ac.

TT.— flolntion by Wai.ter SrmiLT, Oil City, Venango Coonty, Pennsylvania; and 8ih.TE8TEB BouNS,
North Branch Depot, Somerset Comity, New Jersey.

Let a be the width of wide end, h the width of narrow end, I = length of the board
and X ^ width where cut.

Then - _^= length of shorter piece, and -^^£3jl = length of longer piece, which will be rational

when X is rational.

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Ab the pieces are equal, we have --'^z^^ = ~i7i^1o^- ' whence x = Ji/(2i^+2&*).

Put a = m'\-n and b = m—n, and we have afl = m*+n*. Now put m = 2pg and n =p>^g>, and we
haveaj=p^+ff», o = apg-fp»-~g», 6 = apg— pi+g«.
Takingp = 2and9 = l, 2 = 6, a = 7, 6 = 1; p = 3 and9 = 2, « = 13, o = 17, b = 7;eto., etc.

• Another eolation, by Dr. David 3. Bari, M. A., wUl be given in No. 6.

Good eolations receiTeA fkxmi Meaars. Albert, BuBU, Ca»ey, Clark, Drummond, Fleming^ Bomer, FoOardy SeUg, SkUtg, TvUm
snd Wlektr$ham.

05w-~I'iopoeed by T. A. Knrrar, St Albans, Franklin County, Vermont

Given two circles and a point without them to draw through the point a line cutting the two circles so
that the portion of the line intercepted between the two circles shall be equal to a given line. — [From
Chaucenet'8 Oeomeiry.

8<datlon by Marcus Bakkb, U. S. Coast Sanrey Office, Washington, D. C.

This problem which appears as Exercise 85 of Chaunend'8 (Geometry crept into that work through some
mistake. It is there erroneously considered as a problem that can be conetrwied and is stated to admit
of two solutions. It is a constant source of trouble to persons meeting it there without knowing its true
character. It was proposed, in different words, in the Analyst, vol. ii, p. 156, Sept. 1875, by Prof. James
6. Clark of Liberty, Mo., and by a curious coincidence appears as Problem 85. In the next number of
the Analyst, p. 193, the true character of the problem was pointed out by Prof. Scheffer and myself.
Prof. Scheffer stated that the problem admitted otfow solutions while the equation obtained by myself
was of the 8th degree. The following investigation however indicates neither eight nor fimr solutions,
hut six.

It is of interest to note that If we replace the two circles by two right lines perpendicular to each other
we still have a difficult problem and one transcending the elements of Geometry. If however the given
point lies in the bisector of the right angle the problem can be solved by the elements of (Geometry. It
is known as Pappus* Problem. Solutions of it may be found In several books. [See Gatalan*s Theoremes
et Problemee de Geometric elementaire, 8^, Paris, 1872, p. 146.

Let O and O' be the centers of the two given circles whose radii are OM
= R and O'N = r, MN = the given line = 22, P the given point, x the dist-
ance from P to the middle of MN, d = OP, d = OT. / OPO' = a, / 0PM
= ©, /OTN = <?/.

«ow a+g}+<f/ = 19(P, therefore cos ^ = — cos (a -f^) or
cos9>+ooeaco6^ = sinasin^ ;
whence squaring and transposing ^

coeCgf'^-ooe'q/ '\-2iX)Ba(iOB<pooBq/ —Biti»a = (1).

From the figure ooe^=ii=gg=^ an<lcoe^=('+^^+*;--^.
For brevity put <^— B* = m* and ^f*— r« =n?, and substituting in (1) we have

4d'(l-ir)« ■*" 4d'»(l-fa;)« + 4ddf(l^-a^) — sto a-U.

from which equation x may be determined.
A somewhat laborious reduction reduces this equation to

Aa^-J-Baj6-J-Gr<H-Da:»+Eaj*-J-Fa;-J-G = 0,

in which A = <r«— 2Ar 006 a-Hf, B=— 2l(<f'*— <P). C = (2m»— Z*)<l'*-J-{an* — ^)d»+2(3ii —«»«-- n«)Aroo8a

— 4cf»(r«sin»a, D = 4([l^d'«— d*)-fcWXn«— m»)cosa], E = (m^— 4m«Z'— ^)d'« + (n*— 4n«l»— ^)<l*

— 2(m«»»+3*«)Af'cosa+8i«<W«sin«a, P = 2fl:(m^— i*)<r*— (n*-«*)d'-f 2l*(ma— n«)<*r'oo6a).

Gss=P{d'«(P+m»)«+#(J«+n»)«4-2[mHi» + (m?4-n«)?»+^]Afcosa— 4M^'*sin«a}.

This problem was also solved by Prof. Caaey and Walter Siverly. Prof. Caeey employs trigonometric fonctions, and has bad
bis solnuonpablished in the Analytt, vol. vi, p. 189. He remarks— **I do think it very likely the problem Prof. Chanvenet meant
is this: Throagh one of the points of intersection of two given circles, to draw\ Ac., and not that given in his Geometry.'*

96d— 'Proposed by L. C. Walkkr, New Madison, Darke Coanty,Ohio.

A cannon ball, radius r, rolls into the comer of a room whose walls are at right angles, and perpen-
dicular to the floor. What is the radius of another ball Just touched by the cannon ball?

Solution by John 8. Botxb, Ansonia, Darke Ck>nnty, Ohio; Prof, H. T. J. Lunwie, Mount Pleasant, N. C; Prof. W. P. Caskt,
San Frsncisco, Cal.; Prof. J. P. W. ScHsmn, Mercersbnig. Pa.; Prof. E. B. Seitz, Kirksville, Mo.; Prof. Fbaxk Albsbt,
Millersville, Pa.; Hon. J. H. Dbuxxond, LL. D., Portland, Me.; L. P. Shidt, U. S. Coaut Survey Office, Washington, D. C; K. 8.
Putnam, Rome, N. T.; Robins Flbmino, Readington, N. J.; C. H. Tutton, Wilkes Barre, Pa.; W. B. Hbal, Wheeling, Ind.;
J. A. Pollabd, Wbidsor, Vt; and G. H. Lblakd, Windsor, Vt

Let X = radius of smaller ball. The distance from the comer of the room to the center of the larger
ball is the diagonal of a cube whose edge is r, and is equal to ri/3. Similarly, the distance from the

-91-

eorner to the center of the smaller ball is xy/3 ; then X}/S -}- a;+r is the distance from the comer to the
center of the larger ball. .*. x\^S -f aj-f r = r\/3 ; whence x = r{2 — \/3).

Solved also by Messrs. Baker, B<mks, BuUis, Clark, Hart, Beath, Hoover, Nichcls, Norton, Ifatz, Sieerly. Wickersham and

97w— Proposed by Abtbmas Martin, M. A., Member of the London Mathematical Sociely, Erie, Brie Coanty, Pennsylvania.

Solve, by quadratics, the equation x* — 2aa^ — 2ahx + b> = 0.

I^-8olntion bj Prof. W. P. Caskt, San Francisco, California; B. A. Bowser, Professor of Mathematics and Engineering,
Bntoers Scientific School, New Brunswick, N. J.; E. B. Sbitz, Professor of Mathematics, North Western State Normal School.
KlrisviUe Mo.; L. P. Shidt, U. 8. Coast Sorvey Office, Washington, D. C; Prof. U. 8. Banks, Newbnrg, N. Y.; Stlvistbb
BoBiNs, North Branch Depot, N. J.; and W. L. Harybt, Mazfleld, Mahie.

Adding (a«-f 26)aJ« to each side, ar»— 2aa?»+(c^-f 26)a^— 2a6jj-|-ft* = (o»+25>»«.

Extracting the square root, we get «>—ax+6= j-a?i/(a«-f25). and aj«— [a±\/(a* -|- 26) >==— 6;

,-. «=jjo±i/(a?+25)+i/[2a?— 26±2ai/(a?+26)]}.

Similarly solyed by Messrs. Hart, Leland, Bollard and Putnam.

n.— Solution by Waltkb SiyxRLT,Oil City, Venango County, Pennsylvania.

Letic«— 2aa!»— 2a6a?+6« = (a!»-J-fiu:+6)(a!»+iu;-f6). =ar«-J-(m-f n)a'-J-(25+mn)a!8-f 6(m+n)a:-J-6*.

Equating coefficients of like powers of x, we have m-f n = — 2a and mn = —26, which give

m = — o-J-i/(a«+26), n = — a— i/(a«+26),

and we have the two quadratics a^— [n— i/(a^-J-26)]aj-J-6 = 0, aj«— [a+\/(a«-J-26)]a?-f 6 = 0;

.•.aj = l{a=Fi/(a*+26)±i/[2a«-26=F2ai/(a=+26)]|.

Other solutions wHl be published in No. S.

Good solutions given by Messrs. Albert, Baker, Brotm, BuUie, Heal, Hoover, Ludwiff, Matz, Seh^er, Tutlon and Wood.

96^— Proposed by Mrs. Anna T. Sntdrr, Chicago, Cook County, Illinois.

On a circular griddle 12 inches In diameter, 3 equal circular cakes are baked of such size that each
cake touches the edge of the griddle and the edges of the other two cakes. What is the diameter of each
cake?

I«<— Solution by the Profossr.

Let G be the center of the griddle, and draw any diameter AE.
With center E and radius equal C£ describe an arc intersecting
the circumference of the griddle in H. Draw a tangent to the
griddle at E and produce the radius GH to intersect the tangent
at B. Bisect the angle EBG and draw the bisector ; it will meet
the radius CE in G, the center of one of the cakes. Draw 6F
perpendicular to GB ; then GF =GE, the radius of one of the
cakes.

By similar triangles. GF : FG::GE : EB. But, if i2 = radius
of griddle and r = radius of each cake, FB = EB = R\/3 and
GF = 2/2— «v/3; .-. R{2—^/3):r::R:R^/3; whence we get
r = 72(21/3 — 3), and 2r = 12(2v/3— 3) = 5.5692 -f inches.

Elegant constructions also furnished by Prof. Scheffer and L. P. Shidy.

Il.'-Solution by E. P. Norton, Allen, Hillsdale Co., Mich.; J. S. Rotkr: Prof. Frank Albkrt; Prof. E. J. BmcuNDs;
ProfT J. F. W. Schbffkr; William Hoover; J. V. Stewart; Qatin Shaw; A. R. Bullu; William Wilst; J. A. Poltjiro;
Robins Flbming; and Stlvsstbr Rosins.

Let R = (j inches, the radius of the griddle, and r = radius of each caka Then |fV3 = distance from

the center of the griddle to the center of each cake, and we have

r-\-lr\/S = R. .-. r=i2(2v'3 — 3), and 2r = 2i2(2v'3— 3), = 12(2v'3— 3), =5.5692-f inches.

Excellent solutions received from Mei»8rp. Baker, Banks, Bowser, Casey, dark, Drumnumd, Hart, Harvey, Heafh. Ulaml,
Martin, Niehole, Oathout, Seitt and SiveHy.

99«~~P'^opo<«<3<l by W. WooLaET Johnson, Professor of Mathematics, St Johns College, Annapolis, Anne Arundel Co., Md.
The extremities of a line of fixed length which sUdes along a fixed line are joined to two fixed points.
Find the locus of the intorsection of the joining lines.

^.-^oiution by Enoch Bkbut Skitz, Profesfior of Mathematics, North Western State Normal School, Kirlcsville, Adair Co.,
Mii«Fonri.

Let HK be the line of fixed length, OS the fixed line along which HK slides, A and B the two fixed points,
and P the intersection of AH and BK.

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Produce AB till it meets 08 in C; lay off C£ = HK, CD = AB,
and produce CA, making AN = BC. Draw BB, AT, NO, FM
parallel to DE.

Let OT = CB = a, AT = 6, HK = CE = TB = c OM = «and

PM = y. Then we have OB = CT = o-»-c, BB = — r—.

' ^ a-f c

By similar trianglee, BB-PM : OM~OB :: PM : MK, whence

^^^ y(x-a-^c)(a+c) , ^^j j^t-PM : OM-OT::PM : MH.
aft— (a-fc)y

whence MH=fc^\ .>. MK-MH=^^T^yj^> -^^

= c, or by redaction, xy = ab, the equation to the required locus,
which is, therefore, an hyperbola whose asymptotes are OS and
ON.

A Mlatlon bj Prof. Johsuon will aopear in No. 5.

The problem wu abo solved by MesBn. AlbeH, Baker, Bwwn, BuUis, Catqf, Bdmundt, Hoover, Ludtcig, Siverljf and TvUem.

100^— Propowd by Daxhl Kibkwood, LL. D., Profeeeor of Mathematics, Indiana State Uniyenity, Bloomlngton, Monroe
Connty, Indiana.

Bequired to find a number such that when it is added to 15, 27 and 45 there arise three numbers which
are in geometrical progression.

Sointion by Thoxas Baoot, Principal Shelby School, Canaan, Jefferson Connty, Indiana; Trao. L. DsLand, Olllce of the
Becretaiy of the Treasnry, Washington, D. C: Prof. H. 8. Bakks, Newbni^, N. Y.; Mabcus Bakkb, U. S. Coast Snnrey Office,
Washfaigton, D. C: Prof. W. P. Caskt, San Francinco. CalifomU; L. P. Shot, U. 8. Coast Sorvey Office, Washington, D. C;
D. B. OX^OHHVB, Union City, Ind.; W. L. Habtbt, Maxfleld, Me.; WnxiAX Wdjcy, Detroit, Mich.; and J. 8. Bonn, Ansonia,
Daribs Coonty, Ohio.

Let « = the number ; then 15-f a; : 27 +« : : 27+a; : 45 +«• This gives the equation
675+6ai;+a^ = 729+54a:+a:8; whence, 6a; = 54 and « = 9.
^Solutions similar in substance to the above were given by Mesi>ra. Albert. Bower, BulH$. Clark. DnimmomA, Ethnmuie.
Fleming, HoH, Hedl, Hoover, LOand, Matx, Moffrw&r, Norton, OaUumt, PoUard, Putnam, Bobine, Miz, Shaw, BkoeHy and
TuUon.

101^— Proposed by O. H. Mssbill, Mannsville, Jefferson Connty, New York.

Two equal circles intersect, the center of each being on the circumference of the other. From A, one
of the points of intersection, a diameter AB of one of the circles is drawn. Find the radius of the circle
touching AB and the circumferences of both the given circles.

Sointion by K. B. Skits, Professor of Mathematics, North Western SUte Normal School, KirfcsvlUe, MiaaoozL

Let M and K be the centers of the given circles. There may be three
circles drawn to touch AB and the given circles, as shown in the
diagram.

To fhid the radius of the circle O, let AM = r, OP = x. Then OM =
r—x, MP=v'(*^— 2n»), AP = r— y'CrJ— 2rjj), and we have the equation
ir=[r— i/(r'— 2ra;)]tan30o, whence a? = |r(i/3—l).

To fhid the radii of the circles O' and O", let O'B or 0"8 = y, /O'MB
or /0"M8 = ©. ^NMO' or /NMO" = q>. Then + <?> = 180° + 6po ;
hence sin(6-f-^) = ±i\/3, whence by developing and reducing we get
sin3 6 -I- COS'' ^ =F \/3 sin cos <p = }.

ButsinO= ^^-, andoos^=2T~]^Y Substltutingthe values of sin and cos ^, and solving, we and

the radii to be O'B = rh^'vSe — 10^/3), and 0"S = liftr.'Ofi-f 1 V3).

No other complete solnUon received. Partial solutions given by Messrs. Albert, BulUe, Caeeg, Booner^ Sthi§>er, SkverMf,
TutUmvaA WUejf.

108«— Proposed by J. D. Wuxiaxs, Superintendent of Public Schools, Sturgis, Ht. Joseph Connty, Michigan.
Solve the equations ar + y + «y = 75, a^ - y' = 315, by quadratics.

Sointion by William Hoovbr. Superintendent of Schools, and Mathematical Editor of the WUtenberger, Wapakooota, Ohio;
£. P. Norton, Alien, Michigan; and the pBopoeBB.

Prom(l). x'-321 = y2-9 (3). or «;-18 = -^.---^ (4).

x-|-lo aj-f-io

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From (2), x = i^-,-f- (5). or a;-18 = y? -^ (6).

^ -" 1-fy 1-fy ^ '* 1-fy 1-fy "^

From (4) and (6) by reduction we have j^-f ^-?^±A?).y = 9^.51^1?). (7).

Completing the squares by adding ( ^\\ \ ) *^ ®*^^ ^*^® *"^<^ reducing we find y = 3 ; then (5)

gives x= 18.

DUferent ipethods were emirioyed by Meesn. Alberi^ Cat^, Hari, NlchoU, BoMn» and Sivmif/.

108»<— Pvopoeed by Wii,uMi WpOLBtfT Johmsov, Profeeeor of Matbematlcs, St. Johns College, Annapolif^ Mwyland. -
BB is an ordinate, from any point B of a circle, to the diameter passing through the filed point A ;

and T lis the intersection of the tangent at A with the radius produced through B. Find the locos of the '

intersection of AB and TB.

I^-«olation by K A. Bowsdl Profeeeor of Mathematice and Engineering, Rotgers SdenUflc Sdiool, New Bnnuwick, N. J.';
Fbajk Albkxt Profeesor of Matbematice, Pennei^vania State Normal ScboS, MilteriTiUe, Pa. ; Wai.tbb Sitbblt, Oil City, Pa. ;
A. R. BuLLU, Ithaca, N. T.; and C. H. TuTTOM, Wilkee Barre, Pa.

Take the diameter for the axis of X and the tangent AT for the axis of
Y. Call the co-ordinates of B a/ and y', and the co-ordinates of P, the
Intersection of AB and TB, x and y, and r the radius of the circle. Then

AT = ^;^^ ; the equation of AB Is y' = ?(x'. . . .(1) ; the equation of TB Is
-^ = _j^ (2), and the equation of the circle Is y« = (2r— a;>r. ..(3).

Q«M 2T*t/

Solving ( 1 ) and (2) we obtain ixf = —j— and y = — —- , which substituted

^ (3) gives us for the required locus, y> = nr, a parabola, which passe e

through the origin A and the extremities of the vertical diameter, with

parameter = r, and Its focus one-fourth the distance from A to C.

AnalagooB solations given by Meesrg. Edmunds, PoUard, Seh^sr and WUey. This problem wae also Holred by Ijuchu Brown,
the PropOMT and Professor Sem. Mr. Brown*t solntion will be given in No. 5.

104«— Propoee^ by W. T. R. Beu^ M. A., Principal King's Moontain High School, King's Mountain, Cleaveland Co., N. V.

The railroad debt of a certain company is $56400, Interest at 7 per cent., coupons payable seml-annuallv.
What tax must be levied to pay the Interest and create such a sinking fund as will absorb the debt In 15
equal annual payments?

Solution by Robins Flkming, Readington, Hunterdon County, New Jersey.

The yearly rate per cent, of $1 at 7 per cent, interest compounded semi-annually = (1.035)*— 1,
= 0.071225.
Let D = $56400, r = 0.07, /2 = (1 -f Jr)* = 1.071225, n = 15 = number of payments and x = annual

payment. Then ^, p^, ^-, p^ = present worths of the n payments. The sum of these present

XXX X D

worths must be equal to the debt. •*• D+EM+»i + + >»« = A and x= ^i- , , , .

This problem wat* also solved very elegantly by Messrs. Albert, Banks, BuUis, Clark, Seitz, Siverly and Wlckershatn.

105a~PropoB<^ by John Rea, Hiirs Fork, Adams County, Ohio.

Find the co-ordinates a, 6, a', V, of the centers, and the radii r, r' of the two circles
3/«-faJ2 — 2ar— 40 = 0, and ar^-fj^— 40^+50 = 0;
and In case the two circumferences cut each other, what are the co-ordinates of the points of intersection?
— [From Loomis' Analytical Geometry.

Solution by L. P. Shidy, IT. S. Coast Survey Office, Washington, D. C. ; Marcus Bakeb. U. S. Coast Surrey OlHce, Wash-
ington, D. C. ; and Frank Albert, Professor of Mathematics, Pennsylvania State Normal School, Millersville, Lancaster Co., Pa.

The general equations of the circles are (x— a)« -f (y— 6)2 = >^ and {x—a'Y -f- (y — 6')' = ^^ respect-
ively. Expanding and comparing with the given equations we find for the first circle a = 10, 6 = 0,
r = 2i/36 = 11.832-1-; and for the second, a' = 0, 6' = 20, 1^ = 51/14 = 18.708 -f-.

-94-

If the circles intersect, the values for x and y at the points of crossing must be the same for each
equation.

Subtracting (1) from (2) we get 20a;— 40y+ 90 = 0, a; = 2y— 4.5. Substituting in equation (1), and
reducing, y = A[58 + ^/(ISSa)]. = 10.226 or 1.374 and a; = ^<5[71 ± 2^/(1959)]. = 15.952 or —1.752.

Note.— The sign of 40y in the second equation was erroneously printed -f instead of — .

Solntione of the problem w printed were received from MessrH. BuUia, Oatey, Edmunds, Fleming, Bsal, //oor<v. Ijeiand,
liiUard, JShidy, Siverly and Tuiton.

lOftir-Propoced by 8. C. Brack, PhiUuielphia, Pemuylvania.

A circular e#w, one foot in diameter, in cutting off a round log, is stopped when it reaches the center
of the log, when it is found that the wood covers J of the side-surface of the saw. What is the diameter
of the log?

Solution by Framk Albsbt, ProfeaMr of Matbematics, Pennsylvania State Normal School, MiHemvUle. Lancaiiter Co., Pa.;
L, P. Shidt, U. 8. CottBt Survey Office, Washington, D. C; E. B. Sbits, ProfesMir of Mathematics, NortA Mlnsoari State
Normal School, KlrksvUle, Adair County, Mo.; Abram R. Bullis, Ithaca, Tompkins Coonty, New Toik; and W. L. Harvxt.
Vazfield, Penobscot Coonty, Maine.

PutAG = i = a. BC = r, /ACB = e; then r = 200066, the sector BCD
= lr«0 = 2a«9co6se, the sector BAG = Ja«(jr— 26). and the triangle BAG
= Ja«8in2e; hence DBCE = a«(jr— 26-f 46cos«e— 8in2e).

or sin 26 — 20 cos 26 = |)r.

By Double Position we find 26 = 125^ 4^' 5'' and 6 = 62^ 53' ^".

But r = 2aco86 = cos(62o 53' 2i'0 = 0.45579 feet = 5.46948 inches, and
the diameter is 10.9389 inches.

Solved alHO by Prof. Casey, SapH Hoavef, Geo, H, Leland, E. P. Norton, Prof. Sch^er. WcOUr Shmiy and Ch<u. H. Tutton.

107»— PropoMMl by Sylvester Robins. North Branch Depot, Somerset Coanty, New Jersey.

At a Fireroens's Fair a silver truippet is offered to the company exhibiting the ladder that can bo used
in the greatest number of streets and alleys for the purpose of reaching windows on either side without
changing the location of its foot. All bases and perpendiculars must be rational lengths, and a company
may Include in their count dimensions having as many decimal places as their ladder has, but no more.

The *'Hudsons" bring a ladder 65 feet long, the *• Keystones" offer one 321 '®®^ ^^ length, and the
"Delawares" show one of 42 J feet. To whom must the trumpet be awarded, and on what count?

I.^Solution by the Piioposxr.

General expressions for the sides of a right-angled triangle are 2^8, g(r-—«'») and g(r--f«'').

Hypothenuse 65 = 5x13 = 5(3* -f 2^) = 13(2^ -f-l*) = 1(8^ + 12) = 1(7^+43). These values substituU^l
in above expressions give sides of four right-angled triangles. I. 25, 60, 65; II. 39, 52, e>5; III. 16, G3,
65 ; IV. 33, 56, 65. The "Hudsons" ('«,n use this ladder in every street whose width is the sum of any
two of these 8 legs; the number of these combinations is 8x7-^2 = 28. They can also use it in every
sti'eet whose width is tmce the length of one leg — foot of ladder being in middle of street. There are 8
Hueh streets. And they can use it in every street whose width is the length of a single leg — foot of ladder
being against a house — S streets. Count for the "Hudsons", 44 streets.

Hypothenuse 32<5 = 5x5x1.3 = 2.5(3'^-f 2^) = 6.5(2--' -fl^^) = 1.3(42-f3-^) = .5(82-1-1'^) = .5(7*4-4*)
= .1(17- -f-^*) = .l(lS*-f-l'^)- Substituting these values in the general expressions, we have sides of »eveit
right-angled triangles. 1.12.5,30.0,32.5; 11.19.5,26.0,32.5; III. 9.1,31.2,32.5; IV. 8.0,31.5.32.5;
V. 16.5, 28.0, 32.5; VI. 20.4, 25.3, 32.5; VII. 32.3. 3.6, 32.5. In all 14 logs, which taken 2 at a time
give widths of 14x13 -f- 2 = 91 streets. Ladder in middle of street, 2x7 = 14. Foot of ladder againm
a house, 14 streets. Count of "Keystones", 119 streets.

Hypothenuse 42.25 = 5x5x1.3x1.3 = .05(29* -f 2--) = 8.45(2^ -f- 1*) = 5(2.2* -{-1.9*) = .13(18-'-fr^) =
13(1.7*-!- .62) = 3.25(33 -f 2*) = 1.69(4* -f-3*) = .65(8* -f-l*) = 65(.7*-f .4*) = 2.5(1.2* -|- .5*) = 1(6.3^ + 1.6*)
= 1(5.6* -f 3.3*). Substituting these values as before, we have sides of twelve right-angled triangles.
I. 41.85,5.80,42.25; II. 25.35,33.80,42.25; III. 6.15,41.80,42.25; IV. 41.99,4.68,42.25; V. .32.89, 26.-52,
42.25; VI. 39.00, 16.25, 42.25; VII. 40.56, 11.83, 42.25; VIII. 40.95, 10.40, 42.25; IX. 21.46, 36.40, 42.25:
X. 29.75, 30.00. 42.25; XI. 37.13, 20.16, 42.25; XII. 20.47. 36.96, 42.25; in all 24 legs, which taken two at
a time give 24x23 — 2 = 276 streete. Ladder in middle of street, 2x12 = 24 streets. Foot of ladder
against a house, 2x12 = 24 streets. Count of " Delawares", 324 streets. Trumpet must be awardtvi to
the '* Delawares".

The Hohition by Wali^-r Sirerly Ip «hnllar to thin. A general nolution by Dr. Ifarf will be puhliHhod In No. .'».

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IflS^^Propoped by Justus P. W. Schkfpkb, Professor of Mathematics and Gennan. Mercereborg College, Mercfn*buiv.
Franuln County, Pennsylvania.

Three army corps. A, B and C, being engaged in a campaign, have provisions sufficient for 30 weekn.
On these provisions B and C would subsist 9 weeks longer than A and B ; and A and G, 15 weeks longer
than B and .C. After 6 weeks the three army corps encounter the enemy, in consequence of which A
loses I of its troops, B looses ^ and C looses I ; also | of the remaining provisions are lost. How many
weeks will the remainder of the three corps subsist on the provisions left?

Solution by Lucius Bbown, Hudson, Middlesex County, Mass.; and Abram K Bullis, Ithaca, Tompkins County, New York.
Let unity represent the quantity of provisions at first, and x, y, z the proportions for A, B, C, respectively.

Then -. + ,+.==1 (1). ^A^-^l-^=A=ft (3). ,i,-,i, = H W

SubstituUng in (2) and (3) the value of « in (1) gives

— — =-Aj and ^ -^ =1.

Eliminating y by comparison gives

T + 3^~- =3^' ^^ 3a^ + 5a^-71a.+23 = 0.

The only value of x consistent with the conditions is a; = }, therefore y = } and 2 = (.

After the encounter with the enemy the provisions remaining are | of J = 1, and the quantity consumed

in one week is 2 of ^X'\-% of ^y -f } of ^z = i^. Therefore what is left of the provisions will last tht^

remainder of the troops 18 weeks.

Good solutions were given by Dr. Hart. E. P. Norton. J. A. Pollard. L. P. Shidy, Professor E. B. SeUx, Walter Sir^y and tho
Propo$er.

109«~Propoflod by Thomas Baoot, Principal Shelby School, Canaan, Jefferson County, Indiana.

A field in the form of a right-angled triangle has a base and perpendicular of 40 and 200 feet
respectively. What length of rope attached at the vertex of the right angle will permit a horse to graze
upon half the field?

Solution by Julian A. Pollard. Windsor, Windsor County. Vermont; Abbam R. Bullis, Ithaca. Tompkins County, N. Y.;
and Lucius Brown, Hudson, Middlesex County, Massachusetts.

Let ABC represent the field. AB = 200 feet, BC = 40 feet, and
let a; = BE = BF = length of rope, = / EBF.

AEF = EFCB = half the field = 2000 square feet, ABF =
lOOa?sin0, EBF=ia^6, BFC = 20a;cos9.

.'. lOOaJsinO — ia^ = 2000..(l), 2Oa;cos0-f Jar^ = 2000 . . (2).

Adding (1) and (2) we get x = ^^ij^fj\^~^^Q- Substituting

this value of a; in (2) and reducing, 100 = 258in«9 — co8«©, or lOd-f 13cos26 = 12, from which we find
6 = 28° 46' 45" and x = 60.928 feet.

Solved also by Messrs. Albert. Casey. Harvey. Hoover, Leland, Shidy. Seitz. Siverly. Tu(4on and miey.

UO^^Pkusk Problbm. Proposed by Abtkmas Martin, M. A., Member of the London Mathematical Society. Erie. Pii.
Give an expeditious method of approximating to the square root of a quantity, and find by it tli<'
Hquare root of 2 to at least one hundred and fifty places of decimals.

I,^-8olution by Waltbr Siyerlt, Oil City, Venango County, Pennsylvania.

By the usual method extract the root in full to one more than half the number of figures required :
then using double the root found as a divisor, and the remainder as a dividend, contract the division by
cutting off one figure from the right of the divisor for each new figure of the root.

Extracting in full to 78 places I find

^/2 = 1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875;^^
764157273501384623091229702492483605585073721264412149709993583 + .

There are various methods of extracting roots by approximation, but to the best of my knowledge,
when used to extract the square root, in all of them, the work is more than by the above method.

-96-

Hw— Solntion by K. S. Putnax, Rome, Oneida Connty, New York.

1. Let n = the number, and x = itB square root ; then o^ = n. Suppose the root ^r and <^r-f-l, and
let r+p = the root ; p will then be a fraction, and p^ quite small.

... aJ» = n = r*-f 2rp-fp«; or, rejecting i>*, t^+2rp = n, 1> = -^, a; = r-fp = r-f ^~-^ = ^±^.

By constantly substituting the derived value of x for r in the equation we shall approximate to the
value of X,

2. If a^ = n = 2, then r = 1 and we have, from x = r-f p, p = J ; .*. a; = |. Putting r = f we find for
second value, aj = }} ; third value, x = JJJ ; and we see that assuming any value of x thus derived

= ~, the next value wiU be «= ?^^ = ~ J^.
2ab b 2ab

w^Ko^^fi,^ /IN « 3 /ox 2a«— 1 17 ... 2(2a»— 1)«— 1 577

Wehavethen (1)^ = 5^ = 2. (2)^=-^^ = i2. W ^ = 2k6(2a^^) = 408'

. .V _ 2 [2(2 a«— ly — ip~l _ 66585 7 __ 8867310 88897

^*^ * - ^X2:a5(2a^- l)[2(2a^- 1)*- 1] - 470632' ^^^ * " 627013566048*

Substituting the numerator of (5) for a in (4) and the denominator for b we have

JJ = t/2 = 48g»»*<>g»^4g»881fl54B868Qe8 898566945B8492188^^

*^ 84fi9686861691909g7tt5818M5S8001486151788d860071W»34^

This fraction will give the root true to 195 or 196 decimal places, and the next correction is a fraction
having 1 for its numerator and for a denominator twice the product of the numerator and denominator
of the above fraction. By applying this correction the root will be found to over 390 decimals. It is
quite easy to correct the root to 200 places, and I find

"V/2 =1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432
7641572735013846230912297024924836055850737212644121497099935831413222665927505592755799950501152782
060837320076 4. .

If the numerator and denominator of the last-foimd value of x be substituted for a and b in (4) the
root can be foimd true to over 1560 decimal places, and then quite easily corrected to 1600.

The root was not found to the required number of places in any of the other eolutions received.

l^ist of Contributors to the Junior department.

Walter Siyerlt, Oil City, Venango Co., Pa., solved all the problems. Abbaji R. Bullis, Itbaca« Tompkins Co., N. T.,

polved all but 05 and 102. Frakk Albert, Profeesor of Mathematics, Pennsylvania State Normal School, Millersvllle,

Lancaster Co., Pa., solved all but 107, 108 and 110. E. B. Seitz, Professor of Mathematics, North Missouri State Normal

School, Kirksville, Adair Co., Mo,, solved all but 96, 108, 107 and 110. Professor W; P. Casey, San Francisco, California,

solved all but 108, 104, 107 and 110. L. P. Shiby, U. S. Coast Survey Office, Washington, D. C, solved all but «J, 99, 102,

108, 104 and 110. Julian A. Pollard, Windsor, Windsor County, Vermopt, solved all but 95, 99, 101, 102, 104, 107 and 110.

WiLLi/LM Hoover, Superintendent of Schools and Mathematical Editor of the WUUnberg&ry Wapakoneta, Auglaize County, O.,
solved all but 88, 90, 95, 108, 101, 107, 106 and 110. Sylvester Bobims, North Branch Depot, Somerset Co., N. J,, solved SB,

89, 90, 91, 92, 94, 96, 97, 98, 100, 102, 107 and 110. Dr. David S. Hart, M. A., Stonington, Conn., solved 88, 89, 91, 92, 94, 96,

97, 98, 100, 102, 107 and 108. Charles H. Tutton, Wilkes Barre, Pa., solved 91, 94, 96, 97, 99, 100, 101, 108, 106, 106, 109 and

110. Roams Flemino. Readlngton, N. J., solved 88, 89, 90, 91, 92, 94, 96, 98, 100, 104, 105 and 107. William Wiley,

Detroit, Mich., solved 88, 89, 91, 92, 98, 96, 98, 100, 101, 108 and 109. J. P. W. Scheiter, Profeesor of Mathematics and

German, Mercersburg College, Mercersbuig, Pa., solved 91, 98, 94, 96, 97, 98, 101, 108, 106, 108 and 109. K, S. Putnam, Rome,

Oneida Couvty, New York, solved 88, 89, 90. 91, 92, 94, 96, 97, 98, 100 and 110. Gboroe H. Leland, Windsor, Windaor Co.,

Vermont, solved 88, 89, 90, 92, 06, 97, 98, 100, 106, 106 and 109. Marcus Baker, U. S. Coast Survey Office, Washington, D.C.,

solved 89, 91, 98, 96, 96, 97, 98, 99, 100 and 106. John I. Clark, Moran, Clinton County, Indiana, solved 88, 89, 91, 92, 98, 04,

96, 98, 100 and 104. Hon. J. H. Druxmond, Portland, Maine, solved 88, 89, 91, 92, 94, 98, 100 and 110. Prof. E. J. Ed-

munds, B. S., Principal of Academic School No. 8, Nevr Orleans, Louisiana, solved 89, 91, 96, 99, 100, 108, 106 and 110. Datid

WicKERSHAM, Couuty Survcyor, Wilmington, O., solved 88, 89, 91, 92, 98, 94, 96 and 104. Gavin Shaw, Kemble, Ontario,

Canada, solved 88, 89, 91, 92, 98, 100, 102 and 110. E. P. Norton, Allen, Michigan, solved 88, 89, 91, 96, 96, 100,102 and 106.

Professor Huen S. Banks, Instructor in English and Classical Literature, Newburg, N. T., solved 88, 89, 91, 92, 97, 96 and 100.
Walter S. Nichols, Editor Insurance Monitor, New York, N. Y., solved 88, 89, 91, 92, 96, 96 and 102. ^, L. Harvey, Max-

fleld, Maine, solved 92, 96, 97, 98, 100, 106 and 109. Thomas P. Stowell, Rochester, N. Y., solved 88, 89, 91, 93, 100 and 110.

V. Webster Heath, Rodman, N. Y., solved 88, 89, 90, 92, 96 and 98. John S. Royer, Principal Public Schools, Ansonia,

O., solved 88, 89, 92, 96, 98 and 100. Lucius Brown, Hudson, Mass., solved 97, 99, 108, 108 and 109; P. P. Mats, M. S.,

Professor of Pure and Applied Mathematics, Bowdon (State) College, Bowdon, Georgia, solved 88^ 89, 96, 97 and 100. O. D.

Oathout, Read, Iowa, solved 89, 91, 92, 98 and 100. H. T. J. Ludwig, Profeesor of Mathematics, North Carolina College,

Mount Pleasant, N. C, solved 88, 96, 97 and 99. E. A. Bowser, Professor of Mathematics and Engineering, Rutgers Scien-

tific School, New Brunswick, New Jersey, solved 97, 98, 100 and 108. W. E. Heal, Wheeling, Indiana, solved 96, 97, 100 and

106. Theo. L. DeLand, Office of the Secretary of the Treasury, Washington, D. C, solved 88, 89, 91 and 100. D. W. K.

Martin, Webster, Darke Co., O., solved 89, 91 and 96. W. T. Magrudbr, Stevens Institute, Hoboken, N. J., solved 88, 89

and 100. B. F. Burleson, Oneida CasUe, N. Y., solved 88, 89 and 90. J. V. Stewart, Muncie, Ind., solvod 88, 89 and 98.

-97-

Thoxas Baoot, Principal Shelby School, Canaan, Ind., solved 88, 89 and 90. W. W. Johnson, Professor of MathematicH,

St. John's College, Annapolis, Md., solved 99 and 106. J. R. Pagan, Erie, Pa., solved 88 and 89. D. B. O'Conner.

Union City, Ind., solved 89 and 100. G. G. Washburn, North Bast, Erie Co., Pa., solved 88 and 89. I. H. Turrbll, Cum-

minsville, O., solved 98. George A. Joplin, Danville, Kentucky, solved 89. Jambs Q. Brigham, Walton, Kansas, solved.

92. Mrs. Anna T. Sntder, Chicago, 111., solved 96. J. D. Williams, Superintendent of Schools, Sturgis, Mich., solved

102. DeVolson Wood, M. A., C. E., Professor of Mathematics and Mechanics, Stevens Institute of Technology, Hoboken,

New Jersey, solved 97.

The first prize is awarded to K. S. Putnam, Rome, Oneida County, N. Y., and the second prize to
Walter Siverlt, Oil City, Venango County, Pa.

148.~Propoeed by Artemas Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.
A man bought a horse for $100, and sold him for $60, and then bought him back for $75. How much
did he lose by the transaction?

149.— Proposed by John I. Clark, Moran, Clinton County, Indiana.

My grocer sold me a cheese, which he said weighed 32 pounds ; but when it was placed on the other
side pi the scales, it only weighed 18 pounds. He then proposed that I should buy another of the samo
size, and weigh it on the opposite side from the first, to which I consente.l. Did I gain or lose by the
transaction? And what was the true weight of the cheese?

150."*Proposed by W. S. Hinklb. Leacock, Lancaster County, Pennsylvania.

How much more can a bank make in 52S days, with $10000, by discounting notes on 45 days time, than
by discounting them on 30 days, the rate of discount being 6 per cent., and the profits in both cases
retained in bank till the expiration of the time?

J 51«^Proposed by 0. F. Msad, Uniontovni, Fayette County, Pennsylvania.

If the sides of any quadrilateral be bisected, and the points of bisection joined, the resulting figur*»
will be a parallelogram and equal in area to half the qua^lrilateral.

152.— Proposed by Artemas Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

A company of n men were counting their money. The first said to the second, '^Give me your money
and I will have $a" ; the second said to the third, ''Give me J of yours and I will have $a" ; the third said

to the fourth, "Give me | of yours and I will have $a" ; the nth said to the first, "Give me ono-wth

of yours and I will have $a". What sum had each?

153,— Propo«<ed by Sylvester Robins, North Branch Depot, Somerset County, New Jersey.

What are the first ten num>)ers which, when multiplied by 1, 2, 3, 4, etc., will give products containing
the same figures, and in the same order though beginning at a different digit, but when multiplied by 7,
17, — , — , etc., respectively, will give products containing all nines?

154*— Proposed by Joseph Ficklin, M. A., Ph. D., Professor of Mathematics and Astronomy, University of the State of
Missouri, Columbia, Boone County, Missouri.

The area of a triangle ABC is 6 ; the side AB is a, and the angle opposite AB is /3. Bequired the sides

AC and BC.

155*^Propo^^ ^y G- ^' Mead, Uniontown, Fayette County, Pennsylvania.

If from any point in a diagonal of a parallelogram lines be drawn to the opposite angles the parallelo-
gram will be divided into two pairs of equivalent triangles.

156«^l*roposed by Theo. L. DeLand, (^ce of the Secretary of the Treasury, Washington, D. C.

A owes B $1750, payable as follows : 70 notes of $25 each, the first payable in one month, the second
in two, and so on to the last which is payable in seventy months ; each bears simple interest at 10 per
cent, per annum payable with the note. When will he neither gain nor lose by borrowing money at 8 per
cent, per annum simple interest to pay all the unpaid notes and interest, and give new notes, interest
due added, payable, the first one month from the date of the change, the second two months, and the
last on the date when the original notes terminated?

15T-^Propo8ed by Sylvester Robins, North Branch Depot, Somerset County, New Jersey.

It is required to prove that a polynomial of n terms may be found which can be divided by a poly-
nomial of n terms and give a quotient of n terms, the coefficient of every term in all of them being unity.

158«~~Propo^e(i l>y O. F. Mead, Uniontown, Fayette County, Pennsylvania.

If two circles intersect, the common chord produced will bisect the common tangent.

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1M«— Proposed by William Hooyeb, Snperintendent of Schools, and Mathematical Editor of the WiitenderffeTt Wapakoneta,
Auglaize County, Ohio.

August, 1879, had five Fridays, Saturdays and Sundays ; when will the month of August have five of
each of these days again?

160»— Proposed by L. P. Shidt, U. S. Coast Survey Office, Washington, D. C.

Find a point within a given quadrilateral, such that when joined to the middle points of the sides the
quadrilateral will be divided into four equivalent parts.

161«""Propo6ed by Abtbmas Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

What rate per cent, of interest paid in advance is equivalent to r per cent, paid at the end of the year?

168.— Proposed by O. H. Merrill, Mannsville, Jefferson County, New York.

Two equal circles intersect, the center of each being on the circumference of the other. A circle is
drawn touching that diameter of the right-hand circle which joins the centers of the given circles, and
the circumferences of both circles, the right-hand one internally and the other externally ; also a circle
is drawn touching the one last drawn and the circumferences of both the given circles. Find the radii
of these two circles.

163.— Proposed by Thbo. L. DbLan d. Office of the Secretary of the Treasury, Wasington, D. C.

In 1861 a 6 per cent. 20-year coin bond of the U. S., interest payable semi-annually, sold on the market
for $0,891 on the dollar ; what, on this basis, would have been the market value of a 4 per cent. 28-year
coin bond of the U. S., interest payable quarterly?

164«— Proposed by V. Webster Heath, Rodman, Jefferson County, New York.

A field 81 yards square has one of its corners clipped off by a line meeting the sides at 60 and 80 yards
respectively from the comer. A man commences plowing around the field, turning a furrow one foot
wide. How many * 'rounds" will he plow before the unplowed portion of the field becomes a square?

166.— Proposed by lion. Josiah H. Drummond, LL. D., Portland, Maine.

A railroad company has $400000 of preferred stock and $300000 of common stock. The agreement is
that the net income each year shall be applied to the payment of six per cent, on the preferred stock,
and the balance shall be divided on the common stock. The net income is $36000 rental paid annually ;
a debt of $50000 is created pavable in twenty years with annual interest at six per cent, (payable at the
same time as the rent) in such manner that the annual interest is payable out of the current income, but
the prhicipal out of all the Income after the debt becomes due until it is paid. It is agreed to create a
sinking fund. What amount must be carried to the sinking fund annually, assuming that five per cent,
compound interest may be earneii to extinguish the debt when it becomes due, and how shall that
amount be apportioned on the two kinds of stock?

166.— Proposed by Dr. 8. Hart Wright, M. A., Ph. D., Mathematical Editor Yates County Chronicle, Penn Tan, N. Y.

Required the variation v of the magnetic needle, in latitude A = 42° 30', the declination of the North
Star being d = 8b^ 40', and its magnetic bearing being 6 = 8*^ 48' 30" when at its greatest elongation east,
and 6' = 5^ 11' 30 " when at greatest elongation west.

167.- Proposed by B. F. Burleson, Oneida Castle, Oneida County, New York.

Given 3(ar^+y^) = 2a?i/(x-fy) and xy{3fi—^^) = V3{x — y), to find accurate expressions for the values
of X and y.

168.— Proposed by J. F. W. 8cHErpER, Professor of Mathematics and German, Mercersbui^ College, Mercersbarg, Pa.
The position of a ball on a circular billiard table is given. Which path must the ball describe in order
to pass through its original position after touching the cushion twice?

169.— Proposed by II. T. J. Ludwio, Professor of Mathematics, North Carolina College, Mount Pleasant, N. C.

Required (1) the radius of the auger that will cut out one-nth part of the surface, and (2) the radius of
the auger that will cut out one-mth of the volume of a sphere, radius r, the axis of the auger coinciding
with a diameter of the sphere.

ITfO.- Proposed by K. S. Pctnam, Rome, Oneida County, New York.

Traveling recently on a train moving 30 miles an hour and overhauling a freight moving 23 miles an
hour, I endeavored to ascertain the length of the freight. One minute from the time I was opposite the
rear of the freight a third train came between and put a stop to my investigation. At the next station
we stopped, the door of the depot being dirt^ctly opposite my seat." The freight here overhauled us, the
front of the train being opposite me, when our "train started. When we had attained the same rate of
speed as the freight I was opposite the same point in the freight as when I closed my first obsrvation.

It took our train IJ minutas to get under full headway, during which time its motion was uniformly
accelerated. How long was the freigiit train, and how far was I trom the depot when I passed the freight?

-99-

ITlk^Proposed i)j Maboub Baksb, U. S. Coast Survey Offlcef Washington, D. C.

In a plane triangle ABC the center of the circumscribed circle is O, the center of the inscribed circle
is I and the intersection of the perpendiculars is H. Knowing the sides of the triangle OIH, determine
the sides of the triangle ABC.

n2«— Proposed *>y W. W. Johnson, Professor of Mathematics, St. John's College, Annapolis, Maryland.

The base of a triangle Is fixed and the difiTerence between the vertical angle and one of the angles at
the base is constant. Find the locus of the vertex. Discuss the curve and consider the cases in which
the constant difiTerence is and 90^.

173,— Proposed by Dr. Samuel Habt Wright, M. A.. Ph. D., Mathematical Editor TcUss CtourUy CArordcU^ Penn Yan, N. Y.

Beqiiired the variation v of the magnetic needle, in latitude X =■ 43P 30^ the declination of the sun
being 6 = 20° N., and the magnetic baring of its upper limb when rising on a horizon elevated A = 1°
is 6 = 69C 21' 40^'. Badius of sun = r = 16', and refraction = p = 35'.

1'74«'— Proposed by Bsmjamik Pbibce, LL. D., F. R. S., Professor of Mathematics, Har\-ard University, and Consulting
Geometer to the U. S. Coast Survey, Cambridge, Mass.

To find by quadratic equations a triangle of which the angles are given and the distances of the vertices
from a given point in the plane of the triangle.

Solutions of these problems shonld be received by September 1, 1880.

SENIOR DEPARTMENT.

Solutions of Problems Proposed in JY^. 3.

Ill,— Proposed by Oavin Shaw, Kemble, Ontario, Canada.

It is required to divide a given number a so that the continued product of all its parts shall be the
greatest possible.

Solution by L. P. Shedt, U. S. Coast Survey Office, Washington, D. C; W. E. Heal, Wheeling, Delaware County, Indiana;
and Waltxb 8. Nichols, Editor Insurance Monitor, New York, N. Y.

Assume that the number is divided into any number of parts 6, c, d, e, /, g, h, etc., the first being the
greatest, so that 6-fc+d-|-€-|-/+flr+fc-fetc. =a.

Letc = 6— 1>, d=:h—q, e=ib—r,f=^b—8, etc. ; then 6(6— p)(6— g)(6— r)(6 — 8)etc. = a maximum.
It is evident that the smaller the differences p, g, r, 8, etc., become, the greater the continued product
will be, for 6 — (p — 1) is greater than 6— p, and hence the product becomes a maximum when 6=c=rf
=:e=^f=g=h= etc.

Having shown that the number must be divided into equal parts, let us assume that there are x such

parts ; then / - j = a maximum, or a;logf - j = a maximum.

Let y = a;log(-Y DiCferentiating. ^ = log(-^— 1. Placing this differential coefficient = 0, we find
log(- j = 1, which shows that - must be the base of the Naperian system of logarithms, or - = 2.71828
= €. Therefore x = ^ . Substituting this value of a; in the second differential coefficient, the result is

negative ; hence f ^ j is a maximum when x = -.

Solved also by Messrs. Baker, BuUis, Casey, Edmunds, Harvey, Hoot?er, Kummdl, Norton, Oathout, JMlard, SeUz, Shaw,
Stverly and TuUon.

USa-'P'oposed by BoBnrs FLEimrG, Readington, Hunterdon County, New Jersey.
Find 34 right-angled triangles having the same hypothenuse.

Solution by I>r. Dayxd S. Hart, Stonington, New London County, Connecticut.
Let :p, y be the legs and z the hypothenuse of a right-angled triangle ; then d^ -f 2^ = s^. The general

values of a;, y are x =

{p' — q')z

y =

2pqz

-100-

the sum of whose squares is = z^.

The formula 4/ + 1 contains all the prime numbers each of which is the sum of two squares. The four
least of these numbers are 5, 13, 17, 29, whose product is 32045 which put =2. Each of these factors is
the sum of two squares in one way ; the product of every two is the sum of two squares in Uro ways ; the
produ.^t of every throo Is the sum of two squares in four ways, and the product of all four is the sum
of two squares in eight ways ; so that the whole niunber of wavs is forty. — See Barlow's Theory of Numbers.
pp. 176, 177,

The following table contains, in perpendicular columns, the sum of two squares, ;j^-f-9'. equal success-
ively to each of the above factors, and combinations of factors, and number of ways in which each com-
bination may bo varied ; p, q the roots of two s(iuares in each case ; the values of j, y in each case, which
belong to the common hypothenuse z = 32 )45, in all 4() sets of right-angled triangles, any 34 of which
will answer the conditions of the problem. As any four factors derived from the formula 4/ + 1 may be
used, the number of answers to the problem is infinite

p'-hq'

P

q

X

5

2

1

19227

17

4

1

28275

('.5

7

4

16269

85

7

4iK)l

145

12

1

31603

221

14

5

24795

377

1()

11

11475

493

22

3

30875

1105

33

4

31117

"

31

12

23093

1885

43

f.>

30^21

•'

38

21

17051

24G5

49

8

33J>-1

"

44

23

18291

«409

80

3

31955

,,

72

35

19795

20 i5

179

2

82037

"

173

40

27813

...

163

74

21093

••

142

100

^2S3

y

25636
15080

27(:0)

316(58

53«!4

2.no()

2)920

8580

7(>56

2157.;

8772

27132

10192

26312

2400

25200

716

15916

2il24

30956

P'-hq'

P

9

X

13

3

2

12325

29

5

2

23205

()5

8

1

31059

85

9

2

29029

145

9

8

3757

221

11

10

3045

377

19

4

29325

493

18

13

10075

1105

32

9

27347

••

24

23

.1363

18.5

42

11

27931

'•

34

27

7259

2465

47

16

25389

•♦

41

2S

11661

6409

75

28

24205

(1

60

53

3955

(2045

178

19

31323

'•

1C6

67

23067

"

157

86

17253

•♦

131

122

2277

y

29580
22100

7888
13572
31824
31900
12920
30420
16704
32016
15708
31213
19552
29848
21000
31800

07<>4
22244
27001
31964

Solutions were alno reccivtMi from the I^jxintr and Mesyrn. DaiU^ Drummontl, IleuL Pollard, RdAnn, Shidy and Sirerhj.

lis.— Propof*<^ d by Fredliuck S. Sami:ei.!*. Cerro Ciordo, Inyo County, California.

What is the volume of a chip cut at an angle of 45 degrees to the center of a round log, radius r?

Solution by Wiij^iam Hoover, Superintendent of Schools and Mathemaiical Ediwr of the Wittenbtrqir, WaiJakoueta, Ohio:
Waltkk Siverly, Oil Citj, Pa.; and E, B. Seitz, ProfeHsorof MathematicH. North Mi^»^o^ri State Normal School, KirkHville. Mo.

Conceive a plane to cut the chip parallel to the "crease" and perpendicular to the end of the log, at the
distance x from the center. The section isa re<'tangle, sides 2.r, 2i/(r* — a;*-) and area *.x\/{r- — x'^). Thf

A\'^xdxV{r' — x''), =Jr».

volume is oxpresse I by

(iood Holutioni* given by Mei^^rs. Alltif, liull'tx, liowm\ Hart, Harvey, McAdam, Xorton, Putnam, ikhifer and Wood.

114.— Propoj?ed by Dr. David S. Hart, M. A., Stouinj^ton, New London County, Connecticut.

To find two numbers such that their sum shall be a square, the sum of their squares a square, and if
the cube of each be added to the square of the other the sums shall be equal.

I,— Solution by Profe-i^or Asueu B. Evanh. M. A., Principal Lockport Union School, Lockport, Niagara County, New York.
Lt^t X and y represent the required numbers ; tlien must

x-{-y=o (1), x^-\-y'' = D (2), x^-^-y-'^y'-^T'^ (3).

Condition (3) readily reduces to x-\-y = x^-\-xy-\-y^, which on putting x = 7iy gives

n-fl

y =

n^-j-n-fl ■

.(4).

-101-
On making the same substitution in (1) and (2) we obtain

y^'^'^^^^^nii^^ (^)' yKn^+i) = n (6).

It now only remains to make w' + l and n--f »-fl squares. The first of these expressions becomes a

f2 1 r*4-2r^-4-2r- 2r4-l

square when n = - - , and the second then becomes n2-|-n-fl= ^ ^ - ^^— = n .

To make this expression a square put r*-f2r^-f Sr^ — 2r-f 1 = (r--fr + l)^ ; this condition gives r = — 4.
To obtain a positive value of r, put r =p— 4 and let (p^— 7p — -13) be the root of
r*-f2r^-f2r»— 2r-f l=p<— 14p'»-f74p2 — 178p-fl69;
_?^ - A-.^^ __ 2415 _ 101 1945 __ 7395544
tnenp- ^^ , r_p~4-^^. n- ^^^^ ; ^-13227769' 2'-i3227769-

Solved aleo by Messrs. Albert, BvUU, Drummond, Davis, Hart, Hoover and Siverly. Mr. Drummond's solation, and some
others, will be pablished In No. 5.

115«— Proposed by Sylvester Robins, North Branch Depot, Soni?r8et County, New Jersey.

There is a series of parallelopipeds whose dimensions differ from a perfect cube by one unit in only
one of its edges. In every ease the solid diagonal is an into;;er. Calling the one whose edges are 1, 2, 2
the^r«f parallelppiped, it is required to find general expressions for the dimensions of the nth solid, and
compute the length, breadth and thickness of the 30th one.

I,— Solution by the Proposer.

Let X, X and a; + 1 be the dimensions of every solid in the series. Then 3a;3 4. 2a; + l = D = {mx ± 1)-

= m^j^ + 2wix-fl, whence a; = „^ ^ or .2__q* I' V3 be expanded as a continued fraction the con-

vergentsarel. ?, |, ?, «. ff. Ih %h \ll 181. 8??. VA^. l?Si. %%\h \%\^ i5§H. iWg?. J8H?. eto.
These values being substituted for m in the formula above will give the dimensions of the solids.

Using m = 2 we get the edges of our first solid, 1, 2, 2 and solid diagonal 3. The second parallelopiped
has for its edges 6, 6, 7 and for its solid diagonal 11 ; the third has edges 24, 24, 23 and diagonal 41 ; th(»
fourth has edges 88, 88, 89 and diagonal 153 ; the fifth, edges and diagonal 330, 330, 329 and 571 ; the sixth,
1230, 1230, 1231 and 2131 ; the seventh, 4592, 4592, 4591 and 7953 ; the eighth, 17136, 17136, 17137 and 29681.

Now it must bo remarked that the sura of the edges of our solids give the odd numerators in the series
of convergents, 1, 5, 19, 71, 265, 989, 3691, etc., and the solid diagonals are the denominators of the same
odd convergents, 1, 3, 11, 41, 153, 571, 2131, etc., etc. Here we must notice that every even numerator
having twice its square diminished by unity will give the numerator of the convergent standing twice e^
high in the series, and every odd numerator when squared and increased by unity w^ill give the numerator
of the convergent standing twice as high in the series. So much for the numerators of the even con-
vergents. Every odd numerator is the difference of the two even ones between which it stands, and every
odd denominator is J of the sum of the two even numerators between which it stands. By observing
this rule we step from the 15th and 16th numerators to the 30th, 31st and 32d. Then by same law we

U4.U fii* * 196736618251755451 „ ^u- ^u ^. » . o/.^u ,ij ^ ^x 1

reach the 61st convergent, fiQ5u-9395A'-io7(-i • From this the dimensions of 30th solid are fomid to be

05578872750585150, 65578872750585150 and 65578872750585151, and solid diagonal, 113585939507107651.

From the numerators of the 12th and 13th convergents we obtain those of the 25th and 26th ; from

4.U *i. cn^u A MA x^ ' A 4.U * A 4-*^ mw ^4055034964909829777420559539

these the 50th and 52d are obtamed ; another step and we get the 101st, 3i208688988045323113527764971 '

Hence the edges of the 50th solid are 18018344988303276592473519846, 18018344988303276592473519846 and
18018344988303276592473519847, and the solid diagonal is 31208688988045323113527764971.
Using the numerators of the 100th and 101st convergents we easily obtain the 201st fraction,
2139013518315734698265663104770121639087469812968868432585
1234960030599837928682339736709998512373739432964939784153'
From this fraction the dimensions of the 100th block are found to be

width, 713004506105244899421887701590040546362489937656289477528,

thickness, 713004506105244899421887701590040546362489937656289477528,

length, 713004506105244899421887701590040546362489937656289477529,

soUd diagc»nal, 1234960030599837928682339736709998512373739432964939784153.

This problem was also solved by Mei«srs. Albert, BuUis, Davis, Heal, Hoover, Hart, Seifz and Sherly. Mr. Daris' solntioii
will be published In No. 5.

1

-102-

116.— 'Proposed by Mabcus Baker, U. S. Coast Survey Office, Washington, D. C; and Joseph B. Mott, Neosho, Mo.
Solve the equation a^ = a, and find the value of x when a = 300.

I.— Solution by DeVolbon Wood, M. A., C. £., Professor of Mathematics and Mechanics, Stevens Institute of Technology,
HoDokcn, Hudson County, New Jersey.

Let j;= £+2/, e being the Naperian base, then taking the logarithm of the logarithms of both mem-
bers we have ( « + y) log ( « -f 2/) -f log [log ( e -f y)] = log ( log a) = N.
Developing by Maclaurln's Theorem gives

^.,+(>H;%>_(-;y+(«-i>-C;->+*.

Reverting this series gives

»=GH!..)<''-'-^(4iV.$)<''-'>'+C^'«^4"">''-'>'+-

If a = 3 JO, then x = 2.3035 -f .

Solutions of thi* problem were received from Messrs. Albert, Baker, Casey, Kummell, Magruder^ Nichols, Oathout, Shaw
and Siverly. Mr. JkUter's solution will be published in No. 6. '

ll'Y.^Propoeed by Dr. Joel £. Hendricks, M. A., Editor of the Anaiyst, Des Moines, Polk County, Iowa.

Suppose a ball to be projected. In a horizontal direction, and due west, from a point vertically above
the ^h degree of north latituie; and suppose the projectile velocity to be Just sufficient to arrest its
motion in space which resulte.l from the earth's motion on its axis, so that it will descend to the earth
in a vertical plane corresponding with the meridian through the point of projection. And suppose,
further, that the ball, under the circumstances, shall fall to the earth in 5 seconds. How far will the
ball strike the earth north of the 40th parallel?

Solution by G. W. Hnx, Ph. D., Nautical Almanac Office, Washington, D. C.

All we have to do is to find the distance of a point on the 40th parallel from the axis which is

aco8<p

a denoting the equatorial radius of the earth, c the compression and q) the latitude.

Thence tlie centrifugal force at that point is „, • ,, , -- ^ ^^^ . , ^ , where x = ratio of circum-
° T^ -/LI — (2c— c2)smV]

forence to diameter and r= time of a sidereal rotation of the earth.
Next get the component of this in the direction of the horizontal line toward the north, which is done

by multiplying by 8ln<p, and we have >"• ^[i_(S,'!!t<)„„,^y

This is the foree tending to deflect the falling body in the horizontal direction. If we multiply this
by A(5)-' we have the required answer to the problem.

Using Bessel's dimensions of the terrestrial spheroid I get for the answer 0.218286 feet.

Thus it is seen that we do not need to take into account the heterogeneity of the earth's mass, because
it afifects the question only in so far as it influences the value of c, which value is o>)tained directly
from observation.

ii8.— Propo«?d by Winfield V. Jeffeie?, Inttructor in Mathematics. Vermillion Institute, Hayesville, Achland County, O.
How many different combinations, each composed of n letters can be formed from m letters; of which
a are one letter, b arc another and c arc another?

Solution by Walter Sivkhlt. Oil City, Venango Countv, Pa; Professor J. F. W. Scheffeb, Mercersburg, Franldin Co.. Pa.:
and Abram K. Bullis, Ithaca, Tompkins County, New York.

Let p be the letter of which there are a, q the letter of which there are 6, and r the letter of which
there are c.
If we expand
(1 -fpj-f.|?2ar2-f p3aj3+ .... +p«a?»)(l -f qx-f qr^af^-f gSjja-f .... -f g*a*)(l + rj: + r2.i!« + r»ir»+ .... -{-rx'),

the coefficient of x contains all the combinations of the letters taken one at a time ; of ar^ all the combi-
nations of the letters taken two at a time ; of j;«, all the combinations of the letters taken n at a time.

Putting p, g, r each = 1, the number of the combinations of the letters taken n at a time is the coefficient
of x** in the expansion of

(l+.r4-ari+ar»+ .... -fjT'Xl-f-aJ+a^+icS-j- .... -\- x^)(l ^ x -\- x- -\- s^ -^ .... -f-a<).

-103-

Thls expression may be written

C-i^-)('F^)fc'l^) CiC-). =<■— x-^"x.— ) (.-«)-.

when there are t different sorts of letters.

If m = a-f &H~^f ^^ number of combinations of the m letters taken n at a time is the ooefflcient of a^
in the expansion of

(l_a*+i)(l-.a^i)(l— a-4-i)(l_ «)-».

If there are m—a^b—c = d separate letters the number of combinations is the coefficient of af in
the expansion of

IIQ*— Proposed by W. E. Hbal, Wheeling, Delaware County, Indiana.

Show how to trisect an angle, and how to construct two mean proportionals between two given straight
lines, by means of the curves called the conchoid and the dsaoid.

Solution, (1) by Julian A. Pollard, Windsor, Vermont; (2) and (8) by Prof. W. P. Cabby, C. E., San Francisco, California.

(1). Let COM be the angle to be trisected. From any point D in one le;3^ let
fall a perpendicular DB upon the other. Take CB = 21)0, and with O as the
fixed point, XB as the fixed line, construct the arc CB of the conchoid. Erect
DP perpendicular to DB, and draw PO. Then is POC one-third of CX)M.

To prove this, bisect PH at E, and draw DE. Draw also EF parallel to DH.
Since PE = EH, PF = FD, and ED = PE = EH = DO. By reason of the isos-
celes triangles PED and DEO, we have / DEO = 2/ P = 2/ POC. But / DEO
= / EOD ; therefore 2/ EOC = / EOD, or / EOC = J/ COM.— [From Olney'e
OenercU Oeometry.

(2). Let c and d be the given extremes. Construct the rectangle MNOP, having MN = c and MP = d.
On MN construct an isosceles triangle MNG, having NO = to half MP. Make MX = MN and draw NY

Strallel to XG, and through O draw GF (by the conchoid) so that YF = NG, and draw FO intersecting
P produced in Z. Then NF and PZ are the required means.

ForGY : YF :: XN : NF, orGY : id:: 2c : NF;
and by similar triangles ZP : c : : d : NF, .-. ZP =
GY. Since MNG is isosceles, (GF)^ = MFxFN
-f (NG)2. But also (GF)« = the square of the
sum of ZPand half of MP, or to MZxZP-f the
square of half of MP. Taking away the last from
both, wegetMZxZP = MFxFN, and from this
and the similar triangles we have the proportions

ZM : MF :: mp : nf, zm : mf :: nf : zp.
ZM : MF :: zp : mn ; .-. mp : nf :: zp : mn,
ord:MF::ZP:c.

Any number of means may be found between
two ^ven extremes.

(3). In the right-hand figure let AB = c and BC = d. Upon AC describe the semicircle ANC, and let

tY be the cissoid. Find the side of a square = ABxBC, and let CO = side of this square. Make AM
= CO, and draw ON, MX perpendicular to AC. Then from the property of the cissoid A, X and N are
In a straight line, and XM and NO are the required means.

Join C and N. AMxOC = MXxON = (OC)^ = ABxBC, therefore c : XM :: ON : d.

Mr. PoUard also solved the third part. Walter Slverlu sent a good solution, and C. If: Tutton solved the fir.^t part.
An excellent solution of this problem may be seen in Prof. DeVolson Wood's Analytical Geometry, pp. 218 and IKO; also in
Johnton and Rice's Calculus, pp. !a84 and 886.

AXl

1^0«— Proposed by J. J. Sylvbstbr, LL. D., F. R. S., Corresponding Member of
matics, Johns Hopkins University, Baltimore, Mar}'land.

the Institute of France, Professor of Math-

Prove that for all positive values of k less than unity the equation (a:-f-«)(^ + &) = fc(ir-f c)^ has
two real roots.

!•— Solution by the Proposer.

Consider the sequence (j;-|-o)(a;-f 6) — fc(j;-f-c)*; x-\-a\ 1. Let a; be traveled up from — oo to -f-oo ;
when it reaches the vanishing point of a;-f-a the first term becoming — fc(a;-f-c)2 is negative, and conse-
quently no change in the number of continuations of sign in the sequence takes place as x passes through
this point. Any change then that is brought about can only be due to x passing through a vanishing
point of the first term.

Now when a; = -|-oo there are two continuations of sign In the sequence, and when ar = — x there are
none. Hence the first term must have two vanishing points ; i. e. two real roots as was to be proved.

-104-

11,— Solution by Walter S. Nichols, Editor Inmrance Monitor^ New York, N. Y.; DeVolson Wood, M. A., C. E.»
Profe8w>r of MathematicM and Mechaufcn, Stevens Institate of Technology, Hoboken, N. J.; and J. A. Pollard, Windsor, Vt.

Expanding, transposing and dividing by 1 — fc, we have ^+^_i. — i ~jt = ^•

By the Theory of Equations, this can have but two roots, and if one be imaginary so must the other.
Also in order to have two imaginary roots the last term must be positive. But given fc<l, \ — k is
positive ; also since k£<^x, therefore &c2>a6, and the last term is negative. Therefore the equation has
two real roots.

Good Holntione received from Mes^rH. Brown, Bowtttr, HmL, SeUz, Siveriy and 7\ittm.

181,— Proposed by Prof. E. J. Edmundh, B. S., New Orleans, LoolBlana.

Prove that zY^YlY^Yr^^^ . . ^("~-) = (27ry^*.n"^ n being a poslUve integer and T denoUng
the well-known Eulerian integral.

Solution by Charleb H. Kuxxvll, Awistant Engineer, U. 8. Lake Snr\-ey, Detroit, Michigan.
The second theorem in /'' functions,

7X».)r(l-m)=j;'^;^ = ^J^^ (1>.

I \m)i {^L — m)= I
is considered established.

1 Q Q *« Q M 1

Replace in (1) successively to by , , , , - ; then we have

<)<7')-,%-'0-("T^)=3.' " 'n')<)-i,.-.,. <='•

n n n

Taking the product of these equations we have

KK) 't7')]'=,„.,,. -,„<.-. <"■

n n n

We have, factoring jt^** — l by DeMoi^TC's Theorem,

jr^« — 1 = (a:— l)(r3 — 2xco8^-f-lVar2— 2xcos^^-f A ^or^ — 2a;cos^**~^^*-fl\x-f 1).

or "^"^^^ = (a^-2j;cos-^-f l)(^-2a:co8^-fl^ (a^^2xooe^'^~'^^'^ -\-l\ (4).

In this equation let x be replaced by -|- 1 and — 1 and it becomes

n = 2-.(l-co83(l-eoe2^) (l-cos^"-^)*) (5).

«^2-.(l+oo8*)(l + coe?*) (l+cos("-^>*) (5').

Multiplying (5) emd (5^) together and extracting the square root of the product, we have

n = 2«-isin-^sin^''8in^'' shi**"^ (6).

n n n n ^ '

Extracting the square root of the product of (3) and (6) we obtain finally

which is only a particular case of the more general theorem known as Gauss' third theorem in JT fiuic-

tions, viz; r(x\r(x-\-^^\r(x-^^\ r(x-^^~^^\ = {2ii)'^'t^''*r(tx\

The above solution is condensed from Price's Ir^finUesimal Qdculus, vol. i, pp. 164 and 165.
Solved alHo by MenerH. Brawn, Bmoser, BuUig, Heal, Hoover, Matz, Seitz, Sirerly, TuUon and the Propoter.

182,^Propoped by Prof. David Trowbridge, Walerburg, Tompkins County, New York.

When the earth is in perihelion, suppose the sun's mass to be suddenly increased by a half of itself, or
so that m' becomes Jm'. Required the change in the elements of the terrestrial orbit.

Solution by the Propober.
Let 7/1 be the mass of the earth and sun together, and for convenience of calculation let m become |fn.

-105-

Put the present mean distance of the earth equal to 1, and what it becomes when m becomes ]m equal
to a'. Let the perihelion distance of the earth be p = 0.9832249. Put V for the velocity of the earth at
perihelion, which, as we shall see, will >)ecorae the aphelion velocity, and p the aphelion distance, T =
365.2563744 days, the earth's sidereal period before the change, and T that after.
Works on Analytical Mechanics will give the following equations :

V* = ^^— w = ?^ _ 1*^ (1), p = a'(\ -{-€f) (2), e being the eccentricity required,

^ = *^ (3). |m = ^5'f (4). .-. r^=r^x|a'» (5).

From (1), a' = ,^/_f^^= 0.7436. From (2), e'= ^^- 1 =%-^ = 0.3221. ^^' = ^^[(4,)']^

.-. r = 191.2545 days.
The position of the perihelion will be changed 180°, and the perihelion distance = 0.5040.

Solved also by Prof. D. J. McAdam, Prof. E. B. SHU, Walttr Siverly and Prof. DeVolsan Wood.

183«— Proponed by Dr. David 8. Hart, M. A., Stonington, New London Connty, Connecticut.

To find three whole numbers such that the sum of the squares of any two of them increased by tho
product of the same two shall be a rational square.

I.— Solution by Rev. U. Jkmx Kniselt, Ph. D., Newcomeretown, Tuscarawaii County, Ohio.

Let Xj y and z denote the required numbers. Then we must make

a^+^-f y» = a = a« (1). ar3-fa»+2;'= D = 6« (2), ^^yz-\-:fi =0=0^ ^3).

Multiplying (1) by 4 times (2),

4(ar'-f a?y-f y«)(ar'-f«:-f 2?*) = ^a^b^-c' (4V

Subtracting (3) from the sum of (1) and (2),

'^^-{-xy+xz—tpi = a^-\-b^--c^ (5).

Subtracting square of (5) from (4),

3jr2y2+6r^+3jj^-f 6a?2^ + 6a:y2e+3y22rJ = 4a26.' — (o«-f 6-'— r^)^.
Taking square root,

xy'\-xz-\-uz = Jv 12a^6'— 3(a'-j-fr' — <^-)^ = M (6).

Adding (1), (2) and (3), transposing and dividing by 2, we have

= i(«- + &- + c-) — 4v 12a'fri— 3(o- + 6" — r-7^

»0- (6) = l(a'^ + 6» + c") - JA (7).

Adding twice (6) to (7), and extracting the square root,

a;-fy-f2 = J\/2(a' + 6i + c7 + 2v 12a<^fe2.^3(a^-f-6^ — c^*, =Jv 2(a'-f6i-f r^ ±2A, =J/?....(8).

From (6), j; = ^ _7^» and from (8), a; = JB— (y-f z). Hence we have lB--(y-f z) = i^""^.

Clearing of fractions, transposing, and reducing,

2

SB
Subtracting (9) from (8),

x = 4B — ^^"^?- = 3B^-4^--12r3 _ a^ + fr^ — c^ + J^^ 12a-7ii— 3(a-!-f 6* — r^^

^^ ^^ V'2(a'''-f-62 + d) + 2v 12a^6-'-3(a--f 6^~r-)2 '

In like manner, find x-^-z and x+y, and then subtracting from (8) we get
y _ a'-^c^—b'i^U 12a%^-3((7- + 6- — r^i ^ ^ b^^C' — a^ + h 12a«?>J— 3(a^-f6^-r0''

\/2(a3 -f b^ -f c2) + 2 y V2a^¥ — 3(a'' -f b^ — c'-y V2{ai + b- -^-C) ±2y 12a%' — 3(a-^ + 6- — c')-

We must now make

12a'b^—^a^-{-b'—c'y = D = A- (10), and 2(a^-ffri + r-) ^2A = Uz=B' (11).

Put a = (1 — n)6, c = (1-f n)6; then (10) becomes A^ = (9— 3671-')^^ =0 (12). Therefore wc

must satisfy 9— 36n* = Q, or, dividing by 9, 1 — 4w- = D, which put = (1 —jmy and wo get

iB(y+«) = lA + y^ + .V^+2^^ = JA + r^ by (1) ; whence y-f-^ = ^^.."t^"^ («).

2i>

^?. - ^ -.(*:;>

-106-

Substitutiiig ill (11), and taking the upper sign,

^^- + (p-f-4)^"*" P^-f 4 - °' ^^ (p.+4)« - °'

or 3p2 + 16 = D, which put = (4: + g/>)2 and we And ;> = i, ^^ - .

The problem is now resolved, if we find swh values of p, q, b, as will make the trinomials positive.
Since we have treated c as the largest of the three absolute-term roots, we have to make flr2-|-6*>c^ or

2 — 2n-f n2 > l + 2n-fw-, or 1 > 4n, and n < J. Hence ^^^< J, Sp<^p^-\-^, or p«— 8p-f 16 > 12,

p — i > 2\/3 and p > 4+2\/3, or > 7.46-|-. Hence q may be taken = —2, and then p = 16. n = J*^,
a = § J6, r = J|6. A = »«Bft6^ 5 = «^V6. -^^ = il8*. 2/ = f Si^ « = ?il^- Now take 6 = 455, and we
have X = 195, y = 284, z = 325, which are the least values that have yet been found.

This problem wa» aJw) very elegantly 8olved by Heuben Davis and Dr. Uart\ their eolations will be publwhed in No. 5.

Whence

184.— Propo.sed by Arteh ar Mahtln, M. A., Member of the London Mathematical Society, Erie. Erie County, Pa.

The first of two casks contained a gallons of wine, and the second b gallons of water ; c gallons were
dtiawn from the second cask, and then c gallons were drawn from the first cask and poured into the
second, and the deficiency in the first supplied with c gallons of water ; c gallons w^ere then drawn from
the first cask, and c gallons drawn from the second and poured into the first and the deficiency in the
second cask supplied with c gallons of wine. Required the quantity of wine in each cask after w such
operations as that described above.

Solution by Walter Siverlt, Oil City, Venango County, Pennsylvania.

Let M« and i'„ represent the wine in the first and second casks respectively at the end of the nth oper-
ation ; the quantities of wine in each cask at the successive stages of the (n-f l)th operation are

""• ""= (^-«)"- O-ft)*- O-a/"- 0-6)*"+ 6"' =

Also,
Btiminating r„ from (1) and (2),

(i-^)«„+,-^«„+,=(i-^y(i-^y«.-^' <-5^

Eliminating v^^x from (3) and (5),
Let ri, r-i be the roots of the equation

--[(■-n'+:S+(-m»+('-:)'('-0"-»-

The solution of (6) is {Hymer'e Finite Differencee, pp. 54, 55)

i ti'-i) I

]'-(■-:)'-:;-(-:)•+ (-:)'0-n*f

(q (.Vi (^

Put the last expression = S. From (1), Wo = a, Mi = > h- Ci -f- C^ = a— S,

r,C, + r..C, = u,-S: whence C. = »'-«-_:<" -^>. Ci = ri(«-^)_-». + S.

Eliminating u„ from (1), (2), and «„+i from this and (4),

'--[(■-;y+i;+o-o']'-+(*-^)"(-n*-='['-('-:)'-:]-

tt„=C,(rO"-f a(ri)« +

-107-

Whenoe ii,= Ci(r,)» + Ci(r,)« + iSr,. From (2), ii„ = 0, «, = c(2-^V
SolTed also, in an elegant manner, by Prof eseor E. B. Seitz.

185«""Propo0ed by Orlando D. Oathout, Read, Clayton Connty, Iowa.

What is the average thioknees of a slab sawed at random from a rowid log?

I,— ^Solution by Abtxxas Mabtin, M. A., Member of the London Mathematical Society, Brie, Erie County, Pennaylvania.

Let the annexed diagram represent a cross-section of the log, ABC being that
of the slab.

The average thioknees of the slab is the mean value of the ordinate DE.

Let OL = EF = tn, FO = EL = a;, DE = y, J = average thickness of the
slab. Then (ir -f- y)« -f aJ" = r\ r being the radius of the log,

whence y = "i/(r«— a^)— ti».

The limits of a; are and ^/(r*— aJ») = af; of w, and r.

A soluUon by Waltsr Siverl^ will be published in No. 5.

IM.— Proposed by Francis M. Prisst, Bryan, Williams Connty, Ohio.

Divide unity into three such positive parts that if unity be added to each part the three sums shall b<?
rational cubes.

Solution by Rkcbsn Davis, Bradford, Stark Connty, lilinoift.
^^\zj —1. C^~) —1 ^^^ \z) ~~^ represent the required parts.
ThenwUl(^y-l + (^^"y-l + 0y-l = l...(l), o^
or ^-^Hmn^-^t/^ = 4^; therefore n' = ^:^^-t = aw-12m^-6my» .

To make the numerator of (3) a square we must find such values of m-f-n, m—n and y as will fulfill
the condition (2). If m = 18, n= 1, y = 12, (2) is satisfied, z being 15 ; but the required parts will not all
be positive. To find positive values, make the numerator of (3) a square after substituting 18 — or for
m, 12 — go; for y and 15 for z.

Making these substitutions we have

(108)» -f (209304|)-f 46656g)a; ~ (23328p'-f 2592pg-f 3888^)a?»

-f (864p»+216pg«-|-108^)aj» ~ (12p*+6pg«)ar* = D.
Assmne [l08 + (969pH.216g)^ - a06921pM-:^^+5616<j?)a^J ^^^ ^^ ^^^ ^^ ^^^ ^^^^^ ^^^ ^^^^

squaring and reducing we find

_ 144(34538939p» -f 22814712p«g -f5184432p^ -f 40478V)

ii432107153p* -f 1000780560()p3g -f 3391176G72i? ^ + 525661056pq8 -f 31539466g* *

If we take o- land a --^i 23159322 902434176 790757247 3540876766005

irwetakep-landg- ^g. x- ^^^^^. m- ^^^^g^^ . 2( -ggT^-gQ-^ »»= 52708490x73584'

and the required parts of imity are

/m-|-ny_ _ 145291790670531905206919248956783548276049
\ z ) 196909050430895125339325091034151424000000'

/m — ny _ 51520189639391567352035150176008549814959

\ z ) 196909050430895125339325091034151424000000'

(y\^ -.1 — 97070120971652780370691901359325908992
\z) 196909050430895125339325091034151424000000 '

-108-

ISl.—PropoMd by Saxubl Robkbts, H. A., F. R. 8., Member of the London Mathematical Society, London, England.

Two random points being taken within a circle (1) oft opposite sides of a given diameter, (2} on tho
same side, (3) anywhere ; find in each case the average radius of the concentric circle touched by the
chord througn them.

I«— Solution by the PnopoaxR.

Let P, Q be the two random points anywhere in Uie surface of the given
(*ircle whose center is O and raidius r. Draw the diameter AOB and let OP
= a?, OQ = y, 2^ POQ = <», Z POB = 0. Then if i2 is the required average
radius,

jj_./0*'0*'0*'0 1/(3?^+^ — 2lW COB <») "^

-i.rSJl'-'^^-

16r

Let P and Q be both taken in the semicirele above AOB. Then the required average radius

' dOxdtydyd(M)

rwrrrxrw-$ xyslna

Let P be taken in the lower semicircle and Q in the upper, then R: = 2R — Ri = ^ ^ ^.

Elegant solntioiis received from ProfeMor SeUz and Walter SHoeri^^ which will be pnblbhed In No. 5.

m P ropoeed by E. P. Norton, Allen, Hllledale County, Michigan.

There is a circular fish pond surrounded by palisades, to the outside of which a horse is tethered.
The length of the tether is equal to the circumference of the pond. Required the diameter of tiie pond,
supposing the horse to have the liberty of grazing an acre of grass.

Solation by CHASLsa H. Kummxll, Aasiatant Engineer, U. 8. Lake Sorvey, Detroit, Michigan.

Let AC = BC = a = radius of the pond and
A the point where the tether is fastened. Sup-
pose the horse winds his tether around the entire
circumference of the pond ; he will then be at A.
If he unwinds the rope, keeping it stretched, he
will describe an involute to the pond APDE^
From £' he will describe a semicircle, radius
AE' = AE = 2ajr. From E over D to A he wUl
again move in an involute.

The entire area in reach of the horse may be
divided into three heterogeneous parts, viz : 1,
the semicircle E'HE ; 2, the two involute spaces
AODEA and AO'DE'A, which are equal by sym-
metry ; 3. the space BODG'B.

Let C be the origin, AD the a>-axi8 and P any
l)oint ; then we have

CM = « = aoo8^-f-a^sin^ (1),

PH = y = asin^— a^cos^ (2),

where q> denotes the angle which the radius CT,
perpendicular to the radius of curvature PT,
makes with the a>axis.

At the point D where the involutes intersect
on the a>azi8 we have

CD = 2;o = ocos^-f-o^8in^ where ^ = AOTBO; = asin^—o^cos^^b, .*. fito==tan^

*°^*^ = <;os^="^(^+^>

(5).

. (4;. We have then BODG'B = ««^— a*(^— jr) = a«jr .

Because any radius of curvature p=:€up and the area element between two consecutive radii of curva-
ture dA = ifl^dip = ^tflifMip (6), we have the two involute areas

-109-

AGDEA 4- AG'DE'A = a^ (^<fM<p = Ja^(8«« - ^») . . - (7). Finally the semicircle E'HE =2cfiit^: . . (8).
Hence adding (5), (7) and (8) 1 acre = 160 square rods = 43560 square feet = a«(jr-|-Sjr» — J<jDb'+2jr«)

= «=(-+ V""-!^') W- ••• « = V(>r+^w) <'">•

Solving equation (3) we obtain <^ = tan <^ = 4.494039 = 264^ 2T 18.35'\ Using this value in (10) wo
obtain a = 19.24738 feet = 1.1665 rods.

Solved aloo by Mesero. Brawn, BuUU, Casey, Harvey, Hoover, MeAdam, SeiU, Siterly and Tutlan.

129a""Propo«>ed by Reuben Davis, Bradford, Stark County, Illlnoie.

It is required to find three positive numbers, such that if each be diminished by the cube of their sum
the three remainders will be rational cubes.

. I«— Solution by Dr. David S. Habt, M. A., Stonington. New London County, Connecticut.

Let X, y,zhe the numbers, and s = their sum. Then x— «» = m', y— «» = n', «— u* = p* ; whence ye
have a; = m'-f-«", y = Ti»-f-«», « = p"»-|-«»; and by addition, a;-|-y-f-« = iw»-|-»ir»-f-p»-f-3^ = «;
.*. m'*-f-n^-f-p'* = 8— «^. In this expression a must be taken such a fraction that 8 may be greater than
3rf«. Let 8 = 1, then «— 3^9 = J = iVft = ^J J^ -f- ff l^-f- f^, which let = m», n\ p», respectively ;
whence x = jVft. y = /,%, « = iVA-

Let « = f, then «— Sir** = /^ = ^/ft = ,JJ;j -j- ^y»Jij -f- /,»^«^, which let = m\ n», p\ reapeollvely ;
whence a? = rf,. y = ,1/,* z = /A-

Proceeding in the same manner for four numbers, we have w = wHv ^ = Mv V = AVV* * = aWt*

In like manner we can find 5, 6, 7, n numbers.

A general solution by Seuben Davie will be published in No. 6.

ISO.'Proposed by Artex as Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.
A circle is inscribed at random in a given semicircle. Find (1) the average area of the circle and (2)
the chance that the circle does not exceed } of the semicircle. ^

Solution by the Proposer..

1. Let O be the middle of the base of the Semicircle, and C
the center of the inscribed circle. Through G draw the radius
OP, and draw CD perpendicular to AB.

Let r = radius of the semicircle, OD :^ x, CD = y = radius of
inscribed circle and J = average area required. Then OC =
r—y and we have ic«-|-y« = (r— y)«, or a?* = r* — 2rj^, the equa-
tion to the locus of C, which is, therefore, a parabola, axis ver-
tical, passing through A and B, having its focus at Q. The cen-
ters of the inscribed circles are uniformly distributed on this
parabola, and the number of circles is» therefore, proportional to
the length of the parabolic arc AKB.

Let 8 = any portion of the parabolic arc, measured from the vertex, then the average area sought iw

-^= . .= ^ .'^ ^ ,=^'^^0 . since <fe = v/('ii^ + d3^) = ^^^'^^''^

jd8 ^ fd8 jds

__ 2/13_ ^ 1V2 \ ' ^

-'^'*V32 24[>/2-|-log(l + i/2)]7*

2. Put 4 = i and we have ^^^'~^'^' = ^, from which x = r- Yl - ^^ = y. If x is greater
thwi x' the inscribed circle will not exceed one^nth of the semicircle.

^^_j;v..+.^ . (.-;^)'(^-;;:y+.4(.-;;j)'+(-»']

- 1 - V>^^ + log(3-f-i/3)\
-' 2V2 4.1pg(l + >/2);-
Solved alM> very elegantly by ^1. i?. BuUi«, Profeswr H. T. J. Lndwig, ProfeH»or E. B. SHtt and Waifer Sirerly.

-no-

131«*Propoeed by DsVolson Wood, M. A., C. E., Professor of Mathematicfl and Meclianioa, Stevens Institute of Technol-
(igyTHoboken, Hudson County, New Jersey.

A spherical homogeneous mass m, radius r, contracts by the mutual attraction of Its particles to a
radius nr ; If the work thus expended be suddenly changed into heat, how many degrees F. will the
temperature of the mass be Increased, its specific heat being 8 and the heat uniformly disseminated?

I«— Solution by Abrax R. Bullis, Ithaca, Tompkins County, New York.

Let W == the work expended, M = the earth's mass and R = the earth's radius.

Then ^=-jfii-JoJ^3^^^»*'= Mi- " Jo »** = "Sjk^ •

If the foot-pound be taken as the unit of work, then H= f.-^, = oftAQAf" » where H Is the heat-

oquivalent of the work, its unit being the poimd-degree Fahrenheit, and the required increase of tern-

perature= — = oo/i«»>- — •

An elaborate solution by the Proposer will be published in No. 5.

J«*Proposed by Fraickun P. Mate, M. E., M. S., Professor of Pure and Applied Mathematics, Bowdon (State) College,
Bowdon, Carroll County, Georgia.

Three persons, A, B and C, are banished to a level circular island, diameter 2r feet. At the center of

the island is a cylindric fort, diameter 2a feet. During a dark and foggy night B and G stray away from
A, cmd from each other, and lie down to rest. At the first dawn of clear morning, while A and B are
yet reposing, A looks around for them. Bequired the probability that A, without moving from his place
of observation, can see both of his companions.

Solution by Abtxhab Mabtik, M. A, Member of the London Mathematical Society, Brie, Brie County, Pennsylyania.

A can see his companions if they are anywhere in the area EHEPLJE.
PLJF

Then

Let X = AO, u = area EHKPLjE and p = the required chance.

J a
Z AOL = cos-i^^Y /LOJ = cos-i(^ Y JL - ^(r^— a*), area

FKPLGE = (r2— a2)co8-»(^^ ; s^ment GLJ = ir^co&-'(^^ —

iav/(r^ — tt«), and m = (f^~a«)cos-Y'*^ -|-r«cos-»("^ — a\/(H — a«).

••• P = (^-_i.)3;^/;2[(r^-a^)cos-(«^ (2).

/^{.„-.(.)j-..l^-,(:))-_/----<^.),

/-

2atdccos-»( ^^

^(^«a^) - = 2«,/(x«-a».)co6-i(^) - 2anog^;

,-. J2idx I cos-»^^) 1 ' "^ "^ { ^^"'{x) 1 '-■2a^/(a^-o«)oo6-Y^^ + 2a«log«.
The other integrations are easy ; performing them and substituting in (2), we have, finally,

Solved alKO by Professor E. B. Seitz and Walter Siverlp.

-Ill-

•138,— rropo— d bj Abtimas Martin, M. A., Member of tbe London Mathematical Society, Erie, Erie Coonty, Pa.
A duck swims across a river a rods wide, always aiming for a point in ttie bank b rods up stream from
a point opposite the place she started from. The velocity of the current is v miles an hour, and the
duck can swim n miles an hour in still water. Required the equation of the curve the duck describes in
space, and the distance she swims in crossing the river.

I«— Solution by Walter Siterly, Oil City, Venango Co., Pa.; D. J. McAdam, M. A., Profeseor of Mathematics, Waphlngton
na Jefferson College, Wanbington, F ' " - — .. . - . „ . ^ . .,-

School, New Brunswick, New Jersey.

and Jefferson College, Wasbineton, Pa.; and E. A. Bowskb, Piofessor of Mathematics and Ergineering, Rntgers Scientific

Let A be the point of starting, B the point on the other side
of the river opposite A, Q the point toward which the duck
swims, QX, QY the axes of a?, y, QX being measured down the
river and QY directly across it, P the position of the duck at the
time « after starting, AB = a, QB = 6, /^ PQX = ©, x and y the
co-ordinates of P.

Resolving velocities in the directions of the axes x, y.

dx
dt

= «— ncosO

dy 1 fl u dy nsinO

-r = — nsinO; whence -^ = r.

dl * dx t?— ncosG

^ — p ^^ - .

Putting ;=e, ^=.

sinO V

c— cosG ~" ei/CaJ^-fy') — a? ' I

whence xdy—ydx = cV'Ca^ + y*)di^ (1). i

Let x = yz, then dx = ydz-^-zdy and by substitution (1) becomes

^^ ^ Integrating, elogy = log[(l-hO-«)] + ^7= log(^>^^ + 2^^""^) + C.

y " i/(l-h«*)'
When a: = 6, y = a; .

ey'C 1 TJ- z^)dy = — yds, whence

y

whence
Taking the reciprocals of (2) and then subtracting (2) from the result,

which is the equation to the curve.
Let 8 = the required distance ; then

.(2).

(3).

Put (l-e)^(^<«^+r)+^y = p. 2(1+*.) = ,. (l+e)<i^+i')=-7 = r. ^ = «. a. = «..
and « - = m ; then 8 = ^1 ti?"V(p -|- ^ -f- rw^)dw, which can be integrated by formulas of reduction
when m = or any whole number, and when - = any odd whole number greater than unity.

To find the time of crossing, we have ^ = nsinO = -77^ j_ ^y Substituting the value of y/{i^ -f y-)
obtained by adding (2) to its reciprocal,

^^ I [\/(a3-|-6')-f ftlj/'-'a*-! + [\/(a«-f6i) — 6]y»+«a-i-* [ = — 2»kft.
Int^rating from 2/ = a to y = 0, putting (1 = the time of crossing,

2rU = [>^(«y-^') + ^]«' 4. [>^(«'+^*)-_^]«' whence t = >^' + «»l)+«^ = n>/(a^ + 6*) + f>r
* 1 — e "^ l-i-e ' ' 2n(l— 6*) n*— v*

Cor. — If 6 = 0, the equation to the curve becomes 2a; = a«y'-« ■

a-ej«i+* and ^ _ — .

II-— Solution by DeVolson Wood, M. A., C. E., Profetjeor of Mathematics and Mcchanice, Stevens Inetitate of Teclmology,
Hoboken, Hudson County, New Jersey.

Let P and p be consecutive points in the curve, origin at C, C^ = a?, yP = y, P^ = — dx, hp = dy,

V V

PQ = vdl. Op = ndt. From the two last we have Qp = -PQ = mPQ, putting m = - . Draw Pe perpen-
dicular to CJQ, and let c be the point where it intersects hp. The similar right-angled triangles C^P, Phc,
«p and PeQ give hc = -^f. cp = dy-hc =^±'^ , q, = d^(x»+3/»), Pc = -^^J-**?^.

y ' y y *

PQ= gxeQ = >^^^^i^[(l>/(a^+j^) + m.PQ];
... PQ = VS^+^)dVVfi±t)

Also PQ=CP p^^i/(;^+|^M^Hhy^)r-l/(^+3^)d;c'l

' c^ « L y J*

Making these values of PQ equal,

^ \ y ) X X y '

Transpose , multiply by y, divide by \/(afi -f y*), and we find

dVi^-k-t/") = V^-+J^!>^ - ^y^^ which added to the preceding, observing that

X . X

/s^±^\dx xdx _ydx d[|H-V(^+3/*)] _ n_wN^

igrating, observing that when x = a,y==h and CA = c, we find
(i)'^ = '-^f+t^' Which „ay be reduced to 2.= (5+c)(f)'- - ,-$,(-)""•

This problem was also very elegantly solved by Professor E. B. SHU.

134,— Proposed by William Hooveb, Superintendent of Schools, and Mathematical Editor of the Wtttenberger, Wapakoneta,
Auglaize County, Ohio.

Two equal rings begin to move freely from the extremity of the horizontal radius of a quadrant of a
circle, one down the arc, the other down the chord ; If the radius be 200 feet, how long after the motion
begins will one ring be vertically above the other?

I,— Solution by Chablbs H. Tuttok, Wilkes Barre, Lnzeme County, Pennsylvania; Waltkb Sivbblt, Oil City, Venango
County, Pennsylvania; and Abrax R. Bullxs, Ithaca, Tompkins County, New York. ,

Take the origin at the extremity of the horizontal radius, from which the bodies start, x horizontal, y
vertical and positive downwards.

The velocity at any point is by Mechanics v = V(^y)> and we also have ds = vdt, 8 being the space
described and t the time.

TtLc

For arc, y = \/(2raj— a3); ... « = (2i7)*(2ra;— a^)i. d8 ^Vidafi-^d^) z= -^^—-^ whence for

arc dt = 72~wo — "^Si* ^^P*^^ ^y Maclaurin's Theorem and integrate.

Porarc.* = ^J^^ J4(g^y + i(j^,y + A(r6^ay+-} -a)- C=0 because for x=0.i=0.
For chord, y=zx, hence v = a/(2ga:), ds = ^/{d3fi -f- dy«) = >/2dc. dt = ^^-^ and t = ^Ji * V . . (2).

By the question these times are equal, as also the values of A\ hence equate (1) and (2) and divide
by xk ; 1.11712 -f 0.0004306a; -f 0.00000096*2 + ....= 0.3524ri,

whence x\ = 3.3358 nearly, x = 123.822 nearly, and from (2) t = 3.922 seconds nearly.

Ab elaborate solution was furnished by Professor McAdam.

136«*Proposed by Abtexas Mabtin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.
Three equal circles touch each other externally ; find the average area of all the circles that can be
drawn in the space encloised by them.

I,— Solution by Abbax R. Bullxs, Ithaca, Tompkins County, New York,

Let ABC be the- space enclosed by the circles, D the center of one of them, CD = a, DE = r, / CDE
= 6 and x = the radius of a circle having E for its center.

The limits of x are and T^a\ of r, a and — = = r' ; of S, and'^^r, and the required average area Is

-113-

= *^^*^0 Ucos^Q "" 4 oo6^ "*^j«>83e ~ 2co^ "*■ 20/^

Jo VSoo^'"" 2co^ "^ e)^

_ 3raa/6;r4-308— 320i/3 + 2071og3\

~ 6b V ;r-f4— 6i/3-f-31og3 )'

\etj elesnnt solutions received from Meaers. Broum, Seitx and Skceriy. Mr. Brown''9
solution wiU be published lu No. 5.

136**Propo6ed by John W. Bsrry, Pittston, Lozeme County, FennsylTania.

A smooth, straight, thin tube is balancing horizontally about its middle point, and a particle whose
weight is one^th that of the tube is shot into it horizontally with such a given velocity that it Just arrives
at the middle point of the tube. Find the angular velocity communioateid to the tube.

I««8olution by Waltbb Siyiblt, Oil City, Yeiuuigo County, Pennsylvania.

Let 2a = the length of the tube, m its mass, k its radius of gyration, 6 Its Inclination to the horizon at
any time U v the given velocity of projection of the particle, a? the required angular velocity, x and y its
horizontal and vertical co-ordinates at the time t, the origin being at any point in the horizontal line
passing through the initial position of the tube.

By the principle of Vis Vtwi,

gy-

InitiaUy J = r. ^ = 0, ^? = 0. y = 0; .'. C7= ^. and

When the particle reaches the middle of the tube,

£ = »• i- =»■ s = '»•''=»: -^ *"= ^-^^ ••• »'»«"»' = T ■• '^^'^ ~ = «V(»)-

Solved also by Professor DeVolMm Wood.

137«*Propoeed by E. B. Sbitz, Professor of Mathematics, North Missouri State Normal School, Kirksville, Missouri.

Two points are taken at random in the surface of a circle, but on opposite sides of a given diameter ;
find (1) the chance that the chord drawn through them does not exceed a line of given length, and (2)
the average length of the chord.

I«— Solution by the PBOPoesR.

Let AB be the given diameter, O the center of the circle, P, Q the two
random points, and CD the chord th]:pugh them. Draw OM perpendic-
ular to CD.

LetOA = r, EP=a;, EQ=y, EC=u, ED=t4', /OOM = 0, /EOM = ip,
and let 2/S = the angle formed by the radii drawn to the extremities of
a chord equal in length to the given line. Then we have CD = 2rsin0,
u = r(sin6-f-cosStan<p), u' = r(sinO — cosOtan^); an element of the
circle at P is r&inBdBdx, and at Q it is {x-\-y)dg)dy.

1. The limits of B are and /i ; those of <p, —6 and G, and doubled ; of
Xt and u ; and of ^, and u'. Hence, since the whole number of ways
the two points can be taken is JrV*, the required chance is

= ^}f^C (1-C08«98cc»8ln=0d5«l^ := ^ J*(e - 8in6cos6)8ln«0<W = /2/J— 8ln2/jy

-114r-

(1 3i/3\*
3 4«~ ) '

2. The limits of B are and Jtt, and those of <p, x. and y the same as above. Hence the required
average Is

= ^JjJ'*''|^^^(l-co8^G8ec^(?>)8ln»0<Wd(?>= ^^J^^^ = ^^.

The flrat part of thu problem is the eamo ae Qne^t. 1849 in the Educational Times, to which erroneous solutions were given
in vol. vili of the Reprint, pp. 98—4, by StephenWataon and Samuel Bobert^. Mr. Roberta has famlithed a correct aolntion which
will be pnblitthed in No. 6. Walter Siverly tUso sent an elegant solution which we intend to publish.

5^ For want of time, and space, we very reluctantly omit the solutions of Problems 188—147; they will be published in No. 5.

Mjist of Ck^ntributorg to the Senior JDepartment.

Walter Siyiblt, Oil City, Venango County, Pennsylvania, solved all the problems but 181 and 142. E. B. Siitz, Pro-

fuHHor of Mathematics, North Missouri State Normal School, Kirksville, Mo., solved 111, 118, 116, 190, 121, 18S, 124, 127, 198, 180,
182, 188, 185, 1S7, 142, 148. 144 and 147. A. R. Bulus, Ithaca, N. T., solved 111, 118, 114, 115, 118, 121, 128, 128, 180, 181, 184,

135, 188 and 140. DeVoleox Wood, M. A , C. E., Professor of Mathematics and Mechanics, Stevens Institute of Technology,

Hoboken, N. J., solved 111, 118, 116, 117, 120, 122, 181, 188, 186, 189, 145 and 147. W^tlliam Hoovkr, Superintendent of Schools,

and Mathematical Editor of the Wittenberger, Wapakoneta, Auglaize Co., O., solved 111, 118, 114, 115, 121, 128, 128, 188 and 140.
Ohas. H. Tuttom, Wilkes Barre, Luzerne Co., Pa., solved 111, 112, 110, 120, 121, 128, 184 and 133. Cras. H. Kvxmsll,

Assistant Engineer, U. S. Lake Survey, Detroit, Mich., solved 111, 116, 121, 128, 188, 140 and 147. Dr. David S. Hart, M. A..

Stonington, Conn., solved 112, 118, 114, 115, 128 and 120. W. E. Heal, Wheeling. Ind., solved 111, 112, 115, 120, 121 and 146.

D. J. McAdam, M. a.. Professor of Mathematics, Washington and Jefferson College, Washington, Pa., solved 118, 117, 122, 198,
188 and 184. Reuben Davis, Bradford, Stark Co., 111., solved 112, 114, 115, 123, 128 and 129. Frank Albert, Professor

of Mathematics, Pennsylvania State Normal School, MiUersville, Pa., solved 113, 114, 115, 116 and 128. E. A. Bowser,

Professor of Mathematics and Engineering, Rutgers Scientific School, New Bmnsviick, N. J., solved 118, 120, 121, 188 and 147.
H. T. J. LiTDwiG, Prof elisor of Mathematics, North Carolina College, Mount Pleasant, N. C, solved 180, 188, 146 and 147.
Lucius Brown, Hudson, Mass., solved 120, 121, 128 and 185. Prof. W. P. Casey, C. E., San FmncLsco, Cal., solved 111, 116,

1 19 and 128. Walter S. Nichols, Editor iMurartce Monitor, New York, N. Y., solved 111, 118, 110 and 120. Juuan A.

Pollard, Windsor, Vt., solved 111, 112, 110 and 120. J. F. W. Scheffbr, Professor of Mathematics and German, Mercers-

burg College, Mercersbnrg. Pa., solved 118, 115 and 118. Gavin Shaw, Kemble, Ontario, Canada, solved 111, 116 and 196.

W. L. Harvet, Maxfleld, Me., solved 111, 118 and 128. . Henrt Hbaton, Atlantic, Iowa, solved 188, 140 and 147. O. D.

Oathout, Read. Iowa, solved 111, 116 and 188. J. J. Sylvester, LL. D.. P. R. 8., Corresponding, Member of the Institute

of France, Professor of Mathematics, Johns Hopkins University. Baltimore, Md., solved 190 and 141. Samuel Roberts.

M. A., F. R. S., Member of the London Mathematical Society, London, England, solved 187 and 187. F. P. Mate, M. £.,

M. S., Professor of Pure and Applied Mathematics, Bowdon (State) College, Bowdon, Ga., solved 121 and 188. Prof. E. J.

Edmunds, B. S., Principal of Academic School No. 8, New Orleans, La., solved 111 and 121. Hon. Josiah H. Drummokd.

LL. D., Portland, Mo., solved 112 and 114. Syl\'E8TBR Robins, North Branch Depot, N. J., solved 112 and 115. E. P.

Norton, Allen, Mich., solved 111 and 118. Marcus Baker, U. S. Coast Survey Office, Washington, D. C, solved 111 and

116. Benjamin Pbirce, LL. D., F. R. S., Professor of Mathematics, Harvard University, and Consulting Geometer to the

U. S. Coast Survey, Cambridge, Mass., solved 145. Dr. Joel E. Hendricks, M. A., Editor of the Analyst, Des Moinea.

Iowa, solved 117. Willam T. Magruder, Stevens Institute, Hoboken, N. J., solved 116. K. S. Putnam, Rome, N. Y.,

solved 118. Prof. David Trowbridqb, M. A., Waterburg, N. Y., solved 122. Robins Fleming, Readington, N. J.,

dolved 112. Prof. Asher B. Evans, M. A., Principal Lockport Union School, Lockport, N. Y., solved 114. L. P. Shidy,

U. S. Coast Survey Office, Washington, D. C, solved 111. G. W. Hill, Ph. D., Nautical Almanac Office, Washington, D. C,

solved 117.

The first prize is awarded tx) Walter Siverlt, Oil City, Venango Co., Pa., and the second prize to
Chableb H. KuMMEiiii, Assistant Engineer, U. S. Lake Survey, Detroit, Mich.

175,— Proposed by W. L. Harvey, Maxfleld, Penobscot County, Maine.

A man buys a farm for $4000 and agrees to pay it in 4 equal annual installments, interest at 6 per cent,
per annum, compounded every instant. Required the annual payment.

176."P'"opo''<^ ^y Prof. E. J. Edxwds, B. S., Principal of Academic School No. 3, New Orleans, Louisiana.

The sides of an inscribed quadrilateral are the roots of a given equation of the fourth de^ee whose
roots are all real and positive. Find the area of the quadrilateral, and the radius of the circumscribing
circle, in functions of the coelHcients of the equation.

-115-

ITT^-^Pi^POMd ^7 JossPB H. KxBSHNBR, TtoteaeoT of Hathomatics, Mercersbarg College, Merceraborg, Frankltn Co., Pa.
Find three square numbers the ratio of whose sum and product shall be a square.

llS-'Proposed by Prof. W. P. Cabet, C. E., San Francisco, California.

Given the three lines joining the remote angles of the equilateral triangles described on the sides of a
triangle, to construct the triangle and find its sides.

l'X9»*Propo6ed by Stlvbstbr Robins, North Branch Depot, Somerset County, New Jersey.

In a series of rational right-angled triangles where every hypothenuse is unity, each leg of the nth
triangle contains n decimal places. Find the legs of the first 25 triangles, and those of the 50th one.

ISOa^Pi^poMd by Dr. John Bbix, Manchester, Hillsboroogh County, New Hampshire.

One-third of all the apples on a certain tree are rotten, and one-fourth of all the apples on the same
tree are wormy. What are the respective chances that an apple taken at random from the tree will be
(1) sound, (2) rotten, (3) wormy, (4) both rotten and wormy?

181,— Proposed by Marcus Bakbr, U. 8. Coast Survey CMBce, Washington, D. C.

A polygon is both Inscriptible and circumscriptible, radius of inscribed circle r, of circumscribed
circle R and distance between the centers of these circles d ; then if the polygon be a triangle,

R+d^ R-^d " r '

if the polygon be a quadrilateral, (R:fdy "*" (S^==d5» "^ f^ '

What is the corresponding relation for a pentagon?

182«— Proposed by Alexander Macparlanb, M. A., D. Sc., Edinburg, Scotland.
Prove that

log[o«-f-b«-|-2a6cos(a— /?)]* = logo-f- ^cos(a~/J)-l^cos2(<r— /J)-|-l^oos3(a-/?) — etc.

183.""Proposed by Abtsii as Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

A clock which indicates correct time at the level of the ocean is carried to the top of a mountain near
by, the ascent occupying h hours. On arriving at the top of the mountain the clock was found to be m
minutes too slow. Bequired the hight of the mountain.

184*— Proposed by Dr. Saxusl Hart Wright, M. A., Ph. D., Mathematical Editor Yattig County Chronicle, Penn Yan, N. Y.

Given the latitude = A = 42° SO' N., the sun's declination = 5 = 20° N., his radius = r = 16', and a
vertical wall running 8. 10° W., to find when the sun will first shine on the west side of the wall, the
I>oint8 from the sun's north point of ingress and egress on the wall, the altitudes of those points, and of
the sun's center.

185«*Propo(^ by Saxuel Roberts, M. A., F. R. S., Member of the London Mathematical Society, London, England.
Show that if the system of equations

are resoluble by integer values of x, y, u, v, (zero excluded,) not having a common factor greater than
unity, successive solutions can be obtained of the same character.

188.— Proposed by Marcus Baker, U. 8. Coast Survey Office, Washington, D. C.

Inscribe in any plane triangle three circles each tangent to two sides and the two other circles.

187.*Propoeed by Dr. Joel B. Hendricks, M. A., Editor of the Analyst, Des Moines, Polk County, Iowa.

The base of a hemisphere, whose radius is ten feet, rests on a horizontal plane, and a point A on the
surface of a sphere whose radius is one foot is in contact with the vertex of the hemisphere.

If, from a slight disturbance of the equilibrium, the sphere is caused to roll off the hemisphere in the
direction of the plane of a great circle through the points A and B on the surface of the sphere, and if
the point B strikes the horizontal plane at a point D, distant CD from the center of the hemisphere ; it
is required to find the length of the line CD, and the relative position of the points A and B on the
surface of the sphere.

188«""Proposed by Daniel Kirkwood, LL. D., Professor of Mathematics, Indiana State University, Bloomlngton, Ind.

If a mass of matter be projected vertically upward from Jupiter's equator with a velocity Just sufficient
to carry it to a hight of 450000 miles, find the path it will describe, c»r determine the circumstances of
its motion ; the equatorial diameter of Jupiter being 90000 miles, and its time of rotation 9 hours and
55 minutes.

189**Pi^po9®<l by Prof. Hugh S. Banks, Instructor In English and Classical Literature, Newburg, Orange County, N. Y.
How many spheres one inch in diameter can at th« same time touch the surface of a sphere one foot
in diameter?

-116-

190.— Proposed by F. P. Matz, M. E., M. 8., Profe.MM>r of Pure and Applied Mathematics, Bowdon (State) College, Bowdon,
C'anoli County, Georgia.

A given rectangle is divided at random by a line parallel to one of its diagonals, and then two points
are taken at random within the rectangle ; find the chance that both points are on the same side of the
dividing line.

191**~Propofled by Reitbbn Davis, Bradford, Stark Count}', llllnoli*.

Find three square numbers such that if to each its root be added, and from each its root be subtracted,
the three sums and three remainders shall all be rational squares.

192«"*PToposed by DeVolson Wood. M. A., C. E., Profei^sor of Mathematics and Mechanics, Stevens Int<titute of Tech-
nol<^y, Hoboken, Hudson County, New Jersey.

Three concentric shells of infinitesimal thickness and uniform density, radii r, are in contact but
move independently of each other without friction with angular velocities <», oj\ cw". Their axes of rev-
olution are all in one plane but make angles a, p, y with a fixed line.

Suddenly the three shells become one solid. Bequired the position of the new axis, and the angular
velocity.

193**Proposed by Williax W'oolset Johnson, Professor of Mathematics, St. John's College, Annapolis, Maryland.
If we write {x—iyY = A-^-Bi, (f = i/— 1, ) sho>v that the expression A -f aJB, where q is any real nu-
merical quantity, is the product of n real factors linear in x and y, and find these factors.

194«*"Proposed by John W. Bebrt, Pitu^ton, Luzerne County, Pennsylvania.

Two points are taken at random on the surface of the earth, but on opposite sides of the equator ; find
the average distance between them (1) supposing the earth a sphere, radius r, and (2) supposing it an
oblate spheroid, semi-axes a and b,

105«"Pi^PO«^^ hy Oeobob Lillet, M. A., Coming, Adams County, Iowa.
A first integral of the differential equation

Show that the complete primitive is j/ = Jor^ -f &x -f a* -f- &'> a and h being arbitrary constants.

196«— Proponed by W. A. Kite. M- A., Professor of Mathematics. Grceneville and Tusculum College, Tuscnium, Tennessee'.
Required the maximum ellipse that may be drawn within a trapezoid, parallel sides a and b and length
of each of the oblic^ue sides c.

107*"*Proposed by Dr. David S. Uabt, M. A., Stonington, New Liondon County, Connecticut.

To find three numbers such that if the sum of their cubes be either added to, or subtracted from, the
8(|uare of each of them the sums and remainders shall be squares.

198«~PToposed by Arte.mas Mabtin, M. A., Member of the London Mathematical Society, Eric, Erie County, Pa.
At the center of a square park, sides a, stands a square tower, sides b. Two persons are in the park ;
II nd the chance that they can see each other.

199.— Proposed by L. G. Barbour, Professor of Mathematics, Central University, Lexington, Fayette County, Kentucky.

A thousand foxes start from a point in a field and at the same instant a thousand hounds start in pur-
suit each of a fox, from another point situated a rods distant from the first point. The foxes run in
straight lines in divergent directions, at the .same uniform rates of speed, and the hounds at a speed m
times as great. Bequired the equation to the curve upon which all the foxes will be caught, and the
length of the curve.

200.""Pi'oposed by Artexas Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

From one comer of a square field a projectile is thrown at random with a given velocity which Is such
that the greatest range of the projectile is equal to the diagonal of the field ; find the chance of its falling
in the field.

201«*~Proposed by £. B. Seitz. Professor of Mathematics, North Missouri State Normal School, Kirksville, Missouri.
A cube is thrown into the air and a random shot fired through it ; find the chance that the shot passes
through opposite faces.

202,— Proponed by Benjamin Peikce, LL. D., F. R. S.. Professor of Mathematics, Harvard University, and Consulting
<;eometer to the U. S. Coast Snr\'cy, Cambridge. MasKachusetts.

Find a curve which is similar to its own evolute.

203«*Proposed by Artbmas Martin, M. A., Member of the London Mathematical Society, Erie, Brie County, Pa.

A cylinder, radius r, rolht down the surface of another one, radius R, on a horizontal plane, the sur-
faces of both cylinders and plane being rough enough to secure perfect rolling. Determine the motion
of the cylinders, the line of separation, and the path of a given point in the axis of the upper cylinder.

-117-

204.— PropoMd by S. B. Sbite, ProfeMor of Mathematics, North Miflflouri State Normal School, Kirksville, Miseoori.
Find the average area of the triangle formed by Joining three points taken at random in the surfooe
of a giyen ellipse.

2Q5.*Propofled by F. P. 3f atz, M. E., M. 8., Professor of Pare and Applied Mathematics, Bowdon (SUte) CoUege, Bowdou,
CanolT County, Georgia.

Find the average area of the triangle formed by joining three points taken at random (1) in the surface
of a given quadrant, and (2) in the surface of a given semicircle.

20^— Proposed by Abtbvas Martin, M. A., Member of the London Mathematical Society, Erie, Erie Connty, Pa.

A point is taken at random in one of the longer sides of a rectangle and a line drawn from it at ran-
dom to the opposite side, and then two points are taken at random in the surface of the rectangle ; find
the chance that both points are on the same side of the line.

207«""Propoeed by Waltbb Siverlt, Oil City, Venango County, Pennsylvania.

An ellipse, major axis 2a and eccentricity e, revolves about a tangent at the extremity of the major
axis. The major axis increases imiformly from 2a to 2(a-f-^) while revolving through an angle ft, the
ellipse increasing in such a way that e remains constant. Bequired the volume of the solid generated.

208.— 'Proposed by Rev. W. J. Wbioht, M. A., Ph. D., Member of the London Mathematical Society, Jcnklntown, Pa.

Show by actual construction that two plane surfaces lying in one plane can be attached in such a
mannner as to be capable of being spread out into two sheets of a developable surface in which one edge
of the former plane becomes an edge of regression and a curve of double curvature.

Show also how to obtain the equation of the developable surface.

209*'— 'Proposed by E. B. Seits, Professor of Mathematics, North Missouri State Normal School, Kirksville, Missouri.
Two small circles are drawn on the surface of a sphere so as to intersect ; find the average of the
spheric surface common to the two segments cut from the sphere.

210«*Proposed by Abtexas Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.
Three circles, radii a, 6, c, touch each other externally ; find the average area of all the circles that
can be drawn in the space enclosed by them.

211,«Proposed by £. A. Bowser, Professor of Mathematics and Engineering, Rutgers Scientific School, New Brunswick,
Miodlesex Connty, New Jersey.

A sphere, whose equation is or^-f-S^H-^ = &^» ^^^ its center coinciding with that of an ellipsoid whose

3fi 'if^ z^

equation is ^4-?^ + " ~ ^» where a'^h'^c Find (1) the surface and (2) the volume of the solid
common to both by the formulas 8 = \ idxdyl 1 + ^ + s-^) ♦ V = j i jdxdydz,

212**Proposed by B. B. Lakin, Streator, La Salle Connty, Illinois.

A pole 80 feet long was standing upright against a vertical wall. A monkey began to ascend the pole
at a uniform velocity, and at the same instant the foot of the pole began to move out from the wall with
the same uniform velocity, the monkey arriving at the other end at the moment that end reached the
ground. Bequired the equation to the curve the monkey described in space, and the distance he moved.

213«"*Proposed by Chablss H. Kumxbll, Assistant Engineer, U. S. Lake Sunr'ey, Detroit, Michigan.
Find the surface of the solid whose volume is required in Problem 147.

214«""Proposed by D. J. McAdam, M. A., Professor of Mathematics, Washington and Jefferson Collie, Washington, Pa.

In Morin*s apparatus for deducing experimentally the laws of falling bodies, the length of the revohing
cylinder is a feet, its diameter 26 feet, and it revolves n times per minute. Required the equation of the
curve which the pin that projects from the falling body traces on the cylinder, and its length.

216«*^opofied by Enoch Beery Seite, Professor of Mathematics, North Missouri State Normal School, Kirksville, Mo.
Let A, B, C, D, £, F be six random points within a sphere ; find the chance that the planes through
A, B, and D, E, F intersect within the sphere.

216«^PRI2B Problem. Proposed by Artemab Martin, M. A., Member of the J^ndon Mathematical Society, Erie, Pa,
Required (1) the equation to the curve described by a given point in the rod that connects the two

driving wheels, and (2) the equation to the curve described by a given point in the rod that connects the

forward driving wheel of a locomotive with the piston rod.

SlTa^PBizE Problbx. Proposed by Artkmas Martin, M. A., Member of the London Mathematical Society, Eric, Pa.

Three equal coins are dropped horizontally at random, one at a time, into a circular box whose diam-
eter is twice that of one of the coins. Find the chance that only one of the coins rests on the bottom of
the box.

Solutions of these probleme should be received by October 1, 1880.

-118-

EDITORIAL NOTES.

It is with feelings of deep regret we record the deaths of three of our former mathematical correspondents: Hon. Josiah Scott,
Ex-Judge of the Supreme Court of Ohio, Bucyrus, Ohio; Hon. Robert M. DbFrancs, Attorney at Law, Mercer, Pennsylvania;
and James Clark, Wayne, Maine.

"J08IAH Scott was bom on the first day of December, 1803, in Wai«hIngton County, Pennsylvania, on his father's fium, three
miles from Cannonsburg, the seat of Jcfler^on College, where he was educated under the celebrated Dr. McMillan. He boarded
at home, walked daily to and from college, and graduated m IftJl with the highest honors of his class.

He emigrated to Bncyrus, Cra^i'ford County, Ohio, when it was but a hamlet in the wilderness, and there he first engaged in
the practice of the profession of which he afterwards became such a distinguished ornament. He was elected Judge of the
Supreme Court of Ohio, and twice re-elected, remaining on the bench for flilteen successive years. In 1876 he was appointed
Chief Justice of the Supreme Court Commission."

The above is condensed from a biographical sketch published in the Yates County Chronicle of July SI, 1879, in which it is
stated that he died on the 15th, of hemorrhage of the bladder.

Judge Scott was a mathematician of marked ability. He was particularly skillful in the solution of Algebraic and Geometrical
problems, especially those pertaining to the Diophantine Analysis, as may be seen by referring to his elegant solutions in the
Sehoolday Visitor MagaHne and the Tatee County Chronicle.

Hon. RoBEBT M. DeFranos died on the 19th of August, 1879, at the age of 58 years and 28 days. In October, 1880, he was
struck with paralysis and rendered unable to get about without the aid of crutches, but retained his mental faculties and attended
to the duties of his office until in the winter or spring of 1678. Mr. DeFrance was a lawyer by profession, and was elected a
member of the Constitutional Convention of Pennsylvania in 1872.

We are not at present in possession of information for a fuller biographical notice.

In mathematics we believe Mr. DeFrance was chiefly self-taught. He was a valued contributor to the mathematical depart-
ment of the Sclwolday Visitor Magazine, and to the Sehoolday Visitor Annual. His solutions were always ingenious, and usu-
ally original in method.

"J AXES Clark was bom in 1794, at Ayr in Scotland, a place famous in song, and a familiar name to readers of Robert Bums.
He did not in early life enjoy the advantages of education which are, or may bo enjoyed by every one in our highly-favored
country. He learned the trade of a muslin weaver at which he worked till the age of 14, when he was apprenticed on board a
ship. In his 17th year his captain sent liim to school to study navigation. The term occupied about five weeks, and this cloned
his school career.

He left his native country for the last time in 1818, and learned the trade of a cabinet maker with his brother's employer in
Portland, Me. He also learned pattern nmking, which introduced him to the machinist business, and for six or eeven yeans he
was foreman of a large establishment in the city of Alton, Illinois.''— [From the Tate'^ County Chronicle.

Mr. Clark was a self-made mathematician of more than ordinary ability. He was for many years a regular contributor to the
Maine Farmers'* Almanac over the signature of Mechanic. He was also a contributor to the mathematical departments of the
Sehoolday Vitifor Magazine and the Yafee County Chronicle, in the pages of which periodicals many of his fine solutions are to
be found. He delighted in working out the solutions of curious and intricate problems in the Diophantine Analysis.

We are not informed of the date of the death of Mr. Clark, but think it occurred sometime in 1879.

E. B. Seitz of Greenville, O., has been elected Professor of Mathematics in the North Missouri State
Normal School, Kirksville, Mo., and F. P. Matz, M. E., M. S., of Reading, Pa., has been elected Professor
of Pure and Applied Mathematics in Bowdon (State) College. Bowdon, Georgia, where they are now
hard at work.

Our talented lady contributor, Miss Christine Ladd, has been deservedly honored by the Johns Hop-
kins University. The trustees have voted her an honorary stipend and an invitation to continue her
mathematical studies in that Institution.

Miss Ladd is a graduate of Vassar College, and is probably the best mathematician of the gentler sex
in this country.

We take pleasure in acknowledging our indebtedness to Prof. E. B. Seitz, Prof. DeVolbon Wood
and Walter Sivebly for valuable assistance.

The following persons, and many others, have our thanks for the subscriptions procm-ed by ttiem :
John S. Royer, Prof. E. B. Seitz, Prof. W. P. Casey, Prof. E. A. Bowser, Prof. DeVolson Wood,
Sylvester Robins, Reuben Knecht, W. D. McSwane, A. R. Bullis, and Prof. Frank T. Freeland
of the University of Pennsylvania.

A few such **live" mathematicians in every state would soon secure for the Visitor a paying circulation.

This No. of the Visitor has been delayed some months in consequence of the sickness of the Editor,
who has done all the type-setting with his own hands. He is not a practical printer, and never had set
up a stickful of type till last May or June.

The cover, the first six pages and page 9G were printed by the Edljbor on a No. 3 Self-Inker Model
Press ; pages 93, 94 and 95 were printed at the Erie Morning Dispatch Office, and the remaining pages
at the Erie Gazette Office.

We have decided to issue the Visitor semi-annually, at $1.00 a year in advance. Single numbers will
be supplied at 50 cents each. Back numbers can be obtained at the same price.

We hope to have No. 5 ready in July ; it will contain the Solutions omitted in this No., some Additional
Solutions, and a few Papers.

Contributors will please send Problems, Solutions and Articles intended for publication In No. 5 at as
early a date as practicable.

Send subscriptions soon to ARTEMAS MARTIN, Lock Box 11, Erie, Ba.

-119-

NOTICES OF BOOKS AND PERIODICALS.

An Elementary Treatise on the Differential CMciUus, Founded on the Method of Rates or Fluxions. By
John Minot Bice, Professor of Mathematics in the United States Navy, and William Woolsey Johnson,
Professor of Mathematics in Saint John's Ck)llege, Annapolis, Maryland. Bevised Edition. 8vo., pp.
469. New York : John Wiley A Son.

This i0 the most extensive work on the Differential CalculoA yet published in this conntry. It ie well sappHed with interesting
examples, and contains many excellent solutions.

Sections IV and V of Chapter II are worthy the special attention of the student.

A larfre amonnt of space is devoted to the geometrical applications of the Differential Calculus, including Curve Tracing and
the discussion of Higher Plane Curves.

The typography is ver}- fine, and we hetfrtily commend the book to all who want a good text-book on the subject it so fully
treats. It is to be hoped that the authors will soon follow it with an equally exhaustive treatise on the Int^ral CfQcuJus.

The BJlementH of Co-ordinate Geometry, in Three Parts. I. Cartesian Geometry. II. Quaternions. III.
Modern Geometry, and an Appendix. By DeVoIson Wood, Professor of Mathematics and Mechanics in
Stevens Institute of Technology. 8vo., pp. 329. New York : John Wiley & Son.

In Part I, which contains 228 pp., the subjects usually treated in works on Analvtic Geometry are very clearly presented, and
illustrated with a large number ol^ well-chosen examples and solutions, some of which have been selected from the Educational
Times Reprint^ the WUtenberger and the Visitor, and properly credited.

The Second Part, which contains 72 pp., is devoted to Quaternions, and the subject is treated in the most elementary manner.
The examples are of the simplest kind, and intended "to explain and illufttrate the principles and the character of the operations
without taxing the ingenuity of the student in the mere solution of problems,"

The Third rart is very brief— too brief, containing only 10 pp. It is devoted to Tangential and Trilinear Co-ordinates, and
Abridged Notation, &c.

The mechanical execution is first class. We heartily commend the work.

Elements of the Differential Caleulus, with Examples and Applications. A Text-Book by W. E. Byerly,

Ph. D., Assistant Professor of Mathematics, Harvard University. 8vo., pp. 258. Boston : Ginn & Heath.

A good work, based on the Doctrine of Limits, well printed on heavy paper and supplied with a goodly number of examples.

The notation is somewhat peculiar. The differential coefficient is represented by Dxy. Although not in common use it
has the advantage of being easier to print.

Principles of the Algebra of Logic, with Examples. By Alexander Macfarlane, M. A., D. Sc., F. B. S. E.
Bead before the Boyal Society of Bdinburg KJth December lb78 and 20th January 187^. 12mo., pp. 155.
Edinburg, Scotland : David Douglas.

*'It is the object of this little work to investi^te the foundations of the analytical method of reasoning about Quality, with
' * * 'Boole I ' ' ' "' .....

special reference to the principles laid down by Boole as the basis of his calculus, and to the observations which ha\'e been pub
lished by various philosophers concerning these principles.*' P. 2, Art. 5.
It wiU well repay a careful reading.

Principles of Physics, or Natural Philosophy; Designed for the Use of Colleges and Schools. By Benja-
min Silliman, Jr., M. A., M. D., Professor of General and Applied Chemistry in Yale College. Swond
Edition, Be\ised and Bewritten. 8vo., pp. 710. 722 illustrations. Now York and Chicago: Ivison,
Blakeman, Taylor, & Company.

An extensive treatise, beautifully illustrated and well supplied with appropriate examples and problems.

The CoUegiate Algebra; Adapted to Colleges and Universities. By James B. Thomson, LL. D., Author
of New Mathematical Series, and Elihu T. Quimby, M. A., Late Professor of Mathematics in Dartmouth
College. 12mo., pp. 308. New Y^ork and Chicago : Clark & Maynard.

The definitions and rules nn* clear and concise^; the demonstrations simple, rigorous and logical. The examples are numerous
and good, but there is a lack of illustrative solutions. Besides the subjects usually found in works on Algebra, there are chap-
ters on Infinitesimal Analysis and Loci of Equations, and two pages on Probabilities.

The Notes at the close of the book contain brief biographical sketches of several world-famous mathematicians.

Tracts Relating to the Modem Higher Mathematics, Tract No. III. Invariants. By W. J . Wright, Ph . D. .
Member of the London Mathematical Society. 8vo., pp. 75. London : C. P. Hodgson A Son.

A valuable addition to the literature of the subject. In the limited si)ace of 75 pages Dr. Wright has treated in a very ck«r
and forcible manner the important subject of Invariants.

The Modem Higher Algebra is bt^ginnin^ to rceive some attention in this countn', and the thanks of American students ant
due the gifted author for placing within their reach, at so small a cost, the results of the ret^farches of the Great Masters.

Thermodynamics. By Henry T. Eddy, C. E., Ph. D., ITniversity of Cincinnati. 18mo., pp. 182. Bc»-
printed from Van Nostrand's Magazine. New York : D. Van Nostrand.

A New, Simple, and Complete Demonstration of the Binomial Theorem and Logarithmic Series. By J. W.
Nicholson, M. A , Author of the Formula of Bight-Angled Triangular and C'ircular Functions, and Pro-
fessor of Mathematics, Louisiana State University. Baton Bougo : Louisiana Capitolian Print.

TT^e Educaiional Times, and Journal of the College of Preceptors. London : C. F. Hodgson & Son.

The valuable Mathematical Department of three or four double-column qimrto pajros is ably tnlited by W. J. C. Miller, B. A.,
Registrar of the (iencral Medical Council. Among its coniributors are numlx*r<»a many of the leading mathematician]* of thi*
world. About J85 problems are proposed in each No., and about half as many solutions are given.

Mathematical Questions, ^cith their Solutions, Reprinted from the Educaiional Times. Same publishers.
Issued in half-yearly volumes of 112 pp., ^;vo., boards.

The Reprint contains, besides the mathematics publiHhcd in the Ti/n^ff, a Inrgp number of additional solutions and papers.
Vol. XXX contains 12 papers and solutions of 84 nronleins: Vol. XXXI contains (i pajKTs and solutions of «8 problems. Some
of the papers relate to important and disputed nomts in the Theory of Probability. \ ol. XXXII is announcwi as ready.

The Editor of the Visitor can fumii'h the Tinuj* at $2.00 a year, and the Reprint at $1.T3 jwr volume.

-120-

The Analyst. A Journal of Pure and Applied Mathematics. Edited and published by J. E. Hendricks,
M. A.. Des Moines, Iowa. Bi-monthly. $2.U() a year.

NoH. 1 and i of Vol. Vil contain valuable articlcK and elegant rahitiouH. The Analyst it* ably conducted and should receive
I he cordial Hupi>ort of every mathematician.

The Indiana Mathematician. Edited and published by W. DeKalb MeSwane, Superintendent of Schools.
Pett^rsbur^, Indiana. Published semi-annually at $1.00 a year. Single numbers 50 cents. No. 1, Vol. I,
August. Ih7:>. 'Ito., pp. 8.

A local niatliematica] periodical intended for circulation among tJie Htudentt* and teachers of Indiana. Only the first Xo. lia*»
been received; it in well ^^otten up and neatly printed, and contains 87 problemn for solution, 8ome solutions and an article on the
l>octrine of Chances. No. -2 is not out yet, [delayed by nicknesHj but is expected in a month or two. We wish it Bucce^p.

The School ViHiior, Devoted to the Study of Mathematics and Engksh Grammar. Edited by John S.
Koyer and Thomas Ewbank. Ansonia, O. : Published by John S. Royer at GO cents a year.

'I he S<j/iool I'M for in a new periodical, and very ta«tcfully gotten up; No8. 1 and 2 contain a number of interesting problem!*
and i>olutionf) of an elementary character. ItM cheapncHe* and merit should ensure it a large circulation.

The Wittenberger. Spriugfleld, O. : Wittenberg College. 10 Nos. in a year. $1.00 a year.
The excellent Mathematical Department, to whicli »onie of the bent mathematicians contribute, is edited by William Hoover,
Superintendent of Sch'joln, Waiiakoncta, ().

Barnes' Educational Monihly. New York : A. S. Barnes & Co. §1.50 a year.

.\ very valuable educational magazine, and should be read by all teachen*. It \\a» an interenting Mathemadcal Department
odited by Prof. F. P. Maw, now of Bowdon College, (ieorgia.

Educational Noten and Queries. Edited and published by Prof. W. D. Henklc, Salem, O. $100 a year.
i^'sued monthly except in the vacation months of July and Augu**t, and coutaiuH mathematical not«H, probleais and sohitiouH.
I er-.ide!< much valuable gi»neral information.

New- England Journal of Education. Weekly. Boston and Chicago : Published under the Auspices of
the American Institute op Instkucjtion and the Teachers' Associations of tlie New-England States.
$3.iK) a year : $2.5 ) if paid in advance.

One of the l>est educational jonmalj* in the c mntry, and nhonld be liberally patronbEcd.

The Mathematical Department in edited by Pn)f. E. T. (^uiniby, but the publtHhcr has "nqueezed" the life almost out of it.

The Yates County Chronicle. Weekly. Penn Yan, N. Y. : Chronicle Publishing Co. $2.00 a year.
The intercepting and vahuible Mathematical Department continue:* under the able and efficient management of Dr. Wright.

The Canada School Journal. Monthly. Toronto, Ontario, Canada. $1.00 a year in advance.

A larce 2:-page educational monthly crammed full of good matter. It has an able Mathematical Dcimrtment edited by Alfi^d
Baker, M. A.

The Pennsylvania School Journal. Monthly. J. P. Wickershara, Editor. Lancaster : J. P. Wicker-
sham & Co. Sl.fiO a year.

One of the beet of the educational monthHen. It hiv> no department of mathematic:*, but Kcientiflc articlen and examination
problems are occa^onally published.

American Journal of Education. Monthly. Saint Louis. J. B. Merwin. $1.(50 a year.
A "live" i)cri(Klicjil de«<erving a wide circulation among the teaching fraternity.

The Maine Farmers' Almanac. 1880. Hallowell, Me. : Masters & Livermore. Price 10 cents.
This fxcellent little annual contain;* the annwern and nolutions to the Puzzlet*, RiddleR, Mathematical Questions, Ac, pro-
tK>unded last year, and a new lot for t^olulion in the next number.

CORRIGENDA.
No. 2.

Page 28, linen 22, 25 and HfS.for "K(r)" rrad Eu').

No. 8.

Page a3, ^olutiou of I*roblem 5n, line 0, numerator of the fraction,/or "ViPc" rtcui nVe.
Page 77, second line of tlrsi solution,/w "D" rtad B.

No. 4.

Page 90. dilution of Problem »3, numerator of the value of co9^',/or **j'-*' read r^.
i'age 94, solulicm of Problem 107, line H from the bottom, for "2.5(1.2* .5«)" read 25(1.2''' .o").
Page m. Problem lM,/or "81 yards'* read HI rods; Problem 16(5, omit all after "* 8« 48' 30"" and insert "E".
Page 102, solution of Pniblora 117, llne.^ 4 and 7, for "w"' read w*; and in line 4 from bottom rem/ 0.(185760 inrtead of therer^ult
irivcn there.

Piige 104, Problem 121, numerator of exponent of {in), far "1 — »" read n 1; last line of Holution,/<>r "vol. i" read vol. ii.

Pagi» 108. solution of Problem 127, next to the last line. /or **cos<^"' rtad conB.

Page 10J>, solution of Problem 12», Hue H.for "* - *»" read * - 8**.

Page 110, solution of Problem 11,5, line H, numerator of 14th convergent, /or "9" read 4.

Pag.- 114, lim. UM ■;. - (J - ^]^=^)'- no,,,, - (J - |^|)'.

Vol. I. JULY. 1880. Ho. 5.

■.f.pnomo

THE

MATHEMATICAL
VISITOR.

EDITED AND PUBLISHED BY

ARTEMAS MARTIN, M. A.,

Member OF the London Mathematical Society.

I8BUHD SEKI-ANNUAIiliT. TKBMS : ONE DOIiLAB A TBAB IN APVAKOB.
SIKQIiE NUHBEBS, FIFTY CENTS EACH. BACK NUMBEBB SDPPI«IED AT THE SAMB BA^B.

1880.

THE

MATHEMATICAL VISITOR.

[CNTUSO ACCOHDINO TO ACT OP CONOaUS, IH THI TCAM IMS, tV A^TEMAt MARTIN, M. A.. IN TNI OmCI OF TNI LIMAMAN Of OONaiU*. AT WASHINOTON.]

Vol. I. JULY. 1880. No. 5.

JUNIOR DEPARTMENT.

Solutions of Problems Proposed in JY^s. I9 3 4* 3.

1.— Proposed by Prof. Edward Brooks, M. A., Ph. D., Principal of Ponnsylvania State Normal School, MUlersville, Pa.

Two trains, one a and the other b feet long, move with uniform velocities on parallel rails ; when they
move in opposite directions they pass each other in m seconds, but when they move In the same direction
the faster train passes the other in n seconds. Find the rate at which each train moves.

Ill*— Solution by K. S. PtiTNAM. Rome, Oneida County, N. T.; W. L. Harysy, Maxfleld, Maine; and William Hoover,
Sapermtendent of Schools, and Matnematlcal Editor of the WUtenierger, Wapakoneta, Ohio.

Let X = number of feet the faster train moved In one second, and y = number of feet the other moved
in the same time.
n^x-^-y) = distance traveled by both trains while passing when moving In opposite directions ;

.-. m(a;-»-y) = a-»-& (1).

The faster train will gain x—y feet each second, and In n seconds It gains n(x—y) feet ;

.-. n(x^y) = a-\-b (2).

From equations (1) and (2) we have

xA-y = - and x—y = — ;
m n

whence x = j("±* + '•+-\ j, = jW_«+^y

^—Proposed by Miss CHiusTim Ladd, B. A., Baltimore, Maryland.

Show that in any plane triangle

aslnA-f- &slnB- f-cslnC __ /a«-|-& i-f-cg \
acosA-f-bcosB-f-ccosC " \ abc /

II.— Solution by the Proposer.

Since i2= ^~ * = o r^ = n -^^^ a«-|-6«-|-c« = 2fi(oslnA4-6slnB-|-cslnC).
28lnA 28lnB 2smC ^ ^ ^

As shown by J. J. Walker In the Educational Times, March, 1875,

r(a4-64-c) = 12(a cos A -f- 6 cos B 4- c cos C).

Hence «£l5.A + 68inB + c8lnC ^ ay-6.+c' /«.+^+_^X since «6c = 4fl«.

ocosA-|-6co6B-}-ccosC 4r8 \ abc J

X

x^l

8

y

1

1

1

3

4

7

5

20

21

41

29

119

120

239

169

-122-

51.— Proposed by Artemab Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

There is a series of right-angled triangles whose legs differ by unity only. Calling the one whoso
sides are 3, 4, 5 the first triangle, it is required to find general expressions for the sides of the /ith triangle,
and compute the sides of the 100th triangle.

II.— Sohition by K. S. Putnam, Rome, Oneida County, New York.

Let X = shorter leg, x-\-l = longer leg, « = sum of legs and y = hypothenuse.

Then a;3-f(ar-fl)« = y^ from which we obtain (1) x= l['/(2y^ — 1) — 1]. (2) i/ = "v/UC^'-f !)]•

Now by trial the first four values of x that will satisfy the above equations are 0, 3, 20 and 119.

Hence we have

Primary or vanishing triangle
First triangle
Second triangle
Third triangle
An inspection of the above gives the following formulas :

(3) (a:-fl)„ = 4y,_i-f(a?-fl)^, (4) 3^„ = 4«^i + 1/„-2.

From (3) and (4) we obtain values for x, x+l, « and y for the 4th triangle, and as mau^' more as
are desired. We shall then find by inspection (5) «„ = 6«„_i -- 8,»_3.

Representing the sum of legs of the primary triangle by 8, and the sums of the legs of the 1st, 2nd, 3(1,
etc., triangles by «i, a^, «3, etc., we have by (5),

8s = 6«i — 8, 83 = 3581—68, 8i = 23481 — 358, 85=118981 — 2048.

And thus proceeding we may find the sum of legs and thence the sides of each triangle in the series in
succession.

When a single triangle is required the procjess may be shortene<l. In solving for the 100th trianglt^
by proceeding as above we find

(6) 820 = 36178655593983681 — 620727596307718, (7) 8a, = 21086465760082458, — 3617865559393368.
Now 8 = 1 and 8i = 7 and we have *» = 247043313194S081, s.^ = 14389733476117^79.

Substituting these values of a^ and 831 in place of 8 and 8, respectively in (6) and (7) we have

8« = 5055923762956339922096065927393, 841 = 29468083200663558275864384257639.

Substituting these values for 8 and 81 In (6) and (7) we find s^ and 8f,i which substituted in (6) and (7)
give 8^0 and 881, and the values of 8^0 and 8gi substituted in (6) give us

8,00 = 433393862972275766610959594585726143280331055373989306921633S3984677691935393 ;
whence x,^ = 21669693148613788330547979729286307164015202768699465340081691992338845992696.
(j; + l)ioo = 21669693148613783330547979729280307164015202768699465346081691992338845992697.
From (2) we have »/ioo = V \ i[.{^\(*iY + 1] i • Performing the operations indicated we get

3/100 = 30645573943232956180057972969833245887630954508753693529117371074705767728665.

irintendent, Burlington County, Bordcutow-n, N. J., and Mr. Reuben DatUty Bradford.
' -em.

Prof. Edgar Haas, M. A., County Siiperintende]
Illinoifi, fumiBhed elaborate solutions of this probl(

94.— Proposed by Henry Nichols, Hampton, Rock Inland County, Illinois; and Mrs. Anna T. Snyder, Cbicago, ]llinoii<.
It is required to divide a tapering board into two equal parts by sawing it across, parallel to the ends.
Find the width at each end so that the lengths of the pieces will be expre^jsed by rational numbers.

III«— Solution by Dr. David S. Hart, M. A., Stonington, New London County, Connecticut.

Let ABCD represent the tapering board, EF the division line, AI = BK = .c =
length of the board, CI = DK = ax, EG = HF = bx, and AB = GH = IK = y.

From the similar triangles AIC, AGE we find AG = . Then, by Mensuration,

x{ax-\-y) := area of ABCD and ^^^y-^^^^ = area of ABEF ;

, , , hx(y-\-bx) . iai—2b^)x

.'. x(ax'\-y)= ^""J ^ whence 3/= 26_a '

where x may be any number, and a and b such that a > 26^ and 2b > a.

Let a = -,^, 6 = A, and we have y = ^^x. If now x be taken = 36 then will 3/ = 1 = AB, y+2cu =
7 = CD and y-^ibx = 5 = EF.

-123-

97«— Propose ^7 ABTBMA0 Mabtin, M. a., Member of the London Mathematical Society, Brie, Brie County, Pennsylvania.
Solve, by quadratlos, the equation x*—%aafi—2abX'\-¥ = 0.

in.— Solntion by Lucius Bbown, Hudson, Middlesex Co^ Mass.; DsVolson Wood, M. A., C. B., Professor of Mathematics
and Mechanics, Stevens Institute of Technology, Hoboken, N. J.; J. F. W. Scheftbr, Professor of Mathematics and German,
Mercersbnrg College, Meroersbuig, Pa.; Marcus Bakkb, U. S. Coast Survey Office, Washington, D. C; and Chas. H. Tutton. .
Wilkes Baire, Luseme County, Pennsylvania.

The equation may be written (a;^ 4. &>) ^ 2ax(afi 4- 6) = ; adding 2b3fi to each side, and then extract-
ing the square root,

aJ»4-6 = a«±an/(d»4-26), or a5«— [a±i/(a*4-26)]a? = — 6;

whence a? = l{a±i/(tf«4-26) ± ^[2^ — 2ft±2ai/(a?»4-26)]}.

iy«— Solution by F. P. Mats, M. B., M. S.. Professor of Pure and Applied Mathematics, Bowdon (State) College, Bowdon,-
Carroll Co., Georgia; Wzllxax Hooter, Superintendent of Schools, and Slathematical Bdltor of the WtUenbergert wapakoneta,
Ohio; Abrah R. Bulus, Ithaca, N. T.; W. B. Heal, Wheeling, Ind.; Prof. Frank Aubrt, Millersville, Lancaster Co., Pa.:
and Prof. H. T. J. Ludwig, Mount Pleasant, North Carolina.

2fr
By adding -r to each member, the given equation can be put In the following form :
a?

and ' x= l{a±i/(d»+26)±^[2fl{»— 26±2ai/((«»+26)]|.

M«— Proposed by W. Woolsrt Johnson, Member of the London Mathematical Society, Professor of Mathematics, St. John's
College, Annapolis, Maryland.

The extremities of a line of fixed length which slides along a fixed line are Joined to two fixed points.
Find the locus of the Intersection of the joining lines.

II«— Solution by the Proposer.

Taking the line joining the fixed points as axis of x and that on which the line of fixed length slides as
axis of y, and denoting the distances of the fixed points from the origin by a and 6, the fixed length by
c and the distance of the extremity joined to (a, 0) by the variable 2, the equations of the stra^ht lines

are ? + - = 1 and * -|. _4_ = 1.

Eliminating z we obtain

caj* 4- (a — 6>cy — c(a 4- 6)a; 4- oftc = 0,
an hyperbola whose asymptotes are

a? = and caj4-(a— 6)y = c(o-|-6).

The curve passes through the given points and has the given line for an asymptote.

L-— Proposed by William Woolbet Johnson, Member of the London Mathematical Society, Professor of Mathematics,
um^s Oollege, Annapolis, Maryland.

BB Is an ordinate, from any point B of a circle, to the diameter passing through the fixed point A ;
and T is the Intersection of the tangent at A with the radius produced through B. Find the locus of the
intersection of AB and TB.

n,— Solution by Lucius Brown, Hudson, Middlesex County, ICassachusetts.

Let G be the colter of the circle, and P the point whose locus is required .
Draw PS perpendicular to AC, and let AS = a;, PS = y, and / ACB = q>.

Then xiy li r(l--co6<p) :rsln<^ :: 8in<^ : 1-f cos<^ (1)

and X irtAiKp—y ;: r(l — cos^p) :rtan^ (2).

Therefore rtan^— 2/ ly r: tan^:sln<p :: 1 : cos^,

whence rsln^— ycofl<p = y, or 8ln<p : l-f-cos^ :: 2/: r (3).

Comparing (3) with (1 ) gives xly llyir, or y« = nr. Hence the
locus is a parabola.

The methods employed by the Proposer and Professor E. B, Seits are similar to this.

-124-

107.— Proposed by Sylvester Robins, North Branch Depot, Somerset County, New Jersey.

At a Firemen's Fair a silver trumpet is offered to the company exhibiting tlie ladder that can be used
in the greatest number of streets and alleys for the purpose of reaching windows on either side without
changing the location of its foot. All bases and perpendiculars must be rational lengths, and a company
may include in their count dimensions having as many decimal places as their ladder has, but no more.

The '•Hudsons" bring a ladder 65 feet long, the "Keystones" offer one 32 J feet in length, and the
"Dela wares" show one of 42^ feet. To whom must the trumpet be awarded, and on what count?

II,— Sohition by Dr. David S. Hart, M. A., Stonington, New London County, Connecticut.

Let X, y be the legs, and z the hypothenuse of a righ1>angled triangle ; then a;^-f y- = a*. The general

values of j; and y are x = - - --l- , y = . ,^, the sum of whose squares is = a!*.
p-^qi ^ p--\-qi

The formula 4/-|- 1 contains all prime numbers each of which is the sum of two squares. The product
of any two such prime numbers is the sum of two squares in two ways ; the product of any three is the
sum of two squares in four ways ; the product of any four is the sum of two squares in eight ways ; and,
generally, the product of any n such prime numbers is the sum of two squares in 2"-» ways.

Let « = 65 = 5x13 ; then we have, by substituting p and q for the roots of the squares in 5 and 13, and
of the two squares in 65, four sets of legs to the hypothenuse 65, or four triangles for the ladder of the
"Hudsons".

Let z = 325 = 5x5x13 ; then, substituting p, q for the roots of the squares in 5, 5, 25 and of the two
squares each in 65 cmd 325, we have seven sets of legs to the hypothenuse 325, or, dividing by 10, we
have 7 triangles whose legs correspond to the hypothenuse 32.5, the ladder brought by the "Keystones".

In a similar manner when z = 4225 we find 12 triangles for the ladder exhibited dy the "Dela wares".

Each of these triangles can be placed in three streets, having widths 2x, 2y and x+y. Besides, placing
each ladder in a perpendicular position close to the walls of the street, and moving the top so as to reach
windows on the opposite side of the street, we shall have two streets with widths x, y. Then, for each
triangle we shall have streets of the widths x, y, 2x, 2y and X'\-y. Calling the number of triangles n, we
have 5» == number of sti-eets for the triangles taken separately. Combining the first triangle with each
of the n— 1 remaining; the second with the n — 2 remaining, and so on; and, finally, combining the

(» — l)th with th nth, we shall have 1 -f-2 -f-3 -f 4 + -|- (n— 1) for all the combinations. But, putting

JJ, «/, x\ l/ = the legs of any two triangles in combination, we have X'\-x', y-^jf, x-\-yf* y-f ^ ^or the

widths of four streets; therefore we have 4[l + 2-f3-f4+ +(n — 1)] = number of streets for the

triangles taken in combination, and 5n + 4[l-f2-f-3-f4-f -f(n— 1)] = 3n + 2?i^= whole number

of streets for n triangles.

Let » = 4 ; then 20+ 24 = 44 streets in which the "Hudsons" can place their ladder.

Let n = 7 ; then 35 + 84 = 119 " " "Keystones"

Let n = 12 ; then 60+264 = 324 " " "Delawares"

The trumpet must be awarded to the "Delawares".

NOTE ON THE SOLUTION OF 110.

By Walter Siveklt, Oil City, Venango County, Pennsylvania.

The method of solution by Mr. Putnam is Ingenious, and probably as good as any of the methods of
approximation ; but to suppose that any of those methods are "expeditious" for the purpose of extracting
the square root is a mistake. For instance, to extract the square root of 2 to about 200 places, by the
common method, the divisor increases from one figure to 100 figures, then decreases to one figure. If
the figures of the root were all Int^ers, this would give 100 figures in the divisor, (which need be written
but once,) 200 figures in the root, and about 20000 figures in the rest of the work; in all, about 20300
figures.

By Mr. Putnam's method, to form his fraction will require over 8000 figures ; after it is formed he must
divide in full to 100 places, requiring 20000 figures. The contracted part of his division will require
10000 more figures ; in all, upwards of 38000 figures. The work is, therefore, nearly double that by the
common method.

-125-

PROBLEMS.

218.— Propo««ed by Artbhas Martin, M. A., Member of the London Mathematical Society, Erie, Brie County, Pa.

A merchant marked his goods at an advance of 60 per cent, on their cost and then gave his customers
"40 per cent, off', thinking that he was clearing a profit of 20 per cent. Did he gain or lose, and at what
rate per cent.?

219* — Proposed by Thomab Baoot, County Superintendent of Ripley County, New Marion, Indiana.

A rope attached to the top of a vertical pole standing upon a plain just reaches to the groimd. Desiring
to find the hight of the pole I took hold of the end of the rope, keeping it perfectly taut, and pulled it
out a feet from the foot of the pole ; I then found that the end of the rope was h feet from the ground.

How high was the pole?

220.— Prom the Saturday Bvenixo Post, Philadelphia, Pa., June 1, 18T8.

A fisherman began a cast net with 36 meshes. He enlarged it by adding 12 meshes more to the third
row ; the same to the sixth, ninth, and so on to each and every successive third row, termed ''widen-rows".
The last row, being a ''widen-row", contained 3G0 meshes. Required the entire number of meshes com-
prised within the net.

BnierpriM, S. C. Ego Geo.

221.— Proposed by Mies Eliza D. Crake, Brooklyn, New York.

It is required to determine the greatest number of marbles one inch in diameter that can be packed
in a cubical box whose inside dimensions are each one foot.

222«— Proposed by Elijah A. S^^uieb, Le Grand, Marshall County, Iowa.

The top and bottom diameters of a flat-bottom kettle holding an ale gallon are in the proportion of 5
to 3, and its depth is 12 inches. Required the two diameters.

223«— Propo8ed by H. H. Hough, Principal Linden Female Seminary, Doylestown, Bncks Connty, Pennsylvania; and
J. mT TAYLOB, Milton, Umatilla County, Oregon.

Find X and y, by quadratics, from the equations a*-fy = ll, aj-f^ = 7.

224.— Proposed by Christian Hornttno, M. A., Professor of Mathematics, Heidleburg College, Tiffin, Seneca Co., O.

Let A and B be two given points in the diameter of a circle equally distant from the center C. In AB
take any point P, so that PC x radius = (AC)^. Let CP meet the circumference in D ; draw AE and BE
to any point E in the circumference, and take the arc DQ double of DE ; join PQ. Prove that AE X BE
= PQ X radius.

225.— Proposed by D. C. Jones, Saint Charies, Butler County, Ohio.

The product of the length of a field by the width divided by the diagonal will give a quotient equal to
the number of acres in it plus 4J, and 4 times the diagonal divided by the difference between the length
and width will give a quotient equal to one-sixth of the length. Find the dimensions of the field.

226«— Proposed by Professor W. P. Casey, C. E., San Francisco, California.
Solve, by quadratics, the equations ar* — y« = a*, 3^'\-3xi/^ = b^.

227.— Proposed by Professor E. J. Edmunds, B. S., Principal of Academic School No. 8, New Orleans, Louisiana.
A triangle being given it is required to compare its area with that of the triangle formed by connecting
the feet of the bisectors of its three angles.

228.— Pi^oposed by Frank Albert, Professor of Mathematics, Pennsylvania State Normal School, MillersTillc, Lan-
caster County, Pennsylvania.

A ball, radius r, is enclosed in a cubical box in which, it just fits ; there are also eight infinite series of
balls, one in each comer. Each ball in those series touches the three adjacent sides of the box and the
tw(» balls next to it, except the first in each series which touches the large ball. Find (1) the radius of
the nth ball in one of the series, (2) the sum of the volumes of the first n balls in one of the series and
(3) the sum of the volumes of all the bails.

229.— Proposed by Sylvester Robins, North Branch DeiKJt, Somerset County, New Jersey.

In finding the solid diagonals of a series of parallelopipeds, it was noticed that the length, breadth and
thickness are consecutive numbers, and that in extracting the square root, if the sums were only a unit
less every diagonal would be an integer. The sides of the first paralleloplped are 3, 4, 5 It is required
to find general expressions for the sides of the nth solid, and compute the dimensions of the 100th solid.

230.— Prize Problem. Proposed by Artemas Martin, M. A., Member of the London Mathematical Society, Erie,
Erie County, Pennsylvania.

Give an expeditious method of approximating to the cube root of a quantity, and find by it the cube
root of 2 to at least one hundred places of decimals.

Solutions of these problems should be received by Februarj' 1, 1881.

SENIOR DEPARTMENT.

Soluiions of Probleins Proposed in J^os. I9 3 4* 3.

114.— Proposed by Dr. David S. Hart, M. A., Stonington, New London County, Connecticut.

To find two numbers such that their sum shall be a square, the sum of their squares a square, and if
the cube of each be added to the square of the other the sums shall be equal.

II,— Solution by Hon. Josiah H. Drummond, LL. D., Portland, Maine.

Let X and y be the numbers ; then x-\-y= D (1), ar^-f t^ =0 (2), and a^^+y^ = j/'-fa^ (3).

Prom (3), ar^+ary-f y* = •*^+2/ W' ^^^ V = »»*— a; and (1) Is a square and (2) becomes

2ar2— 2m«a;+m4 = D ....(5) and (3), a?«— m«a;-f m"* = Q . . . (6).

Put X = rn^n and (5) becomes by reducing 2n2— 2n-4-l = D = (pn± 1)*;

2(p— 1) , ^, 2m-^(m--l)

whence n= ^ ^^ and then x= - \ ^ .

jyi — 2 p' — 2

u2 2

Substituting this value of x in (6) and reducing we get m = //-4_o«3 i o 2_xn4.4\ *

Assuming the expression under the radical = {p'±p± 2)* and solving we get p = | or J with two
other impracticable values. Then m = /j, and x = ^gg and rrc^ — x = — ^^ are the numbers, both
values of p giving the same result.

To obtain positive values, other values of p must be found. It is evident on Inspection that p = 1, or
2, but both these values give zero as one of the numbers. Take p = g + 1, substitute it in the expression
and solve it as indicated above and we have q = — 4 or J, and p = — 3 or 5 ; but these values of p give
the results already obtained.

Take now p = g— 3, and we find g = 4 or Yt*. a^ci p = 1 or f f . Using the lajst value of p we find
__ 4183 _ 1768x4183 _ 7395544 ^__ _ 10101945

"^"3637' '^~ (3637> "13227769' ^ ^ "^ "" 13227769'

III.— Solution by IiEUBE^' Davi«, Bradford, Stark County, XUinois; and Walter Sivbrlt, Oil City, Venango Co., Pa.

Let x{x'\-y) and y(X'\-y) be the required numbers ; then must
■r{^-\-y) + y{x-\-y) or (ar+y)^ = a (1), x^K^-^yy + ^<x-^yy = (ar^H-y-K^+y)'^ = □ (2),

x\x-^yr-\-ir^{x-\-yy = x^(x-\-yy-^U\^'^yr or (a?»-3^)(a:-f i/)« = (^-|/^)(a;-f2/)«.

Dividing tiie last equation by (ar— i/)(x+i/)3 it becomes a^-|-ar^-f y^ =r 1 (3).

I The conditions will all bo fulfilled if x'^-^j^=D (4), and af^-f a?3/-f 3^ = 1 (5).

Transposing (5), ar'-f-y^ = 1— a:y (6), and a^4-2a^-f y* = 1 -f «!/ (7).

Let 1 —xy = n- ; then xy = 1 — n^, and substituting in (7), (x-\-yy = 2— n^ (8]^

j To make (8) a square, substitute g-f 1 for n and put the result, 1 — 2g— g^, = {inq—iy; then will

! 2(m— 1) , , , m-'-h2m— 1 « . / . xo /fn^—2m—l\^

; ^ = m«^ + l • ^^^ n = g-f 1 = -;;;,^-y-, 2-n« = (x+yy = ( -^^^j- ) ,

I To make the last expression a square, let

[ a^ + 12in3-f 2m2 — r2m-|-l = (w^-f 6m— 17)3;

then m = |. Substituting r-|- 1 for m we shall have to make

Let r- -f-9r + ^f be the root of this square ; then r = — jo and r-f } = 5 J. Substituting this value for to,

,, ^ /2993V 4269720 , , ,, /4183V2 /647 V 2415 1768

•^•+^' = (.3637) ' -^ = (3637)^ ' (^ + 2/)' =13637) » i^-y)' = (3^537) • ^ = gegT' ^ = 3637'

, . , 10101945 , . , 7395544

^(^-f 3/) = 13227769, y(x^y) = 13227769-

-127-

115*— Proposed by Stlvbsteb Robxnb, North Bmncli Depot, Somenet Connty, New Jersey.

There is a series of paralleloplpeds whose dimensions dlflfer from a perfect cube by one unit in only
one of the edges. In every case the solid diagonal is an integer. Calling the one whose edges are 1, 2, 2
theJirHt parallelopiped, it is required to find general expressions for the dimensions of the nth solid, and
compute the length, breadth and thickness of the 30th one.

II«— Solntion by Rkuben Davis, Bradford, Stark Coanty, Illinois.

Letx, X and x— 1 be the edges of one of the paralleloplpeds; then, representing the diagonal of the
solid by A we have 3ic«--ar-f 1 = D* (1).

Assume Sa^ — aaj-f-l = (nx—iy = w«aJ»— 2na;-|-l ; then x = ^^J- To make this value of z in-
tegral, n may be ±2. If n = 2, a: = 2 ; if n = — 2, ar = — 6.

For the edges of the first parallelopiped we have a; = 2, a; = 2 and x^l = 1; and for the second
x = — 6, rc = — 6 and a;— 1 = — 7. But as we use only the squares of these num>)ers in finding the
solid diagonal, the signs may be disregarded after the edges are obtained.

Substituting y-f-r for x in (1), it becomes 3^— 2y-f 1 -f-(6j/— 2)r-|-3r^ = D.
Putthis = [\/(32J« — 2y-f 1) — arp, = 3i/»— 2y-|-l — 2i/(3j/8— 2y+l)«'-f ««»«. and we get

28i/(32/' --2y4-l) 4 -6^-2

If8 = 2. r = 4i/(3i/*-2y-»-l)-»-6y-2; butif8 = -2, r = -V(3!^~22/-hl) -|-6y~2;

.-. X = y+r = + V(33^-2y+l) -♦-72/-2 (2).

Let a = 2 = one of the equal edges of the first solid and & = —6 = one of the two equal edges of the

second ; then, if a%, aj^, a?5, etc., denote one of the two equal edges of the 3d, 4th, 5th, etc., solids, we have

xs = 4v^(3rt«— 2a-|-l) -f7a-2 = c = 24, 3^4 = — 4i/(36«~-26-|-l) -f 76— 2 = d = —88,

X6 = 4v'(3<j«— 20+1) -f7c— 2 = e = 330, a% = — 4v'(3d«— 2d -»-l) -f 7(1—2 =/= —1230.

xj = 4i/(3e2— 2e-|-l)-f 7e— 2 = ^ = 4592, a% = — 4^/(3/*- 2/4-1) + 7/— 2 = /i = —17136,

x^ = 4v'(3</«— 2p-f 1) +7i?— 2 = i = 63954, «,o = — 4^/(3^*- 2ft-f-l) -|-7/i-2 =y = —238678,

a;„ = 4i/[3(aj9)«-2x8 + l] +7a:«-2 a^, = iV'[3(«„-s)'-2a^-8 + l] + 7a:n-.«-2.

In (1) substitute 2 -f tp for a? and it becomes 3(tr -f «)« — 2(«j -\-z)-\-1 = D^,
or 3252— a5-f-l-|-(6«— 2)tr-f 3tt^= □.

Taking (32?— 2a;-f-l)* -\- -i^-'^)^ ^ ^. . for the root, we find

w = — 2(3«— 1)(325«— 22f-f-l) = — (1828— 18253-f 105—2), and x = w-f-« = — (1&^— 18«!8-|-92J— 2) .... (3).
Writing i for z hi (3), we have x» = — (I8t»— 18i2-f-9t— 2) = —4708351225665448 =p.
Substituting p for y in (2), a^ = _4-v/(3p« — 2p-f 1) -|-7p — 2 = —65578872750585150, and the edges
of the 30th solid are 65578872750585150, 65578872750585150, 65578872750585151.

Substituting^ for z in (3), we find xn = 244743685008277952, and this substituted tor y in (2) gives us
a-33 = 3408839784121828674 ; substituting this for z in (3) we find

a;,oo = —713004506105244899421887701590040546362489937656289477528,
and the edges of the 100th solid are

713004506105244899421887701590040546362489937656289477528,
713004506105244899421887701590040546362489937656289477528.
713004506105244899421887701590040546362489937656289477529.
Substituting the value of a^ioo for z in (3), we find

a%,oi = 6524531448049515350118452018470428054703217342336205218854946680046881369087279604040080780818
15833732939213199039629958935050601555098839300287381755249691615536030967002,

and therefore the edges of the SOlst solid are

652453144804951535011845201847042805470321734238620521885494668004688i369087279604040080780818158337
32939213199039629958935050601555098839300287381755249691615536030967002,

6524531448049515350118452018470428054703217342386205218854946680046881369087279604040080780818158337
32939213199039629958935050601555098839300287381755249691615536030967002,

6524531448049515350118452018470428054703217342386205218854946680046881369087279604040080780818158337
32939213199039629958935050601555098839300287381755249691615536030967001.

-128-

To find the diagonals : From (1), a; = J[l±i/(3i^— 2)]; .-. 3X)«--2 = D (4). i

In (4), Bubstitute v-^-t for D and it becomes 3t^--2+6trt-f 3P = a. Assuming \/(3«»— -2) — tu for the I

root of this square we find < = -^^^^^T_ 3^"*^— . If u = 2, « = V(3tJ«— 2)4-6r; but if « = — 2, then

t = — 4i/(3v«— 2) 4-6u ; therefore D = « + « = ^4v'(3i?8— 2) -|-7« (5).

If 1? = — 1. D = —3, but if u = 4-1, D = 11. Let a' = 3 = diagonal of first soUd, ft' = 11 = diagonal
of second solid, I^, D4 D^ etc., the third, fourth, fifth, etc., diagonals ; then

Ih = 4i/(3a'«-2)-f- 7a' = 41 = c', 1)4 = 4i/(36'*-2) + ?&' = 153 = d\

A = 4v'(3c'»— 2) -|-7c' = 571 = e'. Da = 4v'(3d'*— 2) -f 7d' = 2131 =/',

A = 4i/(3e'^-2) -I-7C' = 7953 = ^, A = 4i/(3/'2-2) -f-7/' = 29681 = h',

D2 = 4^/(3y'*-2) 4-7y' = 110771 = i\ Ao = 4i/(3A'*-. 2) 4-7/i' = 413403 =/,

Dm = 4i/[3(A)*-2] -h 7A, A = 4i/[3(Z),^)2-2] 4-7D*-t.

Let 3«»-2H-6irt+3«' = [^(3tr»~2)* + 7^^^^ - ,^^2 )i\' ^^I^^^^^K *°^ reducing we get

i = 2»(3tj'— 2). and D = v+i = fit?* — 3« = 3iK2t?«— 1) (6).

Writing *' for v in (6), D« = 3t'(2t'*— 1) = 8155103642731753; substituting Dn for v In (5), we have

Ao = 4i/[3(Z)4g)*— 2] 4-7Z)« = 113585939507107G51 ; substituting/ for v in (6),

Ai = 3;'(2J'»— 1) == 423908497265970753 ; and then A» = 4v'[3(Ai)*— 2] +7Ai = 5904283700961130691

Writing the 33d diagonal for v in (6) we find the 100th diagonal,

Aao = 1234960030599837928682339736709998512373739432964939784153.

Using the last number in place of « in (6) we have

Aoi = 113008199636026994298326592220955053426348838670618234182627054822531216755603249090056493829
3341593264019376761065450354578681963877363685857880056605581408120057091945003.

III«— Solntion by Abtbmas Martin, M. A., Member of the London Mathematical Society, Brie, Erie Coonty, Pa.

Let Xr, XniXn±l and yn be the lengths of the respective edges and solid diagonal of the nth parallel-
opiped. The length of the solid diagonal is i/[3(aj»)« ± 2a:»4-l] ; therefore 3(ic»)« + 2aj„-f 1 = D = (y,)«,
whence %XnY ± 6x«H-l = D = 3(2/,)«-2 = (w«y .... (1) ; and 3«» ± 1 = i/[3(y«)«— 2] = w„ whence

Xn = J(tu» + 1) (2). From (1), by transposition, (»«)«— 3(y,)a = — 2 (3).

Ifwemultiplytogethertheequationsp^— 3q« = l, t*— 3u« = — 2, weget(pe-f3gu)»—3(pu4- (]<)» = — 2;
hence we may take ir» = pt -f 39U, s/n = l^i -f 9^' But the least values of t and u are obviously t = 1 and
tt = 1 ; therefore w = i>-f-3^, y = p+g.

, , . . (2-f-i/3)»-f(2—i/3)» (2+i/3)« — (2— I/3V*
The general values ofj? and g are Pn=-—^--^—w . ^* ~ ^^2i/3 '

•^ «^ = l[(l/34-l)(2+l/3)» - (1/3- l)(2-i/3)»] and 3^ = J[(3+i/3)(2+l/3)*H- (3-,/^^^^
Substituting in (2) the edges of the nth solid are
i[(l/3H-l){2H-i/3)»-(,/3-l)(2-,/3)- + 2], j[(^3H-l)(24-i/3)--(i/3-l)(2-,/3)«q=2]
and i[(^3 + l)(24-i/3)«-(i/3-l)(2-,/3)« + 2] ± 1.

The upper sign obtains when n is even and the lower when n Is odd.

Edges of second soUd are 6, 6, 7 ; of third, 24, 24. 23 ; of fourth, 88, 88, 89 ; of fifth, 330, 330, 329 ; etc.

(2H-i/3)« = 7+4i/3, (2+1/3)* = 97-h56i/3, (2 + 1/3)8 = 18817 + 10864i/3.

(2 + i/3)w = 262087 + 151316i/3, (2 + 1/3)« = 137379191137 + 79315912984i/3,

(2 + i/3)» = 72010600134783751 +41575339372323900i/3;

(l/3+l)(2 + i/3)w = 196736618251755451 + 113585939507107661i/3,

(l/3~l)(2— i/3y» = — 196736618251755451 + 113585939507107651i/3.

Substituting in the general formulas we find the edges of the 30th solid are

65578872750585150, 65578872750585150, 65578872750585151.

If we expand i/3 by Continued Fractions the convergents are "^ = :i, ^* = -, ^ = 5, ^ = -,
' "^ " gil98lg^3 94 4

^ = ,, , ^^ = - - , ^ = ., , etc. The numerators and denominators of the odd convergents, ^' , ^,
q^ IV qfi 15' gy 41' gi ^i

^*, etc., are the values of w and y satisfying (3). If ?*^* , ?^-'*^, ^'^^ be any three consecutive odd
gs ga«+i ga*+« *«+8

convergents, we have the relations pi^+g = Ap^m^ — Psm-i (-*)» g8-»+8 = 4g8M^ — gsM+i (5)-

-129-

lie.— Propoeed by Mabcus Bakeb, U. S. Coast Stin-ey Office, Washington, D. C; and Joseph B. Mott, Neoeho, Mo.
Solve the equation x^ =zai and find the value of x when a •=. 800.

II,— Solntion by Mabcus Bakeb, U. S. Coast Snn'ey Office, Washington, D. C.

Taking Naperlan logarithms,

a?'loga; = l<^a= & and arloga;-4-log(loga;) = log6 = c.
Now put log a? = 1 -}- y, whence x = c»-^» and (1 -f3/)fi^+» + log (1 -f y) = c.

Developing, {l^y)el+1, = \^y^{l^yy^lil^yf^^\^yY^^^{\^yY^., ..

and log(l-}-2/) = 2/— J^4-Jy'— Jj^-f Jj^— ; and developing still further, and reducing to

decimals, we have a3^+&»®-hci/8-f dj^-f cy^-f/^+i/y^-f =c— e=<p

where = 4-6.4365637, 6 = +3.5774227, c = +2.1455212, cf= +0.3163087, e = +0.3025808,
/= — 0.1402390. Reverting this series we have

^ 5 __ e^ps (2&a~ ac)^ __ (d«d-- 5afec-f-5y)<y?* ( 146^ — ^lab^c ^ 6ct«M + SOtfca ~ a''e)q/ > _

^■"a a?"*" a» a^ "*' ' fl» " '

or loga;=l-fy=14-^^-f59»-f C<^-|-i><p4-f-Jg?^4-.... in which ^ = -f- 0.1553624. B = — 0.0134155,
C=-f 0.0010668, D = -f 0.0000109, ^=—0.0000340; or for logarithmic calculation we have

log«= 1-f [9.1913459]^-f-[8.1276080n]<?.«-»-[7.0281006]93-f [5.0378928]<p*+[5.5321611,]^

When aja^=300 we have Nap. log 300 = 5.7037825 = 6, Nap. log6 = 1.7411296 = c, c = 2.7182818284.
^ = c—£ = — 0.9771523, log «p = 9.9899622, and a: = 2.303511.

120.— Proposed by J. J. Stlvesteb, LL. D., F. R. S.. Corresponding member of the Institute of France, Professor of
Mathematics. Johns Hopkins University, Baltimore, Marj-land.

Prove that for all positive values of k less than unity the equation (ar+a)(x4-6) = &(a;-|-c)« has
two real roots.

in.— Solntion by K B. Seitz, Member of the London Mathematical Society, Professor of Mathematics in the North
Missouri State Normal School, Kirksville, MiseK>uri.

Developing and reducing, we have (l--/c)a:«-f (a-f-ft— 2fcc)a;-f a&— fcr» = (1).

Now we must prove that (a+6 — 2*c)2 > 4(1 — fc)(a6— -frc^) for all positive values of k lees than unity.
When fe = 1, we have (a — & y^ -f 4^a — c)( 6 — c) = (a -f 6 — 2c)-, a positive quantity. Hence for all pos-
itive values of k less than unity, we have (a— 6)^ -|-4/c(a— c)(6— c) > (2).

Develophig (2), adding 4a5— 4/ec2— 4fai6-f 4*2c3 to both members, we hnd

(a -f- &)« — 4fcc(a -f ft) + 4/r^<r2 > 4a6 — 4Ac« — 4to6 -f Al^i?,
or (a-f-6— 2to)*>4(l — Jfc)(a6— *(<).

The second solution published in No. 4 of the Visitor, p. 104, is, in my opinion, defective. In the
next to the last line the "therefore ke^ > ah" does not follow. Take a = 4, 6 = 5, c = 6, and A = J ; then
the equation reduces to a?- H- 6^4-4 = 0, whose roots are real, but the last term is positive. I understand
t^iat a. &, and c may have any values whatever.

Mr. NichoW solntion was selected for publication on account of its brevity, and the defect pointed out above by Prof. Seitt
was not noticed. The solutions by Prof. yVood and Mr I'oUard were t>imilar in method, but not defective.

183.— PropoBed by Dr. David 8. Ha.bt, M. A., Stonington, New London County, Connecticut.
To find three whole numbers such that the sum of the squares of any two of them increased by the
product of the same two shall be a rational square.

II.— Solution by Artexas Mabtin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

Let X, y and z be the numbers, and we must find

«'-f-iry-f3/a= D, ar^-f a»-|-aj2= Q, t^-\-yZ'\-^ = D.

Put a«4-aJ3/-f2/« = (?? — i/V, then * = ?'^^5?- Takep = 2, g= 1; then - = ^. Now let 5m = a:,

3m = 2/ and wm = z ; then the other expressions become, after expunging «i«,

25-f-5ir-f«^ = D (1), 9-h3w-hii^ = D (2).

Put (1) = (mc— 5)2 and we obtain w = ^-C^jH: 1) . Substituting in (2) and reducing,

9n^-f 30n8-|-97n«-|-70n-fl9 = D (3).

Putting (3) = (3n2-f5n-f Ja)« we get 3(24— a)n« -|-5( 14— o)n = Jd*— 19 ^4).

-130-

Let6+19 = a,then(4)beoom€e 3(5— 6)n« — 5(6+6)»» = J(6'+386+285). Now take 6 = and we

57 19 264 264fn

havetne quadratic Sn^-— 5n = ^ , which gives n = ■« ; therefore w = -^-, and z = -gg-- Taking

m = 65 we have x = 325, y = 195, z = 264.

m«— Solution by the Pbofobeb.

Denote the numbers by x, y and 2, and the conditions to be satisfied are

iB*H-a^+3/* = a (1), aJ3-»-aa!H-2f» = D (2), yi-f-y8+«8 = D (3).

If X = m'—n^ and y = 2mn+n^ (1) is a square, and if a = ^^^^^ (2) is a square. Substituting

this value of z in (3), we have j^ -f (^P^^)xy + (^M±^^^tjfi = D .

Expanding the above, and arranging the terms according to the powers of p, etc., we have

2^|>^4.2x2ff>»5 + (4aJ*+ajy-2j^)p«^ + (4x*-2a:y)pg8 + (a^^ (4).

Let (4) = («p» +^ + (3^±^2^)g»y . ^^^^ ^^^^^^^ ^^ ^^^ p ^ __(3^%)?.

In order to get a positive value of p, let p = < — - — ^ J **^®° ^y substitution in (4), etc., we have
^^^(^+53/.)^ + «23a.+38.,+59^)<,=t-(??^±^^

Reducinirwefind ^ _ 3<?(5^ + 11^+11^ -hSj/^) ^j,^^^^ (9 a.^-H3a^+ 27a^H-153/')g .

Keduclngwefind (- 2^(7^+60:^+33^) ' ''^®''*''' ^~ 4y(7x«+6x2/+33^)

.-. we may take p = 9a^-f 13xfiy-|-27a^-|-15jf», and ^ = 42/(7ic«4-6ajy+3^).

Now, in the values of x^y m,n may be any numbers if m]> n, whence p, q are known, and then z is foimd.

Let m = 2, n = 1 ; then a; = 3, j/ = 5, p = ™ ; therefore p = 197, g = 190 and z = — qqq~ I ^J^** ^
integers, a; = 2709, y = 4515, a = 110960.

By this general method, any number of answers may be found, but they are all large numbers.

To find smaller numbers, let to = 2, n = 1 ; then a: = 3, y = 5, whence (4) becomes

i:Cp*-|-3()pSg-|-paQ«-|-CpQ'-f 19q*= D (5). This is so when p = ±Jg, ±q, —dq, Letp = Jg-fr

= i(2r-f g) ; then by substitution, etc., we have from (5)

400r* 4- 1280T«g -f- 1336t<gfi + 672rg» -f 441g* = D = (20t« + 32rg + 21g«)*.

14o 17o 264

Reducing we find r = —^ , P = — 22 * '" ^ ~ '^^''* ^ ~ ^^' * ~ "es » ^^* ^ int^ers, x = 195,

2/ = 325, « = 264.

(180y^\ ^ 7^ 9o

21g«-f 16rg4-~ 7 ~ ) » whence r = -^ , p = ^ ; .-. p = 9,

^ 264 , .

g = 4, « = _„ as before.

DO

Letp = r— 3g, then, by substitution, the expression (5) becomes

25r* — 270f«g -f 1081r«g« — 1890rQ» + 1225g* = D = (5t« — 27rg — 35q«)« ;

whence we getr= -r^^, .-. p= -^, p = 31, g = 13, «= - ; or in integers, a? = 264, j^ = 440, « = 325.

Proceeding in a similar way, we may find other sets of numbers, and also by varying m and n,
A eolntion by Seuben Davis will be publiehed in a fntore No.

126.— Proposed by Oblakdo D. Oathout, Read, Clayton County, Iowa.

What is the average thickness of a slab sawed at random from a round log?

II,— Solution by Waltxr Siveblt, Oil City, Venango County, Pennsylvania; and Abtexas Mabtim, TA. A., Member of
the London Mathematical Society, Erie, Erie County, Pennsylvania.

—131-

Let BLG be the flat side of the slab, AL^x^ its thlokneee in the middle.

The average thiokneee = sum of all the ordinates D£ divided by the number
of them. The smn of all the ordinates may be represented by the area of the
oroes-section ABC, and their number by the width BC of the slab.

BG = 2^/{^rx^afi), r being the radius of the log, and the area ABC is easily

found by the rules of Trigonometry to be r«ooe-»(^^^^) — (r— a;V(2ni;— a?*) ;
therefore, when the thickness at the middle is known, the average thickness of

t^coe->(^^) - (r-»)i/(2ne-aJ»)
theslabis o /.o^ ^n •

But when the slab is sawed at random from a given log the thickness x is not known, and may be
anything from to r. Hence the required average thickness is

J^''[f«ooe-»(*'"~") - (r-x)i/(2nr-a^)](fc

"''2i/(2ne-aj«)dr ^^

'0

^0

127«— PropoMd by Bamuxl BoBnm, M . A., F. R. 8., Member of the London Mathematical Socfetf , London, England.

Two random points being taken vrithin a circle (1) on opposite sides of a given diameter, (2) on the
same side, (3^ anywhere ; find in each case the average radius of the concentric circle touched by the
chord through them.

II.— Solation by B. B. Sbits, Member of the London Mathematical Society, Profeseor of Mathematics North MlMOori
Ktate Normal School, KiriuviUe, MisMmri.

Let AB be the given diameter, O the center of the circle, P, Q, two
random points on opposite sides of AB, P, Q' two random points on
the same side, CD the chord through P, Q, or P, Q', and OM the ra-
dius of the concentric circle.

Let OA =: r, DP = «, DQ or DQ' = y, DE = «. /COM = 6, and
/EOM = ip. Then we have OM = roosO, CD = 2rsine and
u = r(sinO — cosOtanip).

An element of the circle at P is rslnB dfjdx; at Q or Q', {x—y)d<pdy,

1. The limits of are and in ; of <p, --0 and 0, and doubled ; of x,
u and 2rsinO = xf ; and of y, and v.

Hence, since the whole number of ways the two points can be taken
is i«*r«, the required average is

*^' = i^ii^'/^X^iT*^^^®^®*^^^*-^^^ = ^ jj^*'/^*J^[rr«-(:r-t«)*]sinOcosM&^

= ^j^*'/j|^\l~co6«0secV)8in«0cos0<»Ap, = ^J'^*'(0~sinOoosO)sln«OoosO(i©, = ^-^.

2. The limits of are and Itc ; from <p := S to <p = B, the limits of x are u and 2rsinO, and those
of y are u and x, and doubled ; and from 4p = to ^ = |;r, the limits of x are and 2rsinO, and those of
y are and x, and doubled. The result of the integration vrith respect to <p must be doubled.

Hence the required average is

= 5^ r*'(2«8ta»6— 39 + 38ln6oo60)sta«9oo86<«. = ^^ — ^-.

07C* JQ 403t* \bX

-132-

3. ThelimitB of 9 are and |;r ; of <p, and ^n ; of x, and 2r8in9 = xf ; and of y, and o^ and doubled.
Hence the required average is

= 3% JTiT^^^^^'^^^' = SiF/o'^^'^^^^^^' =

16r
15ar*

III,--So]ution by Waltsr Siybblt, Oil City, Venango Connty, Fenn«yl7aiiia.

1. Let O be the center of the circle, AB the fixed diameter,
M, K the random points, CD the chord through them inter-
secting AB in E, OH the perpendicular on CD which is the rar
dius of the concentric circle; OA = r, CM = «, DN = y,
/COH = ©, /KOH = <p, J = the required average.

An element at M = rvAnBdBdx, at N = ydydtp. When H is
above AB, CE = r(8in 04-oo86tan<p) = tt,
DK = r(8in8 — cosOtan^) = v ;
when H is below AB, CK = v and DE = u.

The limits of x are and u when H is above AB, and and
V when H is below AB. The limits of ^ are and v when H is
above AB, and and u when H is below. The limits of (p are
and and doubled, and the limits of 6 are and \k and
doubled.

Since the whole number of ways that the two points can be
taken is JjtV*,

"^ J^ J^*'/T/"«'<»8®8toe(ir4-j^tiJkK)80shiOd«l^

= ^^ J*'J*(sin*6cosO-sin«Ocoe»0tanV)^<*^. =^Jj"J*'(Oshi2ecosO-sin50co6«8)<», =|^ — ^.

2. In this case let M, N' or M'. N" be the two random points, both above AB. CN' or ON" = x, MN'
or M'N" = y. OJy = 2r8infi = w.

\ The limits of y are and x ; of x, and v when H Ls below AB, and and u when H is above and D
below, and and w when the chord CD' is wholly above AB ; of <p, and and doubled when the chord
crosses AB, and and «— and not doubled for the chord C'T>' ; of 0, and \ic and doubled.

= 1^ J**'J*(8ln*flco86 + 38to«eco6»9ooe'9>)<»«i9) + ^ J^*'J^~*oo868ln<ed9<l9>.
= ^^ J*'(2«8ln<ecoee-398ln«eco86 + 3sln»9oo8»9)<i9 = ^ - jg^-

3. In this case the number of ways that the two points can be taken is x^,

16r
15* •

-133-

12fl»— Propoeed by Rbubkm Dayxs, Bradford, Stark Coimty, Blinoifl.

It is required to find three positive numbers, suoh that if each be diminished by the cube of their sum
the three remainders wili be rational cubes.

II«— Solatlon by the Pbofosvb.

Let Xt y and « denote the numberd ; then must
X — (x+yH-«)» = a cube .... (1), y — («-|-y4-«)' = a cube .... (2), «— (aj4-y-f-«)5 = a cube .... (3).

Put «-.(«+y4.«)» = ^(«4-yH-«)» (4), y-(x+y+«)» = 5(«+y+«)' (6).

«-(a?+y+«)» = 5(«4-y+«)' (6).

By addition. (aj4-y+«)-3(a;4-y+«)» = ^ +^+^ (a;-f-y4-g)3 (7), and if m'-»-n8 4-|>»= ga

then (aj+y+i;) — 3{aj+y-f «)» = (aj-f y4-«)», whence aj+y+z = J. Substituting this value of a;-fy+«
in (4). (5). (6) we tod .= '^+«'. y = ^. * = ^^.

243 280 341

If m = 3, n = 4, p = 5, then g = 6 and a? = ^^, y = j^^s' * == 1728*

From (7) we obtain (ar+y 4-«)»(^^^^^ + ») = 1- Take m = 3. n = 5, p = 6. g = 2;
then 49(a;+y+«)s = 1, and x-\-y+z = f .

••^"•V g9 y^7»'"392' ^~"\ ^ /^7»""392* \ ^ J^V

32

III«— Solution by Dr. Datid S. Hart, M. A., Stonlngton, New London Coonty, Connectlcat
Let —'J—, - "t— I — t— be the numbers, and put the sum of the numerators,

CI* O* Clip

c^ 1
aE^>^3^+s^, = D, = fl« (1) ; then the sum of the numbers = - = , the cube of which, subtracted

a^ ^ ifi
from each number, gives the cube remainders - , ^, ^.

Let X = ffti0— 1, y = mr— 1, = pto+2 ; then, substituting in (1), we have

(fl,34.wi»-fi^)w^--3(m«4-n«— 2p»)t»» + 3(mH-n+^)w4-9= D = [3 + 4(iii+n4.^)ipp;

this being reduced we have w = v^+^n- P;jJL_^C^.-r^..r:LfgJ^ j^ which expression m, n must be

taken unequal and positive, and p = 0, or any number that will make 2, if negative, < 1, v being positive.

Let m = 1, n = 3, p = ->2 ; then 10 = }, a; = — }, y = i, z = l, and = 2; whence the numbers
7 9^ 16
"^ 64' 64' 64*

Let 111=: 2, n = 4, p = — 3; then 10 = ^, a; = — i, y = i» « = 1, and a = 2 ; whoice the numbers

11 ii ^
^^ 108* 108* 108'

These two answers are the least possible.

If m = 0, then a; = — 1 whatever the value of tr, and we shall have two numbers answering the
conditions.

Let TO = 0, n = 4, p = — 1 ; then to = f, « = —-1, y = |, « = J, a = 3; whence the numbers are
152 91
729* 729*

131«— l*ropofled by Db Yolson Wood, M. A., C. E., Profeaeor of Mathematics and Mechanics, Stevens Institute of Tech-
noio^, Hoboken, Hndson County, New Jersey.

A spherical homogeneous mass m, radius r, contracts by the mutual attraction of its particles to a
radius fir; if the work thus expended be suddenly changed into heat, how many degrees F. vrill the
temperature of the mass be increased, its specific heat being and the heat uniformly disseminated?

n,— Solution by the Proposbb.

Let g' = the acceleration of gravity on sphere m, then, mUiaUy, g^- = the acceleration of any particle
at distance p from the center. During contraction the force varies as the inverse squares ; hence if a; be

1— n
n

-134-

any distance between p and np, we have ^^ = — y' .Va» which integrated between the limits p and np
gives Jtr' = y' - . - "~ . The mass of a spherical shell will be <im = 3m ^J^- , and the work of contract-
ing the sphere will be } j v^dm = 3my'— ""** f p*dp = fmry'

To And ^, consider the earth a homogeneous sphere, its mass M, radius R, gravity at the surf^e g ;
then y' = %w Za9' ^* ^ = Z^-^* '" = V^* «/^ = Joule's mechanical equivalent of heat, d = density of
the earth compared with water. Substituting the value of gf, dividing by J and also by the weight of a
sphere of water equal in volume to that of the g^ven sphere (or - » )> the result will be the number of
d^rees F. that the temperature would be increased if the body was composed of water ; hence for the
given body we have <o = J f • """. Let R = 20893000 feet, the mean radius of the earth, J= 772
foot-pounds, 5 = 5J, and we have to = 89310 ^ . ""**.

Remark,— It Is not necessary, as assumed above, that the particles fall freely ; the result would be the
same if they met with resistances, but in either case the initial and terminal velocities must be the same
in order that the work done by the attraction through the space considered shall aU be changed into heat.

As a practical example, suppose the given body be the earth, contracted from double its present di-
ameter to its present volume, and that its mean specific heat be 4 (or the same as that of iron) ; then /i
= 1, y = 2, « = J, n = J, and we have to = 357240^ F.

133.— Propoeed by Artev as Mabtih, M. A., Member of the London Mathematical Societ}', Erie, Erie County, Pa.

A duck swims across a river a rods wide, always aiming for a point in the bank b rods up stream from
a point opposite the place she started from. The velocity of the current is v miles an hour, and the
duck can swim n miles an hour in still water. Required the equation of the curve the duck describes In
space, and the distance she swims in crossing the river.

III.— Solution by Enoch Beert Seitz, Member of the London Mathematical Society, Profeattor of Matfaematico in the
North MlMonrl State Normal School, Kirkavllle, Adair County, Mlewuri.

Let A be the point from which the duck starts, B the point directly op-
posite A, C the point for which the duck continually aims, P and p two con-
secutive positions of the duck while crossing the river. Draw FQ parallel
to GB, meeting Gp produced in Q, and draw Pe perpendicular to CQ. Then
PQ is the distance the duck would be carried by the water in the time of
swimming from P to p, and Qp is the distance she would swim in still water.

Let AB = o, BC = 6, AC = c, CP = r, /_ PCD = 9. Then Pc = rd%

pezzzdr, Qe = Petane = rtaned9, PQ = PesecO = rsecQdO, Qp = -.PQ

= '''"sec©<«. Buti)c = Qe — Qp; therefore dr = r tan OdO — **'" sec 0d9.

dr fi

Dividing by r, we have — = tanOdO — - secOdS.

Integrating, and observing that when r^a, cos 9 = , we find

<T') - "■o'CCt-odr-"-.)]- " -« = ■■[(-t-)(.ri.)]=-

the polar equation to the curve.
Passing to rectangular oo-ordinates, taking CD for the axis of x, and putting = m, we have

The distance the duck swims is

-135-

136*— Propofled by Abtxmas Mabtin, M. A., Member of the London Mathematical Society, Erie, Eric County, Pa.
Three equal oirclee touch each other externally ; find the average area of all the circles that can be
drawn in liie space enclosed by them.

II«— Solation by the Pboposbb.

Let B be the center of one of the equal given
circles, a = its radius, = BD = BI, a; = IF =
radius of one of the circles drawn in the space
enclosed bv the given circles. The number of
circles, raaius x, that can be drawn in this space

is represented by the curvilinear equilateral tri- ^^^^^^^^^^^^^m^^xm^mi
angle ^^^^^B^^^^HQ^Hv^^T^

Let G = ^FBO = /FBE, <p = /GBD, r == ^^^^^^^m|^Q^^
BF = a+x, tt = area of the curvilinear triangle ^^^^^^^^^ISI^SIHB^
ETG, and J = the average area required.

Then EG = 2rsin6, area of the rectilinear ^^^^^■OfBH^^B^^^BIM
equilateral triangle ETG = rV38in^, sector ^^^|R«S^^8^^^^^Vwl1
£BG = f«6, segment EFGH = 1^0— r^inScosO. ^^I^^^B^^^^^^HlH
.*. M = »V38in«6 + 3r«8ineco66 — 3r«6....(l). ^^ —

Also, ^+6 = ijr...(2). Butcos^=- ...(3);

therefore g> = coe-»("Y and from (2), B = in—tp, = tir—cos-*^**) (4).

sin© = sin(iHr~ip), = Jco6^-Jv'3sinip.==^?-=i^~"^^^^^ (5).

cose = cos(4jr~<p), = Ji/3co6^H-4sin^ = «1^?+_V^(*:^«') (6).

Substituting in (1) the values of 6, sin 6 and cosO obtained from (4), (5) and (6), we have
« == i r2aV3 — *i< — 6ai/(** — «») -H 6*<co6-» (^^^l .

The limits of r are r = BI = o, and r = BO = ia\/3, = r', being the center of the circle inscribed
in the rectilineal equilateral triangle ETG.

J^ 7c{r^a)hidr nj^ r2aV3 — «»* — 6ai/(»* — «*) + 6i^cos-» (^^"1 (r — afdr

Sa^'^ f^ r2aV3 — ?rr» — 6ai/(/«— o^) + 6r^cos-'^"^1 dr

(^ r2dV3 — ;rr^ — 6ai/(r2 — o^) + er^cos-i^-^l {r-^aydr = rio«*(6^ -f 308 — 320i/3 + 207 log3).

J"*^ r2c««i/3 — jfr^ ~ CaV(»*— a?) + 6»<co6-i(^^l cfr = K(^ + ^ — V3 + 3 logS) ;

A . ^^/ 67r-f 308-320 v^3-f 2071og3\
••• ^=i&'^«'(^ ^^.4^61/3 + 310^3 ;•

III«— Solution by Lucius Bbowk, Hndson, Middlesex County, Maaeachaeetts.

In the figure O is the center of one of the given circles, and A is the
center of the enclosed space. ABC is one-sixth of that space, and is
similar to each of the other five^izths.

Let OC = r, OM = aj; then arc MN = ajfj^r — cos-^r*'^!.

Let any point' in MN be taken as the center of a circle (radius /j) ^^^|^9|^^^^^B1fS|
drawn in the enclosed space. Then we have for the limits of x, r and ^P^S^^^HH^^HS
|ri/3 (=ri), and for those of p, and ar— r ( = pi).

Hence the average area of the variable circle is

«g^[i*-co8->Q]p'dr^ _ »j;;-'H^-r)»[K-C06-.(^)]dr

_ / 6)r+ 308 — 3201/3 + 207 log3\

~"*^V « + 4 — 61/3 + 31(^3 )'

-136-

137.— Proposed by B. B. Bbxts. Member of the IxmdoD Hatbematical Society, Professor of BCatheiOAUes in the North
Miseonri State Normal School, Kirksville, Adair County, Miesoori.

Two polntB are taken at random in the surface of a circle, but on opposite sides of a given diameter ;
find (1) tlie chance that the chord drawn through them dcfes not exceed a line of given length, and (2)
the average length of the chord.

II«— Solntion by Waltbb Sivsklt, Oil City, Venango Coonty, Pennsylvania.

1. Let O be the center of the circle, AOB the fixed diameter, M, N
or m, n the random points, CD or od the chord through them Intersectr
ing AB in K or ft, OH or Ok the perpendicular on CD or cd ; OA = r,
CM or cm = «, DM or dm = y, /COH or /cOfc = 6, /KOH or /^JiOh
= <p; 2rB\nfi the given length of the chord, p = required probability,
J = the required average.

An element of the circle at M or m = r sin 6 d'jcLc, at N or n = ydifd<p ;
CK = r(8in -^coBBtang)) = u, cfc = ^(sinO — cos^ tan <p) = v.

The limits of x are and u for CM and and v for cm; of y, and v
for DN and and u for dm ; of g>, and and doubled ; of 0, and fi
and doubled for p, and and {it and doubled for J.

Since the whole number of ways the two points can be taken is ^x^r*,

= -^ r^r*(8ln<e — 8ln»eco8«9tanV)<M<i^ = ^^ /^(flsln^e — 8in»Cco89)<»,
Oor. — When the given line Is equal to the radius of the circle, 2/S = 60°, = Jr, and P = ( « — a ~) •

= ^J^*'J^ (sin*6-.8in3eco82etanV)d©^<P. = ^^f J^^'(O8ins0-sin*Oco8O)de, = ^J.

III«— Solution by Savcbl Robkbts, M. A., F. R. S., Member of the London Mathematical Society, London, England.

1. Let O be the center and AOB the given diameter of the circle
ABBB'. Take a random point P in the Jower semicircle and through
it draw two chords BPB', B'TB'" equal to the given lino and meet-
ing the diameter in 8, 8' respectively.

If 8' lies between 8 and B, the favorable space for the other random
point Q is B8B — B8'B"' ; if otherwise the favorable space is BSB.

Put /BOP = «p, /OPB = 4ar— Q, /BOB' = 2cr, OP = ^, OB =
r, p' =r rcosa. Then if p is the required chance

lic^^p = 2f^^\^f^'^\mii)d<p - £ \BS'B,"')d<p'] pdp.

Jp'Jo

'a-{-9

[a -f — ^ — J sin 2a -f cos^a tan( <p — B)2d<pf}dp

Jo' Jo

a— 9

[a — — ^ — i sin2a + O06>atan(^+6)]d9jy9(^,

2a — sin2a\«

= »«(2a-sin2a)J']ep(^5. = Jr*(2a— sin2a)«, and p = / ^ 8ln2a y

-137-

2. Put OB = re, SQ = y, SP = e, SB = yt, SB' = Zu Z^^ = ^ ^^^ ^^^ ^^ be the required average.
The element at Q is sinBdxdy and that at P is (y'j'tt)dzdB. Then we have

Butyi, — % are the roots of y' + 2i!.»/oo39 — (r'— a:*) = 0, and therefore

138.— Propoeed by Abtemab Martin, M. A., Member of the London BCathematical Society, Erie, Brie County, Pa.
Two sides of a plane triangle are a and b; find its average area.

I,— Solotion by the Pbopo«kr.

Let X = third side, J = average area required ; then ii/[(o-f 6)' — *»]i/[iB«— (a— 6)«] = area of the

triangle, = A, and J = C^ Adx -h C^^dx, = ^ Pt.V[(a+&)» -«»]\/[a^ -(«-&)* ]d^-

Let (a-f6)«-a!» = y^; then « = i/[(a+6)«-y«]. dx = ^j. -^'^f^ ; the limits are y = (\

I^t (aH:6)i = e^. and y = n/(4a6). then J = ^^/^ V(i-e4 "
gV(l ~g^) _ 1 gV( l— e«2^) _ !-.<* «»

(1).

gV (l— c^) _ 2g«— 1 |/(1 — c3g») l~.2(l-He»)8^-f 3g*g« l~g* 1^

l/(l— «^) 3€« ' -/(I— 2;*) *' -v/(l— >)i/(l — C«ar') "^ 3€« V(l— «-)v'(l— <^2*) '

ifi _1 i/(l— «V) 1 1 .

Subetttntlng in (1) we finally obtain

II.— Solntion by Hbnbt Hbatov, B. S., Atlantic, Case Coanty, Iowa.

Let X = the third side. Then the area of any triangle is l\/[(a + by — a;*]i/[a^ — (o — 6)*]. = A, and
the average area, J= f^^^Adr -h C^dx, = j. r""^i/[(a+6)« — aj'VCa?^ — (a-.6)«]di;.

Put (o-f 6)« — aj«= 4a58in«6, and , irra=«*- Then J= . v I ,,, -^i^v

^* l/(l~e«8in«-©) = «»v/(l-'^l«'<>X« + ^(i-^m^,) + M8ineco66i/(l~e*8in-6)].
Then sin«ecos»e = m(l—e2 sin*©) -fn +p[ 1 — 2(1 -fe«)6in*0-f3e»8in*G]; whence, m-^n-^p = 0,
e«m + 2(H-e»)p = -1 and 3e«p = -1. .-. m = ~^^, n = - - ^^""'^ and p = -3^.

•■' ^ = '\i-[(a«+&^)E(e. iit) - (a-6)*F(6. Jir)].

-138-

III.— Solatlon by Chable^ H. Kummbll, Aiwistant Engiueer, U. 8. Lake Survey, Detroit, Michigan.
Let X = third side, A) = average area required ; then if A = area of triangle (a, 6, x) we have

A = il/[(a + 6)« - ar^VCar^ - (a-l^)«] (1).

and the average area

ra+b , /•«+*, 1 /'«+*. , 1 ra+br -| cLb

We have

_4A'

a— 6 idx
8iil)Btituting this into (2) we obtain

^" = li/a-?[-(-^-*^)' + ('^^+*')^]4A ('>•

Place X = — 6co8^ -|- i/(a^ — ft^sin-^) (<p = exterior angle included by b and a;) ; then
, 1. . J 6*8ln«2>co8a>(i<z) hsin <pd<p F .»/»,., t ^ v t. 1 6xsinG>d<z)

4a = 26a? sin <p. We have now, the limits of q> being and it,

-iiai:n-'-''i.j.'rrJS"'*'""''>-.4[''+'-K(»-)-''-'-<(i)]«'>-

Place X = \/{a^-\-b'-\-2cLbcoB2(f/), = \/[(a-f-ft)'' — 4a6slu^<p'] (2<^' = exterior angle included by a
and 6); then dc = — — r^^ , ,.~- . ? . .-ri, 4 A = ^ab sin <p' cos a/ ^ and since the limits are ijr and
we have, Inverting the limits in (4),

^0 = ,'.ft[(«^ + ^')j;*V[(a+6)^ - i«68lnV]d^ - (— '')^Jrv'[(a+6)''^4«68lnV ]]'
_ a-f 6
126

expre<
V, pages 18, 19 and 97--i()0.)

The two expressions (5), (6) are connected by the so-called Landen's substitution. (See AncUtfst, vol.
100.:

139,— Proposed by M. H. Doolittlb, U. S. Coaat Survey Office, Wa^hingtou, D. C.

Suppose Intinitesimal aerolites equally distributed through all space, everywhere moving equally in all
directions with a given uniform and constant absolute velocity. The aggregate mass Intercepted In a
given time by a given stationary sphere is supposed to be known. Determine the effect upon the
eccentricity of a spherical planet of given mass and volume moving in an eccentric orbit, all of whose
elements are given.

Solution by Db Volson Wood, M. A., C. E., Profeseor of Mathematics and Mechanics, Stevens Institute of Technology,
iloboken, Hudson County, New Jersey.

If the velocity of the planet be less than that of the aerolites, the same mass will be intercepted as if
the planet was at rest. Ck)nsider this case.

The change of the elements of the orbit will be due to two causes. Ist. The Increase of the mass of
tlie planet will increase the attractive force between the sun and planet. 2nd. The aerolites will cause
a direct resistance to the motion of the planet.

Let M be the mass of the sun, m' the Initial mass of the planet, s the distance between them, k a con-
stant; then will the acceleration of one body In reference to the other at the end of time I (the time in

the problem being unity) be - — "^ ' - , hence at distance unity // = k(M+m'-\-mt)y and dju =

knuU. For an elliptical orbit, we have from Mechanics Vq^o^ = /ia(l--e-) ^ c, a being the semi-trans-
verse axis. Since the changes are small, consider two quantities only to vary contemporaneously. If

c be constant, we find da = — „- ,. eft; therefore Je = — ,,, — ,, nearly.

-139-

Similarly, if a be constant Ve find de-= -- ~" - <ft ; therefore i^e = .^ ,/ v-. nearly.

Hence, the major axis decreases and thttieQfie9trj^i):y ii^creases with the tlme,'4Midj.th^ii^i|/(^\mt of
change for one revolution may be found by making i equal to the corresponding time.

2d. The law of resistance is not given. The aerolites ISelhg-inQnUesiinaL we do not consider Uie
impact as be^een finite masses, bu^ that they constitute a medium thrbugn which the planet moves.
Considering the medium as of uniform density, D, and the resistance i^ varying with the square of the
velocity, the case comes within one discussed by La Place, ifcoamgtie C(We8w,T[«925], [8926].

The equations will become for this case da-^f=.^^y .;i\~M± = -dh'^K ))^E||^/^^.c^taiit de-
pending upon the form and jjffos of the planet. If = ^^' ^* ^^ iXd^Vr Q » ^ ~ -/a-l-2ff'' 6^ •

oi and ti being initial. The major axis and elccenfcricity both decrease as the vectoKal angle increases,
and the orbit becomes more nearly circular.

The plane of the orbit will not be changed ; and finally, the longitude of the perihelion will not be
changed, Meccmique (J^kMe,, [891^J. ^'

140-— 'Proposed by Abtemab Mabtin, M. A., Member of the London Mathenuitical ^ietfy, Erie, Eric'Coanty,. Pa..
A circle is drawn intersecting a given circlerUs center being at a given d&ltance frqj^.that of the
given cii^c)^ •^H'^ ^^ average area common tj> both circles.

!•— Solntion by HBkbt fisATo^, B. S., Atlantic, Ca»8 County, Iowa.

Let a = radius of given circle, r = radius of ^Variable circle, b = distance between the centers and
2x = length of the.chpr^. Joining the points of "^ntersedtion.

Then z = ^i/(2a*6« + 2a«r* + Wr^ ^a^ — b* — r*)\ a|id> ihj^ area eommon to the two circles is

If 6 < a, the average area Is ' '

Int^rating by parts, | / ^"^ ^ " - . * "' '•!•'♦

Put.^=,^p+t^ir^;,„,0. ailj m^) = . ; then <ir'^ -^"^^^^^l^l"" . -d x = 2a«ln6eo«6.

' ^L J ■'. . ,

"*" 3&(a + 6j Jo L" " >/(l--<-^.8ln*6) ' J '

If /i>a, thelower limit of integration should be 6— a, and , ;». ^' • .„

J = 4;ro(a+6) + ?+*r(6-o)^F(e, J^) - (7a«+6*)E(e, i^r)].
If U.«^ )h,eijj[^^ ««(«- V)--; v4' ., ->•

-140-

II«— Solatlon by Charlk:* H. Kuxmell, Atwistant Engineer, U. S. Lake Survey, Detroit, Michigan.

Let a = radius of given circle, r = radius of variable circle, b = distance between their centers ; then
we have the area common to both circles, whether 6>a, 6 = a or 6<a,

A =.0^-^-+^--)+. .0.-^-1^-)-.^ (1).

where A = area of triangle (6, a, r),

= Jl/[(6-fay -t^VC*-^- (6-a)^]. = Ji/[_(6--a«)-^ + 2(d-^+a3)r*-r«] (2).

We have then the average area common to both circles

A. = ''"

where r' = 6+a and ri = 6— o if &>a; r' = 2a, n = If 6 = o; r'zsa+ft and r, = a~6 if 6<a.
By partial integration we obtain

- r'LrJr, [-i(6'-«')^ + K2«^ + 6')^ - J^]^ (*)•

Substituting this into (4). and placing C= ^^ r"'^*^^^®"*(**~^^~^) "^ **^^^"*(''*~2C^01'^'
(reserving the determination of its special values in the three cases for the present) we have

Place r = '\/{a^ H- &^ H- 2<i5 cos2<p) (2<p = exterior angle of the triangle 6, a, r at the center of the given

. , X ^rx . ..V, ....<.-. ^1. J 4aftsinfl)COSfl>da?

circle) or r = •[{«+6)*-4«68in-^^]; then '^ = -^Ha+b/-iJZ*<p)y

4a = a/[(aH-6)^ — r^]i/[r^ — (o— 6)*] = ^/{4ab) Bin <p.'\/{4ab)coe<p = 4a!) sin ^ cos ^, and the new
limits ai*e ^' = 0, ^ = }ir for each of the three cases. Therefore in (5), after inverting the limits,

hence if 6 > a,

^ = »*+«.+ 't'[<'-'-F.''(-st')-'"'+'->Erd')]-

If 6 = a, Ao = na^ - Va«E*'(l) = a*(jr— V).

-141-

141*— Proposed by J. J. Stlvbbtkr, LL. D., F. R. S., Corresponding Member of the Institnte of France, Profeneor of
MafhematiCH, Johns Hopkins University, Baltimore, Maryland.

A row of particlee in contact are charged some with positive and some with negative electricity.
Under the effect of their mutual actions or the influence of some external body, these electricities are
8uY)]ect to continuom variation so conditioned that when the electricity of any intermediate particle
becomes neutral the electricities of the particles on either side of it are of opposite signs.

Supposing tlie electricities of each of the particles to be known for one moment of time and for some
subsequent moment, show that a certain logical inference can be drawn as to the Joint changes undergone
by any two extreme or any other two particles in the interval, and state what it is.

!•— Solution by the Pbofoskb.

The number of variations of sign in passing from particle to particle is to be constant for the one
moment of time and for the other : the absolute numerical value of the difference between these two
numbers will be equal or inferior to the sum of the number of changes of sign undergone by the elec-
tricities of the two extreme particles, and ^ill differ from it by zero or an even number ; for the change
of sign of any intermediate particle can not alter the number of variations in the chain of particles,
seeing that it can not take place by hypothesis without the two neighboring particlee having at the
moment of such change opposite signs : the total change brought about in the number of variations of
sign between contiguous particles implies therefore the previous occurrence of at least that number of
changes of sign at the extremities of the chain since it Ls only when one or the other of these undergoes
a change of sign that a change in the total number of the variations of signs in the series can take place :
that change may operate to increase or diminish the existing number by one imit, but a number n can
not pass into the number by additions or subtractions of a tmit without the imit being added or subtracted
at least r—n or n—v times, whichever of these numbers is positive. Gall the positive difference 6 ;
the actual operations must evidently be ^ of one kind /i of the same kind, and /i of the opposite kind ;
.'. 6-^2/1 altogether, where // is or any positive integer.

A particular case of this, that in which one of the extreme particles is incapable of undergoing any
change of sign, lies at the foundation of the proof of 8turm*s Theorem.

The above is as purely a logical process of deduction as the so-called Quantification of the Predicate) ;
and in my opinion the number of such forms of logical process is as far as we can say a priori unlimited.

Observe that into the solution of the problem there enters a pure act of Imagination, (that which
consists in raising up the conception of counting the variations of sign from particle to particle,) such
as Kant correctly observes takes place in the construction appertaining to any Geometrical problem or
theorem ; thus, ex. gr., in £uclid*s proof of the exterior angle being equal to the sum of the interior and
opposite angles of a triangle, the drawing of a line parallel to one of the sides is an exercise of the
voluntary or imaginative and not of the reasoning faculty : in this act resides the real stress of the pr(x>f .

Perhaps according to this view it may be said that the greatest mathematician is not so much tlie
closest reasoner as he who carries the most imagination into matters of the reason.

II,— Solution by Waltbb Sivebly, Oil City, Venango County, Pennsyh-ania.

Let -f denote the positive, — the negative and the neutral particlee, and suppose them to be arranged
and change as in the following tables : —

0, +, 0, -, 0. -f. 0, -, (1) 0, +, 0, -, 0. +, 0. -, (1)

-f , 0, -, 0, -f . 0, -, 0, (2) -, 0, -h, 0, -, 0, -f , 0, (2)

0, — , 0, +, 0, ~, 0, -h, (3) 0, -, 0, -I-, 0, — , 0, H-. (3)

-. 0. -f , 0, -, 0, -f , 0, (4) H-, 0. -, 0, -f , 0, -, 0, (4)

0, +, 0, -, 0, +, 0, -, (5) 0. -f , 0, ~, 0, +, 0, -, (5)

Hence if the electricities in all the particles are the same we may know that each particle haa passe. i
through 4n changes.

If the neutrals are the same, and the positives and negatives occupy different positions, they have
passed through 4n-|-2 changes.

If the neutrals are changed to positive or negative we will know whether they have passeil through
4fi -f 1 or 4n + 3 changes if we know which way the neutrals are changing in the given positions.

-142-

148.— Propoeed by £. B. Ssitjb. If ember of the London Mathematical Society, Professor of Mathematics North Mlssoarl
KtateNormal School, Kirksville, Missouri.

Two points are taken at random in the surface of an elUpee, one on each side of the major axis ; find
(1) the average distance between the points, and (2) the average length of the chord drawn through them.

Solution by the PBoroaER.

Let AA' and BB' be the axes of the ellipse, P, Q, the two random points,
and BS the chord through them. Draw CD' and AN parallel to BS, and
CD conjugate to CD', intersecting BS in M.

Let GA = a, GB = 6, CD = a', CD' = &', CM = a;, CN' = a^, EP = y,
EQ = «, BM = M8 = tt. EM = ». / BEA' = ^ D'CA' = ©, /DCA = 4p,

andZCMB = ^. Then we have 8in^=||,. «^=^£^- = -T^'

"^ = -2-^^ ^ = S(«"-^)' ^ = l^-^B* "^ a'sin^=6V(l-e^)ooee.

An element of the ellipse at P is sin^d«%f, and at Q it is {y+tt)dMz. The limits of 6 are and Jsr,
and doubled ; of x, ^xf and xf ; of y, and u+v; and of e, and u-^v. Hence, since the whole number |

of ways the two points can be taken is JxW&s, the average distance between them is ,

J :s=

8

.rxrir"X"^<v+«)'«^***^'

"" 453rti^ Jo L(l— «*ooe"®)' (1— <!»co0»©)»J •

_ 160^15^+1) 86«(15<i»-.l) /l-fe\

The average length of the chord through the points is

— 6^_ r^r 5g»-»i 1— €» lg,«Q^
■^ iSir^oe^Jo L(l— ^"coe'O)* "^ (1— e»cofl»e)»J®^ '

- 8a(17f +_!) . 4^ (17^-1) /l+e\

1472a 832a

Cor. — In the case of the circle, e = and the results reduce to J = io^-i and Ji = ..— ;i.

loojf 4oif*

See solutions of Problems 29 and 137.

-143-

liSv—Propoeed by Abtkmab Mabtin, M. A., Member of the London Mathematical Society, Erie, Brie County, Pa.

A chord is drawn through two points taken at random in the surf^ioe of a circle ; if a second chord be
drawn through two other points taken at random in the surface, find the average area of the quadrilateral
formed by Joining the extremities of the chords.

I*— Solution by Bnoch Bixbt Sxxte. Member of the London Mathematical Society, Professor of Mathematics in the
North Mieeouri Stete Normal School, Kirksville, Adair Connty, Missouri.

Let AB be the chord through M, N, the first two random points, CD
the chord through P, Q, the second two random points, O the center of
the circle, and ABDC the quadrilateral formed. If the chords AB and
CD intersect, the chords AC, AD, BC and BD will form the quadrilateral ;
and if C and D are both in the arc AEB, the chords AB, AD, CD and CB
will form the quadrilateral. Draw OH and OK perpendicular to AB
and CD.

Let OA = r, AM = w, MN = «, CP = y, PQ = «, AB = «, CD = v,
/ AOH = 0, /COK = <p, /KOH = i>, and oo = the angle AB makes
with some fixed line. Then 8 — 2rsin6, v = 2rsin<p; from ^ = to
^ = 6— <p the area of the quadrilateral is

ut = ir>[sin2^ — sin26-f 2cos^8iu(6 — 9)];
from if = 6— <p to ^ = G + 9 it is ti^ = 2r3sinOsin<psin^; and from
if = Q + tp to ^ = ir the area Is Us = ir»[sin2<p-f 8in2e — 2cos^8ln(0-f <p)].

An element of the circle at M is rBinBd&dw ; at N, xdxdia ; at P, rshi <pd<pdy ; and at Q, zdzdif.

The limits of are and \tc ; of <p, and 6, and doubled ; of ^, as stated above, and doubled ; of co,
and 2n ; of ir, and 8 ; of a;, and tr, and doubled ; of y, and v ; and of «, and y, and doubled.

Hence, since the whole number of positions of the four points is jrV», the required average is

~ ^^ /*' Pj^tC^— 2e)sin6cosG + (jr— 2<p)sln^cos<p + 2sin«© -f 2sin«<p]sin*Gsln><iOd(?^da),
= -^^ f*' r*[(3r— 2e)sinecosG -f (;r— 2<p)sin9)Cos<p -f 2sln«e -f 2slD!^<p'\B[n<esin*<pd'}d<p,

= |~^r*'[64(jr©—2©«) sine cos© - 54jrsin«G + 6(7r--2e)(3sin<e + 10sinfie) -f 1056

35r2
+ 216esln«G — lOSsinOcosO — ITSsin'Ocos© — 128 sin* G cos 6] sin* Od!), = ^ .

-144-

II,— Solntlon by Walter Sivbbly, Oil City, Venango Oonnty, Pennsylvania.

Let M, N be the first two random points, AB the chord through them,
P, Q the second two random points, CD the chord through them and O
the center of the circle. Draw OH and OK perpendicular to AB and CD.

Let OA = r, AM = ir, MN = «, CP = y, PQ = «, /AOH = ©,
/ COK = <?>,/ AOC = if and // = the angle AB makes with some
fixed line. Then we have AH = rsin^, and CK = rsin^p.

An element of the circle at M ia r sin BdOdvp ; at N it is d^idx ; at P it
is r Bin gjd(pdy\ and at Q, di/fedz.

When the chords intersect, the area of the quadrilateral is
ir2[8ln^ + sln(2^— ^) + 8in(2©H-^— 2<p) — sin(2e-f.^)] = u;
the limits of 6 are and ln;ot cp,0 and 0, and doubled ; of ^, and 2<p,
and doubled ; of >i, and 2;r ; of 10, and 2rsin6 = 10' ; of x, and tr, and
doubled ; of y, and 2rsin <p = ^ ; of e, and y, and doubled.

When the chords do not intersect (see Fig. on preceding page), the area of the quadrilateral is
ir^[sin2^ — 8in20 + sin(^— 2«p) + 8in(2e— ^)] = «;
the limits of B are and ^r ; of ^, and 6 ; of ^, 2(p and 2G ; of ^, to, x, y, e, the same as above.

Hence, since the number of ways the four points can be taken is jr*?*, the required average area is

+ ^ r'r*f**[8ln2^ — 8ln20 + 8ln(^— 2.p) + 8in(2e— ^fr)]8ln*«8inV<»<M(fr.

= ^^ j^*"/%in«08in69)d6cl^ + ^J^'J^*[(0-^)(sin9CO8<p-sinOco6O)

H- sin^Ocos^^ — 2sin6fcosGsin^co8^ -|- cos* 6 sin^^jsin^Gsin^^ ddei^.

= ^| P'(150 — 8sin*©cos6 — lOsin'OcosO — 158in©cose)sin^ed© + ^^ f' (15 9 — 64 6« sin 6 cos 6

+ 180sin«© + aoecos^e — ^Ssln^ecose + 2sin»ecose + 26sin30cos6 — ISsinOcosO

25f^ lOr^ 35H

H-488in<0coe'e — 12sin3eco60 — 18sin*0coer'»e)sin<0cM, = 9^" + 9^ » = 9,^-

-146-

144.— Proposed by E. B. Skts. Member of the London Mathematical Society, Professor of Mathematics in the North
Missoori State Normal School, Kirks^ille, Miesonri.

A circle is circumscribed about a triangle formed by Joining three points taken at random in tlie
surface of a circumscriptible polygon of n sides; (1) find the chance that the circle lies wholly within
the polygon ; and (2), a second circle being described in the same manner, find the chance that both
circles lie wholly within the polygon and one of the circles is wholly within the other.

Solution by the Pboposxr.

1. Let ABGDEF be the polygon, FOB the triangle formed

by joining the three random points P, O, B, O the center of the

circle circumscribing PGB. Draw the polygon abcdtf , making

its sides parallel to those of the given polygon, and at a distance
from them equal to MP, and draw ONS and MH perpendicular to
AB and PG.

Now while PO is given in length and direction, and the angle PBG
is given, if MP is less than the radius of the inscribed circle of the

given polygon, the area of the polygon abcdef represents the

number of ways the three points can be taken, so that the circle
circumscribing the triangle will lie wholly within the given polygon.

Let PG = 2a?, OS = r, perimeter of ABCDEF =8, area of

segment PBG = <, area of sector PMG = «, area of triangle PMG

= tt. area of polygon ABGDEF. ...=A, £ PMH = 0, ^ = sin-«(-Y
and ^ = the angle which PG makes with some fixed line.
Then we have PM = «cosec6, ON = r--xco6ec©, area abodef. . . . = (r— rcoosecOy-, r = Oa^oosec^©,

u = aJ«oot©, < = r— tt, and de = <l» — (fw = 2a^coseo«e(l — ©cot©)d9.

An element of the polygon at G is 4xd(axf^, or 4rBsin^coS(^d^xi^, and at B it is dt. The limits of x
are and r ; of ^, and \ic ; of 9, <p and ft—q)\ and of ^, and 2^.

Hence, doubling, since B may lie on either side of PG, we have for the required chance.

= ^,J'''aj»drJ'~*(r — a?oosecO)«(l--OcotG)co6eo'»e«»,

= oyfi Jf. [2(jr—2^)sin29)4-4-- 4co64^ — 38in*24poos2^ + 64sin*<5pco6<plogtan4<p]d<p,

Oor. — When the polygon becomes a circle this part of the problem is the same as Question 1843 in the
Educational TmeSt » = ^^rr, and we have p = j, which agrees with the result of the Editor's solution on
pp. 17 and 18, and Stephen Watson's on pp. 95—99, of vol. xii of the Reprint.

2. While the circle M is fixed, the number of ways the second circle can be described so that it will
lie wholly within the circle M, is, according to the cor. of the first part of this solution, lit^coaec^O.

Hence, doubling, to allow for the number of ways the first circle may be wholly within the second,
we have for the required chance

J J afidxl (r-.arco6ece)*(l — ©cotO)co8ec«6dO.

128ir*

= ^^^ r*'[105(jr— 2<p)8in«pcos<p + 210sin3<p + ITOsinV + 608sin«(p + 6326binV

•f 20334 sinio<p . 27648sini3<p -|- 3C750sin>o<pcos^logtani<^]d<p,
— 8jr«r»o ___ 256^
"" 275>' ~ 275«6 •

Oor, — ^When the polygon becomes a circle, we have Pi = f^^.

-146-

145*— Proposed bv Bbmjamik Pkibcx, LL. D., F. R. S., Professor of Hathenuitics, Harvard Univereity, and Conaalting
Geometer to the U. S. Coast Snrvey, Cambridge, Maasachiuetta.

Prove that if two bodies revolve about a center, acted upon by a force proportional to the distance
from the center, and independent of the mass of the attracted body, each will appear to the other to
move in a plane, whatever may be their mutual attraction.

I,— Qnatemion Proof by tbe PROPoexB.

Let p and fh be the vectors of the two bodies referred to the center as origin.

Let p' = Dtp, pI = Dtp,y pf' = Dtp\ pi' = Ap/, i being the time, and D the sign of differentiation.

If 3f is the central force at the unit of distance and N and N, the mutual attractions divided by the
distance apart, we have \

p" = --Mp'^N{p,^p\ p/'==^Mp, + N,{p^p,), p/'^p" = ^M{p,^p)-\-{N,-j^NXp-p% .
SKp-p.W-p/W-p/'n = (-ftf+A'+JV;)5(p-p,)«(/y-p/) =
which proves that the apparent orbit is a plane.

II«— Solution by Ds Volson Wood, M. A., C. E., Professor of Mathematics and Mechanics, Stevens Institute of Tech-
nology, Hoboken, Hndson Coanty, New Jersey.

Take the plane xy in the plane of the central force and the two bodies at any instant, the origin at the
central force and the axis of x passing through the body a ; the co-ordinates of a being a/ and 0, and of
the other body (5), x" and y", d their distance apart; if the central forceat a unites distance, Fthe force
of 6 on a, and F' that of o on 6 (according to the Newtonian \aw F=^ F'); X, X'\- Y\ Y" the axial
accelerations of a and 6.

Then X = -!£«' + if^-~^ , F = 1?^ ; Z" = -ifo" - i?"^^^^, 7" = -ify" - i?"^' ; .

y// — y _ y"

which gives the direction of the relative accelerations, and which is peurallel to the line ah. Hence,
wliatever be the directions of motion of the two >K>dies or their absolute velocities, their relative po-
sitions (a, &), their relative velocities and their relative'accelerations are parallel to a plane, which was
to be proved.

Solved also by WaUer SiveHy.

146.— Proposed by Bev. W. J. Wbioht, M. A., Ph. D., Member of the London Mathematical Society, Jenkintown, Pa.

If Sit be the coefficient of ou in the determinant D = ^ ± anctn Om, and J denote the determinant

S ± SiiS^. . . . 8nH, prove that J = 1>-K

Solation by W. E. Heal, Wheeling, Delaware County. Indiana; A. R. Bums, Ithaca, Tompkins County, New TortLi
H. T. J. LuDwiG, Professor of Mathematics, North Carolma College, Mount Pleasant, N. C; and Waltxb Sivxblt, Oil
City, Venango County, Pennsylvania.

If we multiply the determinants J = 2 ± SuS^i 8nn* D = 2 ± anOn Oim together we get the

determinant S ± CuC^. . c»» in which the constituents are formed by the rule

Cik = SiiOki + ^fitttt -h -h Sinatn.

Now if i = fc, Cflk = D, and If i does not = /c, c« = 0. Thus the determinant 2 ± CnCat — Cnn reduces
to its first element CuCb {^», that is to i)». .•. JD = i)», and J = D^K

147,— Pbizb Pboblex. Proposed by Abtkxas Mabtin, M. A., Member of the London Mathematical Society, Brie, Fa.

A right cone, whose equation is ar>-f^ = ^^ W* ^^^ & paraboloid of revolution, whose equation

is yi-\-g^ =px (2), have their vertices coincident, the axis of the cone being perpendicular to the

axis of the paraboloid. Find the volume common to both by the formula V = jfjdxdydz.
Solution by Walter Siyeblt, Oil City, Venango County, Pennsylvania.
The limits of integration for z are from z = ^v_±Il) to « = \/(px— y»). Eliminating z from (1),

- 14r7-

(2). the Umlte of y are from y = -"*^^f^'^,'"{f - to y = +~ ^|*^"^f ^ The greatest and least values
of X are in the plane a», hence putting y = 0, the limits of x are from a; = to x = c^.

= AP'CCSe' — 2e' + 3) tan-' e — 3e(l — «»)].
Excellent eolndons were also given by Me«8t«. Boieter, Heabm, Kvmmell, UiduHg snd &Wz.

SOLUTION OF "UNSOLVED"
PROBLEM 40.

By E. B. Skitz, Member of the London Mathematical Society, ProfeBBor of Mathematics in the North Miseonri State
Normal School, Kirksville, Adair County, Mieeonri.

40. The first of two casks contains a gallons of wine, and the second 6 gallons of water. From the
first is poured into the se<x»nd as many gallons as it already contains, and then as much is poured from
the second into the first as was leit in the first. How much wine remains in the second cask after n
such operations?

Let 14. = the number of gallons of the mixture in the second cask after n operations, Vn = the strength
of the contents of the second cask, and Wn = strength of the contents of the first cask.

Then a-j'b'^Un = the number of gallons of the mixture in the first cask, and UnVn = the number of
gallons of wine in the second cask.

After the (n4-l)th operation the nimiber of gallons of the mixture in the second cask is

Un+} = 2tt» — (a+6— 2ii„), whence «„+, — 4m, = — (a-f ft) (1),

an equation In Finite Dlff^ences, whose solution gives

Un = 2««C-f i{a-\-b). When n = 1, tti = 4C-f J(aH-6) = 36 — a; .-. (7= |6— Ja,
and t*„ = J6(2-»+» + l)- Ja(2*»-1) (2).

The (n-l-l)th and (n4-2)th operations give the following relations :

Vn+l = KVn + tTn) (3), Wn+l = i(tJ«+iH-lf«) (4), Vn+^l = i(««+I +ir„+i) (5).

From (3), (4) and (5). by eliminating w, and tfn+i, we find 4iJ„+« — 5t?n+i -f Vn = (6).

Solving (6) we find v, = C, + CiHY^ (7). When n = 1, i', = Ci + iC^ = J (8),

and when n = 2, r^ = CI -f ^d = g (9). From (8) and (9) we find Ci = |, and Cl^ = —f ;

". v« = l(l - ^„). and u^Vn = i(l - ^^ [6(2««+i + 1) - a(2*- - 1)] .

-148-

ON EVOLUTES AND INYOLUTES.

By Abtxmas Mabtin, M. A., Member of the London Ifatbemaiical Society, Erie, Erie County, Pennsylvania.

Ip the inU'lnsic equation to a curve be known,
that of the e volute and Involute can be found.

If fj be the radius of curvature of the curve
at the point determined by s and q>, we have

{Todkwnler'B LHff, Cal., Art 324) /:> = ^ .

"Let AP be a curve, BQ the evolute ; let « be
the length of an arc of AP measured from some
fixed point up to P ; s' the length of an arc of
BQ measured from some fixed point up to Q.
It is evident that (p is the same both for s and
8^ if in BQ we measure q> from BA, which is
perpendicular to the straight line from which q) is measured in AP.

In the left-hand figure, sf = ij—C= ^ — a In the right-hand figure, 8f := C — ^z=i C— , ,

Thus if 8 be known in terms of <py we can find s* in terms of (p. The constant C is equal to the value
of fj at the point corresponding to that for which «* = 0". Todkunter'a lid. Col., Art. 114.

Examples. — 1. Find the equation to the evolute of the tractrix.

The intrinsic equation to the tractrix (see Math. Yisitob, 2d edition of No. 1, p. 11) is 8 = a log sec^ ;

da
therefore ^- = atan<p, and «' = otan<^-— C When «' = 0, <p = 0; .-. C=0, and af = atan<p is the

intrinsic equation to the evolute of the tractrix, which is, therefore, the catenary.

2. Bequired the equation to the evolute of the catenary.

da
We have ._ = asec'^ therefore a' = asec*^— C. When «•' = 0, <?> = 0, and C= a; therefore

af = a(sec«^ — 1), = atan*^ <p, is the intrinsic equation to the evolute of the catenary.

/•2asin^<p(i<p __ osin<p i«io<r/^+^^°^\ -u r/ « - f^asin^pd^ 2a

When <p = 0, a? = 0, y = 0, .-. C = and C" = — 2o; hence

asin<z) , , /l-\-8inq)\ 2a ^ 2a , ^/{t^+^o,y) ,

= , y- — ialogC--^ r -). y= 2a. .-. cos<p= .a-, 8ina>=: .^-^ and

1 — sin^^p " **\1 — sin^p/' ^ cos<jp ^ y-|-2a ^ 3/H-2a

*"" 4a"" ~ *"*''^Vy-|-2a-y(y^+4ay)>

the rectangular equation to the evolute of the catenary.
If y = 2a when « = 0, the equation is

da C

To find the intrinsic equation to the invohde of a curve we have ^ = C ±af \ .-. 8 = l(C ± af)d<p.

Exampka. — 1. Determine the equation to the involute of the circle.

The intrinsio equation to the circle is 8^ = a<p; .-. a = uC ± a<p)d(p =: Ctp ± ^aqjf^ + C, = Ja^ if

we suppose a to begin where the involute meets the circle and <p = at that point.
Since the axis of « is a tangent to the involute where it begins, we have

z = rcosS = jcos<pd8 =Ja<pcos<pd^ = a(<psin^-f-co6^) -f C,
y = rsind = (sin g)da = ta^sin <pdg> = a(—<pcosg}-\- sin (p) -\- C,

X =

When ^ = 0, x:=^a, y = 0; therefore C=i> and C = 0.

.-. ^sln^ + ooe^=~ (1), — ^0O6^ + 8in^ = ^ (2).

Adding squares of (1) and (2),

^ + l = ^+V = 5. ^, ^=^-^) (3).

Multiplying (1) by sin <p and (2) by cos^, and taking the difference of the results,

__ r cosQs in y — r sin Q cosy __ rsinX^— 6).
fp _ _^ . ... _ .. ^ ^

.-. sln(y-0) = ^(rL-z^), whence <p-B = sln-i(^^-^) = eoe-iQ.
and therefore by (3)

. = ^•^r.-L) _ eoe-.(^).

the polar equation to the involute of the circle.

2. Bequlred the equation to the imnohOe of the involute of the circle.

The intrinsic equation of the involute of the circle being ef = \aq^, that of the involute of the involute
is a = \\<i^d<p = ^<f^* supposing both involutes to begin when y = ; and, QeneraXtiy, the intrinsic

equation to the nth involute is « = - ^^ /' • — ^f^fp^^*

^ 1.2.3 (n+Yi)

The axis of 2^ is parallel to a tangent to the involute where it begins,

.*. X = rcoeO = Jsin^cfe = JoT^sinydy = ia(--^co6y + 2ysiny + 3co6y) — (1),

y = rsinO = j— coe^cb = \a\—<f^QORqtd<p = ia(—^ sin 9 — 2^0089 + ^8^9) (2)*

Adding squares of (1) and (2),

a*(^ + 4) = 4r-. and 9= (^-^~— ^ (3).

Multiplying (1) by sin <p and (2) by Gm<p, and taking the difference of the results,

rsinycosG — rCosysind •= a<p (4) ;

whence rsin(y — 0) = oy, and y— © = sin-»(^ j (5).

Substituting value of q) from (3),

6 = (*<^-^)* - 8ln-.(*?^''^--«^>)*.

3. Bequired the equation to the involute of the parabola.

The rectangular equation to the parabola being ^ = ipx, its intrinsic equation Is

and that of the involute is

-;f^sS^+/ii'i<^i)<^ w-

supposing a; = and 2^ = when 9 = 0.
From the preceding equations we get

/l-f"SiUfl)\ X psinc y

-160-

Squaring and solvliig we find

SubeUtnUiig In (3),

^x+py±%^

4. Beqolred the equation to the involute of the ellipse.

The rectangular equattim to tiie eUlpee^ the origin being at the vertex, is

^ = V(2a.-a.); ... * = ,^^^S) = ^^^ W'

Prom(l).ar = a±^^-^^^^- = a- -^^^^___^^ taking tiie lower dgn ;

<fe _ a»ft»8 lny Bt™ ^ .^ -^ - ?

•*' dip "" [a» — (a»— 6«)8ln«<p]l* cia; ~ sin^* •*' dg) ~ [a» — (a?— 6»)8in«<p]l "" a(l— ^8in«^)l"

Integrating. i^sln^poos^ 2

the intrinsic equation of the elllpee.
For tiie Involute we have ds = «'(l^ == aE(«. <p)Ap ~ ^®^^^^ (3),

but E(e, <p)d4p does not appear to be Integrable. We can, however, find the polar and rectangular
equations.
Since the axis of a; is a tangent to the involute where it begins,

X = rooeO = Jeoe^A. = aJ(E(e. <p) - ;,'Ji"l^X'^))«»^d^ (*).

y = r9lne=/eln^*, = «/(E(*.^)-^';«^^^j)8in^d^ (5).

Integrating, and observing that when g> = 0, x = a and 2^ = 0,

X = rcosO = asin<p£(e, <p) -\-ac€e<py/{l'^^^n*<p) (6),

y = reinS = — acoe^E(e, <p) + a8in^(l — e^sin*^) (7).

Multiplying (6) by sin ^ and (7) by cos ^ and taking the difference of the results,

;i;8iu^ — ycos^ = rsln^cosO — rsin^cos^ = rsin(^— 6) = aE(e, g}) (8).

Multiplying (6) by cos^ and (7) by sin gy and taking the sum of the products,

:rcos^ + 2^sin9 = rcos(9>— 6) = av^(l — -^sin*^) (9).

From(8), = <p~sin-i(^"^^ ^)). (10).

Squaring (9) we get

af«8infioos6v'(»in*^ — sin*<p) = cp» — »*co8«0 + (f«cos26--aV)sin«^ (11).

Squaring again and solving the resulting equation, we find

_ /«*(r« + a*«') + «^(*'^--2a« — a«e^)cos*0 + af«8in2G\/[r-(l--<«coe^) — ^(1— ^]\i
8ln<p— ^ r*4-o«6« — 2a«r»e*co826 * - - j,

_ . r* _ a^€«(l--f )^- (g¥»(l— gg g ) ~r»(i^— 2a»H -3a'eg)co8^ — af^sin 2ei/[y«(l--^kx)e»e>- fl«(l— ^) 3\i
«oe^--±^ - r*-|-a^ei— 2a*r»e*cos26 )'

Substituting in (8),

/ a«(*< + a«<?») +t5r«-- 2ai — c^)cos^ + oi^ sin 2et/[f^( l— g>c o6«e) — a«(l— e*)]\*
^^ V r« + tt^-2a%^co626 /

/ r*—a^(l-^ ) — a^{ l--2e«) — rHr»-~2a«-|-3a«e8)co6«e — ar^in2 V[y« (l— g»co6»e) —cfl(l^^)J U
±sino^ - - r*+oVj*-2a«r^c8co826 )

«l?r I ./a«(r»+o*c») + **(t«— 2a8— as^)cos«© + aH8in26v^[r»(l— <«co8»e)--d^l--^)]\n
= -Ji^j^CSin (^ ,^+o^«4-2a^e*'cos2e ^j J'

I am indebted to Professor £. B. Ssirz for simplifying the value of sin <p.

-16^-

COMPUTATION OF STURM'S FUl^CTIQNS.

By W. B. HsAL, Marion, Grant County, Indiana.

The following process (which so far as I know has never been published) seems to simplify, to some
eactent, the calculation of Sturm's functions.

Let " f»l8*(a?) = a,aj» + a8a!«-» + a9a5*-« + «MJ»-8+ H-om-i (1).

t^.ix) = ftiaj^i + M»-^ + 63a?^4-ft4aJ»-^+ +{>* (2).

t, (x) = Ciaf^ + Ciia?«-» + cba!'^ + c*aj'^+ +c»_i (3).

be the (m— a)th, (m— l)th and mth Sturmlan functions.
Multiply (2) by an arbitrary function, pa?+g, of the first degree in x, and we get

(jM?+g)f^i{a;) =p6,a- + (i>«^+g6i)af-' + (p63+g6a)«^ + (p64+g65)«"^+ +qbn .. (4).

If we subtract (1) from (4) and determine j>, q so that the two highest terms shall vanish we get (3).

.-. f«(a;) = (56,+i>6,-a3)«— 8 + (g&s+pft4— a4)«"-«+ +(gft«— a«+i) (5).

in which q = ??^ -«^^^ p^^l.

Multiply (5) by 6iB, and we may take

t^{z) = (rft8+«65-6i«a,)«-^ + (rb3+8b^-bi*a^)af^ + + (r6*-6i«an+i) (6),

in which r = a^ — a,t^, a = Oi6i.

Since a positive factor may be suppressed in any of Sturm's functions the las|i function may always be
taken as one of the numbers +1 or —1.

The computation by (6) may be arranged as follows :

Given F(x) ==«< — 6ir»4-&iJ« + 14aj— 4 = 0, to find Sturm's functions.

F\x) = 2jr»-9aJs« + 5a? + 7. n = (— 6)(2) — (1)(— 9) = —3, «, = 1x2 = 2.

f,(a?) = 17aj* — 57a; — 5. r« = (— 9)(17) — (2)(— 57) = —39, 0s = 2x17 = 34.

U{x) = 152a; — 457. ri, = (— 57)(152) — (17X— 457) = —895.

Ux) = 1.
It is unnecessary to find 8 In computing the last function.

Given F{x) = a;* — 8a« + 14a;2 -f 4a; — 8 = 0, to find Sturm's functions.

F'(a;) =aJ»-6a;*-f 7a; + l. n = (-8)(l)-(l)(-6) = -2, a, = 1x1 = 1.

fi(ar) = 5a;« — 17x + 6. r^ = (-6)(5) — (1)(— 17) = —13, a, = 1x5 = 5.

f,(a;) = 76a; — 103. rg = (— 17)(76) — (;5)(-103) = -777.
U{x) = 1.

EXPAKSION OF PQLYl^OMIALS.

By E. P. Thompson, M. B., Elizabeth, Union County, New Jersey.

The present article is intended, not to set forth any new mathematical principles, but to establish
rules whereby polynomials may be easily and rapidly expanded to any power.

The analysis of the method of obtaining these rules may be briefly stated, thus ; — by expanding any
polynomial to any power by means of the Binomial Theorem, the terms may be arranged in such an
order, as to constitute a formula. From an Inspection of this formula a few simple rules are deduced.

According to tlie Binomial Theorem, the expansion of ( C'\'d'\-e-^x)* may be put In the form

indicated on the left of the brackets and when expanded more completely, the formula becomes that
which is shown on the right of the brackets. The latter formula Is the one from which the rules are
obtained.

[....c+ («!+•+«)]- =

ifi{d+e+xy =

-163-

nc'(d+e+x)»-' = •'

-153

«(»-l)0.-2)^^_ ^ n(«-lj. . . .(»-3)^.^^_

?(5^)e.(rf+e+x)- = .( iT2:i.2 ""^^ + 1.2.1.2 *^*^

' 1.2.1.2.3 «"^»"«««" + 1.2.1.2.3 """"^

+ '"r^iii:?:-?-'^'^*^ + '^f.7.y.i:fi72-^'«^' + + »

+ ni -1^.2.I !2.7'-^-^-' + t2-3.\:2ii^-?3-^-^- + + «

11.2.0 1.2.9

n(n-l) ,^(»-3)^^^^ + !K«ril3^. X»-4)^.,.^_

1 . 2.U

+'^-Tl3Ti%-''«'«^-+''r2'^:l:?^^^^^ +»

+ ^"7.fefl:2'^^-^' + ^2:3.^:2:?.?:?-'-"- + + °

n(n-l)^^(n-7) «(n-l). . . .(n-8)^^_.
^1.2.3.1.2.1.2.8*^^'^ +1.2.3.1.2.3.1.2.3*^**^ + +

+

-154-

3c'd«€PaJ0
3c*'d»€0a!0

As might be expected, this formula is long and appears, at first glance, to be entirely without regu-
larity ; but notice nine laws.

(1.) In each bracket, the exponent of x diminishes by 1 from left to right in each horizontal line and
also from top to bottom in each vertical column, while the exponent of e increases by 1 in each hori-
zontal line from left to right and the exponent of d increases by 1 in each column from top to bottom.
Upon the exponents of these three letters (d, e, x) depend the coefficients ; for —

(2.) If the coefficient of any term is multiplied by the exponent of x and the product divided by the
exponent of e Increased by 1, then the quotient is equal to the coefficient of the succeeding term In the
same horizontal line ; while, by dividing the product by the exponent of d iocreased by 1, ttke coefficient
of the succeeding term in the same vertical column is the result. The coefficients of the first term in
each bracket are the same and in the same order as those of a binomial raised to the nth power, t. e.,

n(n-l) n(n-l)(»-.2)
^' **• 1.2 ' 1.2.3 ' ®^-

(3.) The exponents of c are in the first bracket, 1 in the second bracket, 2 in the third, n in

the(n+l)th.

(4.) The exponents in each term siun up to n.

(5.) The first horizontal line in the first bracket is equal to the expansion of the binomial (e+x) to
the nth power ; all the terms in the first bracket are equal to the expansion of the trinonUal (d+«+^) t»
the nth power ; while all the terms taken together represent the expansion of the letranomial (c-^-d-^-e-^x)
to the nth power.

(6.) In any bracket, the coefficients of the first vertical column are the same and in the same order
as those of the first horizontal line. In performing any example, the terms in each bracket will arrange
themselves in the form of concentric triangles. The three sides of each concentric triangle are made
up of terms containing the same coefficients and in the same order. Moreover, the coefficients of the
first column in the second bracket are the same and in the same order as the coefficients of the second
column in the first bracket ; in the same manner, those of the first column in the third bracket are the
same and in the same order as those of the third column iu the first bracket, and those of the first
column in the fourth bracket are the same and in the same order as those of the fourth column in the
first bracket, etc. This law will facilitate the performing of an example, in that, only a few coefficients
need be found by the second law — ^the rest of them are found by copying. This law will be more easily
comprehended by referring to the example as well as to the formula.

(7.) Let e = exponent of «, = (n— e' — c"— -c"')! ^ = exponent of €, c" = exponent of d, ef" = ex-
ponent of c. Then, according to the formula, the general term

„ n(n— l>(n— 2) (n— e'— c"—e'"-hl) ,..^

^"" 1.2.3....e'.1.2.3....c".1.2.3....e'" ^ ^ ^^

"(.^.-IXn-i)- . ,,-.(e+li__ . ...d^.'e..^

~ 1.2.3. ...c'. 1.2.3. ...e'M. 2.3.... e'"

Thus the coefficient of h-fi^e^z^, in the expansion of (6-f c+d+c-fx)*, is equal to ~-^ - = 30.

(8.) The number of terms in an expanded polynomial of t terms is equal to the number of combi-
nations of (n-f t— 1) quantiUee taken (e— 1) at a time. (See my problem, 299, in the Analyet, and its
solutions. Vol. VII. No. 2, p. 98.)

12 11 10 9 8 7

Thus, the number of terms in {a ±h ±c ±d ±e ±xy is equal to y ^ ' ' ^* " = 924.

(9.) The sum of the coefficients, considering them as all having the same sign, is always equal to t" ;
for, if a: = e = d = c = 1 then . . . .(c-f d+c-f a;)" = 4».

The numerical values of the coefficients in the expanded polynomicd cure independent of the signs.

These last two laws, especially, should be applied to an example performed, in order to test the
accuracy of the work.

In performing an example the zero powers of d, e and x should not be omitted, as th^ are necessary
for obtaining some of the coefficients according to law (2.)

-155-

In order to make the manual labor rapid, observe the following outline : (Reference should be made
to the given example.)

(A). Write the product {bcdex) repeatedly, In such order, as taken together, they form sets of con-
centric triangles as represented by the following dots :

(B). Write the exponents of each letter separately, thus; we give x the following exponents in the
order represented :

543210 43210 3210 210

43210 3210 210 10

3210 210 10

210 10

10

Ac. according to law (1) and then write the exponents of the oUier letters separately.

(C). Annex the coeilldents according to laws (2) and (6).

(D). The signs of the terms of the expanded polynomial are determined by the signs of the powers
and products of the positive and negative terms of the unexpanded polynomial, and may be foimd ac-
cording to the two algebraic rules —

(1). The product of like signs gives plus, and of unlike signs, minus.

(2). If the quantity be negative, make the even powers positive and the odd powers negative.

Intheexpanslonof(o — 6-f-c-fd—e—/-fy — aj)», it is evident that the term — 3380ao6«co<f»c«/>yOx« is
negative.

By becoming familiar with the above laws and rules, the expansion of polynomials may be performed
with wonderful rapidity.

THE DAY OF THE WEEK FOR A GIYEN

DATE.

By Frank T. Fbkbland, B. 8., Iiutructor in Mechanics, Uniyj^isity of Pennsylvania, Philadelphia, Pa.

Since January 1st, 1877, is the second day of the week. It will be found that if <*1" be added to a day
of the month, the remainder obtained by dividing this swn by *'7" will be the day of the week. "1**
may be called the ruling number of January, 1877. In general, if 6 =: the ruling number of any month,
M= the day of the month, n = the quotient obtained by dividing by "7**, and w = the day of the week ;
then 6 -f- if = 7n + IT. (I)

Since February 1st is the fifth day of the week, the ruling number of February, 1877, is "4", which is
three more than that of January. Since March 1st is the fifth day of the week, the ruling number of
March, 1877, is "4", three more than that of January, Ac, &c. Therefore, if a represents the ruling
number of January of any year, which may be conceived to be the ruling number of the year, then

b = a in January, a + 1 in May, a -f 5 in September, i

a + 3 in February, a -f 4 in June, a in October, (

a -f 3 in March. a + 6 in July, a -f 3 in November. C ^ ^

a -f 6 in April, a + 2 in August, a -f 5 in December, j

In leap years the ruling numbers of all the months except January and February will be one more..
According to the New-Style Calendar, every year has 365 days, every fourth year 366 days, every one-
hundredth year 365 days and every four-hundredth year 366 days. Now 365 is of the form of 7n -j- 1.
Hence after every completed common year a will be one more, and after every completed leap year two
more. But a = 1 for 1877 is the result of these additions to the ruling number of the year 1, A. D. If
p represents this number, then

p -f 1877 - 1 -f i(1877-l) - Ti5(l&77-.1) + ,i^(1877-l) = 7n + 1,
which gives p = 1.

According to the Old-Style Calendar, evenr year has 365 days and every fourth year 366 days. Jan-
uary 1st, 1877, Old Style, is seventh day. Hence "6" is the Old-Style ruling number of January, lb77.
If g = the Old-style ruling number of 1 A. D., then

g 4. 1877 — 1 -f J(1877 — 1) = 7n -f 6,
which gives 9 = 6.

-156-

Therefore if F = any year and a = the ruling number of that year, theai for New-Style datee

.. r.... . . i^+i(i^-l)-Tiii(I^-l) + ii»(I'-l) = 7n + a, (III)

and for Old-Style dates

I'+5 + i(I'-l) = 7» + a. (IV)

l)ecause the New<Style ruling number of the year F is the result of these additions to **!", the ruling
number of the year 1, A. D., and the Old-Style ruling number of the year F is the result of these ad-
ditions to «*6*\ the Old-Style ruling number of 1 A. B.
In finding the value of the fractions the decimals must be n^lected.

Emmple.-'BuTr fought the duel with Hamilton July 11, 1804. What was the day of the week?

Substituting 1804 for Fin (III) we have

1804 + 450 — 18 + 4 = 2240 = 7n -f a.

Dividing 2240 by 7 the remainder is 0, therefore a = 0.

Substituting in (II) for July we have 6 = -f 6 -f 1 (for leap year) = 7. Substituting in (I) we have
7 4- 11 = 18 = 7n -f W. Dividing 18 by 7 the remainder is 4, therefore W = 4 = Wednesday.
This result may be compared with pp. 347 and 857 of Farton's Burr, Vol. I.
The years in any century in which February has five Sundays are given by the formula

F=4/7n-f l+3(C-l)-3-^^)

in which C = the century. The decimals must be neglected and n so chosen that the year falls wittun
the century.
The years in which February has five of any other day of the weelL are given by the formula

C-l>

'= 4(^7n -h 3(TF-1) + 1 + 3(C-.l) - 3 -^-)

in which W = the given day of the week.

The years next after leap years, or the inauguration years, in which March 4 comes on Sunday are
given by the formula

F

= 4(7n + 1 -h 3(C-1) - 3.^^) -

MfemongtratioH of two JHophanHne Theoremg.

By Aktbmah Mabtim, M. A., Member of the London Mathematical Society, Brie, Pa.

1. The sum of the squares of three consecutive whole numbers can not be a square number.

Any three consecutive whole numbers may be represented by a?— 1, z and «-|-l ; the sum of their
squares is 3ar^ -f 2, which is of the form 3m -f 2.

We must now show that no square number is of the form 3m -f 2. All whole numbers are included in
the forms 3n, 3n + l and 3n-f2; and all square numbers are included in the squares of these forms.

(3n)* = 9n^ = 3(3n*) ; (3n-f 1)« = 9?i?-f 6n-f 1 = 3(3n«-f 2n) + 1, and (3n-f 2)« = 9n« -f 12n + 4
= 3(3n^4-4n-f 1) + 1. AU square numbers are therefore included in the forms 3m, 3m +1, and 3m -f 2
can not be a square number.

Cor, — In a similar manner it may be shown that the sum of the squares of four consecutive numbers
can not be a square.

2. The sum of five consecutive int^ral square numbers can not be a square number.

Let a;— 2, a;— 1, x, x-pl, 2-f-2 be the roots of the squares; then the sum of the squares is

5a^-flO = 5(aJ»+2).

If this is a square the square must be divisible by 5. Put, therefore, 5(a^+2) = ^^ ! then a?^-f2 =
by^, and 3fi-\-2 must also be divisible by 5. All numbers divisible by 5 end with or 5. All square
numbers end with 0, 1, 4, 5, 6 or 9. Hence o^ + 2 ends with 2, 3, 6, 7, 8 or 1 and therefore is never
divisible by 5.

-167-
SotuHmt 0f the ^^ Three-nicety M'robtem.

By E. B. Skits, Member of the London Mathematical Society. Prafefleor of Mathematics in the North Mlnoori State
Normal School. Kirksrille, Adair Coonj^, MiseonrL

If three dice be piled up at random on a horizontal plane, what is the probability that the pUe will
not fkll down?

Let the squares ABCD and EFGH (Fig. 1) be the projections of
the first and second dice on the horizontal plane, and O and M the
centers of the squares. If the sonare EFGH be moved paraUel to
itself about the fixed square ABCD so ns to be exterior to, but in
contact with it, M will describe the octagon KLNPQBST whose
curea r^resents the number of positions the second die can have.

Let AB = 1, and let r^resent the angle A6H which is equal to
the angle formed by AO and EG produced ; then area KLNPQBST
= area abed — 4area KoT = (1 + sinO + cosd)< — 2slnOeosO
= 2(l + sin© + cose).

If ^ be the angle which a side of the third die makes with that of
the second, the number of positions of the third die for each position
of the second will be r^resented by 2(1 -f sin^ + oos ^).

Let N (Fig. 2) be the proJectioQ of the center of the third die.

The pile will not fall down, if M and the middle point of MN are in the square ABCD, and N is in
the square EFGH.

If EFGH be moved parallel to itself so that MN will be constantly bisected by a side of ABCD, M will
move over the line IKL, and the area of the rectangle IKLC will represent the number of favorable
positions.

Let MN = a;. MQ = x", /AME = 0. /MQF = <p, and area
CIKL = u. Then we have ac' = Jcosec^; if ^ is lees than Jjr— ©,
the number of favorable positions is

u = [l-ja:sin(0 + <p)][l-ia:cos(0 + <p)],
and If ^ iB greater tlian })r— 0, the number is

ui = [l-Ja;sin(OH-<p)][l+Jxcos(© + <p)].

The limits of B are and ^jt; those of <p, ^Jt and lie— 6, Jir— G
and })r ; of a;, and ac' ; and of ^, aod J^r.

The results of the integrations with respect to ^ must be mul-
tiplied by 4 to allow for the cases in which N falls on the other
parts of EFGH.

Hence the required probability is

ttx { /rr "-^ +XL r- -^ } <^'^

P'f ** 4(1 + sine -f co66)(l + sin^ -f co6^)<»d0
P'(ir -f 4)(1 -f sine -f cosG)(i9

192(jr+4)«

"*»-*/

r**! r**'~"(96 + iesinG — 16cose — SsinGcose — irslnOcot<p — 16cosecot<p

4-3co626cota>-f 3shieco6ecot*^)cosec«^d<p -f- f^ (96 — IGsinG — 16cose-|-3slnOco8e — irsin6cot<p
+ 16cosecot^— 3cos2ecot^— 3sineco6ecot*<p)co8ec-<pd<p [ d^
= 100/ *iA\« P'(192 — 32co6e — 16secG-f 2co320-f 8ec*G)de

= (,+4).[i + i - W + iVo8(^2_l)].

-158-

PROBLEMS.

231.— Pi^p(M^ by P. F. Manob, Alamm, Sonom, Mexico.

If from any point in a circular arc perpendiculars are drawn to the bounding radii, the distance of
their feet is mvarlable.

232*— I^P<Mod by Hon. Josiab H. Dbummomd, LL. D., Portland, Maine.

From the expressions x*-\'i^-^l = D and x< — ^—- 1 = a And a series of values for x and y in
positive whole numbers, each pair of which shall be the initial terms of an infinite series of values.
Also, the law of the latter series.

233.— Prop<M«<i by Dr. A. J. Puboib, Otselic, Chenango Coanty, New York.

A cone is 30 inches in circumference at tne base and 120 feet in hight. What length of inch ribbon
will wind around the cone, from bottom to top, leaving a space of 5 inches between?

834,— Proposed by Prof. W. P. Casbt, C. E., San Francisco, California.

Let a, 5, c be the three sides of a triangle, r, r", r", r"" the radii of the four circles which touch them ;
prove by a geometrical demonstration that

od -h oc -f- ftc = ly -f rr" -f rr"' -f r'r" H- r'r'" -f r'V".

836.— PropO0ed by Db Vouon Wood, M. A., C. E., Profeenor of Mathematics and Mechanios, Stevens Institute of Tech-
nology, Hobo&en, New Jersey.

A surveyor at latitude 40^ north sights due east along a true level at a vertical staff 3 miles distant;
how far south of the parallel of latitude passing through the position of the surveyor will the foot of the
staff be?

836«- Proposed by Woostbr W. Bbman, Assistant Professor of Mathematics, University of Michigan, Ann Arbor, Mich.
From the equations

tlnd the values of x, y and z by determinants.

837.— I^roposed by D. J. McAdax, M. A., Professor of Mathematics, Washington and Jefferson College, Washington. Pa.

A street bears N. 12^ W. At right angles to the street is a wall 22 feet high and 61 feet long. How
long will the sun shine upon a shelf 4^ feet high, at a point 27 feet from the street and 36 feet from the
wall, on the 22d of December?

838«- Imposed by F. P. Mats, M. A., Professor of Pore and Applied Mathematics, Bowdon (State) College, Bowdon,
Carroir County, Georgia.

At what angle with its axis must a vertical cyllndric corn-stalk be struck with a sharp, wedge-6hai>ed
blade so as to sever the stalk with a blow of minimum force?

239.— Proposed bv Dr. 8. H. Wbioht, M. A., Ph. D., MathemaUcal Editor of the Yatet OowUy ChronkU, Penn Yan.
Yates Coanty, New York.

In what latitude will two stars cross the prime vertical at the same time, their declinations being S
and d', and the difference of right ascension PT

840*— Proposed bv W. W. Johnson, Member of the London Mathematical Society, Professor of Mathematics, St. John'8
CoUege, Annapolis, Maryland.

The center of a circle which passes through a fixed point moves on a given curve ; prove that the
envelope is similar to the pedal of the given curve with reference to the fixM point

Ml*— Proposed by Miss Christinb Ladd, B. A., Baltimore, Maryland.

Given three circles ; four pairs of circles may be drawn tangent to them, and four nairs of circles may
be drawn through their six points of intersection. Show that the product of the lengths of tangents
from the radicfii center of the given circles on any pair of tangent circles is equal to the product of the
lengths of tangents from the same point on any pair of circles through the six (real or imaginary) points
of intersection.

242*— Pi'oposed by De Volson Wood, M. A., C. S., Professor of Mathematics and Mechanics, Stevens Institate of Tech-
nology, Hoboken, New Jersey.

A homogeneous sphere, mass m, radius r, having an angular velocity lo, is dropped ; if the volume
increases at a uniform rate, how far will it have descended when the angular velocity becomes cd'?

243.— Proposed by Rbubbn Datis, Bradford, Stark Coanty, Dlinois.

It is required to find three positive integral numbers, such that the stun of their cubes is a cube, and
also the sum of the cubes of everj- two of them a cube.

244.— Proposed by C. A. O. Rosbll, A. B., Teacher of Qerman and Mathematics in the High School, Beading, Pa.
A cup has the volume v. What are its dimensions when its exterior surface 8 is a minimum?

245.— Proposed by Artemas Mabtxn, M. A., Member of the London Mathematical Society, Brie, Brie Coanty, Pa.
Two rodi^, lengtlis m and n, rest with their ends against each other in a fixed hemispheroidal bowl,
radius of top a and depth b. Find the position of equilibrium.

-169-

240*— Proposed by Charlb? H. Kummeix, Ai*8i8tant Engineer, U. 8. Lake Sun-ej, Detroit, Michigan.
Find the equation and lengtti of the shortest cur\'e connecting two points on the helicoid, or winding-
stair-case su^Cace.

241.— Propoeed by Abtexas Mabtin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

A sportsman shot a duck in the middle of a river 2a yards wide, and his dog started after it when it
was directly opposite him, swimming continually towards the duck. The velocity of the river is v miles
an hour, and the dog can swim m miles an hour in still water. Bequired the equation to the curve the
dog describes in space, the distance he swims to get the duck and the time occupied.

248*— Proposed by E. B. Sbitz, Member of the London Mathematical Society, Professor of Mathematics North Missouri
State Normal School, Kirksville, Missonri.

Find the average area of the triangle formed by joining three points taken at random in the surface
of a parabola whose base is b and altitude h.

249.— Proposed by Abtemas Martin, M. A., Member of the London Mathematical Society, Erie. Erie County, Pa.
A point is taken at random in the surface of a given circle and two random chords drawn through it.
Find the average area of the quadrilateral formed by joining the extremities of the chords.

250*— Proposed by £. B. Sbitz. Member of the London Mathematical Society, Professor of Mathematics in the North
Missouri State Normal School, Kirksville, Missouri.

A plane triangle is formed by joining three points taken at random on the surface of a sphere ; find (1)
the average area of the triangle, and (2) the average area of its inscribed circle.

251»— Proposed by Artemas Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.
Two coins, radii a and 6, are lying on the bottom of a circular box, radius r, >(a -f- 6) ; find the chance
that both are on the same diameter of the box.

252.— Proposed by B. B. Seitz. Member of the London Mathematical Society, Professor of Mathematics North Missouri
State Normal School, Kirkt*ville, Adair County, Missouri.

A triangle is formed by joining three random points within a circle ; find the chance that its circum-
scribing circle is less than the given circle.

253«— Prizk Pboblbb. Proposed by Abtbmas Mabtin, M. A., Member of the London Mathematical Society, Eric, Pa.

A sportsman saw a duck in a river a yards wide which dodged behind a rock directly opposite him, b
yards from the shore; he ran down strt«,m, along the edge of the water, at a speed of m miles an hour,
but the duck kept all the time ''behind the rock" and swam for the "other shore". The velocity of the
river is v miles an hour, and the duck can swim n miles an hour In still water. Require<^l the equation
to the curve the <luck describes in space.

Solutions of these problems should be received by March 1, 1881.

EDITORIAL NOTES.

Professors W. W. Johnson of AnnapoliH, Md.. and E. B. Sbitz of Kirksville, Mo., have been elected Members of the London
Mathematical Society.

The Honorary Degree of Master of Arts has been conferred upon Professor F. P. Matz by Ursinus College.

This No. of the Visitor has been delayed some weeks by the ill health of the Editor, who has done all the type-setting him-
self: the press-work was done at the office of the Erie Gazette.

We hope some of our contributors will take more paint< in the future in writing out their solutions; many of the errors which
disfigure the pases of the present No. are not tvpographical. Read over your solutions several times after copying, and he tmre
they are "O. K.^' before sending them for publication.

Solutions of the problems proposed in No. 4 will be given in No. 6, and solutions of those proposed in this No. will be given
in No. 7.

Contributors will oblige us by sending in their solutions of problems in No. 4 as soon as practicable.

Send subscriptions soon to ARTEMAS MARTIN, Erie, Pa.

NOTICES OF BOOKS AND PERIODICALS.

An Elementary Treatise on Analytie Geometry, Embracing Plane Geometry and an Introduction to Geometr}- of Three Di-
mensions. By Edward A. Bowser, Professor of Mathematics and Engineering in Rutgers College. 12mo. pp. 287. New York:
D. Van Nostrand.

An excellent elementary work, containing a large number of well-selected examples, and very neatly printed.

Mathematical Queffiontf. with their Solutiow. from the ''Edvcational Timee", with many Papers and Solutions not pul>llKhed
in the ^'Educational Times". Edited by W. J. C. Miller, B. A.. Registrar of the General Medical Council. Vol. XXXII. From
July to December, 1879. 8\'o, boards, pp. 112. London: C. F. Hodgson & Son.

This interesting volume contains 4 papers and solutions of 111 problems. Several of the most difficult Probability Holutionn
are by our valued contributor. Professor Seitz.

The "Educational Times" is published monthly: the Remint. half-yearly. Vol. XXXIII is announced as nearly ready.

The Editor of the Visitob can furnish the "Times" at 5J2.00 a year, and the Sej)rint at $1.75 per volume.

H&jp to ika/re and Retain Attmtion. By James L. Hughes, Inspector of Public Schools, Toronto, Canada. 18mo. pp. 80.
Toronto: W. J. Gage & Co.

An admirable little book that should be read by every teacher.

-160-

AtMiican Journal qf Mathematics Pure and Applied. Editor in Chief: J. J. Sylveflter. AMociate Editor in Ctutrgc: William
E. Story. With the Co-operation of Simon Newcomb, H. A. Newton, and H. A. Howland. Pobliahed under the Aospices of the
Johns Hopkiiu Univen*{tv. Baltimore: Printed for the Editon* by John Murphy A Co. Quarterly; 4to, pp. 100. $5.00 a year;
tiingle uumben*, $1.50. Vol. II.

No. I, Vol. II, contains a very interesting and valuable paper on the Paacal Hexagram by Miae Chrietine Ladd. This volume
hIho containri many other articleri of special Interest, but wc have not room to mention even their titles.

The world-wide reputation of the diiitinguii<hed Editors is a sufficient guaranty that the pages of the Journal will always be
filled with matter of the highest order.

Tht Analyst: A Journal of Pure and Applied Mathematics. Edit49d and Published by J. B. Hendricks, M. A., Des Molnea,
Iowa. Bi-monthly; 8vo, pp. 82. $8.00 a year. Vol. VII.

Devoted to valuable mathematical papers and solutions of interesting problems, is ably edited and should be liberally supported.

The School VisUor, Devoted to the Study of Mathematics and English Grammar. Edited by J. S. Royer and Thomas Ewbank.
Monthly; pp. l± $0.60 a year; single numbers, 10 cents. Ansonia, O: Published by John B. Koyer.

This handsome young!(ter increases in interest with its age. The Julv No. contains 7 pages of excellent solutions in elemen-
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matter of general interest to teachers. We wish it abundant success.

The WUtettberger. 10 numbers in a year. $1.00 a year. Springfield, O: Wittenbeig College.

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Barnes' Educational Monthly. New York: A. S. Barnes & Co. $1.50 a year,

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Ten numbers in a year. Contains, besides other matter of general interest, mathematical notes and problems and solutions.

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CORRIGENDA.

No. 2.

Page 38, line *7,/or "4miia" read im*a.

No. 3.

Page 68, solution of Problem 59, line 5,/(W "A =- sin-i(sinwisin^'' read A ^ sin-i(9in.lcosa).

No. 4.

Page 92, solution of Problem 101, the letter S is omitted in the diagram.

Page 108, solution of Problem 187, next to the last line, /tor *V[«*+y* Jftrycos-]" read y[x* \ y* +2arycosr|.

Page 110, solution of Problem l&i, the letter P is omitted in the diagnm.

Page 111, solution of Problem 188, in the denominator of the value of ^ at the end of the last line /or "«'' read v*.

Page 116, Problem 199, Prof. Barbour's address should be Richmond, Madison Co., Ky.

No. 5.

Page laj, second solution of Problem 114, line 3, /br "(8)" read (4), and /or " - Q" read =-m^; line 6, numerator of the
value of X, for "(w- 1)" read {p 1).

Page 187, line 19, the subscript 8's should be 9's; in the values of the edges of the aoist solid insert a "4'' between the 88d and
Mth figures.

Page 188, in the value of the 901st diagonal insert an "8'' between the 38d and 88d figures from the end of the line.

Page 189, solution of Problem 116. line 9. /or '•6*«'' read bit>^; and /or "80o*c«" read 8a«c«.

Page 188, second solution of Problem 189, line 2, for "ar*+y*+«*, ^ " read x*-{-y* i-«' f 8, = .

Page 187, line 5, after "cW" insert "djr".

Page 138, Une 7, denominator of last integral, for ''dx'' read A; last line, ftir "A«'' read Aa.

Page 148, line 4, for "CN'" read CN. The letters D and 8 are omitted in the diagram, and "Q" is put for Q.

Page 148, "E'' is omitted in the diagram.

l^age 144, 6th integration, in the denominator of the first integral, far "v-*'' read «>, and in the denominator of the second

integral for ''r*'' read «>; in the last integration, first integral, the factor sin^ outside the parenthesis should be sin*^. For
the second integral substitute the following:

J (336 — 54G*8ln9co80 + nOco&'B - 488iii'eoo86 — 28iii*eeo8e + 26sin3 9coeO

— ISslnOcoeQ — 488in*9co8»9 — 60&\n^BcoePe — 908lneco636)8ln*©(«.
Page 145, at the end of line 8 of the solution insert * Inscribed circle of the polygon, and M the center of the''.
Page 147^ line 7, /or " A J»[*/K1 -«')-*«]" read ^p[ip{l~^) -ix]y(p+x)V(^^P-x).
Page 168, first line below the second brace, insert ''+'' before the last term.
Page 154, line 8 from the bottom, after e insert "±/'\
Page 156, line 25, /or " -8880" read -8880.

Ko/r/.

JAVUAftr. 1881.

l-Ff]l5lB83

Uo. 6.

THE

I t

EDITED AND PUBLISHED DY

ARTEMAS MARTIN, M. A.,

Member of the London Mathematical Society.

ttBUSB sEiu-ANmrALinr. tbrhs: onb j>oiaJukb a ybab in advanob.

snrOltB NUHBBBfl, FIFTY OBNTS BiACH^ BACK NUMBBBS 8VFPI<IBD AT 7HB flAMB BATB.

^

1881.

NBW MATHEUATJCAL TEXT BOOKS.

MATHEMATICAL BOOKS

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GEOMETRY OF THREE DIMEmOHS.

BY EDWARD A. BOWSER,

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OONTfeNTS,

Pakv 1.— Amalttic OaomTBT of Two Dihenbions: V
Chap. 1— The Point. Chap. 0— The Ellipse.

S^The Right Line. " 7— The Hyperbola.

** d— Transformation of '* S— Cleneral Bqaafton of

Co-ordinates. '* the )id Degree.

•* 4— The Circle. . " 9>-Higher Plane Curves.

" 5— The Parabola.

Pakt 2.— Analytic OaoMCTBT oy Taim DixBRSiom:

Okap. 1— The Point. Cbap. »— The PUme.

** »~The Right line. " i-^urfaces of Jlevolu-

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PART I.-DIFFKRKNT1AL OALCULUS!

L

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1 Diflerentiatum of Algebraic Sb Transcendental Functions.

8. Limits— Detived FuucUons.

4. Successive Differentials and Derivatives.

5. Development of Fhnctinns.

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10. Direction of Curvature— Singolar Points— Tracing of

Curves.

1 1 . Radint» of Curvature— B volutes and In volutes^En velope«.

PART IL-INTBGRAL CALCCLU8.

1. Blementaiy Forms of Integration.

2. Integration of Rational Fractions.

8. Integration of Irrational Functions by Rationalisation,
i. Integration by Successive Reduction.

5. - Integration by Scries— Successive Integration— Integration

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FED15188J

THE

/

MATHEMATICAL VISITOR.

[IHTIIIKD ACOOWINO TO ACT OP CONORUS. 1

N THI TiAR ISM. •» AUrEMAI HARTIM, H. A., IN THI OfflOt OF TMt UMUUAN W OONORIt*. AT WAMINOTOH.]

Vol. 1.

JANUARY. 1881. No. 6.

JUNIOR DEPARTMENT.

SoiutioHS of the Probiemg Proposed in J^. 4.

14&— Proposed by Abtmmab Mabtin, M. A., Member of the London Mathematical Society, Brie, Brie County, Fa.
A man bought a horse for $100, and sold him for $60, and then bought him back for $75. How much
did he lose by the transaction?

I«— €k>lotion by J. M. Quraoz, Alamos, Sonora, Mexico; Jossfh Tubrbull, East liyeipool, Ohio; Bdwin Fulcb, Cln-
dnnatna, New York; and J. W. Donovan, Anaonia, Ohio.

If he sold a horse which cost him $100 for $60, the man lost $40. By buying the same horse again for
$75 he gained $25. Therefore the difference between $10 and $25, or $15, is what he lost.

a~Solntion by Dr. David S. Habt, M. A., Stonington, Connecdcat; and L. P. Shidt, U. S. Coast Survey Office,
ngton, D. C.

If the man had repurchased the horse at the same price for which he sold it, he would have lost
nothing. Therefore he lost the difTerence between the two prices ; that is, 75 — 60 = 15 dollars, his loss.

m^^Solntion by Professor W. P. Cabbt, C. B., Ban Frandsco, California.

As the man had to add only $15 to the $60 to make the second purchase, he therfore lost only $15.

IV^-Solndon by Oblando D. Oathout, Luana, Iowa; and Hon. J. H. Dbummons, LL. D., Portland, Maine.

Having the horse, no matter what it cost, the man sold it for $60, and bought it back for $75 ; therefore
he lost $75 — $60 = $15.

V*— Holntion by H. T. J. Lunwio, Professor of Mathematics, North Carolina College, Moont Pleasant, North Carolina.

By selling and then buying back again the horse stands him 100 — 60 -f 75 = 115 dollars ; 115 — 100
= 15 dollars, his loss.

Good solndons from Messrs. Brown, Clark, Hoover, NichoU, JMktrd, Sylvetter RotAnt, C. C. RotAnt, BotOl and Skceri^.

149«— Proposed by John I. Clabk, Moran, Clinton County, Indiana. *

My grocer sold me a cheese, which he said weighed 32 pounds ; but when it was placed on the other
side of the scales, it only weighed 18 pounds. He then proposed that I sliould buy another of the same
size, and weigh it on the opposite side from the ilrst, to which I consented. Did I gain or lose by the
transaction? And what was the true weight of the cheese?

Solution by A. E. Hatnks, Professor of Mathematics, Hillsdale College, Hillsdale, Michigan; C. A. O. Bosbll, A. B.,
Teacher of Mathematics at the Carroll Institote, Reading. Pa.; L. P. Sbivt. U. S. Coast Sur\'ey Office, Washington, D. C;
J. M. Tatlob, Milton, Oregon; Chablbs Gilpin, Jr., Philadelphia, Pa.; and C. C. Robins, Princeton, New Jersey.

Let X = the true weight of the cheese, a and h the false weights, and c and e the arms of the scales.

Then, from the principles of Mechanics, we have the equations : cx = be (1), ex = cu: (2).

MulUplylng (1) by (2), ceafi = abce ; .-. sfi = ab, and « = i/(a6).

-162-

Therefore the true weight Is a mean proportional between the false weights', and the weight of each
cheese was 24 lbs. ; and since the purchaser paid for 3*2 -f 18 = 50 lbs. and obtained only 24x2 = 48 lbs.,
he lost the price of 2 lbs.

The problem was aleo solved by Messn. Baaot, Oatev, Clart, Donovan, HawUy, Hoover^ LuduAg^ Maiz^ NkhoU^ (kUhomt^
PoUard, 3ylve$Ur Bobins, ;MU and Sit^rfy,

150«— Proposed by W. S. Himklx, Leaoock, Lancaster Ckmnty, Pennsylvania.

How much more can a bank make in 528 days, with $10000, by discounting notes on 45 days time, than
by discounting them on 30 days, the rate of discount being 6 per cent., and the profits in both cases
retained in bank till the expiration of the time?

Solution by Juuak A. Pollarb, Qo«hen, N. T.; C. A. O. Bosbll, B. A., Teacher of Mathematics at the Carroll Institate,
Reading, Fa.; ^ylvesteb Bobims, North Branch Depot, N. J.; and C. C. Bobins, Princeton, N. J.

By discounting notes on 45 days time, the bank could make 528 -i- (45 -f 3) = 11 transactions ; and by
discounting them on 30 days time, 528 -r (30 -f 3) = 16 transactions.
In the first case the discount, at 6 per cent, per annum, is $0.06 x ^ = $0,008 on the dollar. The

profit of each transaction would be j^^^ — ^10000 = $90.6451612^+, and of the 11, 11 times as
much, or $887.096774.
In the second case, the discount is $0.06 x ^ = $0.0055 on the dollar. The profit of each transaction

SI 0000
would be **7^^, — $10000 = $55.3041729+, and on the 16, $55.3041729 X 16 = $884.866766.

$887.096774 — $384.866766 = $2.23, what the bank would gain by discounting on 45 days time.
Solved In a similar manner by Messrc. Clark, Donovan, Drummond, Matt, Seitz and Siveriy.

151.— Proposed by Q. F. Mkad, Uniontown, Fayette County, Pennsylvania.

If the sides of any quadrilateral be bisected, and the points of bisection joined, the resulting figure
will be a parallelogram and equal in area to half the quadrilateral.

Solution by J. M. Quiroe, Alamos. Sonora, Mexico; William Hootkr, Soperintendcnt of Schools, Wapakoneta, Ohio;
C. A. O. BosKLL, B. a.. Teacher of Mathematics at the Carroll Ini*titute, Beading, Pa.; Prof. W. P. Caset, C. E., San
Francisco, California; P. F. Mamoe, Alamos, Sonora, Mexico; Sylvkstbr Bobims; Julian A. Pollard; Edwin Place;
and Orlando D. Oathout.

Let ABGD be any quadrilateral, £, F, G, H the middle
points of the sides. Join £ and F, £ and O, F and H, and O
and H. Draw the diagonals AD, BC,

In the triangle ABD the sides AB, BD are bisected in F
and H, therefore FH is parallel to AD ; in the triangle ACD
the sides AC, CD are bisected in £ and O, therefore £0 i^
parallel to AD, and also parallel to FH.

In like manner it is shown that £F is pcurallel to GH, and
therefore £FOH is a parallelogram.

Draw AL and DM perpendicular to CB. Area of triangle
ACB = AL X iBC, area of parallelogram £FIJ = KL x EF
aEJULL X JBC ; therefore parallelogram £FIJ = | of triangle

Area of triangle DCB = DM X ^BC, area of parallelogram GHIJ = MN X GH = IDM x IBC ; therefore
parallelogram GHIJ = | of triangle DCB.
Adding, we get, parallelogram £FGH = J of quadrilateral ABCD.

Solved al;o bv Messrs. Clark. Donovan, Drummond, Hart, Hawley, Ltidwig, MaU, NiehoU, Putnam, C. C. Bobint,
Sch^er, SHU, Siverly and TumbuU.

162«— Proposed by Artbxas Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.

A company of n men were counting their money. The first said to the second, "Give me your money
and I will have $a" ; the second said to the third, "Give me } of yours and I will have $a ; the third said

to the fourth, "Give me ) of yours and I will have $a" ; the nth said to the first, "Give me one-nth

of yours and I will have $a". What sum had each?

I,— Solution by Hon. Joslah H. DRvmiOND, LL. D.. Portland, Maine; L. P. Shidt, U. 8. Coast Survey Office, Wash-
ington, D. C; and Sylvester Robins, North Branch Depot, N. J.

Let X = amount the first man had ; then a~ x = amount second had, 2[a — (a — x)"] = 2x = amount
third had, 3(a — 2aj) = 3a — 2.3a; = amount fourth had, 4[a — 3(a — 2aj)] = 4o — 3.4a -f 2.3.4c = what

-163-

fifth had, 5[a — (4a— 3.4a+2.3.4e)] = 5a — 4.5a + 3.4.5a — 2.3.4.5a; = amount sixth had ; from which
it is evident that the nth man had

(n— l)a — (n— l)(n— 2)a + (n— l)(n— 2)(n— 3)a— ± 1.2.3.4. . . .(n— l)a?,

which by the terms of the question = a — . Reducing this equation we readily find

_ aln — n(n — 1) + n (n — l)(n— 2) — n(n— l)(n— 2)(n---3) + ± 3.4.5. . . .n]

^ — - - I 4. i. 2.3.4... '.n

The upper signs obtain when n is odd and the lower when n is even.

•

II.— Solution by Lucius Bbown, Hudson, Masftachusetts; J. F. W. Schetper, Professor of Mathematics and German,
Mercersbur^ College, Mercersburg. Pa.; C. A. O. Rosell, B. A., Teacher of Mathematics at the Carroll Intftitntc, Reading,
Pa.; and Walter 8. Nichols, I^itor Insurance Monitor, New York, N. Y.

Let Xu Xi, x^, Xn denote the several sums. Then by the given conditions we have

Xx-\-x^i = a (1), a^H-Jaj3 = a (2), X3H-Jj?4 = a (3), XA'\-\xt, = a (4),

Xn-\ -h ^ ~^Xn = a (n— 1), a^ + ^x, = a (n).

Multiply the first, second, third, fourth, (n— l)th equations byl, —1, H-r-;;, — ioq»

± 1 o~o / o\' ^^^ *^^ ^^^^ results, and we get

Xi '\-Xi-X2 — lXi + ix:, + ix^^ix4-^\xs'\- ± ^ 2 jj .(n— 2)('"-' "^n-l^) ^

% "^ 1.2 1.2.3 "^ i.2.3.4 1.2:3.4.5 "^ i 1:2.3. . . .(n— 2)7 '

^»± 1.2.3.. ''"(^r.-l) = «^'-' (^+^>'

where Sa-i is put for the series on the right, which is the reciprocal of the base of the Naperian system
of logarithms to n— 1 terms.

« /N ^/ .IN Alt u A tt[(1.2.3....n)5f„_i ±n] a(1.2.3. . . .n)(n— 5„_,)

From (n) and (n+1) we readily find x, = ^ ^.3. . . .n ± 1 ' ^- = n(1.2.3. . .n ± 1) '

III.— Solution by Walter Hiverly. Oil City, Venango County, Pennsylvania; and William Hoover, Superintendent
of Schools, Wapakoneta, Auglaize County, Ohio.

Let til, Uj, ttj, tt„ Un be the sums the first, second, third, xth, and nth

persons have respectively.

Then u, -|- '"^^ = a, which gives the difference equation

tt,+, + xu, = ax (1).

Integrating,

«. = (-l)-i[(1.2.3....x-l)(C-;« + ^2«-l.L3«+ •• +(~^)^S.2.3".TI-1«)]

= (-l)-i[(1.^.3....x-l)(c~;a+;a..^V + ii3«- ± 1.2.3: .'. .x-2«)]

= (--l)'-'a.2.3....x-l)(C-aS,_,).
When X = I, C = Wi ;
.-. «, = (-l)'-H1.2.3....x-l)(M,-a^^._,) (2), u« = (-lr-i(1.2.3....n-l)(u,~a^\_,)....(3).

By the problem, u^^^^ = a (4). Eliminating u„ from (3), (4), U| = "'■^^' ^'2 3* ' ' ^^^'""'i ~ ""^ •

Substituting this value of U| in (2) and (3) we gc't

«-n/19<l ^ ../ (1.2.3.. nVS._,-Ln x «(1.2.3. . . .n)(n - Sf„-,)

Solved alM> in an elegant manner by Met«t<n<. Cawi/j Uoftoran, JIaU, Bollard and Putnam.

-164r-

163«— Proposed by Hyiykbtvb, Bqbins, North Branch Depot, Somerset Coanty, New Jersey.

What are the first ten numbers which, when multiplied by 1, 2, 3, 4, etc., will give products containing
the same figures, and in the same order though banning at a different digit, but when multiplied by 7,
17, — , — , etc., respectively, will give products containing all nines?

Solution by the Pboposbb.

Dividing a number consisting of 9's by 7 we find the quotient terminates with six places, and is 142857.
This is the first number, aiid when multiplied by 1, 2, 3, 4, 5 and 6, respectively, gives products con-
taining the same figures, and in the same order but beginning at different digits. The second number
is obtained by dividing a number composed of 9*s by 17. This time the division terminates with 16th 9
used, giving as a quotient 588235294117647. Multipiiftd by the numbers 1 to 16 inclusive, the several
products are seen to observe the same law aj9 before.

*' Every prime number except 2 and 5 toiU exactly divide the number expressed by as many 9'« as there are
units, lss8 one, in the divisor.*' — Leqendbe.

When the division does not terminate before reaching the last of the dividend there will stand In the
middle place or places of the quotient aj9 many 9*6 as there are figures in the divisor, less 1 ; and all the
figures in the latter part of the quotient may be found by subtracting those in the former half thereof
successively from 9's ; and the entire quotient will possess the remarkable property of producing other
numbers containing the same figures as itself, in exactly the same order although beginning at a different
digit, when multiplied by any number less than said prime divisor.

The other numbers result from the division of numbers composed of 9's by 19, 23, 29, 47, 59, 61, 97,
109, 113, 131, 149, Ac., and are:

3. 52631578947368421; 4. 434782608695652173913; 5. 344827586206896551724137931;

6. 212765957446808510638297872340425531914893617 ;

I. 169491525423728813559322033898305084745762711864405779661 ;

8. 16393442622950319672131147540983606557377049180327868852459 ;

9. 103D9278350515463917525773195876288659793814432939690721649484536082474226804123711340206185567;

10. 91743119266055045871559633027522935779816513761467889J08256880733944954128440366972477064220183
48623853211 ;

II, 88495575221238938053097345132743362831858407079646017699115044247787610619469026548672566371681
415929203539523 ;

12, 76335S77862595419847328244274809160305343511450331679389312977099236641221374045801526717557251
908396946564885496183206106870229 ;

13. 67114093959731543624161073825503355704697986577181208053691275167785234899323859060402684563758
389261744966442953020134228187919463087248322147651 ; etc. ; etc.

Excellent solutions received from Professor Sch^fer, Professor Botett and Walter Siveriy.

Ift4«— Pn>poi*ed by Joskph Fickltk, M. A., Ph. D., Professor of Mathematics and Astronomy, University of the State
of MioHoari, Columbia, Boone County, Missouri.

The area of a triangle ABC is h ; the side AB is a, and the angle opposite AB is fi. Required the sides
AC and BC.

Solution by J. F. W. Schevter, Profcwor of Mathematic<« and German, Mercersbnrg College, Mercersburg, Franklin Co.»
Pa.; L. P. SuiDT, U. S. Coaut Survey Office, Wa^hint^ton, D. C; and Walter Sivbrlt, Oil City» Venango Co., Pa.

Denoting the side AC by x and BC by y, we have

xysinfi = 2ft (1), a:« + 3^ — irycos/J = a« (2).

Substituting xy = 2bcoaec/3 in (2), we have

j4-f i/« = a«-f 4ftcoty5.
Adding 2xy = 46 cosec /S to, and also subtracting it from, the last equation we obtain

x-^y = V(a* -f 4fecot J/?), x — y = -i/(a» — 46tan J/?) ;

.-. X = J;/(a« + 46cotl/5) -f J;/(a« — 46tanJyJ?), y = J;/(a«-f 46cotiytf) — li/(a*~4&tanl/J).

Solved aUo by Messrs. Banks, Casey, Drummond, Hart, Uower, Ludwig, Jfatz, OaUumt, Putnam, Ra&eU, C, C, Bobins^
Seilz and TumbuU.

-165-

15ft.— Propoeed by G. F. Mbad, Uniontown, Fayette Coanty, PeniwylTania.

If from any point in a diagonal of a parallelogram lines be drawn to the opposite angles the parallel-
ogram will be divided into two pairs of equivalent triangles.

Solution by P. F. Manov, Alamofs, Sonora, Mexicio; E. J. BDMinrDs, B. S., Ancien Elere de TEcole Polytechnlqne, Pro-
feseeor de mnsais d'Anglaitf et de Mathematiqaeis ParU, France; H. T. J. Ludwio, Profeeaor of Mathematics. North
Carolina College, Mount Pleaaant, N. C; Prof. W. P. Cabbt, C. E., San Francisco, California; Edwim Plack, Cincinnatns,
N. T.; OsoBGB Hawuet, San Francisco, California; and K. S. Pctmam, Rome, N. T.

Let ABGD be a parallelogram, AG a diagonal and P any point in It. Join
PB and PD. Draw BH and DE perpendicular to AC. As the triangles
ABC, ADC are halves of the same parallelogram, they are equal in all their
parts ; therefore BH = DK. The triangles APB and APD have the same
base and equal altitudes, therefore they are equivalent. By the same
method of reasoning the triangles CPB, CPD are shown to be equivalent.

Similar solntions given by Mewrs. Bimkt, Clark, Drummond^ Hart, J7ayfM», Hoover,
MaU, Oaihtmt, liUlard, Quiroz, SylveUer JSobku, C. C. Bobint, BomU, Sch^er, SHtg,
SMdf and Siverly.

Ift6«— Proposed by Thbodork L. DxLand, Office of the Secretary of the Treasnry, Washington, D. C.

A owes B $1750, payable as follows : 70 notes of $25 each, the first payable in one month, the second
in two, and so on to the last which is payable in seventy months ; each bears simple interest at 10 per
cent, per annum payable with the note. When will he neither gain nor lose by borrowing money at 8
per cent, per annum simple interest to pay all the unpaid notes and interest, and give new notes, interest
due added, payable, the first one month from the date of the change, the second two months, and the last
on the date when the original notes terminated?

I«— Solution by J. W. Donovan, Ansonia, Darke Coonty, Ohio; and J. A. Pollard, Ooshen, Orange County, N. Y.

The interest on $25 for one year at 10 per cent, is $2.50. The notes must run a sufficient length of
time that the interest at 8 per cent on the amount will be $2.50. $2.50 h- 0.08 = $31.25, principal at 8
per cent. $31.25 — $25 = $6.25, interest accrued at 10 per cent. If it requires one year to gain $2.50
interest, to gain $6.25 will require 2 years and 6 months, or 30 months, from date of first notes.

II«— Solution by John L Clark, Moran, Clinton County, Indiana.

The last note, which will be due in 70 months, is sufficient for our purpose. If we find the time at
which we can borrow money at 8 per cent, and pay the last note of $25 with interest, and neither gain
nor lose, it will be the time for all the other notes.

We first want to find the amount of money that will accumulate the same amount of interest at 8 per
cent, ^at $25 will at 10 per cent, in the same time, thus :

8 : 10 :: $25 : $31.25.

Now, if $25 will amount to $31.25 at 10 per cent, in a certain time, then that is the time we wish to
find. $25 will gain $2.50 in 12 months at 10 per cent. ; then

$2.50 : $6.25 :: 12 months : 30 months,
which is the time required.

Solred algebraically by the Propom; and Messrs. Ckuey, Putnam, SomU, 8eUt and Siverly.

157.— Proposed by Stltkstsr Robins, North Branch Depot, Somerset County, New Jersey.

It is reqidred to prove that a polynomial of n terms may be found which can be divided by a poly-
nomial of n terms and give a quotient of n terms, the coefficient of every term in all of them being unity.

Solution by F. P. Matz, M. A., Professor of Mathematics. Military and Mathematical School, King's Mountain, N. C.

There may be two cases : — I, n may be odd ; II, n may be even.
I. We readily perceive that

(a?*-f ar-f l)(aJ« — « + 1) = x^ + aj'-f 1,

(x4^-ai9 + a?» + a;-f 1)(«* — ai9 + a?« — a;H-l) = aj»-f ««-f «*-f a?»-f 1,

(aj^» +«— «-f aj"-»-f . . . . -f l)(a*-i — a*-«-f aj— »— .... -f 1)

= s^in-l)^^(n-^)^^(n-9)^ ^1 (A).

Either factor of the first member of (A) will divide the second member in accordance with the re-
quirements of the problem.

-166-

II. In this case,

(*-f !)(«— 1) = afl — h (aj» + aj«+ir-f l)(aj» — *«-f « — 1) = a^B-f a:< — a^— 1,

(a5»+«*H-a' + a:8-f « + !)(«» — x«-fa^ — aJ»-fa; — l) = ajw + aj»+a!6 — a;* — aJ* — 1.

(af«~»+a!*- «-f aj»-5+ . . . . -f l)(aj^i — aj"^+af^— .... —1)

= af^t^-D-f a!«(«-2)-faJ»(»-'»)-h....+«» — aj»^ — jf»-<— ....— 1 (B).

Each factor of the first member of (B) will divide the second member.

Solved alflo by the Propoter^ Mewiv. Drummond, Hart, Putnam, Eotell, Sch^er, Seltz and Siverly.

158*— Propoeed by O. F. Mkad, Uniontowiif Fayette County, Pennsylvania.

If two circles intersect, the common chord produced will bisect the common tangent.

I.— Solution by Edwin Place, Cinclnnatos, Cortland County, N. T.; J. F. W. Schbvfbb, Profeseor of Mathematics and
German, Mercersburg College, Mercersburg, Pa.; and Waltbk Sn'BRLT, Oil City. Pa.

Let O and O' be the centers (»f two Intersecting circles, PF their com-
mon tangent, BS their common chord AS produced to meet it. From
the point P' <iraw FT parallel to 00' the line passing through the centers
of the circles, and from P and F drop perpendiculars to the line through
Oand O'. Call OP = R, OF = r and 00' = a.
. From the similar triangles FTP. POE, FO'F we have
FT _ OE _ O'F . R—r _ OE _ O'F
FT ~ OP "" OT' * ^^ a ~ J^ "" r '

therefore OE = ^i^^^^) and OT = "i^^^) ,

a a

In the triangle AOO' we have (OA)* — (0C)« = ( AC)« = ( A0')« — (C0')« ;

.-. (0C)» — (C0')« = (0A)«-(A0')« = B» — r« (1). Also, 0C-fC0' = a (2).

Dividing (1) by (2), OC — CO' = ^ ^ - (3). From (2) and (3) we easily find

00 = »« + «•-- = - + £--. CO' = ia-^--- = --£-±--

Ec = oc-oE = '^+f-.^- ■«(«-'•) = '^-(f=r)*.

2a a 2a

CP = CO' + OF = «• - f +!• + '^«— ) =, "^(^-Ji'.
^ 2a ^ a a

.-. EC = CF and consequently PB =s BF,

II.— Solution by C. C. Robins, Princeton, New Jersey; Gborob Hawlbt, San Francisco, California; J. M. Quibos,
Alamos, Sonora. Mexico: C. A. O. Rosell, A. B., Teacher of Mathematics at the Carroll Institute, Beading, Pa.; and
William Hoovbb, Superintendent of Schools, Wapakoneta, Ohio.

BA X BS = (PB)« and BA X BS = (BF)*, therefore (PB)« = (BF)*. PB = BF and PF is bisected
by AS produced.

Solved in a similar manner by Messrs. Putnam, Svltfester Robins, SHU, Shaw and Shidy; and otherwise by Messrs.
Edmundt, Hart, Haynet, Mange, Mais, Siveriy and TiimlntU.

159*— Proposed by Wiluak Hootbr, Superintendent, of Schools, Wapakoneta, Auglaize County, Ohio.
August, 1879, had fire Fridays, Saturdays and Sundays ; when will the month of August have five of
each of these days again?

I.— Solution by Jomr I. Clark. Moran, Clinton County, Indiana; C. A. O. Rosbll, B. A.. Teacher of Mathematics at
the Carroll Institute, Reading, Pa.; Dr. David S. Hart, M. A., Stonington, Conn.; and J. A. Pollard, Goshen, N. T.

In order tliat August can have five Fridays, Saturdays and Sundays, the first day must be Friday and
the last Sunday. This would take place every seven years if there were no leap years, because there
are 52 weelcs and one day in a common year, consequently the first day of August will come one day
later in the week each year ; but it will come 2 days later each leap year, and as there will two leap years
intervene between 1879 and 1886, said event will take place two years sooner, or in 1884.

Nearly thus were the solutions by L. P. Shidy, C. C. Robins and Sylvester .Robins.

II.— Solution by the Proposer; and Waltbr Sfverlt, Oil City, Venango County, Pennsylvania.

Any year in which the first of August falls on Friday will satisfy the problem.

The Dominical Letter for Friday, August 1, 1879, is £, and the remaining years of this century iiaving
£ for their Dominical Letter are 1884 and 1890, the years required.

-167-

Tlf-— Solntlon by K. S. Putnam, Rome, Oneida Co., N. T.; and Bluah A. Squixb, Le Grand, Marsha)! Co., Iowa.

The month of August will have five Fridays, Saturdays and Sundays in any year when August 1st
oomee on Friday.

To ascertain when the same day of any month in any given year will next come on the same day of
the week, divide the last two figures by 4 and note the remainder. For January and February if the
remainder is 3 it will occur 11 years afterwards ; if the remainder is 1 or 2, 6 years ; if remainder is 0, 5
years afterwards. For other months, if the remainder is 3 it will occur in 5 years ; if 2, in 11 years ;
and if 1 or 0, in 6 years.

Should any centennial year not a leap year (as 1900) intervene, add 1 to the result obtained above.

In the problem given, 79 divided by 4 gives 3 as remahoider, and the year required is 1884.

Following the rules the same will occur in 1890, 1902. 1913, 1919, 1924, 1930, 1941, 1947, 1952, 1958,
1969, 1975, 1980^ Ac.

An elaborate general liolation wae famished by Prof. Frank T. Fneland, of the University of Pennsyh'ania, which we
relnctantly omit for want of space.

160.— Proposed by L. P. Shidy, U. 8. Coast Survey Office, Washington, D. C.

Find a point within a given quadrilateral, such that when Joined to the middle points of the sides the
quadrilateral will be divided into four equivalent parts.

I.— Solation by K. S. Putnam, Rome, Oneida Coonty, N. Y.; and John I. Clark, Monin, Clinton County, Indiana.

Let ABCD be any quadrilateral.
Through A and D respectively draw
£F and OH pcurallel to the diagonal
BC; through B and C, EG and FH
parallel to the diagonal AD. O, the
intersection of the diagonals of the
circumscribed parallelogram, will
be the point required.

I, J, E and L being middle points
of the sides, IJLK is a parallelo-
gram (Problem 151).

AAIK = JAABC,
AJDL = J/:^DBC,
and AAIK 4- AJDL = } of area of
quadrilateral, = ^ area of parallelo-
gram IKJL. But, drawing perpen-
diculars, AIOK -f AJOL = JIK X N 0-f iJL X OP = JL X iNP = I area IJKL.
So AAIK -f AJDL -f AlOK -f A JLO = AIOK -f JDLO = area of parallelogram ; also CKOL -f OIBJ
= area of parallelogram ; therefore AIOK -f JDLO = CLOK -f IBJO. But AIOK = JIKx( AM -f NO)
= JIK X 80 and JDLO = JJL X (RD -f OP) = JJL X OT. Also, IK = JL and OT — OS, therefore
AIOK = JDLO. So CKOL = OIBJ. Hence, AIOK = KOLC = OLDJ = OJBI.

II.~^lntion by C. C. Robins, Princeton, New Jersey; J. M. Quiboz, Alamo«, '^nora, Mexico; and William Hoovbr,
Superintendent of Schools, Wapakoneta, Ohio.

Let ABCD be any quadrilateral. Draw the diagonajs AC and
BD, and bisect them at E and F respectively. Through each
of these points draw a line parallel to the other diagonal. The
point O where these lines intersect is the point required.

iVoc>f.— The triangles A ED and AEB are equal ; also triangles
CDE and CBE are equal, because on equal bases and between
same parallels. Therefore AECD is one-half the figure ; but
AECD = AAEC -f AADC = AAGC -f AADC. In the same
way DGBC, CGAB, BGDA are each on^-half the figure. Sub-
tracting from equation AOCB = OBCD the common triangle
BGC, we have AAGB = ACGD, and in similar manner AADG = ABGC. Each of these triangles is
bisected by the line drawn from G to the middle of its base, and adding halves of equals together we
obtain HGLC = KGLB = KGIA === GHDI.

Solved also by Meeer*. Brown, Casey, Edmunds, HaH, HawUy, Malx, Place, Pollard, Rosell, Sch^er, SHtz. Shidy and
aiverly.

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161.— Proposed by Abtexaa Martin, M. A., Member of the London Mathematical Society, Erie, Erie County, Pa.
What rate per cent, of interest paid in advance is equivalent to r per cent, paid at the end of the year?

I,— Solution by the Pboposkr.

Let X = the rate per cent, required expressed decimally.

If the interest is to be equivalent to the lender to r per cent, paid in advance it is the present worth

of r discounted at r per cent. ; .-. x = ^ , .

1-f r

But if the interest is to be equivalent to the borrower to r per cent, paid in advance it is the present
worth of r discounted at x per cent. I •'• ^ = i i_ » whence x« -f x = r and x = \/{r -f J) — }.

II,— Solution by Walter Siverlt, Oil City, Venango Co., Pa.; and Sylvester Robins, North Branch Depot, N. J.

Let X = the required rate per cent, in advance expressed decimally.
The interest on x is ar*; on a^, a:»; on t\ x*; on x<, x^; etc. ; etc., and

a;4-a^-fa:» + aT*-fa:*-f etc. = r, or , = r: whence x = , — .

' ' ' ' ' 1 —X 1 -I-*"

Solutions of the first case received from Messrs. Bowser, Casey, Clark, Donovan, Drummond, Haynet, FoUard, Putnam.
C. C. Robins, Shidy and Seitz; and of the second ca«e from Messrs. M<Uz and Roeell.

162.— Proposed by O. H. Merrill, Mannsville, Jefferson County, New York.

Two equal circles intersect, the center of each being on the circumference of the other. A circle is
drawn touching that diameter of the right-hand circle which joins the centers of the given circles, and
the circumferences of both circles, the right-hand one internally and the other externally ; also a circle
is drawn touching the one last drawn and the circumferences of both the given circles. Find the radii
of these two circles.

Solution by Artexas Martin. M. A., Member of the London Mathematical Society, Erie, Erie Co., Pennsylvania; and
LirciUH Bhown, Hudson, Middlesex Co., Massachusetts.

1. Let r represent the radius of one of the
e lual circles, and x the radius of the required
I'ircle touching the diameter AJ.

TluMi AC = r -h jr,

AE = -j/[(r-|-.r)«- x«] = -i/(r«-f 2rx),
BE= i/L(r— a;)«--x«] = ;/(r«— 2rx);
AB = AE — BE, therefore

r = -j/(r«4-2rj;) — ;/(r«— 2ra;) (1).

Squaring we get

2;/(r«— 4dB«) = r;
squaring again and reducing we find x = lr\/3.

2. Draw DF perpendicular to the diameter
A J, and CG perpendicular to DF.

Let y = radius of the second required circle,
AF = uj and DF = z.

ThonBF = r— IT, BE = V[(r-ir;/3)« — (irv?3)«] = ir(i/3-l). FE = BF + BE = lr(l-f i/3) — w,
DG = « — ir;/3, CD = y-|-iri/3, BD = r — y, AD = r-f j/, and we have

(AF)« + (DF)« = (AD)«. (BF)« -f (DF)« = (BD)«, (GC)« -f (GD)» = (CD)«;

or ir« -!-£« = {r^yY (2), (r-,r)» + z» = (r-y)« (3),

(ir-f ln/3- tr)« + («-lr^3)« = {y^{r^/^y (4).

Subtracting (3) from (2), we get ir = Jr -f 2j/ (5).

Substituting in (2) and (4),

(irH-2y)«-|-2r« = {r^tyy (6). (iri/3-2j/)«-|- (2- iri/3)« = (y-f Jn/3)« (7).

Expanding (6) and (7), and reducing,

«« + 3y« = Jr» (8), {z - itV3)« = %ryV^ -3y«-ftr« (9).

Subtracting (8) from (9), we get z = ri/3 — 5^ (10).

Substituting in (8), 28y« — lOre/v'3 = — »r« (11) ; which quadratic gives y = ftri/S.

Solved also by Met*sn«. Hooter, Putnam, C. C. BoMns, Sch^er and Sirerly. Prof. Casey, J. W. Donoran and Edwin
Place each t<olved the flr»t part.

-169-

183«— Propoeed by Thsodojix L. DkLai^d, Office of the Secretary of the Treasury, Washington, D. C.

In 1861 a 6-per-cent. 20-year coin bond of the U. S., interest payable Bemi-annually, sold on the market
for $J.B91 on the dollar ; what, on this basis, would have been the market value of a 4-per-cent. 28-year
coin bond of the U. S., interest payable quarterly?

!•— Solotion by the PiiopoaKR.

Let r = rate per annum paid on a bond, n = the number of interest-payments in a year, / = the
number of years the bond runs, v = the market value of the bond (or one dollar of it), and R = the rate
per annum the investor realizes (paid n times a year).

Then ^ is the income of the Investor during each of the n intervals, which he will receive w/ timo8.
receiving with the last the face of the bond, $1.00.

The payment preceding the last Is at interest - of a year at R per annum ; and ^( 1 -f ) is the
value of that payment when the bond matures ; and the sum of all the payments improved at the rate
R (- -annually j will equal v so improve^l.

Therefore we have the equation

Summing, transposing, and reducing, wo have

(12t-r)(l-f JJ)*" = «-r (2);

or by logarithms,

««log(l + ^)-f log(«t»~r) = log(/i-r) (3).

Taking the elements of the 6-per-cent. bond, we find from (3), by Double Position, R =. 0.070226.
For the constant rate per annum when paid n and n' times a year during any t years we have,

('+jr-('+?r ">■■

or after reduction R' = n'(l -f -Y' — n' (5) ;

that is to say, equation (5) gives the rate the investor realizes when paid n' times a year instead of n
times ; therefore R' = 0.069620, in the four-por-cent. bond.
In equation (1) let R become R' and n become n', or take the elements of the four-per-cent. bond, and

we have v' = — ^'—^, -:.-. -f ti = 0.6361467 ;

or a $100-bond (at 4 per cent.) would sell for $63.61.

II.— Solution by H. H. Doouttlb, U. S. CooBt Survey Office, Washington, D. C.

I assume that in the case of two bonds having the same rate of interest payable at the same intervals,
and differing only in respect to the time at which the principal is payable, if either sells at par the other
sells also at par. I know that bonds having a long time to run usually sell for more than similar bonds
having a short time to run ; but I suppose the dilTeronce to be fluctuating and uncertain ; and unless
data for its determination are given in such problems, it can not be taken into consideration.

I interpret the problem thus: At a certain rate per cent., not explicitly stated in the problem but
determinable therefrom, upon a principal of $0,891, partial payments of 3 cents each are made at regular
intervals of 6 months ; and at the end of twenty years the amount due is $1.03. At the same rate per
cent., upon the required principal, partial payments of 1 cent are made at regular Intervals of 3 months ;
and at the end of twenty-eight years the amount due is $1.01.

Let X = amount of $1.00 in three months at the above-mentioned rate per cent., and y = the required
principal.

Then 0.891a?^« — 0.03x7* — 0.03j;« — 0.03xt* — — O.Oar« — 1.03 = (1),

and yri»« — 0.01a;»" — 0.01x1 w_o.01a:»w— — O.Olx — 1.01 =0 (2).

Multiplying (1) by a^ — 1 and (2) by x — 1,

0.891(x« — xw) — 0.03XW — ar» -f 1.03 = (3),

y(xiw — x"«) — 0.01x» w — X -f 1.01 = (4).

From (3), x = 1.017405, and then from (4) y = 0.6361467.

Mr. DooHitle solved this problem at the request of the Proposer who placed his solution in our hands for publication.
Excellent solutions were also given by F. W. Lantz, Washington, D. C\, and Walter Siverly^ OU City, Pa.

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164.— Prop«»»cd by V. Webster Heath, Rodman, JcfferM>n County, New York.

A field 81 rods square has one of its corners clipped off by a line meeting tiie sides at (>0 and 80 yards
respectively from Uie corner. A man commences plowing around the field, turning a furrow one foot
wide. How many "rounds" will he plow before the unplowed portion bei'omes'a square?

Solution by J. F. W. Scheffer, Proferaor of Mathematics and German, Mercereburg CoUege, Meroeivbuig. Pa.

Bisect the angles A£F and CFE and produce the bisecting
lines to meet in O. Draw GH perpendicular to EF.

There will be as many "rounds" as the width IH of the
furrow is contained times in 6H.

Denoting EB by a, BF by 5, EF by c, the width of the
furrow by d, /BEF by a, /BFE by /3, we easily obtain
c
tanja-flanj/;'

GH

But cosa = - , cos/^ = ,
c ' c

sin a = - , sin ^ =

and from the formula

- 1 — COS<Z> V X 1 c

tania>= , ^ wehavetanla =
*^ sin <p ■

. , tanjy^ =
6 *' a

GH =

Substituting in the foregoing value of GH we get

. , , and number of "rounds" = ,, . ,
a^-o — c a{a -|- 6 — c)

In our case a = 60 yards, b = 80 yards, c = 100 yards,

d = l foot = J of a yard, and the number of "rounds" Is 360.

(>ood 8o]ution8 al!*o given by the Proposer, Lucius Brown, J. A. PoUard, L. P. Shidy and Walter Sivtrly.

\W^—VTO^oifeA by Hon. Josiah H. Druxxond, LL. D., Portland, Maine.

A railroad company has $400000 of preferred stock and $300000 of common stock. The agreement is
that the net income each year shall bo applied to the payment of six per cent, on the preferred stock,
and the balance shaU be divided on the c;ommon stock. The net income is $36000 rental paid annually ;
a debt of $50000 is created payable in twenty years with annual interest at six .per cent, (payable at the
same time as the rent) in such manner that the annual interest is pa3'able out of the current income, but
the principal out of all the income after the debt becomes due until it is paid. It is agreed to create a
sinking fund. What amount must be curried to the sinking fund annually, assuming that five per cent,
compoimd interest may be earned to extinguish the debt when it becomes due, and how shall that
amount apportioned on the two kinds of stock?

Solution by Walter Siterlt, Oil City, Venango County, Pennsylvania; and the Profosbb.

Let J- = the amount laid by infthe sinking fund each year, 1.05 = R, 50000 = a. Then

x+Rx'\-R^X'\-E^x+ J^W^x = a, ''^^^^^- = a, x = "jj^Zp = $1512.127.

The preferred stock under the original arrangement would pay $24001) at the end of 20 years and what
the rental lacked of paying $24000 after paying off the debt at the end of the next year.

At the end of 20 years there is due, principal and interest, $53000, from which subtracting $36000
leaves $17000, plus interest = $18020 due at the end of the 21st year. $24000 -|- $18020 — $36000 = $6020
which the preferred stock would pay at that time, which is equivalent to $3679.244 paid at the end of the
2()th year ; this addeJ to $240rH) gives $29<>7ih244, the portion of the $30(K)0 paid by the preferred stock.

$50000 : $1512.127 :: $2l>(:79.244 : Srt97.575,
sura put in the sinking fund by the preferred stock each year, which subtracted from $1512.127 leaves
$614,462 to be put in by the common stock.

18^— Proponed by Dr. S. Hart Wright, M. A., Ph. D., Mathematical Editor Yates County Chronicle, Penn Yan, N. Y.
Required the variation v of the magnetic needle, in latitude A = 42^ 30', the declination of the North
Star being 6 = 88^ 40', and its magnetic bearing 6 = 8^ 48' 30" E. when at its greatest elongation east.

Solution by Edward A. Bowber, Professor of Mathematics and Engineering, Rutgers College, New Brunswick, N. J.;
and the Proposer.

Let z = azimuth of North Star at its greatest eastern or western, elongation.

Then, by works on Astronomy, we have z = sin-'(<*ort'5secA) = 1^ 4s' 30" which addeJ to the mag-
netic bearing of the North Star when at its greatest elongation west, or subtracted from the bearing
when at its greatest elongation east, gives us tlie variation of the magnetic needle.

.-. r = 6 — 2 = (8^ 48' 30") — (1^ 48' 30") = 7^, the variation west of north.

Solved also by K. 8. Putnam and Waiter Sirerly.

-171-

10nf«— Propoeed by B. F. Bitrlbson, Oneida Castle, Oneida County, New Yorlc.

Given 3(a^ + j/*) = 2xy{x + y) and a;y(x* — |^) = lS{z — j/), to And accurate expressions for the values
of z and y.

Solution by William Hoover, Superintendent of Schools, Wapakoneta. Oliio; the Proposer; E. B. Seitz. M. L. M. S..
Professor of Mathematics, North Missouri State Normal School, Kirksville, Missouri: and Dr. David S. Hart.. M. A..
Stonington, Connecticut.

Let a^ = p, x + y z= 8, and tlie given equations readily become

«» — 2|> = J(26) (1), and iM = 13 (2). These give «3 — J(26)« = 20 (3).

Put 8 = ^w -|- , and (3) becomes ir« — 26tr = — ( j ; whence ir = 5;- or x= ;

. •. « = J[v^(676) -f ^(26)]. Hence p = J(3«« - 26) = ,iJ26^(26) -f ^(676) - 26] = x^/.
Having x-\-y and a;j/ we And

X = 4[«-f l/(«*-4p)] = i{^(676)-f ^(26)-f /[4(26)-26^(26)~^(676)]|.

y = l[«~l/(«*-4p)] = 4{^(676)-f ^(26)--i/[4(26)-26^(26)-^(676)]|.
Solutions received also from Messrs. CoMy, Edmunds, McUz, Putnam, RoteU and Sivetiy.

168.— Proposed by J. F. W. Scheffer, Professor of Mathematics and German. Mercersburg College, Mercersbuig, Pa.
The position of a ball on a circular billiard table is given. What path must the ball describe in order
to pass through its original position after touching the cushion twice?

Solution by William Hoover, Superintendent of Schools, Wapakoneta, Auglaize County, Ohio.

Let P be the given position of the ball, O the center of the circular table,
A and C the points in which the ball strikes the cushion, and B the middle
point of AC.

Let OP = a. OA = OC = r, /OPA = ^OPC = q>\ then

ZPAO = sin-("^|.''?), = 450- J^
since the angle of Inciilence equals the angle of reflection ;

.-. "-^^ = 8ln(460-l.p) = v'[4(l-8in<p)].

\,

V

V

. ... r«sina> r* , , fV(8a«-f r«) — r«

whence sin*a> + - « . = s-i. and 8in<z> = ^- -. .—
^ 2a« 2a« ^ 4a*

Good solutions given by the Proposer and Messrs. Casey, Clark, Place, PuOutm, C. C. Robins, SHtz, Shidy and Sirtrly,

109*— Proposed by H. T. J. Ludwio, Professor of Mathematics, North Carolina College, Mount Pleasant, N. C\
Required (1) the radius of the auger that will cut out one-mth part of the surface, and (2) the radiuH

of the auger that will cut c»ut one-nth of the volume of a sphere, radius r, the axis of the auger coinciding

with a diameter of the sphere.

Solution by J. F. W. Scheffer, Professor of Mathematics and German, Mercersburg College, Mercersburg. Pa.: and
E. B. Seitz, M. L. M. S., Professor of Mathematics, North Missouri State Normal School Kirksville, Missouri.

1, Let X = radius of the auger. The auger cuts out two segments, the surface of each of which is
27rrh, h being = r — i/(r« — x^) ; .-. ^nrlr—y/ii^—jfl)'i = ^^^, whence x = rfl — (l— Vl*.

2. Let y = radius of the auger. The volume removed is composed of two equal segments, radii of
bases y and hight of each = h' = r ~ ;/(r*-— j/"), and a cylinder radius y and altitude 21/(1^—^^). TUv
volume of each segment is JflrV«(3r — h') ; the volume of the cylinder is 2w2^(r«— ^) ;

... iyrh\3r~h') + 27ty»y/if^-y) = ^f .
Substituting value of k' and dividing out n,

[2r«-3/»-2V(r2-y«)][2r + V(,^-2/«)]-f 3yV(r«-y») = ^~;

whence (r*—y»)y/(r^—y) = r^(l^^\, and y = rfl - (l — ^Vl*.

Solved also by the Proposer and Me;«sn$. Casey, Hart, Plac.\ Putnam, C. C, Robins, Roseli, Shidy and Sirerly.

-172-

170«— Proposed by K. S. Putnam, Rome, Oneida Connty, New York.

Traveling recently on a train moving 30 milee an hour and overhauling a freight moving 20 miles an
hour, I endeavored to ascertain the length of the freight. One minute from the time I was opposite the
rear of the freight a third train came between and put a stop to my investigation. At the next station
we stopped, the door of the depot being directly opposite my seat. The freight here overhauled us, the
front of the train being opposite me, when our train started. When we had attained the same rate of
speed as the freight I was opposite the same point in the freight as when I closed my tirst observation.

It took our train H mintes to get under full h^idway, during which time its motion was uniformly
accelerated. How long was the freight train, and how far was I from the depot when 1 passed the freight?

Kolndon by L. P. Shidt, U. 8. Coast Snrver Office, Washington, D. C; E. B. Sbitz, M. L. M. 8., Professor of Mathe-
matics, North Missoari State Normal School, KirksvUJe, Missouri; Lucius Brown, Hudson, Middlesex Co., Massachusetts;
and C. C. Robins, Princeton, New Jersey.

Let P denote the passenger train, and F the freight train. Since P moves 30 miles an hour, it will
move i mile per minute ; and since F moves 20 miles an hour, it will move ) mile per minute. Hence
P gains on F i — ) = I mile in 1 minute, when the observation was broken off by third train.

After stopping it took P 1 minute from the start to attain the same velocity as F ; during this minute
P*8 average velocity was 10 miles an hour, or | of a mile per minute, while F traveling ) of a mile in the
same time, gained I — ii = 1 of a mile o^ P. By the problem this brings us again to the point where
our observations were broken off, so we have now seen the whole of F, uid its length is | + i = i mile.

8ince P took U minutes to attain full speed, its average speed was 15 miles an hour, and during this
time it traveled | of a mile. During the same time F traveled J X J = i mile, having gained 1 — | = t
of a mile on P. After attaining full speed P gains on F at the rate of ^ of a mile per minute, hence to
gain i of a mile it will take J of a minute, during which time it will travel } X } = | of a mile. Hence
the entire distance from the depot when P passed F was | -f 8 = J <^' * mile.

Solved also by the Propooer and .Messn*. Dcnotxtn, Bo$eU and Sir^rly.

171*— Prop<x^ by Marcus Bakvr, U. 8. Coast Sorvey Office, Washington, D. C.

In a plane triangle ABC the center of the circumscribed circle is O, the center of the inscribed circle
is I and the intersection of the perpendiculars is H. Knowing the sides of the triangle OIH, determine
the sides of the triangle ABC.

Solution by Professor W. P. Caskt, C. E., San Francisco, California.

Let OH = a, HI = ft, 10 = c, r = radius of inscribed circle,
R = radius of circumscribing circle.

As the lines OH, HI and 10 are given, therefore the points
O, I and H are given in position. Bisect OH in N, then N is a
given point, and is the center of the nine-point circle. Join N
and 1 and produce NI to meet the circumference of the inscribed
circle in M. As the nine-point circle and the inscril>ed circle
are tangent circles, therefore M is the point of tangency and NM
is the radius of the nine-point circle, = \R. Now NI becomes
a known line being = J[2(HI)«-f 2(0I)« — 4(NH)«1*

= J(2/>«-f 2c«-a«)* =p = NM — IM = i«-r,
and (01)* = B^ — ^Rr; whence R and r become known lines,
and therefore the three circles are given in position and mag-
nitude. It only reomins to find a point C in the circumference
of ABC so that the t^tngent CA to circle I may be bisected by the circle N in the point S, which is easily
done. But the angles of the triangle ABC may be foand from the following well-known properties, viz ;

^ = cos A -f cosB -I- cosC — 1, or cos A -|- cosB -|- cosC = 1 + p (1);

(0H)« = 2R^{\ -f COS2A -h cos2B -f co82C), from which we get

co»«A -f cos«B -f cos»C = ^-t^^ (2) ; cos AcosBcosC = ^li^-^' = ^"i,"* (3).

4/1* oil* oiv'

From (1) and (2) we have cosAcosB -f cosAcosC -|- cosBcosC = 'l^^ — ^1^- - (4),

Ate* litO

From (1), (4) and (3) we have immediately by the Theory of Equations the cubic equation

COS3A- (n.-)cos*A + (^ 2^* - 8ie*->^'^ = S/e*-'
the three roots of which are the values of cos A, cosB and cosC.
The sides of the triangle ABC are AB = 2iJsinC, BC = 2/2sin A and AC = 2/2 sin B.
Excellent solntions received from Lvciw Brmon, Professor Sch^er and Professor SeUz.

-173-

178»— PropoMd hj W. W. JOBHBOH, M. L. M. 8., ProfeMor of Mathematics, 8t Jolin*a College, Aonapoliii, MtryUnd.

The base of a triangle is fixed and the difference between the vertical angle and one of the angles at
the base is constant. Find the locus of the vertex. Discuss the curve and consider the cases in which
the constant difference is and 90^.

Solotion by the Pbofosbb.

Let OAP be the triangle ; take O as the origin, and the base OA as the
axis of a;.

Let POA = e, PAX = <p, the constant difference POA — OPA = a, and
OA = a. The vertical angle is then <p—B and the condition of the prob-
lem gives 20 -- (p = a (1).

Now tanO = ^, and tana)=— ^— (2).

X ^ x—a ^ '

X—i

and (1) gives tan2e = ten(^-ha). threfore ^_^^^ = i^^^^^.

or substituting from (2) J^- = ^^l^^^^}^ reducing to

(^•H-2/*)(y— ictana) -f a(x«—2/«)tana — 2ttxy = (3).

the equation of the locus required.

Writing the equation in the form

2a;i»f — a(2* — 1/*) tan a
v — a; tan a = — ^ —^ .

we have when y = a;tan a = ao ,

2^ — a;tana = atana (4),

which is therefore the equation of the asymptote.

Equating to zero the terms of the lowest degree in (3), we have

y = a:(— cota ± coseca),
the equation of two tangents at the origin, whose inclinations to the axis of x are therefore \a and
90o-f Ja.

These conclusions can also be derived geometrically thus ; when P is at infinity = 9>, hence from
(l)0 = ^=:a the inclination of the asymptote. Now the line AP rotates with twice the angular ve-
locity of OP, since from (1) d<^ = 2cl9 ; hence wlien OP and AP are parallel the distance of P from AP
is double its distance from OP ; therefore the asymptote cuts OX produced at a distance a on the left of
the origin, which agrees with (4).

When <^ = 0, n, 2n, Ac., P comes into coincidence with O, and OP becomes a tangent to the curve.
Now (1) gives for these values = ^ia-^kir), (where Jlc Is an integer,) which gives the direction of the
two tangents as before.

When a = 90« we have xijfi-^j^) — a(a^--yt) = (5),

of which the asymptote is a; = —a, and the two tangents at the origin y = ^x.

The general case (3) Is known by French writers as the oblique sirophoid, and the case (5), or the
right atrophoid, has been discussed by Dr. Booth under the name of the logoqfdic curve. — {Booth's New
Oeometrical Methods, Vol. I, p. 295. See also Rice and Johnson's Differential Calculus, p. 290.)

When a = (3) breaks up into y = and a:*-f y« — 2ax = 0; that is, the axis of x and the circle
whose center is A and radius a, as is geometrically evident should be the case.

Oood solutionB by Messrs. Bowmt, Catty, BdmwuU, Hoover, Ludwig, Jtoteli, Sehtffer, SeU* and Mverl^.

173.— Propelled by Dr. S. H. Wright, M. A., Ph. D., lute MatbemAtical Editor Fo/sf Oounfy CkronkU, Penn Yan, N. T.

Bequired the variation v of the magnetic needle, in latitude A = 42^ 30^, the declination of the sun
being 6 = 20^ N., and the magnetic bearing of its upper limb when rising on a horizon elevated ^ = 1^
Is 6 == 69^ 21' 40"'. Radius of sun — r = 16', and refraction = p = 35'.

Solution by the Proposvh.

The zenith distance of the sun's center = 9(P — /i-|-p-|-r = 8?K>*51' = «. Put J(« -|- A -f 5 ) = «. The
true azimuth is 2cos-V[co8(0— ^)8in(8~A)secAcosec2] = 62^ 30^ 58'' = a. The magnetic azimuth
is h, and > a, therefore 6 — a = 6^ 50' 42", the variation west of north.

SolTod also by K, 8. Putnam, Professor K B. SeitM snd Walter mverlf.

-174-

174«-'Propo8ed by the late Benjamin PsntoB, LL. D., F. R. S., Professor of Mathematicfl, Harvud UniTerBity, and Goo-
salting Geometer to the U. H. Coast Sarvey, Cambridge, Massachiuetts.

To And by quadratic equations a triangle of which the angles are given and the distances of the vertices
from a given point in the plane of the triangle.

I,— Solntion by William Hoovkr, Superintendent of Schools, Wapakoneta, Anglaixe County, Ohio.

Let A, B and C be the angles and a, &, c the given distances
from the vertices to the point P in the triangle.

With vertical angles 2A« 2B and 2C, and including sides a,
6 and c, construct the isosceles ti'iangles AP1P2, BP1P3, CP^Pa,
and arrange them so that their bases shall form the triangle
P1P2P3, their vertices lying without if the given angles are all
acute ; but if one of them is obtuse the corresponding verte^t
will lie within.

Join A, B and G, and ABC is the required triangle ; AP = a,
BP = 6, CP = c.

We easily And P,P^ = 2asin A, PjPa = 26sin B

and P«P3 = 2csinC;

^ \ 2a68mAsmB / \ 2a6sinAsinB /

Then from the triangle APiB we have

Au f • i ... o I. fri . ,/a»8in«A4-&«8in«B — c«8in2C\"| ) i
AB = |a. + 6.-2a6cos[c + cos-( :,;^sinAslMB )J | '

= {a*-\'bf— cosec A cosec B \ cos C(a«3in«A -f b*sm*B — c«sln«C)

— sinCi/|:(2aftsin AsinB)« — (a«sin«A-f 6«6iii«B— <<8in«C)«] \ )*.
Similarly, we find

Ar. I • • ^ o Ft, . ,/fi«ain«A4.c«8in«C — 6«8in»B\"| )i

AC = ja. + c.-2accos[B + cos-i(-— ;^^.^-^^^^^ - -)J } ,

= { a« -f <? -- cosec A cosec C -j cos B(a«sin«A -f c«8in«G — 6«8in«B)

— sinBi/l(2ac8inAsinC)« — (a%in«A-f c«8in«C— 6«8in«B)«] [

]*.

un (k^.w o.. Fa . ,/5«8in«B-|-c«8in«C~a«sin«A\-| ] »
BC= |^ + c^-26ccosLa + cos-(---2^.^3^.^^^ -)J j ,

== { ftt -|. cs _ cosec B cosec C -j cos A(b««in«B + c«sin«C — a«sin«A)

— sinA;/[(26csinBslnC)« — (6«sin«B-f c^insC— a%in»A)«] \ j*.

II,— Solution by Walter Sivbrlt, Oil City, Venango County, Penn&iylvania.

Let ABC be the required triangle, P the given point, PM, AN perpendiculars on BC, and PQ perpen-
dicular to AN ; BP = a, CP = 6, AP = c, BM = x, PM = y, BC = z. Then will BN = 2 sin C cos B cosec A
= mz, AN = 2: sin B sin C cosec A = nz, and we have

ai*^t/» = a^ (1), (z-x)* -f »» = 6« (2), (mz-x)* -f (m-yy = c« (3).

SubsUtuting a« for x« -f y« In (2) and (3),

2« — 2aa:=6« — a« (4), (m«-f n«)2« — 2mxz — 2nyz = c^ — o« (5).

Eliminating xz from (4) and (5),

(tn«-fn« — m>3!« —- 2113^ = <^ — a« — m(6«— o«), or y = ^ — — - „ •

From (4), * = ~ 2« ~'

Substituting these values of x and 2^ in (1),

[(m«-f n«— m)*-|-n»]2* -f [2(iii«-f n«— m)(m5«--ma«-f a« — c«) — 2n«(a«-f 6^)>«

= — (m6» — mo^ -f a^ — c«)« — n«(a^ — 6»)«.
a quadratic for determining z.

For a geometrical construction, see Simpaon'a Algebra, p. 367.
Solutions received also from Messrs. Casey, Eastwood, JSosell, Sch/tffer and S«ikU.

-175-

Xt#f of iJontributorg to the Junior JDepartn^ent.

Waltbb Sitsrlt, oil City. Venango Co., Pa., solved all the problems bat 171. C. A. O Roenx, B. A., Teacher of Math-

ematics at the Carroll InstUnte, Reading, Pa., all but lOi, 108, 164, 165, 166, 171 and 178. E. B. Skits, Member of the Lon-

don Mathematical Hoclety, Professor of Mathematics, North Missouri State Normal School, KiiksviUe, Mo., all but 148, 168, 158,
160, 16S, 168, 164, 165 and 1G6. K. S. Putnam, Rome, N. Y., all but 148, 149, 160, 158, 168, 164, 165, 171, 178 and 174. « Prof.

W. P. Casxt, C. £., San Francisco, Cal., all but 160, 158, 157, 15B, 160, 8d part 168, 168, 164, 165, 170 and 178. J. F. W.

SCBSFFBR, Professor of Mathematics and German, Mercersburg College, Mercersburg, Pa., all bnt 148, 140, 160, 166, 150, 161, 168,
166, 166, 167, 170 and 178. C. C. RoBixs, Princeton, N. J., all but 158. 158, 156, 157, 168, 165, 166. 167, 171, 178, 178 and 174.

L. P. SeiDT, U. S. Coast Suney Office, Washington, D. C, all but 160, 158, 166, 157, 108, 168, 165, 166, 167, 171, 178, 178 and 174.
WiLUAX HooTiR, Superintendent of Schools, Wapakoneta, O., all but 150, 158, 150, 157, 161, 168, 164, 166, 166, 100, m and 178.
Dr. David S. Habt, M. A., Stonington, Conn., solved 148, 149, 151, 154, 156, 157, 168, 160, 100, 167 and 160. F. P. Mats, M. A.,
late Professor of Mathematics, Military and Scientidc School, King's Mountain, N. C, 149, 150, 151, 168, 154, 166, 157, 156, 160,
161 and 167. Stlvbstxb Robins, North Branch Depot, N. J.. 148. 149, 160, 151, 158, 158, 156, 157, 158, 150 and 161. John

1. Clabk, Moran, Ind., 148. 149, 160, 151. 155, 156, 150, 160. J61 and 168. J. A. Pollard, Goshen, N. Y., 148, 148, 150, 151,

168, 155, 156, ISO, 160, 161 and 164. Hon. Josiah H. Drummond, LL. D., PorUand, Me., 148, 151, 158, 164, 166, 107, 160, 161

and 166. J. W. Donovan, Ansonia, O , 148, 149, 160, 151, 155, 156, 159, 161 and 1st part of 168. Edwin Placx, Cinclnnatua,
N. Y., 148, 151, 155, 158, 1st part of 108. 168 and 169. Lucius Brown, Hudson, Mass., 148, 168, 160, 108, 164, 170 and 171.

H.T. J. LuDWio, Profe8M>r of Mathematics, North Carolina College. Mount Pleasant, N. C, 148, 149, 151, 154, 166, 109 and 178.
B. J. Edmunds, B. S., Professor of Mathematics, Southern University, New Orieans, Louisiana, 150, 165, 168, 160, 167 and 178.
JosBPH Turnbull, East Liverpool, O., 148. 151, 158, 154, J65 and 168. Prof. H. 8. Banks. Instructor in English and Classical

Literature, Newbnig, N. Y., 148, 149, 151, 158, 154 and 155. A. E. Hatnbs, M. Ph., Professor of Mathematics and Physics,

Hillsdale College, Hillsdale, Mich., 149, 160, 166, 158 and 161. O. D. Oathol-t, Luana. Iowa, 148, 149, 151, 154 and 155.

J. M. QuiRos, Alamos, Sonora, Mexico, 148, 151, 155, 158 and 160. Okorob Hawlbt, San Francisco, Cal., 148, 151, 166, 166

and 160. Waltkr S. Nichols, Editor Jntvrance Monitor, Now York, N. Y., 148, 149, 151, 158 and 169. P. F. Mani»x,

Alamos, Sonora, Mexico, 151, 166 and 158. Edward A. Bowskr, Professor of Mathematics and Engineering, Rutgers College,
New Brunswick, N. J., 161, 166 and 178. Thbodorx L. DbLand. U. S. Treasmy Department. Washington, D. C, 156 and

168. J. M. Tatlor, Milton. Oregon, 149 and 160. Dr. S. H. Wright, M. A., Ph. D., Penn Yan, N. Y., 166 and 178.
M. H. Doolittlb, U. S. Coast Survey Office, Washington, D. C, 168. Gkoroh F.astwood, Saxonville, Massachnsetta, 174.
W. W. Johnson, Member of the London Mathematical Society, Professor of Mathematics. St. John's College, Annapolis, Md.,
178. F. W. Lantb, Washington, D. C, 168. Frank T. Frbbland, Instructor In Mechanics, University of Pennsylvania,

169. B. F. BuRLXsoN, Oneida Casde. N. Y., 167. V. W. Hbath, Rodman. N. Y., 164. E. A. Squibb, Le Grand,
Iowa, 150. Charlbs Gilpin, Jr., Philadelphia, Pa., 149. Thomas Baqot, County Superintendent, New Marion, Ind., 149.

PROBLEMS.

854«— Proposed by Artbmas Martin, M. A., Member of the London Mathematical Society Erie, Brie County, Pa.
It is required to compute the number indicated by 4'*^*

255*— Proposed by Rbubbn Knbcbt, Photographer, Easton, Northampton County, Pennsylvania. *

A and B are equal partners in a house and share equal profits. A receives $986.67 for rent and pays

for repairs $854.12, also pays B $112.50. B pays $72.95 for repairs. Now does A owe B or B owe A, and

how much?

256«— Proposed by John I. Clark, Moran, Clinton County, Indiana.

Mr. B offered a certain railroad company 60 rods square of ground, on condition that they would
make a station at a certain crossing. The company claimed 78 feet in width for the road, and agree to
accept of Mr. B*s offer, if he will give that amomit of ground exclusive of the 78 feet. The road is to
enter said plat 15 rods teeal of the south-east corder, and leave it 20 rods east of the north-west corner.
If Mr. B agrees to this, how many rods square must he donate?

26>1»~-^^^ McLbu.an*s Tbachbr's Hand-Book of Algebra, p. 8SS, Ex. 8.

A cask. A, contains m gallons of wine and n gallons of water ; and another cask, B, contains p gallons
of wine and q gallons of water. How many gallons must be drawn from each cask so as to produce by
their mixture b gallons of wine and c gallons of water?

SSS*— Propos<^ by !>• !•• Wright, Mallet Creek, Medina County, Ohio.

What per cent, of income do U. S. 4|-per-cent bonds at lOS yield in currency when gold is 106?

250»— l*ropo8cd hy Williax Hoovbb, Superintendent of Schools, Wapakoneta, Auglaize County, Ohio.
Show that in any plane triangle the angle included between the perpendicilar from the vertical angle
and its bisector equals half the difference of the angles at the liase.

M0»— Propo"^ hy Artbmas Martin, M. A., Member of the London Mathematical Society, Erie, Erie Co., Pa.

Four men. A, B, C and D, start at the same time from the same point in the circumference of a cir-
cular island 200 miles in circumference, and travel the same way around it ; A going 16 miles a day ; B,
23 miles a day ; G, 37 miles a day ; and D, 44 miles. In how many days will they all be together agisdn,
and where? A complete general solution is desired.

-176-

9II1«— PropoMd by A. E. HATinn, M. Ph., Profeflsor of Mathematics and Vhjaics, Hillsdale College, Hilbdale, Mkh.
If a, & and c be unequal and poBiiive quantities, prove that a' + &> + c^ > Sabe.

I,— Proposed by L. P. Shtdt, U. S. Coast Survey Office, Washington, B. C.
A man deposits D dollars every year in a Bank, which allows him compound interest at the rate of r
per cent, per annum. What sum will the Bank owe him at the end of n years?

263«— Proposed by Stlvistbr Robinb, North Branch Depot, Somerset County, New Jersey.

How many acres of prairie sod can a farmer break up in all the years that he can turn over a rectan-
gular piece of different size and integral dimensions every summer, surrounded by an outside furrow
100 rods in length?

964.— Proposed by Joskph H. Kebshmbb, Professor of Mathematics, Mercersbarg Collie, Meroersbnig, Pa.
If A ■£ B = 90^, in a plane triangle, sides a, &, c,

2c^* = (o + &)** + (a— 6)=»=«,
the upper and lower signs being taken together.

Ml^— Proposed by J. F. Bonn, Jndge of Probate Court, Tiffin, Seneca Coonty, Ohio.

Determine the relation between the product of the sides of a given triangle and another triangle
formed by joining the centers of the escribed circles of the given triangle.

B.— Proposed by W. F. L. Sandbb-^, New Albany, Floyd County, Indiana.

Two boys, John and James, ran a foot-race of 200 yards. At first, John gave James 2 seconds and 8
yards the start, and then beat him 2 seconds. The next time, he gave James 16 yards and 5 seconds the
start, and was beaten 20 yards. In what time can each run the distance?

•

26T«— I'fopo'c^ by Thbodobk L. DxLano, Office of the Secretary of the Treasury, Washington, D. C.

U. S. 4-per-cent. bonds with 26 years to run, interest payable quarterly, are worth on the market 112.
Consider this the measure of the National credit. The Secretary of the Treasury wishes to place on the
market a new loan to refund maturing bonds — the 6's ; the loan to run 40 years. Interest payable tri-
annually. Required the rate the new bonds must draw in order to sell at par.

288«— ^^poe^ by Abtbmas Marthv, M. A., Member of the London Mathematical Society, Erie, Brie Co., Pa.
The distance between the centers of two circles, radii R and r, is a ; find the radius of a circle touching
them and their common tangent.

9.— Proposed by P. F. Mangs, Alamos, Sonora, Mexico.
Let a line be drawn from the center of a circle to any point of any chord ; then show that the square
of this line, plus the rectangle of the segments of the chord, is equal to the square of the radius.

270«— I'roposed by E. J. Edmunds, B. S., Professor of Mathematics, Southern University, New Orleans, Louisiana.
Circumscribe a circle about a triangle ABC; draw AD, BE perpendicular to BC and AC, and lnter>
secting in P ; produce AD* to meet the circle in F, and then show that DP = DF.

271«— Proposed by Marcus Baker, U. S. Coast and Geodetic Survey Office, Washington, D. C.
From any point P in the periphery of an ellipse lines are drawn to the foci F and F'. Bequlred the
locus of the center of the circle inscribed in the triangle FPF'.

272«— I^roposed by Benjamin Hbadlbt, DiUsboro, Dearborn County, Indiana.

Divide a board 10 feet long and 2 feet wide into four pieces, so that they can be put together to form
a square.

SHfS*— Proposed by J. M. Quiroz, Alamos, Sonora, Mexico.

The area of any triangle is equal to the radius of the circumscrit)ed circle multiplied by half the pe-
rimeter of the triangle formed by joining the feet of the perpendiculars of the given triangle.

2'74«— Prom Dr. Morbison's Trioonoxbtbt, p. 221, Ex. 27.

If the middle points of the sides of a triangle be joined with the opposite angles, and Ru i2s, R^

R^i be the radii of the circles described about the six triangles so formed, and rj, r^, rs r^ the radii

of the circles inscribed in the same, prove that

(1) R,RA = R^R,Rt and (2) 1 + . I + J = J + ^ + ^.

ri rj rs r^ r4 r^

8!75«— Proposed by Wiluax Woolset Johnson, Member of the Liondon Mathematical Society, Professor of Mathematics,
St. John's College, Annapolis, Maryland.

If AB is the transverse axis of a hyperbola whose asymptotes make angles of 60 degrees with its axis,
and D a point so taken that BD = lAB ; then if P be taken on the branch passing through B the angle
PDA is double the angle PAD, and If P be on the other branch, the supplement of PDA is double the
supplement of PAD.

Solutions of these problems should be received by September 1, 1881.

SENIOR DEPARTMENT.

SoiuHong of the JProbiemg JRropoged in Jf^. 4.

176.— Propoeed by W. L. Habykt, Maxfleld, Pcnobecot Coantjr, Maine.

A man buys a farm for $4000 and agrees to pay it in 4 equal annual Installments, Interest at 5 per cent,
per annum, compounded every Instant. Required the annual payment.

I«— Solatlon by Chablm H. Rummbll, U. S. CoMt and Geodetic Survey Office, Waahiugton, D. C.

Let a = $iOOO, r = 0.05 and x = annual payment. If the interest is compounded n times a year we
have the present value of 1st installment = x{l -|-n-ir)-» ; of the second, = a?(l -|-n-»r)-*» ; of the third,
= aj(l -f-n-V)-*» ; and of the fourth, = x(l +n-»r)-^.

Summing we have, since the sum of the present worths of the payments must be equal to the debt,

atl - (Ij-n-^r)-^ . _ a[(l-Hi-»r)» - 1]

"■ (l+n-*r)» — 1 • . . * — i_(i^^-.iy)_4, •

Since (l+«-ir)- = l + ]J.n-ir + ]^.^^n-^ + ^.'^^'^^n-^+

«4 t4 '

= ! + »' + J 2 + 172.3 "*" (if n = 00 ), = r (f = Naporian base),

we have if interest is compounded every instant,

' = ?S^- = MOOO X f^-^, = «131.38.

n,— Solation by B. B. Sxitz, Member of the London Mathematical Society, Professor of Mathematica, North Missomi
Bute Nonnal School, Kirksville. Miiwoari.

Let a =£ $4000, p =. the annual payment, r = 5 per cent., and n = 4, the number of payments.
To find the amount of a fc»r n years, let x = the amount at the end of t years ; then we have

dx = rxdl,
whence by integration, observing that when i = 0, :b = a, we find

When < = n, we have logf - j = nr, or a: = af***, the amount of a for n years.

The amount of the first payment for (n— 1) years is pet*-!)', that of the second payment for («— 2)
years is p£(»-s}>', and so on. But the sum of the amounts of the payments is equal to the amount of the
principal ; hence we have

p(g(»-l)r ^ g(iM-t)r ^ £(»-8)r ^ + 1) = 0«^»

whence by summing the series, and solving for p, we find

= -"■X^-A) = f --J[) = ,n31.37+.
Solved also by the Propoter and Messrs. Eastwood, HoofBer, FoUard, EoteU, Sch^er, SiveHy and Wood.

-178-

n^— Proposed by E. J Bdmvmds, B. 8.. Professor of Mathematics, Southern Unlvereity, Nfew Orleaus, Louisiana.

The sides of an inscribed quadrilateral are the roots of a given equation of the fourth degree whose
roots are all real and positive. Find the area of the quadrilateral, and the radius of the circumscribing
circle, in functions of the coefficients of the equation.

Solution by Walter 8iverly, Oil City, Venango Co., Pa.; Prof. W. P. Casbt. C. E., San Francisco, Cal.; E. B. Skits.
M. L. M. S., ProfesHor of MathematlcH, North MfsHonri State Normal School, Kirkeville. Mo.; J. F. W. ScuBrrsK, Pro-
fet»or of Mathematici* and German, Mercersburg College, Mercernburg. Pa.; and the Pkoi*oi«er.

Let a, 6, c, d be the sides of the quadrilateral, A its area, R the radius of the circumscribing circle
and X* — nu^ -\- rui* -— px -f ^ = the equation.
Then m = a + ft + c + d = 2», » = a6-fac-fa<i + bc-fM + cd,

p = abc + abd -(- acd + led, q = abed.

Also, A = '\/i{H — a){H — b){a—-c){H^d)'\ = ^(#r» — yna^-l-na*— p«+g). But 8 = Jm, therefore

-A = ii/[i(Pg — 4iwp + 2m«n — TO*)].
Again. R = .//(««' + ^rfHm.^-M)(ad+6c)Y

But (ab -f od){ac + bd){ad -j- be) = cflb*d* -f a«6«d« -f a^ifld^ -j- 6«c«d» -f a6cd(a« + &«-f <?+#).
and a«6*c» -(- a«6«d» -j- a*<^d» + b*<fldf = P' — 2ng, a6rd(a«+6«-hc*-fd») = q(m^~2n) ;

Solved aim by Mesens. Softwood, Hoover. KumnusU, Roeell and Wood.

VXtt—VropoaeA by Jocibph H. Kxrsiinbb, Profe«iflor of Matheroatlco, Mercerttburg College, Mercershurg, Franklin Co., Pa.
Find three square numbers the ratio of whose sum and product shall be a square.

I,— Solution by Gborob Eastwood. Saxon ville, Middlesex County, Mai^sachasetts.

Designate the required numbers by jfl, 3^, «» ; then, per question, J^^Jr~ is i>o be a square. The
denominator is a square already, so that it only remains to make the nimierator a square.

Put ^ = p*a«, s» = g«a^ ; then aj« -f p«a^ + g«a^ = ar«(l-fy« -j- q«) = 3fi{l +p« -h Ip") = J:»(l + iP»)*.
when g = ip*.

Ex. — Take p = 4, and j- = 1 ; then 1, Ifi and 64. are the squares that will satisfy the conditions of
the question. But an infinite number of other squares may be found.

II,- Solution by W. E. Hbal, Marion, Grant County, Indiana.

All we have to do is to find three square numbers whose sum shall be a square. This is aatisfled by

(x« -1-3^+2!')*. 'ir*-:^. 43/«2^, where x, y, z are any rational numbers. Taking x = 3, 1/ ^pf4, « = 1 we

get 144, 36 and 16 for one set of the required numbers. -c = 4, ?/ = 2, z = 3 gives 12^^76, 144.

Good Holutiohu received from MeMnrn. Drummond, Hart, Hoover, PMard, C. C. Robins, Sylreftei' Robins, Sch^er, Shidy
and Siverly.

ITO*— Proponed by Professor W. P. Casey, C. B,, San Francisco, California.

Given the three lines joining the remote angles of the equilateral triangles described on the sides of a
triangle, to construct the triangle and find its sides.

I,— Solution by K. B, Sbitb. M. L. M. S., Prof elisor of Mathematics. North Missouri State Normal School, Kirksville,
Mii^Huuri; and Chaklkh 11. Kuxmell, V. S. Coast and Geodetic Sur\ey Office, Washington. D. C.

Let DEF be the triangle formed by the throe given lines. It is known
that the throe lint^ joining the vertit^es of the equilateral triangles to the
opposite vertices of the original triangle are equal, that they meet in one
point, and intersect at anglas of 120^. Hence to determine their common
point construct the isosi'eles triangles DHF and EKF, making the angles
DHF, EKF each equal to 120^ ; with H and K as centers and radii equal to
HD and KE dascrilie arcs intersecting at O ; then ^ DOF = / EOF =
^ DOE = 120^ ; therefore O is the common point of the three lines above
named.

Again, it is known that the sum of the distances from the intersection of
thej^e three lines Xx^ the vertices of the original triangle is equal to one of
the lines ; hence the sum of the linc-i DO. EO and FO is twice one of the
three equal lines.

Produce DO, EO, FO, making DC = BE = AF = i(DO + E0 -f-FO);

-179-
Join AB, AC, BC, AD, AE, BD, BF, CE, CF ; then ABC is the original triangle.

To compute the sidea of ABC, let EF = a, FD = 6, DE = c, o + & -f c = 28, area DBF = u, and
^DFE = F. Then we have FK = Jai/3, FH = J6i/3, u = ioAsinF = i/[«(«-a)(a — 6)(«— c)],

OPxHK = 2FKxFHXBta(60o + F). or OF = ^^^-^^^^J_^^^+2^e^-^r^r

__ 2a68in(60o + F) __ _ (a«-|.M;-<^)V^3 + 4u

l/[3a«+36«— 6afcco6(60"-f-F)]' "" -/[6(a» + 6«+c«) + 24tti/3]'
Similarlv we find OE (a«+c«-&«V3 + 4u_ _ (b«-f c«~a«V3 + 4u

From the construction we have OC = J(OF -f OE — OD), OB = 1(0F + OD — OE) ; hence, since the
angle BOC = 12(P, we have

BC = \/[(0C)« -h (0B)» + OC.OB], = li/[3(0F)« -f (OE- 0D)«],

= it/ { 6a« + 6M — 2c« — 8i/[3«(« — a)(« — 6)(« — c) ] j ,
and by symmetry,

AC = Ji/ j 6a« + 6c« — 26« — 8v'[3«(« — a)(« — 6)(« — c)] } ,

AB = J^{66«-h6c« — 2a« — 8i/[3«(« — o)(«— 6)(«-.c)]l.

The solution in Vol. XXXII of the Educational Times Reprini, p. 60, (of Quest. 5811,) is mine, although
credited (by some mistake I suppose) to others.

II« — Solution by William Hoover, Superintendent of Schools, Wapakoneta, Auglaize County, Ohio.

Let ABC be the required triangle, and DEF the triangle formed by Joining
the remote vertices of the equilateral triangles constructed on the sides
AB, AC, BC. Draw the circumscribing circles of the equilateral triangles ;
they will intersect in the point O. Draw DC, BE and AF ; they will inter-
sect in O and be equal, and the angles DOE, EOF and FOD will each be
equal to 12(F.

Put EF = a, FD = 6 and DE = c. Let FO = u, EO = v and DO = w.
Then, since cos 120° = — i, we have

,4*^_titj-f v« = o« (1), t«» + uic-hw« = 6« (2),

t?« -f wc -f w« = c« (3).

These equations are the same as (1), (2) and (3) in the solution of Prob-
lem 123, No. 4, p. 105 ; therefore

_ a» -f- 6« —^ ± Ji/[12a«6« -- 3(a«-|- &•--€•)•]
" "" v/{2(a? + 6«-f-c«) + 2i/[12a«t« — 3(a«-f6«— c^)*])'

_ a«-f-c« — 6«-f Ji/[12a«6« — 3(a»-|-6« — <^)»]

* " V^|2(o« + &«-fc>«)±2i/[12a«6s--3(a«-|.62-c*)«]|*

5«-|-c« — a« + ii/[12a«5« — 3(a» + 6« — (?«)«]

^ — v/12("a«-|-6«-i-c«) + 2i/[12a«6« — 3(o«-i-6«— c«)«]}*

Now let CO = a;, BO = y and AO = z. Then, since each of the angles at O = 60°, we have

3fi — xu-{-v^ = 2^ — yM-|-u«, or a^ — y* = u{x^y); whence a;-|-y = t« (4).

Similarly, we find x-^-z = v (5), and y-|-2 = w (6).

X = l(u-|-«— tr), =

3a«--6« — c« + Jv'[12a«6« — 3(aH-6« — c«)«]
v/ {2(a«-f 2^-i-c«) + 2i/[12a«6« — 3(o«-f fe«--c2)«] j *

_ _ 3fe« — a« — c» + J-i/[12a«6«--3(a«-f6«— <<)«]

y - i(w-f«« «). - ^j2(a«+6«+c«) + 2i/[12a«6« — 3(o«-f6«--c«)«]}'

— i( J. _ 'i = 3c«--a« — 6« + J-i/[12a«6« — 3(a«-f 6«--c«)«]
« — ii'v-hv) u), — ^j2(a«-i-&«+c»)±2t/[12a«6« — 3(a«-|-6«— c«)«]p

From the triangles AOB, AOC and BOC we have

AB = -/(3^ + 3^-f ««), = Jv/{66«-f 6c« — 2a« — 2i/[12at6« — 3(a«-f6«— c«)«][,
AC = i/(a:«-|-a»-f 2!«), = lv/{6o«-f 6c«- 26«--2i/[12a«6« — 3(a«-f 6«— c»)«]j,
BC = -/(a^+a^y + S^). = l>/i6«* + 6&*--2c« — 2i/[12os6« — 3(a«-|-6«— c»)«]|.

-180-^

III*— Solatlon by J. F. W. Sciuimt, ProfeMor of Mathematice and German, Meroenbiug College, Meroenkmrg, Pa.

Denote the sides of the required triangle ABC by x, y, z; and those of
the given triangle A'B'C by a, h, e; and the angles of the former by
A. B, C.

In the triangle AB'C, we have a« = y» -f «» -h 22^ cos (60° — A).

Since cos (60° — A) = cos 60° cos A -f sin G(P sin A. we have
2y8Cos(60^-A) rrj/BCOsA-fi/SyssinA = g^cosA +21V3,
denoting the area of the triangle ABC by F. But since

cosA = ^-+2**^^, we have 2a« = 3(y»+««) -aj« + 4iV3 .... (1),

and by analogy,

2ft» = 3(aj»H-««)-~y» +41^1/3.... (2). 2c« = 3(aJ«+y»)~.«»-f 4iV3 . • • (3).

From these we derive by subtraction.
j(at_5*) = y« — aj«, whence y* = i(a»— 6«+2;C»); and f{cfl—<fi) =«• — ««. whence a^ = J(or« — <?4-2as«).

Substituting these values of 2^ and 2^ in the well-known formula

16 JJ'* = 2(xV + a?^ + 3^«») — a:* — y* — «*
we get after some reductions which present no difficulties,

82?'= i/[12«< + 4(2a;i—6«—c«)«« — (&«—<?)«].
Substituting this value and those of 3^ and 2* in (1), we obtain a quadratic equation whose root is

x» = J[3(&«+c«) — a« ± i/(2o«6«-h2a»c«+2d«C«— a* — ft*— c*V3].
Denoting the area of the given triangle A'B'C by A. we have, choosing the lower sign,

« = IV'CCC&'+c^) — '^o* — «Ai/3]. and by analogy y = iV'[6(a«4-<J») — 26« — 8A\/3].
z = Ji/[6(o« + 6«) — 2d» — P Al/3].
Solved alM> by the I^ropoMr, Oeorge Ecutwood and Profewor Edmundi,

I'XB*— l^v^P^x^ ^y Sylvbi^tbr Kobinm, North Branch Depot. Somerset County, New Jersey.
1 n a series of rational right-angled triangles where every hypothenuse is unity, each leg of the nth
triangle contains n decimal places. Find the legs of the tirst 25 triangles, and those of the 50th one.

Solution by the Pboposbr.

Three conditions must be fulfilled in the solution of this example : (1) the square of the hypothenuse
must equal the sum of the squares of the legs of every triangle ; this condition is satisfied by writing
2r8, rs— s< and r*-fs< as the general expressions for the sides of any triangle in the series: (2) the
problem requires that every hypothenuse shall be unity ; this condition is met by dividing the above

expressions by f*+«*, and our general expressions become r, , ^. ^^ ^ and 1 : the last condition

(3) insists upon each leg of the nth triangle containing n decimal places ; this demand is complied with

by making r* -j- «• = 5».
Since r^'\-^ = 5», it is plain that (2r4-«)« + (2«-r)» = 5»+» and (2r8)« -i-(r«— 8«)« = 5*». We now

have a key by which the different values of r and s for the several triangles may be obtained.
Setting out with the values r = 2, « = 1 we find the successive values to be :
(2) r = 3, « = 4; (3) r=ll, « = 2; (4) r = 24, « = 7; (5) r = 41, « = 38; (6) r = 44, « = 117;

(I) r = 29, « = 278; (8) r = 336, « = 527; (9) r = 718, « = 1199; (10) r = 237, « = 3116;

(11) r = 6469. 8 = 2642; (12) r = 11753, = 10296; (13) r = 8839, a = 33802;
(14) r = 16124, 8 = 76443 ; (15) r = 136762, 8 = 103691 ; (16) r = 164833, 8 = 354144;

(II) r = 873121, 8 = 24478. (18) ; r = 922077, 8 = 1721764; (19) r = 2521451. 8 = 3565918;
(20) r = 1476984. s = 9653287 ; (21) r = 6699319, s = 20783558 ; (22) r = 34182196, s = 34867797;

(23) r = 35553398, s = 103232189 ; (24) r = 32125393, a = 242017776 ;
(25) r = 306268562. s = 451910159; Ac.

The legs of the first 25 triangles therefore are :
(1) .6, .8; (2) .28, .96; (3) .352, .936; (4) .5376, .8432; (5) .07584, .99712; (6) .668944, .752192;

(1) .2063872, .9784704; (8). . 42197248, .90660864; (9) .472103424, .881543168;

(10) .1512431616. .9884965888; (11) .70db5l37408, .71409248256; (12) .131585609728, .991304810496;

(13) .8719952142336, .4895143985152; (14) .40388753227776. .91480864735232;

(15) .225775162589184. .974179437248512; (16) .7651277924204544, .6438784522452992;

-181-

(IT) .99842930528354304, .06602608634396672; (18) .564236714094952448. .832369096038214464;

(19) .3333452483696001024, .9428043288958906368; (80) .29900669664185430016, .95425101213847257068;

(81) .583996790525665476606, .811755966196566962656;

(88) .0198561472974078063072, .9998028472726528720896;

(83) .61576662620151796989952, .78792858943967761268736;

(84) .260882895830831308210176, .965370454625020943532032;

(86) .3705159561103475195510784, .9288261011985155397517312.

Legs of 50th triangle,

.e88a607818Q10887747O7Ot701M88S8B8797Ban^^ .7»48665a6a8«7OO6118O6OOB88O0OO«714O6MaOO79tiO^

LegB of 100th triangle,

.fiaM201604857»761ia»'l«883a8aW1754S»^

18O«~l*n>po0ed by Dr. John Bbll, Manchester, HUlsboroiigh Coantjr, New Hampehire.

One-third of all the apples on a certain tree are rotten, and one-fourth of all the apples on the same
tree are wormy. What are the respective chances that an apple taken at random from the tree will be
(1) sound, (2) rotten, (3) wormy, (4) both rotten and wormy?

I,— Solution by Abtemas Martin, H. A., Member of the London Mathematical Society, Brie, Brie Coonty, Pa.

1* The least part of apples sound = 1 — (| + J) = ^^ ; greatest part = 1 — | = |. All parts between
these limits are equally likely, hence the chance of taking a sound apple is }(! + /f ) = IJ.

8. The least part of apples rotten only = ) — ) = y>f ; greatest part = | ; therefore the chance of
taking an apple that is rotten only =^ l(t^+i) = /|.

3. The least part of apples wormy only = ; greatest part = I ; therefore chance of taking an apple
that is wormy only = Jxl = I*

4« The least part of apples both rotten and wormy = ; greatest part =. J ; therefore the chance of
taking an apple that is both rotten and wormy = Jx^ = i.

II«— Solution by Abtemas Mabtim, M. A., Member of the London Mathematical Society Brie, Brie Coanty, Pa.

Let 12m = number of apples on the tree. Then in = number rotten, and Sn = numlier wormy.

Letp = probability of taking a sound apple at random from the tree, pi =. probability that the apple
taken is rotten only, p^ = the probability that it Lb wormy only, and p^ = the probability that it is both
rotten and wormy.

1. The number of sound apples can not be lees than 12n — (4ii+3n) = 5n, nor more than 12fi — 4n
= 8ii, and all numbers from 5ii to ^n are equally likely.

If the tree contains 5n sound apples the probability of getting one of them is -;- ; but the probability

that the tree contains 5n sound apples, or any other number from 5n to £n. Lb /~«-~ v^^nr^ = a-. TT »

therefore the probability that the tree contains 5n sound apples and one will be taken is ^ X »- ^^ .

The probability that the tree contains 5n + 1 sound apples and one will be taken is ^' X o^rri •

Similarly, the respective probabilities that the tree contains 5n -f 2, 5n + 3 ^n sound apples,

•ndoneofthem win betaken, are j^3+^jj, ^^^^^y la^^gj^+i) fi^Jr+T)-

The total probability of obtaining a sound apple at random from the tree Lb the sum of these separate
probabilties.

_ 5n-h(5n-hl) + (5n-|-2)-|-(5n-f3)+ .^ . ^. . + 8n _ (^-h8n)xi(3n+l) ^^ la
" ^ "■ r2n(3n+l) "" 12nx(3n4.i) '*'

8« The least number of apples rotten only = 4n — 3f» = n, and the greatest number = 4n.

. « _ n-Kn-fl)4-(n-f-2) + (n+3)- h -1-411 __ (n+4n)Xi(3nH-l) ^

"P'" 12n(3n-|-l) "* 12n(3n-|-l) "'

-182-

3. The least number of apples wormy only = 0, and the greatest number = dn.

__ + 1+2 + 3+ +3n _ 3nxi(3n+l) _ .

•'Pi— 12n(3n + l) ~ 12n(3n + l) "" *'

4. The least number of apples both rotten and wormy = 0, and the greatest number = 3n.

_ + 1 + 2 + 3+ +3n _ 3nxi(3fi + l) _

••^~ • 12n(3n+l) "" 12n(3n + l) "" **

III*— Solution by Abtbxab Martin, M. A., Member of the London Mathematical Society, Erie. Erie County, Pa.

Put . = part rotten and , = part wormy. Let p ^ probability that an apple taken at random from

the tree is sound, pi = the probability that it Is rotten only, p^ = the probability that it Lb wormy only
and ps = the probability that it is both rotten and wormy.

Thed we have pi + jsj = ^ (1), pt +p3 = ^ (2), p +pi +p« +i>3 = 1 (3).

From (1), (2) and (3), p^ = 1~^-P (4). p, = 1-^-P W-

1. When ;, + J < 1. ^c loast part of apples sound is ^ — ( ;, -h ^ )• *"^<i ^® greatest part is 1 — .
If > , ; all parts between these limits are equally likely,

From (3), (4) and (5) we now readily find, by common rules,

^' = 6 ""2d "= A' ^^ = 2d = *• ^' = 2d = *•

2. I' 1^ <C j> the greatest part of apples sound is 1 — ,, and

p = ^-<Us)'

2c\ a c a a

P» = 26' ^* = d-26' ^ = 25*

3. When |^+. = lor]>l, the least part of apples sound is 0, and the greatest part is 1 — . if

Gl C C CI C

. > ' and 1 — , if , < , ; therefore in this case
o a a a

P = 4(1-6) ''hen *>^ and = i{l-'^) when °<^, p, = i(l + °_ J) or l(l-|).

P. = i(i-t) or Kx+^'^-)• "' = c+';-o - iff +:-o-

Solved also in an elegant manner by CharUs H. KummeU of the U. S. C(»a0t and Geodetic Survey, Washington, D. C.

181a— Proposed by Marcus Bakek, U. S. Coast and Geodetic Snrvey Office, Washington, D. C.
A polygon Lb both inscriptible and circumscriptible, radius of inscribed circle r, of circumscribed
circle R and distance between the centers of these circles d; then if the polygon be a triangle,

R-\-d^ R—d r'
if the polygon be a quadrilateral, ^^^^^, + ^^^^^, = ^.

What is the corresponding relation for a pentagon?

Solution by Charlbs H. Kummsll, U. S. Coast and Geodetic Survey Office, Washington, D. C.

The general solution of this problem can, I believe, only be effected by means of Inverse Elliptic
Functions, as has been done by the illustrious Jacobi. — (See Durege, Elliptlsche Functionen, pp. 164 — 186.)

I shall condense out of Jacobi's highly interesting discussion that which pertains directly to the prob-
lem with some modifications of my own.

-183-

Lemma. — If there are two circles, one wholly within the other and the
diameter AB = 2i2 is drawn through their centers C, c. whose distance
Ce = d, and if the radius of the inner circle = r ; then if from A and
likewise from any point F on the outer circle chords AA' and FF' are
drawn touching the inner circle, and arc AA' = 2a, arc AF = 2(^, arc

(1)

l/[(H+d)«-»«]-^'^-

'5^

ir

A ^

J JL-^ .

r-B

! 1

AP* = 2^, we have

where Jg>^ 1/(1 — /c'*sin><p) and the modulus k

If T is the point of tangency of chord FF' then

FT = i/[(Fc)«-r»] = -/(iJ«-|-f|i+2/Wcos2<p— r»)

= \/[(i^+<*)* — r»~iiWsin«<p] = aJtp (3)

if we denote the tangent At = i/[(^ + d)* — r«] by a (4).

Similarly we have F'T = aJ(f/ (3') and A'< = aJa (3").

Draw CH perpendicular to cT ; then

(y = HT = RQoe{(f/ — q,) (5), <-H = dcoB{<pf -{- <p) (6);

therefore cT = r = /2coe(<p' — <p) + dcos(<;p' + <p) (7).

We have Ff= 1(FT + FT) = la(J<?/ + J«p) = R&in{<f/ — (p) (8),

CH = KFT-FT) = ia(J(p-^Jg>') = d&ln^g/ -^ <p) .'. (9);

■• sinC^?/ — <p) "■ a ^ '* sin(^ + <p) «

On the chord AA' we have similarly, placing ^ = 0, ^ = a,

sina a ^ '' sina a ^ ^' \l + ^a/

., ,r> . jx 2Jflcosa

Also. r = (iJ-hd)cosar = i_^j-

These values being used in (7) we obtain, clearing of fractions,

cosa = J(l-h^a)cos(<p'--<p)H-l(l — Ja)cos((^ + <p) = cos^cos*;?/ -f siu<psin^Ja (14).

In a spherical triangle in which €p/ is an exterior angle and the sides x, oif and A are respectively op-
posite the angles ^, n^qif and a we have coscr = cos^cosip' + sin^sln<^'cos-A (14').

» X ,- sina; sin«' sin-A , 2i/(/M) ,--.

But if we assume , = , - = , = * = -^^-^^ — ^ (15)

sin^ sin<p' sina a

we have cosoj = J<p, cosa;' = J<p', cos A = da (16)

and (14') becomes identical with (14), and as I have proved in that case (Analyst, Vol. V, pp. 17, 18) we

r^' dq) _ C'^dqt

J^ Jq) "" Jo ^tp

The integral of this equation is either (14) or the forms

.(11).

(12).
(13).

have

.(17).

I _ sip <P^ CO B <y) i^ y — sin y cos €pl AqJ
sina— 1 — fctsin'^sinV

cosa =

cos q)' cos q> + sin <pf Jq>'%\n q{A$

Ja =

1 — fc»sin*<psinV
Jtp ' J(pj\- Ifi sin q)' cos (^ sinjp cos ^
1 — fc«8in«<p8in«<p'

which are given 1. c. or in any treatise on Elliptic Functions.

If arc AF = 2q>, arc AF' = 2<^, arc AF" = 2q,'\ etc., and arc AA'
= 2a, then we have by (17)

C^'dq>

Jq)'
f ♦" dq>

^q> ~" J*

Ja(

[•♦<•) dq}

y^(»-i) j(p

. 'dq> _ /•♦<"> dq>

^<p

Aq*

-184-
n 9>(*> = n + tp tbe polygon will doee and we have

nfj*?. = r+*-*^-= r'_^?. = 2r*'4?- = 2^ (19).

Jo Jg> J^ J<p Jq Jg> Jo Jq> ^ '

(using Jaoobi's notation for the complete elliptic integral of the first species).
We have then

arc \kk' = a = am -^ (ampUtudo ^^ (20).

This equation or (19) evidently gives Implicite the radius of the inner circle if its center is given such
that a polygon of n sides may be inscriptible in the outer and circumscriptible to the inner circle.
The integrals of equations (17) are by (14) and (20) for ^ =

2 2
oosa = cos am -iT = cos am - IT (21),

4 2 4 2 2

= cos am -iTcos am -K 4- sin am -ITsin am -KJam -K (21').

n n ' n n n '

= oosam2iroosam2(l — ^W+sinam2irsinam2(l — -^irJam^ir....(21(»-»));

or, writing (J» = co6am-2£; S» = sinam -2ir, T)i = jAm -K,

i n -an s «

a= aa + SiS.d.= aa+s,s^B, =

= C,C<t-s + S,S'-iDi (21'. 21" 31(-»).

By the relations sinam (2£^— «) = sinam« .-. Sl = Oi_» (22),

coeam(2£^— «) = — cosamt .•. C£ = — Ci_» (23),

equations (21') become if n is even

a=^aa+s,s,D, = ac,+s.SiD,= = (2i'.2i")

• •

-C^C^_, + S^S^_,I), = S^_,D, (2i«-.)

because (_/. = cos am AT = and 1^. = 1,

= ^4+i^l+'S^i+i'S^i-0i = >Si_iDi by (22) (31K-+*))

by (22) and (23).

= C,,C,_, + S,_.S,_.D,= aC, + S,SrD. (2K-')

by (22) and (23),

= C\C\_L + SiSi_i.DL= Ci because Ci = 1. /Si = (21{— )).

There are then only ^n really different equations. U n is odd there will be one middle equation, viz :

Ci = C+i C-i + S'+xS'^-i D. = - C'^ + S^-i D. (2i«-»).

by (22) and (23).

-185-

Because (y^ -f ^, = 1 we can express Oi and Oi ^om (21'), then C » »nd ^ t ftrom (21"). eto.,

UN n n it ft

in terms of (Ji, Si *»<! Jji- If n is even (21*( »-«)), and if n is odd (21*(«-»)) becomes an equation

• m u

between Oi > Si '^^^ Jji > ^cbt is, between oosa = cos am ~ K, sin a and J a. But solving (12) for

Jaweobtain ^^ = -^l = f^' '• <><»«= Ci = ^^ by (13) (24).

and sin a = Si = RXd ^^^ (*^' ^® have, finally, an equation in R, d, r and a = i/[(iJ+d)* — **]•
!• For n = 3 we have

r r« / ^^ \//^~^\ _ ^^ <^ __ / 2ig \/ r«^ \

^' iJ+d ~" .(/2 + d)* "*" \{R'\-d)*)\R-\-d) ~ /2 + d VJB+d A(^ + <*)»/ *

that is, r = « — d— -^^'*-. Solving this quadratic we find r = ^^-[—« — di (3«—d)]. We

have to reject the lower sign because r = — (i2+d) makes the radius of the Inner circle greater than
that of the outei^ and n^ative which is against the supposition. There remains only

^=-2^--' ^' 7 = :B-f-d+lB3d (25).

2« For n = 4 we have

squaring. ^ = ({^y[(fi+d)._^], ... 1 = ^^^^ = ^^, + ^- (36).

3. For n = 5 we have

From the second equation we find
Substituting in the first equation we obtaia

. - c,V(i>, + ■) - q V(i). - c.) - s,i),V(C. + .).

" \R-}-d)M\R+d) \R+dJ'\\ /e-t-d J \(>e+d)«;\V iJ+d r

= r(fi-|-d)[i/(2r) — -/(H-d— r)] — (i2~d)(«-fd+r)i/(Jfl+d— r) (27).

. Progressing in this manner we find for the hexagon

0==i2«— d» — (Jfl-fd)i/[(Jfl — d)« — r«] — (R — d)i/[(«-fd)«-t4] (28);

for the heptagon,

= r[4fidr«-f-(i2«— d»)«]i/(2i2) — (JfJ + d)[2(i2«4-d«)r«— (K« — d«)«V(fl-d-r)

— 2r(«+d)(i2 — d)«(i2H-d-f r)i/(fl-fd-r) (29);

for the octagon,

= r(i2 — d)[4i2dr« + (fi«— d»)«] - (i2 + d)[2( R«-hd«)t^ — (Jfl«-d«)«]i/[(Jfl — d)« — f^]

— 2^*{R — d){R-\-d)*^/l{R'^d)* - r«] (30) ;

and so on.

-186-

tSS,— PropoMd t^ Alixaiisbb MACFABLAm, M. A., D. Sc., Edlnbarg, Sootiaod.
Prove that

!•— Solution by Waltbb Sivsblt, OU Clcy, Venango County, PennaylTiniA.
Put a'-p=i'y and, applying the well-known formula

log(c+^) = logo + 1 - .g + ^ - 5 + etc..
log(«.+b.+2a6eoer)* = lo8« + (> '+^<^^) _ ( V-^^r^ + (M+agcoey). _ ^
Expanding and arranging terms in ascending powers of -,

log(or«-f 6«-f-2a6co8^)* = loga-f -cos;^ — 5--(2coe^;'--l)-f ^r— (4oo^— 3ooe^) — etc.,

b lA Ifi b*

= l^go + -cos;' — 5^^0062;^ + 3S*^^>' ~ ^*^*^ + ®**' *

.-. log[a«4-6«+2c*coe(tf— /tf)]* = logo + ^C06(a—/J) — ,^cos2(a—>ff) + ^oos3(a—>J) — etc.

II«— Solntlon by W. £. Heal, Marion, Grant Connty, Ind.; and Wilxiax Hooysb, Superintendent of Schools, Wap»-
koneta, Auglaise County, Ohio.

Put f = (a— /tfV—l ; then, by Euler's Theorem, 2cosn(tr — yff) = c"* + c— ♦.

ii» + 6« + 2aftoo6(a-/:^)=^a»[l + 5 + a<<'^-*-^"*)] = ""O +i^)0 + i^"0 '
log[a« + b» H- 2a6cos(a->tf)]* = logo + 1 riog(^l + ^ «*) + log(^l + S'"*)] •
Peveloping by the formula log(l -f «) = » — Jj^ + Ja? -- Jx* + J«» — etc.,

log[a« + 6« + 2a6cos(a~y5)]* = loga4.^[l(£^ + €-»)] -J^,[j(£* + «-^^^ + et«..

= loga + ^cos(a-/:^) ~ l^coe2(a-/tf) + 4^coe3<a~/J) -etc.
Solved also by Profetaor Ludwig and Profewor SeJk^er.

183.— Proposed by Abtsxas Martin, M. A.. Member of the London Mathematical Society, Erie, Brie County, Pa.

A clock which Indicates correct time at the level of the ocean is carried to the top of a mountain near
by, the ascent occupying h hours. On arriving at the top of the mountain the clock was found to be m
minutes too slow. Required the hight of the mountain.

Solution by Edwabd A. Bowsbb, Professor of Mathematics and Engineering, Bntgers College, New Bninswick, N. J.

Let a = the hight of the mountain, r = the radius of the earth at the level of the ocean and i the time
of ascent to any point P. Put auOOA = 6.

Then from the principles of Mechanics we have o^tjtji ( = r) = hight of P above the level of the
ocean, i^\(-) = time of vibration, = one second at the level of the ocean by hypothesis.

^\\^) = (l -f p^ = time of vibration at the point P. 1 — . = number of vibrations lost at

the point P in one second.

.*. r (l — . / )<ft = number of vibrations lost in ascending the mountain in the time b (=• 9600^
seconds), which == COm by hypothesis.

-187-
Hence Integrating we have

l--log(i + -)= J .
Developing by Maolaorin's Theorem we have

Reverting this series and restoring the value of &, we have

„ / m \ , 8r/ m \« , 28r/ m \» , 4Mr/ m \* . ^

3\60A7 ^ 9 \60A7 ^ 13o\60Ay
Solved in a similar maimer by Pnifeasor Afto and TfU/iom Hooter.

184.— Proposed by Dr. S. H. Wbioht, M. A., Ph. D., late Matbemadcal Editor YaUs County Chronicle, Penn Yan, N. Y.

Given the latitude = A = 42^ 30" N.. the sun's declination = <$ = 20<^ N., his radius = r = 16', and a
vertical wall running S. 10^ W., to find when the sun will first shine on the west side of the wall, the
points from the sun*s north point of ingress and egress on the wall, the altitudes of those points, and of
the sun's center.

Solndon by B. B. Sbitz, M. L. M. S., Professor of Mathematics, North Missouri State Normal Scbool, Klrksrille, Mo.

Let Z be the zenith of the given place, N the north celestial pole, ZQ the
vertical circle coinciding with the plane of the wall, S the place of the sun's
center at the required time, and T the point of contact with ZQ. Draw the
arc NM perpendicular to QZ, and produce it till it meets TS produced at P.
Join NS and ZS.

Let ZN = 90O — A, 8N = 90° — <5, ST = r, 8P = 90^— r, MN = >-, ZM = f
PN=e, Z8=K, /NZM = yC^=l(F, /ZN^ = ^, /SNP = >u, Z8PN =
MT = CO, and / PSN = p.

Then in the triangle MNZ we have y = sin-»(8in/tfcosA) = 7° 21' 20",

f = tan-»(oos/tf cot A) = 47^ 3' 46", and ^ = cot-^tanytfsin A) = 83^ 12' 24" ;

.'. e = 9(F+;^ = 97^21' 20".

In the triangle NPS we have

u - 2tan-i(' ^^"^^^-'-^^^ ^-^-*^ ^-^^"^" ^>V - 870 1' 14-
M - -fiuui \^co8j(6-r-(5)sini(eH-r-(5)y "" *

a, = 2ten-i(«^i(|+^-+4)«|4(^-^ = 690 47' 24",

\co8l(fii--r— tf)8inl(6i — r + d)/

p = 2tan-.(-«^"lf^'-+-«>«'"M9;t!:^l)« = 97c 65' 43".
^ \co6l(6 — r— d)co8l(6+r + <5)/

Hence the hour angle of the sun = // — ^ = 3^ 48' 50", and the required time is ^(fi — if) = 15 min.
15} sec. after apparent noon. The distance of the point of ingress from the north point of the sun is
180O — p = 82<^ 4' 17", and the altitude of this point is 90^ — (w— ^) = 67^ 16' 22".

In the triangle STZ we have r = cos-i[(co8rco8((»— ^)].= 22^ 43' 44" ; hence the altitude of the sun's
center is 90" — r = 67^ 16' 16".

For the point of egress we must substitute — r for r in the expressions for oo and p ; hence we have

\co8 1(6 -I- r — d ) sin i(9 + r -I- (5 ) /

^ \cosl(6i+r — <5)cosl(6 — r+^)/ '

which is the distance of the point of egress from the north point of the sun. The altitude of this point
is 90" — (gj' — 0) = 67° 11' 58", the zenith distance of the sun's center is k' = cos-'[cosrcos((w' — 0)] =
22° 48' 8", and hence its altitude is 90" — k' = 67" 11' 52".

Solved also by the Propoeer and Professor De Voleon Wood.

-188-

186.— Propoeed by Samukl Robsbtb, M. A., F. R. S., President of the London Mathematica] Society, London,

Show tliat if the system of equations

cui^-^-bxy-^Ci/^ = u*, ojfi — bxy — cj^ = »•,

are resoluble by integer values of x, y, u, v, (zero excluded,) not having a common factor greater than
unity, successive solutions can be obtained of the same character.

Solntion by the PaopociBfi.

We suppose a, 6, c to be integers and for the satisfied pair of equations we may write

where A stands for axi*, B for bx\yi^ and Cfor cy{*.

Then if Ac« -f Bj^y + Cy = t/*, Aafi — Bxy — Ci^ = V^

It may be assumed that

x = fc«-f^, U=ux'\-2k(vl—vk), V=vx-'2}{vl—uk)

and by substitution •

C:y« + B(&«-fe»)y= [(fc»-f ^)u-f 2fc(i?i— mA)]«^A(*« + ^)«.
or [2Q/-fB(;k«+^)]«===4C[(fc»-h^)ii-f 2/f(iJ«— uJk)p--(4AC— B«)(fc»-fe»)«,

= (B-f 2(7)«(Ae« + i«)« 4- 16C^«(fc»-h««) — ^2Cavm — 16C(t«»— »«)*«^.
and y is rational and generally integral if

[2(7M»r4-f (B-f 2C)»(B-f C)]i -f (B + 2C)«i*tJfc =

so that

y = « +

B + 2C*

186«— Propoecd by Marcus Baker, U. S. Coast and Geodetic Survey OfDce, Washington, D. C.

Inscribe in any plane triangle three circles each tangent to two sides and the other two circles.

!•— Solution by £. J. Edxunds, B. S., Professor of Mathematics, Southern University, New Orleans, Louisiana.

Let M, N, P be the centers of circles inscribed in the
angles A, B, C. Draw MD, NF perpendicular to AB.

Assume AD = x, BF = y, CH =2; a = pbin«a,
6 = p sin*/tf, c = p sln*;^, with p = \{a -f 6 -f c)
and a, 6, c lengths of sides of given triangle ABC ; a, p, y
are then determined.

Let us again assume A. = 4(a-f^-|-;^); we shall prove
that X = psin2(A — or). y = p8in*(A — /3),

«=psin«(A — x)-
In order to do so let us remark that the radii of the two
circles tangent to AB are expressed as fbllows :
MD = a;tan JA, NF = ytanJB.

Evidently we have

DF = c-a;~iy = i/CCMD + NF)* - (MD- NF)«] = 2i/(MDxNF) = 2i/(a?ytaniAtaniB).
But by Trigonometry

-i'-vcKpi-ro-i'-vr

— a^^p — r^

and we have

; .-. tan JAtanJB = 1 = cos«x,

PCP-6)

x + y-f 2cos;^\/(x!/) =psin«x. J

j?-f-2H-2cos/^l/(a») =psin*/^, ^ (I)

y + «-f-2cosa'v/(y5) = psin'cr. ;
In a circle, radius = 1, construct a triangle having angles A — a, A — ytf and n^y (their sum = ;r,
referring to the value of A). The sides of that triangle are expressed by sin(A— a), 8in(A— /^) and
sin y^ and we have

sin«(A — a) + sin«(A--y5)-f 2cos;^8in(A — (r)sin(A— /?)= sln«^, j

8in»(A--a)-f sln«(A — 7^) + 2cosytfsin(A--a)sin(A— r) = sin'/?, \ (H)

sln«(A — ytf ) -f sin«(A — 7^) -j- 2 cos cr sin( A — fi) 8in(A — r ) = 8in«a ; )

and if we take x = psin«(A--a'), y = psin«(A — /?). « = psin«(A — ;') (I) and (II) are identical,
therefore the problem is solved.

The substance of this solution Is taken from Prof. Housel's Introduction a la Geometric Superieure,
page 144.

-189-

II»— Solntlon by Rev. IT. Jsms Knisblt, Ph. D., Newcomerstown, TuBcanwai) County, Ohio.

Designate the sides of ABC, as usual, by a, 5 and c. Suppose the circles Inscribed. Lot DP = x,
HN=y, ST = «. WeknowthatBK = i(o-hc— 6), CV = l(a-f-6— c), AK = l(&-f-c— a); designate

OK = r = J('(_« + &-c)(a-6-hc)(&-a + c)\

these segments by m, n, p. Also,

'*""\'V 4(a + 6-|-c)

AP =

pa?

r r r ■ r

(PD - HN)9 + (PN)« = (DH)«. (HE - ST)» + (SE)« = (TH)«, (DL — Mt)« -f (ML)» = (TD)» ;
or, by substitution,

(y — «)• -f- ( a — "^ J =(y -{-«)«. Hence we have

2v/(«j,) = c-?^+-'»» (1). 2V(x.) = 6-?"|'« (2), 2v'(j^) = «-""+•«.

From (2) and (3) we have

n\/2-|-r^x = \/[(m6 — (pn — r«)x], n^2 + r\/y = !/['"»»« — (w*w — r«)y].

Hence Kl/-c — l/y) = i/[m6 — (pn — r«)a;] — ^[r?ia — (mn— f«)i/].

Squaring, transposing, collecting, and putting for px-f mj/ its value re — 2rv/(ir3/), found from (1), we

26(6-fc— a)(a + 6— c)
4(a + 6 + c)

.(3).

K I -^ 2tt(o + c-6)(a--6-fc)

have, since mn — r«=-^ .^ '\ -^ ^ and pn — r« =

m(a-f 6-c) + (2r«+2mV(xy) = 2J^rHi«a6--

2m»o6

2r'rMi6xy

(px + mi/) +
-1.

)■

- ^'\ L"*U+ft+c+ a + 6 + c + a + 6+c)J

after some transformations ; hence

m(a + 6-c) + (2r»+2m)i/(a^) = 2r^'(j^-^'^^[n» + 2ni/(xy)+a;2/])

Therefore

2*2/ =

-[V(.+n.)-"] -[V(.?r+e)-"]['-+%(.:t;,)]

.( 2aJm \

'*-[\(a+t;.c)-»]'

_ 2CV(0C — CV)(r + OC— CV)
r» — (OC — CV)«

(X).

-190-

2CV(0 C— CV) _ 2CV( 0C~CV) 2CV( QC— CV ) OC--CV _

■""^ ,4_-(oc::_cv)« "" 2cv.oc — (CV)« — [(0C)»— t^l "" 2cv.bc — 2(cv)« - oc— cv " '

hence (X) becomes W{xy) = PN = r -f OC — CV.

Similarly 1V{xz) = ML = r -f OB — BK, W{ys^) = SE = r + OA — AK.

Now, to find AP, BN and CS :

AP-fBN = c — PN (4), AP-fCS=:6— ML (5). BN-fCS = a — SE (6).

Subtract (6) from (5), add to (4) and divide by 2 ; then

AP = i(fe-fc— a— ML — PN + SE), = i[AK — (OB-f r— BK) — (OC-f r—CN) -f (OA-f r— AK)].

= J[OA-hAK — (OB-fBK)--(OC — CV)— r],

= 1[20A — (O A — AK) — (OB — BK) — (OC — O V) — r],

= 0A — i[r-f(OA — AK)-f(OB — BK)4- (OC-CV)],

= 0A— {r — l[r-(OA — AK) — (OB — BK) — (OC— CV)]}.
So, BN = OB— {r — i[r — (OA — AK) — (OB — BK) — (OC— CV)]],

CS=OC — {r — J[r — (OA— AK) — (OB— BK)-(OC-CV)]}.
If a, p, y denote the lines OA, OB, OC and « = l(a-f6-fc),

AP = l(«_r-f a-yiJ-r), BN = l(8_r+/5r-a-r). CS = i(8_r-f ;^-a-y5).
From the similar triangles AOK, ADP we have

AK:OK :: AP:DP; or 8— a:r :: J(«--r-f a — /?— x):aj;

whence x = -^ J- — i ^^. Similarly, y = -^^^ — J ' . , ^^ « = ^ hT x -

2(8— a) ^ ^ 2(«— 6) 2(8- c)

Construction. — Let ABC be the triangle, F, V, K the points of contact of the Inscribed circle, radius r.
From centers A, B, C describe the arcs FK, VK, FV, intersecting AO, BO, CO in the points G, Y, W.
From the center O, on OA, take GZ = OW ; then ZR = OY. Take OH' = r, and through the point of
bisection of RH' describe a circle having its center at O. With A, B, C as centers describe arcs touching
the circle last drawn and cutting AB, BC in the points P, N and 8 which will be the points of contact.
From these points erect perpendiculars to AB, BC intersecting OA, OB, OC in the points D, H, T ; these
last points are the centers of the required circles. For
AP = OA — [r— |(r— OW — OY — OG)]. = OA— {r- i[r-(OC-CV)- (OA-AK)- (OB-BK)]}.

Ill*— Solution by E. B. Seitz, Member of the London Mathematical Society, Professor of Mathematics at the North
Missouri State Normal School, Kirktiville, Mijaeourl.

Let ABC be the triangle, M, N, P the centers of the cir-
cles, D, E, F, G, H, K the points of tangency.

Let MD = a?, NF = y, PH = «, and r = the radius of
the Inscribed circle of the triangle. Then AD = AE =
ajcotJA, BF = BG = ycotJB, CH = CK = acot^C,

DF = W{xy\ EK = 2v'(a»), GH = 2y (yz),
and we have the equations,

arcotJA -f 2-v/(a:y) + ycotiB = c (1),

arcotJA + 2-v/(a») -fscotiC = 6 (2),

2/cotiB -f 2i/(2^) -f2:cotiC = a (3).

By Trigonometry we have

sinJBcosJB sinJCcosJC sin JA cos J A sin^CcosJC

~~6— = ~c ^^)' a -=—~c (^^'

and from these two equations we can deduce the following :

6(cottA— tan^B) = c(cot^A— tan^C) (6), a(cottB— tan^A) = c(cottB— tanJC) (7).

Dividing (1) by (2) and (3) respectively, and clearing of fractions, we have

6[a;cotJA-h2i/(«2/)-f ycot JB] = c[a;cotiA-f 2i/(a2;)-f «cotJC] (8).

a[a:ootJA-f2-v/(a^) + ycot JB] = c[ycotiB-f2i/(2/K)+2Cot JC] (9).

-191-

Sublracting x times (6) from (8), and y times (7) from (9), we iiave

6[ajtanlB-f 2y(a^)-f ycotJB] = c[a; tan ^C -f 2y(a»)-f« cot tC] (10),

a[a?cottA-f 2i/(a^)-f ytantA] = cCs/tan^C-f 2y(y!;)-f «eot JC] (11).

Muldpiying (10) by (4), and (11) by (5), and extracting tlie square root, we iiave

l/«8in JB-f i/ycosJB = i/arsin^C-f i/«0O8 JC (12),

l/«C06 JA+i/ysin^A = i/ysin^C+i/ecostC (18).

Subtracting (13) from (12), we find

Vy __ si ntC-fcoB^A—sintB _ co8jC+co8j(2B+C) _ 0O8jBco8j(»r— A) _ 1-f tanJA

Vx sin JC— sintA-fcosiB "" cosJC-f co8j(2A-f.C) "" eo8jAco8j(;r — B; "" 1-f-taniB ' ^ ^'

StaUarlyweflnd 7x = ij^l^ .^^^>-

Substituting the value of y/y from (14) in (1), we have

or ri-tan«iA JU:^B,u\k^ /lr-t»5!iB\/L+*^i^Ylx - c

L 2tanJA ^ \l-f tanJB/ ^ \ 2taniB Al + tanJB/ J "" '

whence x- 2ctanjAta niB(l+ taniB)

(l+tanJA)(taniA-ftanJB)(l — tttnVAtanJB)[i-ftani(A-fB)]*

C8inlA8inlB(H-tan^^) !;(^_±^"i?^

■" 8inl(A + B)(i-f tanJAyci-HwoKjr— C)]' ~" (r+tanJA)[l-f tani(jr— C)]'

— JK^+tan iB)(l-htan iC)
~" 1-ftanJA

Substituting the value of x in (14) and (15), we find

__ lr(l-ftaniA)(l-ftan4C) _ i r(l+taniA)(l- f tanjg)

^ " 1-ftaniB • "^ * ~" " l+tanJC

Aa excellent eolution was famished by WUliam Bdover, and a very brief one by Profeasor CViMy. We hope to publbb
other solutioiiB in a fatare No.

Thin celebrated problem haa a hialory, and ia known to mathematiciana aa '*Malfatti*e Problem/*

: c,

181«— Proposed by Dr. Jobl E. Hendricks, M. A., Editor of the Analytt, Dea Moinee, Iowa.

The base of a hemisphere, whose radius is ten feet, rests on a horizontal plane, and a point A on the
surface of a sphere whose radius is one foot is in contact with the vertex of the hemisphere.

If, from a slight disturbance of the equilibrium, the sphere is caused to roll off the hemisphere in the
direction of the plane of a great circle through the points A and B on the surface of the sphere, and if
the point B strikes the horizontal plane at a point D, distant -CD from the center of the hemisphere; it
is required to find the length of the line CD, and the relative position of the points A and B on the
surface of the sphere.

Solation by William Hooysb, Saperintendent of Schools, Wapakoneta, Anglaixe County, Ohio.

Put 1 foot = a = the radius of the sphere, and 10 feet = 6 =
the radius of the hemisphere ; let O be the position of the center
of the sphere at the point of separation, v its orbital velocity, t
the time in moving from O to O^ when the point B strikes the
horizontal plane at D, H the vertex of the hemisphere, and ^ the
angle OCH.

We have (Routh'tt Rigid Dynamics, 2d Ed., pp. 103, 104)

COS0 = i? . . (1). and « = ■•[ ^^^^(a-f 6)(l--co8^)] . . . (2).

We also find - = angular velocity of the sphere, and - ^

r= the angle through which the point A has revolved when the center of the sphere is at O. Also

«esin^-fJpf« = (o-f6)co8^ — a (3), and CD = (a-f 6)8in^-f ulcxwf (4).

-192-

From (8), t = -(ii/[%Ko+ft)<508^ — 2a^+««8inV] — «8in^^ (6).

Substituting the value of « from (2) in (4) and (5), and the resulting value of i in (4), and reducing,

CD === (a+6)8in^±coeV{V(a+&)(l~ooe^)[(a+^)<»B^ + f{a+5Xl-<»s^)«lnt^-oJ

— y(a+&)(l— ooe^)ein^ooef
The angular distance described by the point A is

"i"~^+ ? = ir(o+&)f ± i/{ V(«+&)(l-ooe«[(a+6)cos^

** ^^L 1 lOsin^

+ f{a+6)(l~oos^)sint^-a]}J - ^-^(a+5)(l-.cos«.

Restoring the numbers, CD = 11.646+ feet, and ^±^^ + - = 16.043+. which divided by 2* gives

2.39416+, the number of revolutions of the sphere before impinging on ti^e horizontal plane at D. Henoe
the angle AO'D = 2)r x 0.39416 == 141^ 63' 51".

Solyed also 1^ the Propoter,

pr The other eolntioiia dae in thii No. are very relnctlantly deferred till the pabUcatiofi of the next No.

XMf of CkmiributOTB to the Senior JDepartment.

Solutions have been received from the following persons to the Problems indicated by the numbers :

E. B. SnTS. Member of the Loodon Mathenutticel Society, ProfeMor of Marbemetles, North lllwniri State Normal School,
KIrksville, Jliasoari, ]?», 178, 177, 178, ISS, ISi. 188, IftI, IM, 190, 106. 901, 908, 9M, 906, 808, 9(?7, 900, 910, 916 and 910.

Waltsb SiYBBLi:, Oil City, Venango Ca. Pa., 176, 170, 177, 188, 180, 100, 101, 108, 106, 100, 106. 100, 908, 908. 907. 211, 919, 918,

914 and 918.
WiLUAX HooviB. Superintendent of Schools. Wapakoneta. Anglalae Co.. Ohio. 176. 170, m, 178, 170, 188, 188. 186, 187, 109,

101 and 908.
J. F. W. ScHnTBB, late Profeeeor of llathematica and German. Meroefshug CoUeg^ Mercersboig, Fkanklln Co., Pa., 175,

176, 177, 178. 18*, 106 and 106. • ^

Cbarlbs H. KumKLL. U. 6. Coast and Geodetic Suroy Office. Washtaigton, D. C, 17^ 178, 178, 180, 181 and 918.
DbVoliov Wood, II . A., C. E., Profeeeor of Mathematics and Mechanics, Stevena Instltnte of Technology, Hoboken, N. J.,

176, 178, 188, 184, 186 and 188.
Bdwabo a. Bowseb. Professor of Mathematics and Bngliieering, Hateera College, New Bmnswick, New Jeceoy, 188, 196,

907, 911 and 910.

F. P. Mate, M. A, UUe Professor of Mathematics, Military and Scientiflc School, Khig*s Moontain. North CaroUna, 180, 180,
190, 901 and 919.

GioBon Eastwood, SaxouTille, Middlesex Co., Maasachnsetts, 176, 176, 177, 178 and 919.

L. G. Babboub, Professor of Mathematics, Centra] University, Richmond, Kentocky, 909, 919, 914 and 910.

W. B. Hbal. Marion, Grant Co., Indiana, ITT, 1&), 101 and 106.

Dr. Dayid 8. Hart, M. A., Stonington, New London Co., Conn., 177, 101 and 107.

Professor W. P. Cakxt. C. B., SanFrancisco, California, 176, 178 and 188.

E. J. BDMUirDs, B. S., Professor of Mathematlos, Sonthem University, New Orleans, Louisiana, 178, 178 and 188.

H. T. J. LuDwra, Professor of Mathematics, North Carolina CoUege, Moont Pleasant, Cabaims Co., N. C, 189 and 100.

Luoros BnowN, Hudson, Middlesex Co., Mass., 108 and 810.

L. P. Shtot, U. 8. Coast and Geodetic Survey Office, Washington, D. C, 177 and 101.

STLvnTKR BoBiHS, North Branch Derot. Somenet Co., N. JT, 177 and 170.

K. 8. PUTHAX, Borne, Oneida Co., N. T., 100 and 101.

C. A. O. BosBLL, B. A., Teacher of Mathematics at the CanoU Institute, Beadhig, Pa., 176 and 170.

Hon. JoeiAH H. DRtnoiOirD, LL. D^ Porthmd, Maine, 177 and 101.

Samusl Bobbbts, M. A. F. B. 8., President of the London Mathematical Society, London, Bngland, 186 and 108.

Bbubbn Davis, Bradford, Stark Countr, Illinois. 101 and 107.

JvuAM A. Pollabd, Go»hen, Orange Co., N. T., 176 and 177.

W. L. Habvbt, MaxOeld, Penobscot Co., Maine, 176 and 170.

Bev. W. J. Wbioht. M. A., Ph. D.. Burlington, Chittenden Co., Vermont, 908.

Professor Obmond Stoxx, M. A~ Astronomer at the Cindnnati Observatory, Mt Lookout, Hamilton Co., Ohio, 810.

Dr. 8. H. Wbiobt, M. A., Ph. D., late Mathematical Editor of the YcUm OowUh ChronicU, Penn Tan. N. T., 184.

Professor Hugh 8. Banks, Instructor in Ibiglish and Classical Literature, Newburg, N. T., 180.

Mabcus Baksb, U. 8. Coast and Geodetic Survey Office, Washington, D. C. 188.

Hbnbt Hbatob, B. 8., Perry, Dallas Co., Iowa, 196.

Bev. U. Jbssb Kkisklt, Ph. D., Newoomerstown, Tuscarawas Co., Ohio. 186.

Gborsb H. Habvill. Colfax, Grant Co., Louisiana, 919.

C. C. BoBixs, Princeton, New Jersey. 177.

D. J. MoAdam, M. a.. Professor of Mathematics, Washington and JeiferBon College, Washington. Pa., 914.

WiLLiAv WooLSBT JoHNsoN, Member of the London luthematical Society, Professor of Mathematics St. Jolm*s CoU^

Annapolis, Maryland, 108.
J. M. Abnold, South Boston, MassachuPetts, 918.

Alxx. S. Chbistib, U. S. Coast and Geodetic Survey Office, Washington, D. C, 908.
Isaac H. Turbbll, Principal 4th District School, Cincinnati, Ohio, 186.
Dr. JoBL E. Hbmdbicks, M. A., Editor and Publisher of the AnalyH^ Dee Moines, Iowa. 187.

The first prize for solution of Problem 216 is awarded to Professor E. B. Beztz. KlrksvlUe, Ho., and
the second to J. H. Abnold. South Boston. Mass.
No solution of Problem 217 has been reoeived.

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PROBLEMS.

8A3»— Proposed by Bxubsh Datm, Bradford, SUrk Ooimtj, Olioois.

It is required to find three positive integral numbers, such that their sum is a cube, and also the sum
of any two of them a cube.

Bepropoied to ooirect enon; m stated in No. 6, it is impossible.

816*— Proposed by Abtbxas Mabtot, M. A., Member of the London MathematlciU Society, Brie, Brie Ooonty, Pa.
If -- be the last convergent in the first period of \/A expanded as a continued fraction, and r the
integral part of i/il, show that p^ = rq^-{~ g,^i.

9fnf«— Proposed by Samuil G. CAOwiir, New London, Oneida County, N. T.

A fflven volume v of metal is to be made into a rectangular vessel of uniform thickness a. Determine
the dimensions of the vessel so that its capacity shall be a maximum, (1) when it has no cover, and (2)
when it has a cover.

278*— Proposed by B. J. Bdmumds, B. S., Professor of Mathematics, Sonthem University, New Orleans, Louisiana.
When two parabolas having a common tangent and a common focus cut each other imder a constant
angle, find the locus of their point of intersection.

279»— I*roposed by Hon. Jociah II. Orumiiono, LL. O., Portland, Maine.

Find three positive whole numbers in arithmetical progression whose sum shall be a square, and such
that if 1^ be added to each the several sums shall be squares ; the numbers varying when different values
are given to p, and p being any number.

880«— Proposed by Abtsmas Martin, M. A., Member of the London Mathematical Society, Brie, Brie County, Pa.

If the feet of the perpendiculard of a plane triangle be joined two and two a new triangle is formed,
which call the first pedal triangle; if in like manner the feet of the peri)endicular9 of the first pedal tri-
angle be joined anotner triangle i^ formed, which call the eecond pedal triangle; and so on. It Is required
to find expressions for the sides and angles of the nth pedal triangle in terms of the fimctions of the orig-
inal triangle.

S81*— Proposed by SrLvssTBR Bobina, North Branch Depot, Somerset County, New Jersey.
Investigate that series of paralielopipeds in which the law requiring each edge and solid diagonal of
the nth term to be expressed by n figures holds to the greatest possible number of terms.

L— Proposed by Professor W. P. Casbt, C. E., San Francisco, California.
If the lines AO, BO, CO from the angles of a triangle ABC to the center O of the inscribed circle be
produced to meet the opposite sides m a, fi, y, and 'from these points as centers with the respective
distances oO, /HO, yO as radii three circles be described the sum of the reciprocals of the radii of all
the circles respectively touching those circles is equal to four times the reciprocal of the radius of the
inscribed circle.

U— Proposed by Professor David Tbowbridob, M. A., Waterburg, Tompkins County, N.
If (l-2pc-f C*)— = 1 -f-cP, -f.C«PaH- C^Pi-^-

m being Integral or fractional, and P< a function of p, prove that

j^l PiPn(l-p»r-^dp = 0.
t and n being different integers.

884>— Proposed by Abtbmab Mabtik, M. A., Member of the London Mathematical Society Brie, Brie County, Pa.

Two equal circles, radii r, intersect, tlie center of each being on the circumference of the other. A
circle is drawD touching that diameter of the right-hand circle which joins the centers of the given cir-
cles, and the ciroumfer.mce') of both circles, the right-hand one internally and the other externally ; a
circle is drawn touching the one last drawn and the circumferences of both the given circles ; and so on.

Find the radius of the nth circle.

885*— Pi^po^^ hy Dr. Dayid S. Habt, M. A., Stonlngton, New London County, Connecticut.
Find thirty-four biquadrate numbers whose sum shall also be a biquadrate numlier.

88B*— Proposed by Artexas Martin, M. A., Member of the London Mathematical Society, Brie, Brie Co., Pa.
Find the average distance between two points taken at random in the surface of a given ellipse.

S87«— Proposed by Chaflks Qilpik, Jr., Philadelphia, Pennsylvania.

A cup of wine is suspended over a cup of equal capacity full of water ; through a small hole in the
bottom the wine drips into the water cup, and the mixture drips out at the same rate. When the wine
cup is empty, what part of the contents of the lower cup is water?

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888»— PropoMd hy AammAB Martin, Iff. A., Member of Uie London MatbenuitScal Society, Erie, Erie Co., F&.
Three points being taken at random in the surface of a circle, find the chance that the triangle formed
by joining them is acute.

889«— Proposed by Miss Chbistiks Ladd, B. A., Fellow of Johns HopUns UnWertity, Baltimore, Maiylaad.

If A is the radius of the circumscribed circle of a triangle ABC, r that of the inscribed circle, p that
of the circle inscribed in the orthocentric triangle, I the center of the inscribed circle of the tnanffle
ABC, and Q that of the circle inscribed in the triangle formed by Joining the middle points of the sides
of ABC. show that (QI)t = l(7«p— 6JBr+3f<+2Ji»).

890.— Proposed by AnniiAS Mabtim, M. A., Member of the London Matbematicsl Society, Brie, Erie Co., Pa.
A circle, radius r, touches another circle, radius R, internally. Find the average area of all the cir-
cles that can be drawn touching the two given circles.

291.— Proposed by J. C. Olabhax, Ottawa, Ontario, Canada.

From one solution of the equation a^ + y<--fl^ = a an infinity of others may be derived thus : Write
the given values in order x, e, y in a vertical column. Make a second colunm thus : take the sum of the
two upper numbers for a new top number ; the sum of the two lower numbers for a new bottom number,
and the sum of all three for a new middle number. Repeat the operation on this second column so as
to get a third ; the numbers in this third column will give a new solution of the proposed eduation. Oon-
tinue the process to any number of columns, the odd columns will give new solutions. The sign of any
jdumber in an odd column may be taken either positive or negative.

The following is an example of the process applied toa^ + y<— 8^ = 10:

I II III IV

6 13 33 81 197 477 765

8 20 48 116 280 287 836 etc.

7 15 35 83 —199 81 359

(Adapted from Quest 4102, by T. T. Wilkinson, in the EduoaHonal Tmea MathemaHcal Reprint, Vol.

Generalize the above so as to obtain from one solution of xl +2| +a;{ + +;b| — y^ = a an in-
finity of solutions.
jBrample.— From 0»+0»-f 1«— 1« = obtain 2«+2«+l«— 3« = and from this 2«+3«+6«— 7« = 0.

2M<— P'opo"^ hy E. B. Sbite, Member of the London Mathematical Society, Profesor of Mathematics at iho North
MissoQii State Normal School, KlrksviUe, Adair Conn^, Missoari.

Two points are taken at random within a circle on opposite sides of a given diameter, and a third
point is taken anywhere within the circle; find the average area of the triangle formed by Joining the
three points.

|»— Proposed by Gbobob Bastwood, Saxonville, Middlesex Connty, Massarhneetts.
On what days of any year in a given latitude will the contour of the shadow swept out by a vertical
pole, on a horizontal plane, have the greatest length?

894.— Proposed by Abtbmas Martix, M. A., Member of the London Mathematical Sode^, Brie, Brie Co., Pa.
A dog se^ng a fox at the center of a circular field leaped over the fence and "went for him." the fox
running in a straight line and the dog directly towards him.
Find the chance that the fox will escape from the field.

8M»— Proposed by Professor Anmi B. Btahs, M. A., Principal Lockport Union School, Lockport, Niagaim Co., N. T.

Four circles of given imequal radii n, rg, r^ u have their centers at the vertices of a quadrHateral
ABGD whose angles are variable. If AB = n -f rt, BG = rt+rs. CD = r3+r4. DA = r4+*'i» what re-
lation must exist among the angles A, B, G, D that that part of ABGD exterior to the four sectors may
be a maximum?

896.— Pnq^osed by B. B. Shts, M. L. M. S., Professor of Mathematics, North Missonri State Normal School, Kirksrflle, Mo.
Three points M, N, P are taken at random within a triangle ABG, and lines AMD, BN£, GPF are
drawn to meet the sides of ABG in D, £, F, Internecting each other in B, S, T.
Find the average area of the triangle BST.

297*— Proposed by Samukl Bobkrts, M. A., F. R. S., President of the London Mathematical Society, London, Bngland.

Three circles A, B. G are inscribed or escribed about the same triangle ; show that In the triangle
formed by the three double tangents aE touching A and B, fie touching B and G and CA touching G and
A, an infinite number of triangles can be inscribed such that the sides opposite the vertices on aB, 6c,
CA touch the circles G, A. B respectively.

296.— Proposed by AnniiAs Martin, M. A., Member of the London Mathematical Society, Brie, Brie Ooonty, Pa.

There is a cylindrical tower, radius a, at the center of a circular park, radius jR. A himter on the
outside of the fence saw a rabbit in the park which *'scud" around the tower out of sight and "made for
the fence." The hunter ran around by the fence, but the rabbit kei>t **ju8t out of sight" all the time.

Bequired the equation to the curve the rabbit described, and the distance it ran to reach the fence.

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2M»— Pf^ipoMd hy DBVoLflOH Wood, M. A., C. &, Profeaoor of MAttaematies and Mechanics, Sleveivi lutitnte of Tech-
Dolo^, Hobokeu, New Jeney.

If each of n vesselB oloeely oonnected in a circuit contains a dlfTerent liquid, each o gallons, and the
liquids circulate by flowing uniformly in one direction at the rate of a gallons per minute, mixing uni-
formly, how much of each liquid will there be in any one of the vessels at the end of time if

Q/QOm—PtopoBtd by ABTKMAa MABTiif, M. A., Member of tbe London Mathematical Society, Erie, Brie Co., Pa.
During a heavy rain storm a circular pond is formed in a circular fleld. If a man undertakes to cross
the field in the dark, what is the chance that he will walk into the. pond?

301«— Propoeed by B. B. Sbits, Member of the London Mathematical Society, Profoacior of Mathematics at the North
MlMoiui State Normal School, KirkayUle, Adair County, Miaeoori.

Let M, N, P, Q, K, S be six random points within a circle ; find (1) the chance that each of the chords
through M and N, P and Q, B and S, intersects the other two ; and (2) the chance that each of the chords
intersects the other two, and the triangle formed contains the center of the circle.

30S«— Proposed by Abtsxab Mabtxn, M. A., Member of the London Mathematical Society, Brie, Erie Co., Pa.

Having g^ven the polckr equation r =/(6) of a curve, deduce therefrom by a general method, without
changing to rectangular co-ordinates, its mkiMic equation ; and by it find the intrinsic equation of the
spiral of Archimedes, and of its involute.

Solutions of these problems shonid be received by March 1, 1888.

EDITORIALi irOTES.

We deeply regret that it is our sail duty to record the death of the eminent mathematician Professor
BENJAMIN PEIBCE, LL. D., F. B. S., of Harvard University, which occurred October 6, 1880.

The following notice is extracted from a newspaper cutting kindly sent us by Mr. 8. C. Qovuy of Man-
chesier. New Hampehii'e :

**He was bom in Salem in 1809, and from his earliest years manifesUd a taste for mathematical studies, a taste which was de-
veloped by the formerly well-known teacher, Dr. Bowditeh, and which led him, while at college, into complete absorption in
his tavorite pursuit. It is related of him that, dnnng the last year of his course, he neglected all his other studies and ai tended
no reciiation^ wliatever, devoting fourteen hours a day to maihemalic-*. After grsdnating, he continued his mathematical stud-
ies at the same tension, and in IMS he became Perkin^s professor of astronomy and mathematics ai Hatvard college. In 1887 he
aocoeeded Prof. Ba<-he as superintendent of the Unitt^ Staten coast survey, resign mg the latter position in March, 1874. As a
mathematician he attained the first ratik, and he had few, if any, compeers in h.s highest intellectual hibors. He was noted for
his directness and conciseness of demonstration, and by the intuiti%e insight with wbich he approachtni the most difflcult prob-
lems. His published work^ are few, although his contributiontf to the science of muthematic» are most important, and his text-
books and elementary treati«es are widely circulated. Bo abstruse are some of h s works thai, as he himself taid, only one
man l)esides himself has been able to understand tht-m. To Prof. Peirce belongs the dis.iuciion of being one of the founders of
a new brsnch of mathematics, the final form of which is not yet determined, but which may prove to be the great event in the
mathematical history of this country— 'Linear Associative Algebra.' This book has been published in an edition of eome fifty
copies. As an astronomer Prof. Peirce's record is high, although he has written no work on the subject.''

Our thanks are due Prof. £. B. Seitz for valuable assistance in correcting the proof sheets of this No.

Prof. Gaset, Prof. Wood, Prof. Beman, J. 8. Boteb, StijYEBTEB Bobins, and many others, are en-
titled to thanks for their success in procuring subscribers.

Piof. W. W. Beman of Michigan University, Ann Arbor, has been elected a Member of the London
Mathematical Society.

Miss Chbistine Ladd, B. A., has been elected a Fellow of Johns Hopkins University.

Prof. Edwabd Bbooks, Ph. D., Principal of the Pennsylvania State Normal School, Mlllersville, Pa.,
has returned from abroad and is again hard at work among his classes.

prof. £. J. Edmunds, B. S., has returned from France, and is now Professor of Mathematics in the
Southern University, New Orleans, Louisiana.

Prof. F. P. Matz, M. a., late of King's Moimtain Military and Scientific School, is a student at the
Johns Hopkins University.

The Mathematical Department in the Yates C&unty Chnmide, so ably conducted by Dr. S. H. Weight,
M. A., Ph. D.. for more than eight years, has been very abruptly discontinued. The Dr. did not even
have an opportunity of saying a parting word to his contributors. It was begun February 29, 1872, and
ended October 22, 1880. The 2500 problems and their solutions published would fill three or four large
octavo volumes and constitute a most valuable collection.

The Mathematical Department in the WiUenberger has also been didcoutinued. The October Number,
1880, contains Mr. Hoover's valedictory.

The typo-setting and printing of this No. of the Visitob has been all done by the Editor, and he deeply
regrets that tlie delays occasioned by the breaking of his press and various other causes, and having to
move, compel him to "cut it short.**

The present No. closes the first volume of the Visitob. In order that the Editor may get the rest he
so much needs and tnusi have, no Visitob will be issued in July this year. The next No., It is hofied,
will be ready in January, 1882. All who have paid for another No. will receive No. 1 of Vol. II.

A limited number of bound copies of Vol. I can be supplied at $3.50.

-196-

As some of our contributors may desire to procure copies of their solutions published in the Vibitob,
we will state here that we can supply them h^eafter at $1.50 per hundred, printed on good paper the
size of a ViaiTOB leaf, if the solution does not occupy more than one page ; if more than one page, $1.00
per page for 100 copies.

The Editor having had so many applications for his photograph has decided to have his portrait en-
graved. It will be printed on heavy paper, the size of a Visitob leaf, suitable either for framing or
binding with the Visitob, and supplied to subscribers at 25 cents. We will be pleased to receive orders
for the portrait at once so that we can determine the number to be printed. Say whether you want it
bound with the Visitob or not— or order Uco, one for framing and the other to be bound in the Visitob,
and the two will be furnished for 40 cents.

Contributors are requested to send in their solutions of the problems proposed in the Junior Departr
ment of No. 5 as soon as practicable.

We must ask our subscribers not to send us their individual checks nor drafts on private or country
banks, as we are subjected to a "shave" on them. Postal Money-Orders and Drafts on New York pre-
ferred. Our Canada subscribers will please not send Canada postage stamps — we can not use them.
English postage stamps are always acceptable : they can be used in making remittances to London.

NOTICES OF BOOKS AND PERIODICALS.

An Ehmentary Treatise on the Differential and Integral Cfalcuhis. With Numerous Examples. By Ed-
ward A. Bowder, Professor of Mathematics and Engineering in Rutgers College. 12mo, pp. 395. New
York : D. Van Nostrand, 23 Murray street.

An elementary work designed u a text-book for colleges and scientiflc schools, wel. supplied with appropriate examples, some
of them selected from the Visitor.

A iHf€renlkil is defined as the deference bettoeen two eonsecuOve vahut of a variable or function ; and conMCiMve vahtet of a
variable or function, as values which differ from each other by Uu than any ataignabU quantity.

It is a work of rare excellence, beautifully printed on fine paper, and we can heartily recommend it to those in search of a
clear exposition of the principles of this (to many minds) seemingly mysterious subject.

Prof. Bowser*s Mathematical Works (Analytical Geometry, and Treatise on the Calculus,) have been adopted already as the
tex^books in the following Institutions: University of Penna., Philadelphia, Pa.; Free Institute, Worcester, Mass.; Wesl^an
University, Middletown, Conn.; University of Oeoigia, Athens, Ga.; Yale College, New Haven, Conn.; Rutgers College, NeW
Brunswick, N. J.; North CaroUna College, Mt. Pleasant, N. C; Madison University, Hamilton, N. Y.; Polytechnic Instltnte,
Brooklyn, N. Y.; Racine CoUege, Racine, Wis.; University of N. C; Chspel Hill, N. C; State College, Austin, Texas; Syra-
cuse University, Syracuse, N. Y.; Hanover College, Hanover, Ind.; Rensellaer Polytechnic Institute, Troy, N. Y.; Washington
Uiilversity, St. Louis, Mo. ,

For table of contents, and price, see advertisement on second page of the cover of this No.

A Sifnopsia of Elementary ResuUs in Pitre and Applied Maihematic8: Containing Propositions, Formulie,'
and Methods of Analysis, with Abridged Demonstrations. By G. S. Carr, B. A., late Prizeman and
tjjcholar of Gonville and Caius College, Cambridge. Vol. I, Part I. 8vo, pp. 280. London : C. F. Hodg-
son and Son.

Part I of Vol. I contains sevrn sections: Mathematical Tables; Algebra; Theory of Equations and Determinants; Plane
Trigonometry; Spherical Trigonometry; Elementary Geometry; Geometrical Conies. Part H of Vol. I, which is in the Pre«a,
will contain the following sections: Differential Calculus; Integral Calculus; Calculus of Variations; Differential Equations;
Plane Co-ordinate Geometry; Solid Co-ordinate Geometry. Vol. n is in preparation and will be devoted to Applied Mathemat-
ics and other Branches of Pure Mathematics.

Chiefly valuable as a work of reference; it is well printed, and profusely illustrated with excellent disgrams.

An Elementary Treatise on Trigonometry, with Numerous Examples and Applications. Designed for
th« use of High Schools and Colleges. By J. Morrison, M. D., M. A., Principal of the Walkerton High
School. 12m<., pp. 332. Toronto : Canada Publishing Co. (Limited).

A very full and complete treatise on Plane Trigonometry, contaii.ing great store of interesting and well-chosen examples, antf
many elegant solutions.

The Teacher's Hand-Book of Algebra; Containing Methods, Solutions and Exercises Illustrating the La-
test and Best Treatment of the Elements of Algebra. By J. A. McLellan, M. A., LL. D., High School
Inspector for Ontario. 12mo, pp. 229.

Key to the Teacher's Hand-Book of Algebra. By the same Author. Second Edition — Revised. 12mo,
pp. 199. Toronto : W. J. .Gage and Company.

The "Hand- Book" is really a book of examples and problems in Algebra^Hind a very good one, too. It contains, however, a
large number of solutions, and gives complete explanations and illustrdtions of important topics. A valuable Utile work for
both teacher and student. In the **Key" full solutions are given of all the dlfBcult problems.

We regret that we can not praise the priming of these excellent books. Poor printing is a lamentable fault we have noticed
in sU the Canada books that we have seen.

Ray's New Higher Arithmetic. A Bevised Edition of the Higher Arithmetic. By Joseph Bay, M. D.,
Late Professor in Woodward College. 12mo, pp. 408. Cincinnati & New York : Van Antwerp, Bragg A Co.

The work of revision has been very efficiently done by Prof. J. M. Greenwood, M. A., Superintendent of Public Schools, Kan-
sas City, Mo., and Rev. U. Jesse Knisely, Ph. O., of Newcomerstown, O., and this is now one of the best Higher Arithmetics
**in the field.''

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MathemaiiooU Queetiona, with their SobOums, from the ^^Edueationai Times" with many Papers and Solu-
tions not published In the '*£ducaiional Times." Edited by W. J. C. Miller, B. A., Registrar of the
General Medical Ck>uncil. Vols. XXXIII and XXXIY. From January to July 1880, and from July to
December, 1880. 8vo, boards, pp. 116 and 120. London : C. F. Hodgson & Son.

Vol. XXXin contaioB 5 pspen tnd solatione of 138 problems. Vol. XXXIV contains 1 paper and solndons of 147 problems.
Many of the flneet ** Average'' and "Probability" solations are by our valued contribaior FrofeMor Ssm.

The BdveaHonal Times is pabliehcd monthly, and contains a most valoable Hathematical Department. The Rqjrint is is-
sued in half-yearly volomes.

The Editor of the Visitor will be pleased to famish the Tbnea at $&00 a year, and the SeprirU at $1.75 per Vol.

The Analyst: A Journal of Pure and Applied Mathematics. Edited and Published by J. £. Hendricks,
M. A., Des Moines, Iowa. Bi-monthly ; 8vo, pp. 32. $2.00 a year. Vol. VIII.
As interesting and valoable as ever, and deserves the continued support of mathematiciaiis.

The School Visitor, Devoted to the Study of Mathematics and English Grammar. Edited by John
S. Boyer and Thomas Ewbank. Monthly ; pp. 16. $0.75 a year ; single numbers. 10 cents. Ansonia,
O. : Published by John S. Eoyer.

Has grown some, and each No. this year will have a nice cover. The Mathematical Depnrtment contains well-selected and
interesting Problems and Solutions in the Elementary Branches, and is illustrated with beautiful diagrams. Its merits deserve
to be crowned with suocess^-every teacher should have a copy.

Eduoationdl Notes and Queries. Issued monthly, except in the vacation months, July and August.
Edited and Published by Prof. W. D. Henkle, Salem, Ohio. $1.00 a year.

Contains a vast fund of curious and valuable mformation, besides Mathi matical Problems and Solutions.

The Canada School JoumaL Monthly. Toronto, Canada : W. J. Oage A Go. $1.00 a year in advance.
An excellent educational journal. The Mathematical Department is under the able management of Alfred Baker, M. A.

Barnes* Educational Monthly. New York : A. S. Barnes & Go. $1.50 a year.
A live periodical, containing an interesting Mahematical Department.

The Canada Educational Monthly and School Chronide. Toronto : Canada Educational Monthly Pub-
lishing Co. $1.50 a year.

A very readable journal. The Arts Department, edited by Archibald MacMurchy, M. A., is devoted to Problems and Solu-
tions, and Bzamination Papers.

Oage's Sdwol Examiner and Monthly Review. Vol.1. Toronto : W. J. Oage ft Co. $1.00 a year.
A new candidate for public favor, "containing full sets of model Examination Papers, and a survey of current thought and
art, with special reference to the teacher's point of view.'*

. The Pennsylvania School Journal, Monthly. J. P. Wickersham, Editor. Lancaster : J. P. Wicker-
sham ft Co. $1.60 a year.

One of the best educational journals in the country.

New-England Journal of Education. Thomas W. Bicknell, Editor. Boston : 16 Hawley street. $3.00
a year ; $2.50 in advance.

Issued weekly, contains a vast fund of educational intelligence, and should be read by all teachers.

The CaWomia Architect and Building Review. James E. Wolfib, Editor and Manager. Geo. H. Wolfe,
Business Manager. San Francisco : San Francisco Architectural Publishing Co. Monthly. $2.00 per
annum in advance. Single copies, 20 cents.

An excellent periodical devoted to the interests of the architects and builders of the Pacific Slope. In ''Everybody's Column'*
questions of intererest are discussed, and problems solved.

It contains also much valuable information of interest to the general reader, relating to house-drainage, sinks, Ac. These
I of vital importance can not receive too much attention, and their neglect is almost sure t.) result iu a visit from Death.

Discussion of a Oeometrical ProMem: With Bibliographical Notes. By Marpus Baker. U. S. Coast Sur-
vey, Washington, D. C. Extracted from the Bulletin of the Philosophical Society of Washington. Phil-
adelphia : ColUns, Printer, 705 Jayne St.

The problem discussed is the followhig:
"In a right-angled triangle there are given the bisectors c/t the acute angles: required to determine the triangle.'* Six solutions
are given, and two constructions. This problem was proposed in the Ladies' Diary for 1797, and lias appeared in many places
since that time.

Mcromdrical Measurements o/1054 Double Stars, observed with the 11-inch Refractor from January 1,
1878, to September 1, 1879, under the superintendence of Ormond Stone, M. A., Director. Cincinnati :
Published by authority of the Board of Directors of the University.

On (he Extnu-Meridian Determination of Time, by means of a Portable Transit-Instrument. By Ormond
Stone, M. A., Astronomer of the Cincinnati Observatory.

Huree Approximate Solutions of Kepler's ProMem. By H. A. Howe, M. A., Assistant at the Cincinnati
Observatory.

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CORRIGEITDA.

No. 5.

Page 141, Problem 141, last line, first "any" should be "the"; line 1 of solution, for "constant" read
counted ; line 12, after "number" insert v.

Page 147, lines 3 and 4, /or "y(pa;— «•)" read \/(imj— y»).

Page 149, line 14, In the denominator /or "(n-fn)" read (n-f 1).

Page 155, line 1, after "outline" insert "in the expansion oi(fi±c±d±t± xy and omit the sentence
in().

Page 158, Problem 237, the latitude should be given = 40° 10' N.

No. 6.

Page 178, solution of Problem 176, line 6, for

*'A = l\/[4(89— 4mp+2m«fi— m*)]" read A = Ji/(16g— 8inp + 4wi«n— m<);

last line, far "R = J J(--^±^±!i_Y. ^ fl = J(_P^'^-Z^__).
*\\8g— 4mp-f-2m«n— m*/ \\16g— 8mp-f 4m«m— wi*/

End op Vol. I.

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