Mathematical questions and solutions

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4UTH HM ATIC AL QUESTIONS,

Wn-H TTtEIR

SOLUTIOKS.

tTtOM THE -EDUCiTiUXAt. TOiH!^/

VOL. XLIX.

r

Mocih 33*g.S4»,

SCIENCE CENTER LIBRARY

Digitized by VjOOQ IC

-.-, .^

Digitized by VjOOQ IC

Digitized by VjOOQ IC

MATHEMATICAL

aUESTIONS AND SOLUTIONS,

FEOM THE "EDUCATIONAL TIMES,"

WITH MANY

PAPEES AI^D SOLUTIOlsrS

IN ADDITION TO THOSE

PUBLISHED IN THE "EDUCATIONAL TIMES,"

AND POUR

APPENDICES.

EDITED BY

W. J. C. MILLER, B.A.,

BBOISTBAB OP THE GENEBAL MEDICAL COUNCIL

VOL. XLIX.

^ LONDON:
FRANCIS HODGSON, 89, FAERINGDON STREET, E.G. •

1888.

Digitized byCjOOQlC

M(i:Ht^%%si> /<^"'i::r'-'^,

^ I

*J^ Of this series there have now been published forty-nine Volnmes,
each of which contains, in addition to the papers and solutions that
have appeared in the Bducational Times, an equal quantity
of new articles, and comprises contributions, in all branches of
Mathematics, from most of the leading Mathematicians in this and
other countries.

New Subscribers may have any of these Volumes at Subscription-prices.

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LIST OP CONTEIBUTOES.

AiYAB, Prof. SWAMIB^ATHA, M.A. ; Madras.

Allen, Rev. A. J. C, M.A. ; College, Chester.

Allman, Professor Geo. J., LL.D. ; Galway.

Andebsox, Alex., B.A. ; Queen's Coll., Galway.

AWTHOWT, Bdwtn, M.A. : The Elms, Hereford.

ABMENAifTE, Professor; Pesaro.

Ball, Sir Eobt.Stawbll,LL.D.,P.R.S. ; Dublin.

Barton, W. J., M.A. ; Highgate, London.

BAsO.Prof. AbinashChandba.M.A. ; Calcutta.

Babu, Professor Ealipada, Dacca College.

Battaglini, Prof. Giuseppe ; Univ. di Boma.

Batliss, Geoegb. M.A. ; Kenilworth.

Beevob, E. J., B.A. ; Regent's Park, London.

Beltrami, Professor ; University of Pisa.

Berg, Professor F. J. van den ; Delft.

Besant, W. H., D.Sc., F.R.8. ; Cambridge.

Beyens, Professor Ignacio, M.A. ; Cadiz.

Bhattachabya, Prof. Munindbanath, M.A.

Bhut, Professor Ath Bijah, M.A. ; Dacca.

Biddle, D. ; Gough Ho., Kingston-on-Thames.

BiBCH, Rev. J. G., M.A. ; Bunralty, co. Clare.

Blackwood, Elizabeth, B.Sc. ; Boulogne.

Bltthb, W. H., B.A. ; Egham.

BoBCHABDT, Dr. C. W. ; Victoria Strasse, Berlin.

BoBDAGE, Prof. Edmond ; Coll. de Nantua.

BouLT, S. H^ M.A. ; Liverpool.

BouBNE, C. W., M.A.; College, Inverness.

Bbill, J., B.A. ; St. John's Coll., Camb.

Bbocabd, H., Chef de BataiUon du Genie;
Grenoble.

Bbooks, Professor E.; Millersville, Pennsylvania.

Bbown, a. CBC7M, D.Sc. ; Edinburgh.

BucHHEiM,A.,M.A.,Ph.D.:Gov.Sch.,M'chester.

Buck, Edwabd, M.A. ; Univ. Coll., Bristol.

BuBNSiDE, Professor W. S., M.A. ; Univ., Dublin.

BUBSTALL, H. J. W. ; Sch.of St.John*sColl.,Cam.

Camebon, Hectob C, M.A. ; Glasgow.

Capbl, H. N., LL.B. ; Bedford Square, London.

Cabmodt, W. p., B.A. ; Clonmel Gram. School.

Cabb, G. S., M.A. ; 8 Endsleigh Gardens, N.W.

Casey, Prof., LL.D., F.R.S. ; Cath. Univ., Dublin.

Casey, Prof. W. P., M.A.; San Francisco.

Catalan, Professor ; University of Lifege.

Ca VALLIN, Prof., M.A. ; University of Upsala.

Cave, A. W., B. A.; Magdalen College, Oxford.

Cayley,A., F.R.S.; Sadlerian Professor of Ma-
thematics in the University of Cambridge,
Member of the Institute of France, &c.
Chakeavabti, Prof. Byom., M.A. ; Calcutta.
Chase, Prof., LL.D. ; Haverford College.
Chbistie.R.W.D.; MerchantTaylors' S.,L'pool.
Clabke, Colonel A. R., C.B., F.R.S.; Redhill.
CocHEz, Professor; Paris.
Cockle, Sir James, M.A., F.R.S. ; London.
Cohen, Abthub, M.A., q.C., M.P. ; Holland Pk.
CoL80N,C.G., MJL.j University of St. Andrews.
CoTTEBiLL, J. H., M.A. ; R. N. Coll., Greenwich.
Cbbmona, Prof. LuiGi; Rome.
Cboeton, Professor M. W., F.R.S. ; Dublin.
Cboke, J . O'Bybne, M.A. ; Dublin.
CuLLEY, ProfyMA. ; St. David's Coll.,Lampeter.
CuBTiB, R., M.A., S.J.J Univ. Coll., Dublin.
Dabboux, Professor; Paris.
Data, Prof. Pbomathanath, M.A. ; Calcutta.
Davieb, D. O. S., M.A. : Univ. CoU., Bangor.
Davis, R. F., MJL. ; Endsleigh Gardens.
Dawson, H. G., B.A. ; Christ Coll., Camb.
Day, Rev.H. G.,MA.; RichmondTerr.,Brighton.
De Longchamps, Professor ; Paris.
De Wachtbb, Prof., M.A. ; Schaarbeck.
Dey, Prof. Nabendba Lal^M.A. ; Calcutta.
DiCK,G. R., M.A. ; Victoria University.
DoBSON, T., B JL. ; Hexham Grammar School.
D'OCAGNE, Maubicb ; Cherbourg.
Dboz, Prof. Abnold, M.A. ; Porrentruy, Berne.
DuPAiN, J. C. ; Professeur au Lyc6e d'Angoulftme.
£ ASTON, Belle, B.Sc. ; Lockport, New York.
Edwabd, J., M.A. ; Head Mast., Aberdeen Coll.
Edwabdeb, David, B.A. ; Dublin.
Elliott, E. B.. M. A. j Fell. Queen's Coll., Oxon.
Ellis, Alexandeb J., F.BJS.; Kensington.
Emmbbich, Prof.. Ph.D. : Miilheim-am-Buhr.
Emtage, W. T. a. } Pembroke Coll., Oxford.

EssENNELL, Emma ; Coventry.

Evans, Professor, M.A. ; Lockport, New York.

EvEBETT, Prof. J. D., D.CL. ; Qu. Coll., Belfast.

PiCKLiN, Joseph ; Prof, in Univ. of Missouri.

PiNKEL, B. F. : Ada, Ohio.

FOBTEY, H., M..A. ; Clifton, Bristol.

FosTEB, F. W., B.A. ; Chelsea.

FosTEB. Prof.G.CABEY,F.R.S.; Univ.Coll.,Lond

FoPT-pn. \V, S., Hodrtoi^don.

FbaskliNi UJUiibii> £ LiDiJ,M.A.j Prof, of Nat,

Ssei. aiKl Mjilh,, Union Springs, ?few York.
FrniiTES, E, ; Uiiivorsity of NapJeMH
GALEiKUTjj;HivJ.3f .A, , FulLTrin.ColL.Dubliii.
Ga iKh Kat£: Worctoiter P^rk^ i^mroy-
G.ii i.ATLY. W.. M.A,; Highr.iti^ Loridon.
Ga r r EKRi?, K^>v.T., M.A- ; Church Eectoryj^^iilop,
G.u hj.m, Fa.*.Ncia, MA., KKiiS.; Londoiin
GiiNii^E-, Pn>r^ M.A.; Univ. ColU Al>ery$tirith.
G?;^;iM\=^, 11. T„ B.A.; Mtud.of Ch.Ch-.OKfbni.
Gj u r. 1 1 ] : Fi , J . W. L,, l\R.S.i Pall .'TrltuCo] I . , Camb
Geo JicNijERa, Frore*Hiort M.A.; Moscow.
Gru;i>uN, Alic% Ri!l<^; GlourtisttdT.
G< ■ r 1 1 E It, W. P< , I rvinfi Hoi i>o j Bt^rhy.
Gi, uiAM, a. A., il.A.i Trimty Colloiro, Dublin,
G]r r s: ^r hiEtU, \lr.\\ W, J,, M,A. ^ Uulwich Colk^ffB.
Gin: KM STREET. W, J., B,A. ; Hull.
Givi i-yw^aou, James M*; KtrksvillOp Missouri.
Gil I PFimtJ. G. J.. M.xi. ; Foil. Ch. Uoll., Cu-iub-
G]^ [ 1 1 ITHS, Jm -^I a. ; Fellow or Jesus Coll., Oiom
Gnuvt;, W. B„ B.A. J Ferry Bitr, Birminffham.
H.v h I MAR[>, FrofeJ^sfir, M.A. j Fftri?t,
H.uuu,E„B,A.. B,Sc-.; King's Sch,, Wurwick,
H.^ 1 i.r Profosaor Asapu, MA. ; Washington,
R,\ \[ vt<tsTi» J„ 5LA. ; Buokliurst Hill, Esaei,
Ha> umasta Hatt, Fr<jfeswir, B*A,^ Mail res,
H.^ i: r* E >t A, C. ; Univtf raity of St. Peternburg,
H.ui[i>:R, Alfrep, M,A.; Chdtenliam,
Hu;r.i;r, Rtiv. Robeut, FkH.S, ■ Osfonl.
H.^jiias, tl, W.J B.A*; Trinity CoDtigw, Dublin.
H.u;l, IJi*, Davii! S,i ytoTiiTiffton, V^lomicctlout,
Ha i; 1 , 11. ; R. ftl. Acudemv, Woolwieli,
H.\ I (TiiTONt liciv.Dr*. F.It.S.; Trill. CoU,, DubL
H.y.y [»t;ic^a, J. E., M.A* 5 DeM Malnai^ lowav
Hi' i r kIj^ G., M.A. ; The Grovo^. HummcniTnitbi.
H] iMAN, R. A., M.A. ; 'frin- Coll., Cambridije,
H] ir ^E [ TK, Cii . i Mtimbrf] di? rinMUnt, FuiiA,
Hj uvi-Y, F. K.d.. if. A.; Worthinpf.
HiiL, lU'V. K, M.V '^v T ' ■ sCnltt^pp, Cnmb
HlN'lON, C. H., M..A . : '.; ■ .; liliara Colletre.
H1B8T, Dr. T. A., F.K.iS. ; London.
Holt, J. R., M.A. ; Trinity College, Dublin.
Hopkins, Rev.G. H., M.A.; Stratton, Cornwall.
HOPEINSON, J., D.Sc., B.A. ; Kensington.
Hudson, C. T., LL.D. ; Manilla Hall, Clifton.
Hudson. W.H.H.,M.A.;Prof.inKing'8Coll.,Lond.
Inoleby, C. M., M.A., LL.D. ; London.
Jenkins, Moboan, M.A^ London.
Johnson, A. R., M.A. ; Wesley Coll., Sheffield.
Johnson, Prof., M JL. ; Annapolis, Maryland.
Johnston, J. P., B.A. ; Trin. Coll.. Dublin.
Johnston, W. J., M.A.; Univ.CoU., Aberystwith.
Jones, H. S., B.A. ; Llanelly.
Kahn, a., B.A. ; St. John's Coll., Camb.
Kennedy, D., M.A. ; Catholic Univ.. Dublin.
KiBKMAN, Rev. T. P., M.A., F.R.S. ; Croft Recfc.
KiTCHiN, Rev. J. L., M.A. ; Heavitree, Exeter.
KiTTUDaE. Lizzie A. ; Boston. United States.
K:. . . ■^ V. i;. J.. Xt ■■•■.. . ■■.: . o]i[a,

K><.iW[,E5S, It., i*.A.T M-F.:; '^O! ((.■nllEIML

K0EIJ1.EII, J. ^ liu« St. JaoqnEis, Pnrb,
liAtutAN, R„ B.A,; Lowiflham.
Lampb, Prof., Ed, ^f Jahrb.dsr Math.-^ Berlin.
LANQLTTTt E. M„ B,A. ; BerJford.
La V E HTY, W H a „ Si .A, 1 1 ftto KxiLiju i u IJn iv.OiIbrd.
La^vhkmce, K. J. ; Ei-F( 11. Trill. ColL, Catub,
Li; itniJi, G*>T>t?nil : 2S Rui^ CarotZr Brusaells.
Lemuine, E. ; &, Rue Littfd, Ppj-ih.
Lh4abi^kb« Prarcussor ; Dtil^'t,
LfiuiIOUj, R., M.A. ; Firiabuiy Park,
LkcdebbosPpO., M^A.s Fel.PembrokeUoll.jOlon,^
Ll:v£!TTf R.. ilJL.; Kinx£dvp% Skib.^BlruiiUf^htua*
LojfDOW, RcVh H.,Bi,A,^ PoeklingiQ]i.
LawBT, W. H,| Jt)i|Kit^l#ckrQ«*yD)a^lt

IV

McAlisteb, Donald, M.A., D.Sc; Cambridge.
McCat, W. S., M.A.; Pell. Trin. Coll., Dublin.
McCLBLLAur, W. J.,B.A.; Prin.of SantrySchool.
McCoLL, Hugh, B.A. ; Boulogne.
Macdoxald, W. J., M. A. ; Edinburgh.
Macfarlaxe, Prof. A., D.Sc. j Univ. of Texas.
MclNTosff, Alex., B. A.; Bedford Bow, London.
Mackenzie, J. L.. B.A. ; Gymnasium, Aberdeen.
McLeod, J., M.A.; R.M. Academy, Woolwich.
MacMahon, Capt. P. A. ; E. M. Academy.
MacMurcht, a., B.A.; Univ. Coll.. Toronto.
Madison, Isabel, B.A. ; Cardiff.
Malet, Prof., M.A.; Queen's Coll., Cork.
Mann, M. F. J., M.A. ; Kensington.
Mannheim, M. ; Prof, k I'Ecole Polytech,, Paris.
Mares. Sarah, B.Sc. ; London.
Martin, Artemas, M.A., Ph.D.; Washington.
Mathews, G. B., B.A. ; Univ. Coll., N. Wales.
Matz, Prof., M JL. ; King's Mountain, Carolina.
Merripibld, J., LL.D., P.BuA.8. ; Plymouth.
Merriman, Mansfield, M.A.; Tale College.
Meyer, Mary S. ; Girton College, Cambridge.
Miller, W. J. C, B.A., (Editor);

The Paragon, Richmond-on-Thames.
MiNCHiN, G.M.,M.A. ; Prof, in Cooper's Hill Coll.
Mitcheson, T.,B.A.,L.C.P. ; City of London Sch.
MoNCK, Prof. H. St., M.A. ; Trin. Coll., Dublin.
MONCOFRT, Professor ; Paris.
Moon, Robert, M.A. ; Ex-Fell. Qu. Coll., Camb.
Moore, H. K., B.A. ; Trin. Coll., Dublm.
Morel, Professor ; Paris.
Morgan, C, B.A.; Salisbury School.
Morley, Frank, B.A. ; Bath Coll., Bath.
Morrice, G. G.,B.A.; Mecklenburgh Sq., Lond.
MuiR, Thomas, M.A., F.R.S.E. ; Bothwell.

MUKHOPADHYAY,Prof.A8UT08H,M.A.,P.R.S.E.

Mr?rT--r.v -^rAT^ Ti vt. V-iL MA.. LL.E.

NBUBBftt*. Proffisiajors Hmv. uf Li^ee.
Kbwcojilb, FntL &i^oti^UA.\ Wiinhinjctim.
CrCorfNKLL, Mnjpr-Gt^i^eral F, ; Utttb»
OrKASUAW, Eev. T. W., M.A.; Clifton,
ORCHARD, H. L,, M.A.. K.Sc; Hauipstead,
O'Ri^aAir, JouN^ New iSti^ot, Liiit^nck.
Owen, J. A„ B.S6. ; TonnysL>n St., Liverpool.
Paktdn.A. W,,M,A.; FfU.ia'TchjJ-v.ll.JiuhlLn.
Pendlebdry, C, M.A. ; London.
Perrin, Emily, B.Sc. ; Girton College. Camb.
Phillips, F. B. W. ; Balliol College, Oxford.
PiLLAi, C. K.. M.A. J Trichy, Madras. •
PiRiE, A., M.A. ; University of St. Andrews.
Plamenewski, H., M.A. ; Dahgestan.
PocKLiNGTON. H. C.,M.A. ; Yorks CoU., Loeds.
PoLiGNAC, Pnnce Camill]^ de ; Paris.
PoLLEXFEN, H., B.A. ; Windermere College.
Poole, Gbrtrc7DE, B.A. ; Cheltenham.
Potter, J., B.A. ; Richmond-on-Thames.
Pressland, a. J., B.A. ; Brecon.
Prudden, Frances E.; Lockport, New York.
Purser, Prof. F., M JL. ; Queen s College, Belfast.
Putnam, K. S., M.A. ; Rome, New York.
Rawson, Robert ; Havant, Hants.
Read, H. J., B.A. ; Brasenose Coll., Oxford.
Rees, E. W., B.A. ; Penarth.
Reeyeb, G. M., M.A. ; Lee, Kent.
Reynolds, B., M.A.; Netting Hill, London.
Richards, David, B.A. ; Aberystwith.
Richardson, Rev. G., M.A. ; Winchester.
Roach, Rev. T., M.A. ; Clifton.
Roberts, R. A., M.A.; Schol.of Trin.Coll.,Dublin
Roberts, S., M.A., F.R.S. ; London.
Roberts, W. R., M.A. ; Fell, of Trin. Coll., Dub.
RoBSON, H. C, B.A.; Sidney Sussex Coll., Cam.
Rosenthal, L. H. ; Scholar of Trin. Coll., Dublin.
Roy, Prof. Kaliprasanna, M.A. ; Agra.
RuoGERO, Simonelli ; Universitd. di Roma.
Russell, Alex., B.A. ; Ass.Lect. C.Coll..Camb.
Russell, J. W., M. A. ; Merton Coll., Oxford.
Russell, R., B.A.; Trinity College, Dublin.
RuTTER, Edward ; Sunderland.
Salmon, Rev. G., D.D.,F.R.S.; Regius Professor

of Divinity in the University of Dublin.
Sanders, J. B.; Bloomington, Indiana.
Sanderson, Rev. T. J., M. A. ; Royston, Cambs.

Sabadaranjan Rat, Prof., M.A. ; Dacca.
Sarkar, Prof. Beni Madhav, M.A. ; Airra.
Sarkar, Prof. NiLKANTHA, M.A. ; Calcutta.
ScHEFFER, Professor; Mercersbury Coll., Pa.
ScHOUTB, Prof. P. H. ; University, Groningen.
ScoTT, A. W., M.A. ; St. David's Coll., Lampeter.
ScoTT, Charlotte a., D.Sc. ; Professor in

Bryn Mawr College, Philadelphia.
Scott, R. F., M.A.; Pell. St. John^sColl., Camb.
Sen, Raj Mohan; Rajhasbye Coll., Bengal.
Seymour, W. R., M.A. ; Tunbridge.
Serret, Professor ; Paris.
Sharp, W.J. C, M.A. ; Greenwich.
Sharpe, J. W., M.A. ; The Charterhouse.
Sharpe, Rev. H. T., M.A. : Cherry Marham.
Shepherd, Rev. A. J. P., B.A. ; Fell, Q.Coll.,Oxf
Simmons, Rev. T. C, M.A.; Grimsby.
SiRCOM, Prof., M.A. ; Stonyhurst College.
SivBRLY, Walter; Oil City, Pennsylvania.
Skrimshire, Rev. E., M.A. ; Llandaff.
Smith, C, M.A. ; Sidney Sussex Coll., Camb.
Stabenow, H., M.A. ; New York.
STEaGALL, Prof. J. E. A., M.A. ; Dundee.
Stebde. B. H., B.A. ; Trin. ColL, Dublin.
Stein, A. ; Venice.

Stephen, St. John, B. A.; Caius Coll., Cambridge
Stewart, H., M.A. ; Framlingham, Suffolk.
Storr, G. G., B.A. ; Clerk of the Medical CounciL
Swift, C. A., B.A. ; Grammar Sch., Weybridge.
Sylvester, J. J., D.C.L., F.R.S.; Professor of

Mathematics in the University of Oxford,

Member of the Institute of France, &c.
Symons, E. W., M.A.; Fell. St. John's ColL.Oxon.
Tait, Prof. P. G., M.A.; Univ.^dinburgh.
TANNER,Prof, H.W.L.,M.A.; S.Wales Univ. CoU.
Tableton, F. a., M.A. ; Fell. Trin. CoU., Dub.
Taylor, Rev. C, D.D. ; Master of St. John's

College, Cambridge.
Taylor, H. M., M.A.; Fell. Trin. Coll., Camb.
Taylor, W. W., M.A. ; Ripon Grammar School.
Tebay, Septimus, B.A. ; Farnworth, Bolton.
Terry, Rev. T. R., M.A., Fell. Magd. Coll.,Oxon.
Theodosius, A. F., M.A. ; Bath College.
Thomas, A. B., M.A., Merton College, Oxford.
Thomas, Rev.D., M.A.; Garsington Rect.,Oxford.
THOM80N,Rev.F.D.,M.A.; Ex-FeLSt.J.Coll.,Cam .
TiRELLi, Dr. Francesco ; Univ. di Roma.
ToRELLi, Gabriel; University of Naples.
TORRY, Rev. A. F., M.A. ; St. John's Coll., Camb.
Traill, Anthony, M.A., M.D.; Fellow and

Tutor of Trinity College, Dublin.
Tucker, R., M.A. ; Mathematical Master in Uni-

versity College School, London.
Turriff, George, M.A. ; Aberdeen.
ViGARiE, Emile ; Castres, Tarn.
ViNCENzo, Jacobini; University di Roma.
VosE, Professor G. B. ; WashiuKton.
Walenn, W. H. ; Mem. Phys. Society, London.
Walker, J. J., M.A., P.R.8. ; Hampstead.
Walmsley, J., B.A. ; Eccles, Manchester.
Warburton- White, R., B.A. , Salisbury.
Warren, R., M.A. ; Trinity College, Dublin.
Watherston, Rev. A. L., M.A. ; Bowdon.
Watson, D., M.A. ; Folkestone.
Watson, Rev. H. W.; Ex-FeU. Trin. Coll., Camb.
Wertsch, Franz ; Weimar.
Whalley, L. J., B.Sc. ; Leytonstone.
Whapham, Rosa H. W. ; Cardiff.
White, J. R., B.A.; Worcester CoU., Oxford.
White, Rev. J., M.A. ; Royal Naval School.
Whiteside, G., M.A. ; Eccleston, Lancashire.
Whitworth, Rev. W. A., M.A. ; London.
Williams, C. E., M.A. ; Wellington CoUege.
Williamson, B., M.A^ F. & T. Trin. CoU., I)ub.
Wilson, J. M., M.A. ; Head-master, Cliftoh Coll.
Wilson, Rev. J., M JL.; Rect. Bannockbum Acad.
Wilson, Rev. J. R., M.A. ; Royston, Cambs.
WoLSTENHOLME, Rov. J., M.A., Sc.D. ; Profossor

of Mathematics in Cooper's HiU College.
Woodcock, T., B.A. ; Twickenham.
WooLHOUSE, W. S. B., F.R.A.S., Ac. ; London.
Wright, Dr. S. H., M.A. ; Penn Yan, New York.
Wright, W. E., B.A.; Heme Hill.
Young, John, B.A.; Academy, Londonderry.

o

CONTENTS.

iMatftematiial ^apersf, ^r.

Page

Note on a Rectangular Hyperbola. (R. Tucker, M. A.) 116

** Something or Nothing ?'* (Charles L. Dodgson, M.A.) 101

i)iophantine Analysis. (R. W. D. Christie.) 159

Resolution of Squares. (R. W. D. Christie.) 162

Resolution of Cubes. (R. W. D. Christie.) 170

1010. (The Editor.) — Find integral values of a>, y, z which will make
a;3-2y2,y2_3g?^a^_ 5^2 all squares 174

1014. (The Editor.) — Find the least integral value of x which will
make the expression 927a;2— 1236a; + 413 = a square 174

1042. (The Editor.) — Find values of x which will make each of the
expressions dx^-^l, x^ + l, 2a;*— 3x^+2 a square number 174

1898. (Hugh MacColl, B.A.) — Find the number and situation of the
real roots, giving a near approximation to each, of

a;4 + 4-37162a;3-24-964235876U2 + 34.129226840859882a;

-14-63442007818570452204 « 0. ... 70

2144. (Professor Wolstenholme, Sc.D.) — If from the highest point
of a sphere an infinite number of chords be drawn to points uniformly
distributed over the surface, and heavy particles be let fall down these
chords simultaneously, their centre of inertia will descend with accele-
ration i^...... 125

2146. (Professor Nash, M.A.) — D, E, F are the points where tiie
bisectors of the angles of the triangle ABC meet the opposite sides. If
ar, y, z are the perpendiculars drawn from A, B, C respectively to the
opposite sides of the triangle DEF ; j?,, ^2> Pz those drawn from A, B,
respectively to the opposite sides of ABC : prove that

PI.+ Pi ^^«ii + 8 8inAsin J^ sin-^ 125

x^ y^ v' 'lit

a

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VI CONTENTS.

2173. (Professor Wolstenholme, Sc.D.) — The quadric
ax^^by'^^cz^^ 1
is turned about its centre until it touches a'x^ + A'y' + c'z^ _. \ along a
plane section. Find the equation to this plane section referred to the
axes of either of the quadrics, and show that its area is

ir(a + A + <J-a'-*'-c')V(«*^-«'*'0* 126

2352. (Professor Sylvester, F.R.S.)— We may use VJ^^i to denote the
third point in which the right line PQ meets a given cubic ; P#Q«R to
denote the third point in which the line joining the one last named and
R meets the cubic, and so on. Thus P^P will denote the tangential or
point in which the temgent at P meets the given cubic, and [P#P]#[P#P]
will denote the second tangential, i.^., the tangential to the tangential at P.

1. Prove that [P#P]*[P«PJ = I«P«[P»P]#P«I, where I is any point
of inflexion in the given curve.

2. Obtain a function of P, I which shall express the point in which the
curve is cut by a conic having five-point contact with it at P 21

2353. (The late Professor De Morgan.)— The q
late Dr. Milner, President of Queens' College, Cam-
bridge, constructed a lamp which General Perronet
Thompson remembered to have seen. It is a thin
cylindrical bowl, revolving about an axis at P, and
the curve ABCD is such that, whatever quantity _ ^
of oil ABC may be in the bowl, the position of Bx *^
equilibrium is such that the oil just wets the wick yjv^^^^^
at A. Fiud the curve ABCD 64 &

2396. (W. S. B. Woolhouse, F.R.A.S.)— Let ABCD be any convex
quadrilateral, having the diagonals AC, BD intersecting in E ; and let
p, p' denote the ratios 2AE . EC : AC^, 2BE . ED : BD^ respectively.
Then, if five points be taken at random on the surface of the quadrilateral,
prove that the probabilities (1) that the five random points wiU be the
apices of a convex pentagon, will be ^^^ (11 + 5pp') ; (2) that the pentagon
will have one, and one only, point reentrant, will be f ; (3) that it will
have two reentrant points, will be -^ (I— pp') 41

2437. (The late Rev. J. Blissard, M.A.)— Prove that

-J_ + _±_ + -J^ + ... = -^taii^ 40

\^^x^^^^-x' h^^x'^ ix 2

2448. (J. S. Berriman, M.A.)— Let AEB, CED be two lines of railway,
whereof AB is perfectly straight, and CD curved as far as F, the
remainder being straight ; then, if FE be 25 feet long, and the curve CF
have a radius of 3000 feet, and the angle BED = 25^ 26' ; show that the
distance from B to E, so that a curve BC may be struck with 1000 feet
radius is 342*765 feet 49

2814. (The late Matthew Collins, B.A.) — Can the common differ-
ence of three rational square integers in Arithmetical Progression be ever
equal to 17? 161, 174

3419. (Artemas Martin.) — The point A, is taken at random in the
side BO of a triangle ABC, Bi in CA, and C, in AB ; the point Ao is taken
at random in the side Bfii of the triangle AiB,Ci, Bg in CiAi, and C2 in
AjBi, and so on ; find the average area of the triangle AmB„C,» 85

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vu

. 4043. (For Enunciation, see Question 1898) 70

4251. (Colonel Clarke, C.B., F.R.S.)— If A, B, C be three circles, B
being within A, and C within B ; prove that the chance that the centre
of A is within C is f ; 61

4721. (Professor Sylvester.) — Prove that every point in the plane
carried round by the connecting-rod in Watts' or any other kind what-
ever of three-bar motion has in general three nodes, and that its inverse in
respect to each of them is a unicircular quartic 127

4828. (The Editor.) — ^If the corner of a page of breadth a is turned
down in every possible way, so b& just to reach the opposite side ;
(1) show that the mean value of the lengths of the crease is

j{7A/2 + log(l+ V2)}a,

and (2) the mean area of the part turned down is -|-|a3 128

5440. (R. Rawson.) — ^Prove that the general solution of the equation
is K = <?3 [* e't*(»^-^>/*(») . ^ [zy,^' (2) dz-\-e (1)

dx^ \ Xx \dx I dx^ ) dx dx^ x^xdx }
X Xy \dx I dx^ 5 dx Xi \dx J

where a, jS, Xi are given functions of x^ and

N = ^3^ ^ €'.♦(•)-'■»'*(•) . 4) (a)^. <^'(a)-^ e^iiP)-c,!iK$) , 4, (3)^-. ^' (3)] ,
i^dx dx )

dx (^dx

- ^ 6'.«(/9)-f2<'«(^). ,^(j8)<'»*i <p' (3) ] .

dx )

80

6391. (J. J. Walker, F.R.S.)— If 0, A, B, C, D are any five points in
space, prove that lines drawn from the middle points of BC, CA, AB
respectively parallel to the connectors of D with the middle points of C)A,
OB, OC, meet in one point E, such that DE passes through, and is bisected
by, the centroid of the tetrahedron OABC. [Quest. 6220 is a special case,
in two dimenbions, of the foregoing theorem in three dimensions.] ... 129

6911. (W. R. Westropp Roberts, M.A.) — Let H and H' be the
Hessians of two binary cubics respectively, © their intermediate co-
variant ; then, using the notation ot Salmon, prove that

9e2-36HH'=6PJ + H(6J) 74

6931. (For Enunciation see Quest. 2396.) 41

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Vlll CONTENTS.

7131. (W. J. C. Sharp, M.A.) — ^Prove that the vector equations to
the centrodes of a three-har motion, which are easily derived from one
another by a linear substitution, are of the third degree in the vectors,
and reduce to the second where the algebraical perimeter of the iig^e is
zero 130

7178. (W. J. C. Sharp, M.A.) — ^If three concyclic foci of a bicircular
quartic, or circular cubic, be given, and also a tangent and its point of
contact, determine the curve 23

7244. (D. Edwardes.) — The circles of curvature at three points
of an ellipse meet in a point P on the curve. Prove that (1) the normals
at these three points meet on the normal drawn at the other extremity of
the diameter through P ; and (2) the locus of their point of intersection
for difEerent positions of P is 4 {a^x^ + b^^ « {a^-b'^)'^ 68

7384. (Professor R6alis.)— Etant donn6e la serie illimit^e 7, 13, 26,
43, 67, 97, 133, 137, ..., dont le terme general, celui qui en a n avant lui,
est An = 3 (n2+ w) + 7 : demontrer les propositions suivantes : — (1) sur cinq
termes consecutifs, pris a volonte dans la serie, un terme est divisible par
5 ; (2) sur sept termes consecutifs, deux sent divisibles par 7 ; (3) sue
treize termes consecutifs, deux sent divisibles par 13 ; (4) aucun terme
de la Berien'est egal a un cube; (5) une infinite de termes, tels que
A2 = 25, As7 = 4225, etc., "sent des carr6s divisibles par 25 ; (6) la
douxi^me et la troisi^me proposition sont comprises, comme cas particu-
liers, dans la suivante : si N est un nombre premier, de la forme 6m + 1 ,
sur N termes consecutifs de la serie, deux sont divisibles par N ; (7) on
pent affirmer aussi que, k Texception de 5. aucun nombre premier de la
lorme 6;n—l ne pent diviser aucun terme de la serie 140

7759. (Professor Hanumanta Kau, M.A.) — From one end A of the
diameter AB (= 2a) of a semicircle, a straight line APMN is drawn
meeting the circumference at N, and a g^ven straight line through B at
M, at an angle a ; show that the locus of a point P, such that AP, AM,
AN are proportionals, is the cubic through A,

r = 2a sin2 a sec cosec^ (a -6), or 2a sin' a {x^ + y^ ^ (jx sin a — y cos o)',
which, when a = ^t, iir, becomes

2a^ {x^ -i y^) = n^y 2rv2(j;2 + y2) = a;(a?--y)2 69

7949. (R. Knowles, B.A.) — Prove that the sum of the series
ix-'ix^'^ix3-...ad. inf.

.3-l^Uog(J^^*±^^3-Mftan-l?^+cot-l3i]... 84

7986. (J. Brill, B.A.)— ABCD is a quadrilateral, AB and DC when
produced meet in E, and AD and BC when produced meet in F ; prove
that AB . CE . DF cos (ABD + CEF + CAF)

+ AD . CF . BE 003 (ADB + CFE + CAE)

- BC . AF . DE cos (CFE « ADB + DCA)

- CD . AE . BF cos (CEF + ABD + BCA) = AC . BD . EF 02

8020. (Asp iragus.) — A conic circumscribes a given triangle ABC
and one focus lies on BC ; prove that the envelop of the Corresponding

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CONTENTS. IX

directrix is a conic with respect to which A is the pole of BC ; and, if A
he a right angle, the envelop is the parabola whose focus is A and direc-
trix BC. [If (0, 0), («, b)y (a, — c) are the coordinates of A, B, C, the
equation of the envelop will be
Ue (bc-oA x^ + ia (b + c) {be-aP) xy + a^ [4a2+ (i-<.)2]y2

+ a^^b + c)^2ax•~a^) - 0.] 82

8095. (H. G. Dawson, B.A.) — If a, A, c be the axes of a quadric
having the tetrahedron of reference for a self- conjugate tetrahedron,
(if »?» C> ^) the tetrahedral coordinates of the centre of the quadric, and
(A-i, Mi» ''i» »i)i (^2> /*2» ^iy *2)» (^3* i"3» ^3* '3) *^® tangential coordinates of
its principal planes ; prove that (1)

and hence (2), if a tetrahedron be self-conjugate with respect to a sphere
of radius R and centre O, show

-R2 (ABCD) = x«<OBCD) +fi- (OCDA) + v2(0DAB) + ir^ (OABC),
where A, B, C, D are the vertices of the tetrahedron, X, /u, i^, ir the per-
pendiculars from A, B, C, D on any plane through O, and (ABCD;, &c.
are the volumes of the tetrahedra 31

8132. fW. J. Johnston, M.A.) — Prove that, if the section of a
quadric by a plane is given, and also a straight line in that plane ; then,
if through this line a plane can be drawn to cut the quadric in a circular
section whose radius is also given, the locus of the centre of this circular
section is a circle in a plane perpendicular to the given plane 25

8177. (Professor Hanumanta Rau, M.A.) — The images of the circum-
centre of a triangle ABC with respect to the sides are A', B', C ; prove that
the triangles A'B'C and ABC are (1) equal, (2) have the same nine-point
circle ; also find (3) the equation of the circum-circle of A'B'C and the
angle at which the two circum-circles cut each other 95, 131

8270. (D. Edwardes.) — Let ABO be an acute-angled triangle, and
L, M, N the points where the angle bisectors meet BC, CA, and AB re-
spectively. Prove that (1) the circles ALB, ALC cut one another at an
angle A, the circles ALC, ANC at an angle ±i (0— A), and the circles
ALC, BNC at an angle 90° — JB ; (2) the centres of the pair of circles which
pass through L are equidistant from the centre of the circle ABC, and
similarly ior the other two pairs ; (3) if p^, pj^ ^® t^® Ttidii and 8j^ the
distance between the centres of the circles which pass through L, and
similarly for p^, p'^, &c., Pj^PmPn = Pl^m Pn ^ KKh *» W ^^ ^1 ^^ the
distance of the circle ALB (or ALC) from the centre of the circle ABC
(radius R) , and similarly for e^, d^^ R' — R (rf,c?2 + ^i<^3 + d-M ~ '^dA/^z = ^ J
(5) if the base BC and the ciicum-circle BAC be given, the envelope of
the line joining the centres of the circles ALB, ALC is a parabola whose
focus is at the centre of the given circle and latus rectum 4Rsin2JA. ... 66

8300. (Professor Hanumanta Rau, M.A.) — From any point P on the
circle described about an equilateral triangle ABC, straight lines PM, PN,
PR are drawn respectively parallel to BC, CA, and AB, and meeting the
sides CA, AB, BC at M, N, and R. Prove that the points M, N, R
are collinear 60

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X CONTENTS.

8315. (Professor Booth, M.A.)— If

tan"» (^ir + ^i\,) = tan** (Jir + \(p),

8329. (D. Edwarde8.>— Prove that (1) the squares of the lengths of
the normals drawn from a point xt/ to the ellipse b'x^ + a-t/'^ = a^^, are
given by the equation {p-r* - ( U + p^V + 9q*) t^ + U V } 2

- 4 {r4-(2V + 3j02)r2 + 3U + V2} {(j»4-3^*)r^-(2j»2U-3^4V)r2-hU2},
where U = H^j^ -^ a^i/^ - a-b'^, V== x^ + t/^-a^-b^y p^^a^-^V^, and ^=a2^;
and (2) if on the normal at P, a length PQ be measured inwards, equal
to the semi-conjugate diameter, the squares of the lengths of the other
three normals drawn from Q are given by the equation

+ {4(a-*)2PQ2(2a2+2*2 + a*)-4rt«*-^(2«2 + 2*2-7a*)} r'

-4 {(a-*)2PQ2-a2^j2 ^ 0. 99

8331. (H. G. Dawson, B.A.)— Show that the solution of

^^ZJL^^^^ax, y^^y^^by, ^^l^^'^^cz (1,2,3),

yU f^H rpH gtl *" -jjH yU \ J » /

depends on the solution of

fl(p-a)«-i + *(p-A)»-i + <j(p-<j)~-»= (4).

51

8333. (Professor Hanumanta Rau, M.A.) — Prove that the equations
x^ + 19j?- 140 = 0, and 7a:^- 12a:3 + 46a;2^. 12ar+ 7 = 0, have two common
roots 96

8337. (Professor Mukhopadhyay, M.A., F.R.S.E. — Extension of
Question 8107.) — If 0y <^, i// be the angles of inclination of any two tan-
gents to a conic, and of their chord of contact, to a directrix, show that,
if e be the eccentricity of the conic,

^^ Ar^jin^+ji^liin^ eit ^ hz^l ^ hu^ 64

X-icos0 + /4-*cos4) sin^d sin^^)

8344. (R. Knowles, B.A.)— AB, BE, CF are drawn from the angular
points of a triangle ABC, so that the angles BAD, EBC, ACF are each
equal to the Brocard-angle of the triangle ; show that their equations are

bcy-a^z^O, *2^-a<Jz=0, abx-c^i/^0 87

8461. (F. R. J. Hervey.) — Find in how many ways n lines of
verse can be rhymed, supposing that (1) no line he left unrhymed, and
(2) the restriction as to unrhymed lines be removed ; and show that, in
the case of the sonmty the respective numbers of ways are 24011157 and
19U899322 91

8463. (J. C. Stewart, M.A.) — Solve completely the equations
x-^'ly-xy'^-k- VZ(\-.2xy-y'^) = y + 2a?-a;V + (2+ ^'6)(\-2ry-x^ = 0;
and show that one system of values is a; = ± J\/3, y —I and V'6 — 2.

30

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CONTENTS. XI

8503. (N'Importe.)— A rod of length a^l rests in fquilibrium in a
vertical plane within a rough sphere of radius a, one extremity of the
rod being at the lowest point of the sphere ; show that the coefficient
of friction is J1 — \ 73

8640. (Rev. T. R. Terrj-, M.A.)— Show that the series
1 + w X + ^(^+^) y(y-fO ^ w(w-H)(w4-2) g(g-fr) (g + 2r) ^
p 1.2 p{p-^r) 1.2.3 i? (it? + r) (i? + 2r)
is convergent if i? > q-\-mr 67

8677. (B. Hanumanta Rau, M.A.) — Prove that the arc of the pedal
of a circle, of radius a, is equal to the arc of an ellipse (^ ^ f )> the origin
being at a distance Ja from the centre of the circle 33

8592. (Professor Mathews, M.A.) — Through a point P are drawn
three planes, each parallel to a pair of opposite edges of a tetrahedron
ABCD. Prove that the 12 finite intersections of these planes with the
edges of the tetrahedron lie on the same quadric surface ; and that, if
BC2 + AD2= CA2 + BD2« AB2 + CD2 (i.^., if each edge of the tetra-
hedron is perpendicular to the opposite edge), there is one position of P
for which the quadric surface is a sphere 132

8647. (R. W. D. Christie, M.A.) — If «= 13 + 23 + 3^+ ... + n8,
S « l6 + 2» + 3« + . ..+««, 2 = 17 + 27 + 37+. ..+«7j prove that 5 + S= 2«2.

17.8

8667. (N'Importe.) — Two equal perfectly elastic balls, moving in
directions at right angles to each other, impinge, their conmion normal
at the instant of impact being inclined at any angle to the directions of
motion : show that, after impact, the directions of motion will still be at
right angles 66

8668. (Alpha.) — The ellipse whose eccentricity is 4 ^^2 is referred to
the triangle formed by joining a focus to the extremities of the latus
rectum through the other focus : prove that its equation is

72+9(i87 + 7a + ai8) = 84

8701. (A. RusseU, B.A.) — Resolve into quadratic factors
(a2-^c)«(A + c)*(*-c) {a2 + 2a(* + c) + ^
+ (*2- (.a)8 {c + ay (c- a){b^ + 2b {c + a) + ea}
+ {(^^aby{a+by(a-d){c'-^2c{a-t-b) + ab} 68

8742. (R. Knowles, B.A. Suggested by Quest. 8521.)— The circle
of curvature is drawn at a point P of a parabola, PQ is the common
chord ; if 0, O' be the poles of chords of the parabola, normal to the
parabola at P and Q respectively, and if M, N, R, T be liie mid- points of
00', OQ, O'P, PQ respectively, prove (1) that the lines MT, NR intersect
at their mid-points in the directrix, (2) that OP, O'Q are bisected by the
directrix 30

8743. (C. Bickerdike.) — Prove that (1) the length of a focal chord
of the parabola is /cosec2^; (2) when the chord is one of quickest
descent, cos ^ = (|)* ; and (3) the time of quickest descent down the

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Xll CONTENTS.

chord then is ^/(2\t)lg, where / is the latus-rectum, and 4> the angle
made by the choni with the axis 63

8752. (Professor Genese, M.A.) — If AL, BM, CN he perpendiculars

from the vertices of a triangle ABC upon any straight line in its plane,

then, three letters denoting an area, and signs being regarded, prove that

AMN + BNL + CLM = ABC 35

8766. (S. Tebay, B.A.)— If AX, BY, CZ be opposite dihedral angles
of a tetrahedron, show how to construct the solid in order that

{tani(B-Y)-tani(C-Z)}tani(A + X)

+ {tan i (C -Z) - tan HA- X) } tan J (B + Y)

+ {tani(A-X)-tani(B-Y)}tanl(C + Z) -0 123

8771. (W. J. Ureenstreet, M.A.) — Prove that the series

Ussin«Ji+^sin2« + ^^-3_8in^«+ J^-^sm^^

63

8781. (Professor Hanumanta Rau, M.A.) — If She the sun, and A
and B two planets that appear stationary to one another, show that
tan SB A : tan SAB = periodic time of A : periodic time of B 98

8782. (A. Russell, B.A.)— Prove that, if

a3(3 + <j) + ^(<j + a) + c3(a + *) = 2a*tf (a + i + <j), then

'>)(^"-)/(lrH-(^-»)/(,^-')

(2) (a8'»-i*»»(?*")(a2-3r)(* + (?)2(^2 + (?')(i« + c^)...(*2« + c2«) + ... + ... =0;
(3) (62-c2)^a-^j'{3a2 + a(4 + (r) + ^

+ (c2-a2)fi-^y{3i2 + i((? + a) + ca}

+ (a«-i«}^<j-^)'{3c2 + c(a + i) + fl*} = 121

8784. (R. W. D. Christie.)— Prove that, if
«= 1 + 2 + 3 + .. .+«, S2 = 12 + 22 + 32 + ... +w2, S3 = 13+ 23 + 3»+. ..+««,

2 » 14 + 2^ + 3^+...+^, <r- l» + 2« + 3« + ... + n6,
then (3(r + 2«8)/55=S3/S'^ 178

8818. (Professor Mukhopadhyay, M.A., F.R.S.E.)— Show that, (1)
the equation of the directrix of the conic which is described having the
origin for focus and osculates ¥x'^ + a-y^ = a^b^ at the point ^, is

(a-2-3-2) («a?cos3 4>-3ysin3 4>) = 1;
(2) the envelope of this for different values of <p is the quartic
*2.i;-2 + «2y.2 ^ (aA-i_ia-»)^

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CONTENTS* XIU

Which curve is also the reciprocal polar of the evolute of the conic
a^x^ + l^y^ = a^^s with respect to a circle whose radius is a mean pro-
portional between the axes of the ellipse 40

8826. (Professor Sircom, M.A. Suggested by Question 2845.) —

Show that l+i-^+i4;^+|4_6^+...».»Ii:l^ 77

3 3.5 3.5.7 a;(l- x-)^

8850. (W. J. Greenstreet, M.A.) — Prove that the sum of all the
harmonic means which can be inserted between all the pairs of numbers
whose sum is w, is ^ («^— 1) 59

8852. (J. Griffiths, M.A.) — If a, jS, 7, 5 be the roots of the quartic
ax* + ^bj^ + Qex^ + ^dx + d = 0, and it q ^ ^^^ -i- ^^^ ; show that

(2-^)2(1-2^)2(1+^)2 108 J2'
where 1 ^ ae-Ud-i-Zc^, J == ad'' + eb^ + (^-ace—2bcd, 69

8853. (A. Russell, B.A.)— Prove that

Jo Jo Jo \ 4a2«3' 4^2^' ^cYf

is a solution of the differential equation

^==a.^+i.^+,2^ 119

dt dx^ rfy2 az^

8855. (Professor Mukhopadhyay, M.A., F.R.A.S.)— Prove that (1) the

solution of the system ^ . = a, -21 . => ^ is given by

X l+y2 a^ l+y6

X OK- 1 «A-1

where \ satisfies ( ~^, ] « — ^ — ; and obtain (2) all the solutions

by the transformation A + \-i=/i 33

8868. (Professor Schoute.) — If ABC and A'B'C are two positions of
the same triangle in space ; if A", B", C" are the centres of the segments
A A', BB', CC, and if the planes through A", B", C" respectively perpen-
dicular to AA', BB', CC, intersect in P, the tetrahedrons PABC and
PA'B'C are not congruent, but symmetrical 39

8930. (R. W. D. Christie.)— Prove that, whether (n) be odd or even,

sinwa = sina f (2cosa)"-i-(«-2)(2cosa)'»-3+ (n-3)(;»-4) ^^cose)^-^

^(^-^H^-^^(^-6)(2cos0)n-7 4....| 176

8935. (For Enunciation see Quest. 2396.) 41

8940. (W. J. C. Sharp, M.A.)— If

S = aj:- + Ay2 + cz^ + aic2 ^ 2lyz + 2mzx + 2nxy + 2pxw + 2qyw + 2rzw,

h

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XIV

CONTENTS.

and Pi . 2 s MiX^ + hy^y^ + ez^z^ + dw^Wi + / (y,a, + y^^ + &c. ;

show that 818,83+ 2P,.aP2.,P3.i-Sili.3-S2Pj.i-S3P5 2

= A

yii .vj, .vs

«ii

+ &C.,

+ ... + 2L tr,, «?2, t^a arj, arj, a-j
|a?i, a-2, iTal |yi, y„ y,

where A, u are the first minors of the discriminant of S 134

8941. (W. J. C. Sharp. M.A.)— Prove that the conditions that the
binary quantic (a, *,<;... J x, yY should be reducible to a binomial form,

tty bf Cf d ...
bf Cf df e ...
Cy d, e, /...

0.

[This is a generalisation of the catalecticant of the quartic ; those of
quantics of higher order admit of similar extension .] 113

8954. (W. J. C. Sharp, M.A.) — If seven tangents to a cuspidal cubic
(or tricuspidal quartic) be given, and a conic be described to touch any
four of those, the conic which touches the other three given tangents and
the two remaining common tangents of the first conic and the curve,
will always touch a fixed tangent to the curve '. 29

8968. (W. J. C. Sharp, M.A.)— If (a:,, y„ r„ m?,), (x,,, y^, z^, «?j),
(^3, ^3, «s, 1^3) be any three points, and A, /x, v the artial coordinates of any
point in their plane referrea to the triangle of which they are vertices ;
show that the equation to the section of any surface U = by the plane
will be obtained by substituting for Xy y, «, tv from the equations

(A. + /i + y)ar = \Xi + ftX2 + vX3, (A + /i + K) y = Ayj + /xyj + J^a,
{\ + fx + p) z = A2i + /iz, + v^s, (\ + fi + y)w = MVi + fiW2 + ytP3,

86

8969. (W. J. 0. Sharp, M.A.)— If the temiary «-ic be written
aign + ± (3^y + f,^) ^*-i + ^^^^-^) (c^y- + 2c^^ + e^^ x^-^ + &c.,

and ax + b^y + b^ be written for a,

b\X + c^y + e^ be written for ij,
b^ + c?2y + <'zz be written for ig? J^nd so on,

in any invariant or covariant ; the result will be a covariant of the

(w + l)-ic

aa:»*>+ 'i±i(% + *23)a^'*+&c. .

135

8970. (W. J. C. Sharp M.A.)— If X, Y...U denote the deter-
minants :r|, y„ ^j, «?i, 1

^2» .V2> ^2» ^2> «2
^8» y3» Hi <*'3. <
^4» Piy -Ay ^4* <

and Vi, V2, V3, V4 be the valuer of the quinary quadratic V when

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CONTENTS. XV

(^1* yi» «i, w'o «i)» (^2. y2» «2, «^2» «*2)> &c. are put for {r, y, 2, tr, w), and

81,2, &c. stand for if a?, -^ +yi-r +...)V2, &c.,
V dx2 dt/i I

Vu S,.2, S,.3, S,.4 =AX2 + BY2 + &c.,

3i.2» Vj, S2.31 S2.4

Si. 3, S2.3, Vj, S3. 4

Si. 4, Si .4, S3. 4, V4

where A, B, &c. are the first minors of the discriminant of V 136

8989. (Professor Wolstenhohne, M.A., Sc.D.) — In a tetrahedron
OABC, OA = fl, OB = A, 00 « tf ; BO = x, CA « y, AB = «, and the
dihedral angles opposite to these edges are respectively A, B, C ; X, Y, Z.
Having given the equations b == y = ^{a + x)t <?— « = a—x, B = Y,
C + Z = 180°, prove that B = Y = 60°, C-A = Z-X - 30°; and find
the relations between a, d, e 67

9006. (H. L. Orchard, M.A., B.Sc.) — Inside a hemisphere (of radius
p) a luminous point is placed, in the radius which is perpendicular to the
base, at a distance from the base = ip ^^3 ; show that the illumination of
the surface (excluding the base) is = 3irC 89

9018. (W. J. Greenstreet, B.A.) — If the Earth and Jupiter are
in heliocentiic conjunction at the same time as Jupiter and one of his
satellites, show that the times when the satellite will appear to an
observer to be stationary are the roots of the equation

^+£.+ fi + >(* + ,)cos2,r(i-i-V--(« + ^)c08lw(l-i)<
a b e be \ b c I ac \ a c I

- ^^ia^b) cos2,r f J- - 1^ < - 0;
ab \ a b I

where <?, /, s are radii of the orbits of the Earth, Jupiter, and the satt^llite,

«, by c their periodic times, the orbits circular and in one plane 97

9042. (H. L. Orchard. M.A., B.Sc.)— Prove that 13 + 23+33 + ... + a:3
is a factor of the expression Zjfi-^Vlx'^ -^l^jfi^la^-k-^s^ 178

9044. (S. Tebay, B.A.)— If A bo the area of one of the faces of a
tetrahedron ; X, Y, Z the dihedral angles over A ; and

M = (1 - cos^ X - cos= Y- cos'^ Z - 2 cos X cos Y cos Z)* ;
show that A/M has the same value for all the solid angles 99

9087. (H. Fortey, M.A.)— Show that, when the cards are dealt out

at whist, the probability that each player holds two or more carls of each

• suit is 2062806, «&c. ; or the odds are about 4 to 1 against the event. K3

9089. (Emile Vigarie.) — Par Ips sommets A, H, C d*un trianale on
mene des paralleles aux cdtes opposes qui rencontrent le cercle circonacrit
en A', B', C. Les droites A'B', A'C, C'B' roncontrent respectivement
AB. AC, BC eit a, /8, 7. Demontrer que Torthrceiitre du triangle a6y
est le centre du cercle ABC 65

9092. (A. E. JollifPe, MA.)— Prove that
{1n)\ (2m-1)! (2w-2W * / ,% ^

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XVI CONTESTS.

9102. (H. L. Orchard, BJ8c., M.A.)— Show that the weneB

1^ + 27+37 + 47+. ..+97iadiviableby27 178

9122. (Professor HadBon, ILA.)— Prore that the locus of the feet of
perpendiculars from the vertex of y* = 4nx on chords that subtend an
angle of 45" at the vertex ia r*— 24 ar cos 6+ ISo^ cos 2^ = 99

9128. (M. F. J. Mann, M.A.) — Find the sum of all numbers less tiian
n and prime to it is divisible by n 29

9140. (Emile Vigarie.) — Si R, Rj, R, designent respectivement lea
rayons du cercle circonscrit du premier cerde de Lemoine {triplieaU ratio
circle) et le deuxi^me cercle de Lemoine {eohne)^ demontrer la relation

B?«4Ri»-B,« 34

9142. (B. W. D. Christie. See Quest. 8700.)— H

2r = l'' + 2^ + 3*' «^

provethat (92,i + 3029 + 9X7)/ 2, « (112m, + 3028+7^6) /2i ^78

9146. (B. Lachlan, M.A.) — If two circles (radii p, p') intersect in
A and B, and any straight line cut them in the points (P, Q), (R, S) re-
spectively, show that

(AP . BP . AQ . BQ^/p2 = (AB.BB. AS . BS) /p'',
(AP . BP . AS . BS) /SP« = (AQ . BQ . AR . BR) / QR^ 36

9149. (Charlotte A. Scott. B.Sc.)— If ABCD be a quadrilateral, in
which the sidtts BA, CD meet towards A and D in H, and the sides
BC, AD meet towards C and D in K ; and if from a point L in HK,
LAG, LFC be drawn meeting BC in G and AD in F, respectively ; show
that BF and GD meet in HK 75

9164. (Professor Nilkantha Sarkar, M.A.)— Prove that

— 1 c^^^' Bm(eQiiix)Buxnzdx = — 75

IT Jo nl

9183. (A. R. Johnson, M.A.) — Investigate the induced magnetisation
of an ellipsoidal shell composed of any number of strata bounded by con-
focal surfaces 117

9195. (Sir James Cockle, F.R.S.) — Integrate

— = ^ , when m = 1 or when w = 2 53

9200, (Professor Neuberg.) — On casse, au hasard, une barre, de
longueur 3«, en trois morceaux. Demontrer que la probabilite que le"
procliiit des longueurs de doux quelconques des morceaux soit moindre
quo «' est :

f loff„ [i(3+ y5)] + 2->v/5 :;= 0123 (atres-peuprfes) 80

9215. (8. Tobay, B. A.) — The growth at any point of a blade of grass
v»ri«'ri directly as its distance from the root. The respective heights of
grass in thrcf) meadows, of 2, 3, and 6 acres, are 3, 3^, and 4 inches. The
gruHS in 'the lirHt and second meadows is cut in 32 and 30 days, respectively.
If 12 oxon consume the produce of the first meadow in 56 days, and 16
oxen consume the prpduco of the second meadow iu 63 days, find when

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CONTENTS. XVU

the grass in the third meadow must he cut so that 18 oxen may consume
the produce in 80 days 32

9217. (Major-General P. O'Connell.) — In using either the French
or English Arithmometer, any two numhers each containing less that
nine figures can he multiplied together, and the sum of a series each term
of which is the product of two such numhers, whether positive or
negative, can he obtained without writing down any figures. It is
required to find a formula for the product true to, say, thirteen figures on
two numbers each of sixteen figures, so that the result may be obtained
by the use of the Arithmometer alone, i.e., without intermediate record.

96

9226. (J. White.)— Prove that

l3+2H33...M3isafactorof (l»+2*+3»...M«)x3 178

9227. (W. J. C. Sharp, M.A.)— Show that (1) 1.2.3...n'' is divi-
sible by (w) to the power of (/*'*- 1) / (»— 1) ; and (2) when (n) is a prime
this is the highest power of {n) which will measure it 29

9229. (Professor Sylvester, F.R.S.)— Prove that the points of inter-
section of any given bicircular quartic by a transversal, will be foci of a
hyper-cartesian capable of being drawn through four concyclic foci of the
given quartic 37

9250. (Major-General P. 0*Connell.) — If s = the length of an arc
of a circle, v = the versed sine of half the angle subtended by the arc,
c = the chord of the arc ; required a series for the value of s in terms
of V and c 60

9256. (E. Vigari^.) — Dans im triangle ABC si (a) est le pied siu: BO
de la symediane issue du sommet A, et si (a') est le point conjugue
harmonique de (a) ; demontrer que Ao' est egale au rayon du cercle
d'Apollonius correspondant k BC 122

9259. (Professor Sylvester, F.R.S.) — Prove that, if one set of four
colli near points are the foci of a hyper- cartesian drawn through a second
set of the same, the second set will be the collinear foci of a hyper-
cartesian that can be drawn through the first set 37

9264. (Professor Hudson, M.A.)— Prove that y = >/2 {x—ia) is both
a tangent and a normal to 27ai/^ = 4 (x — 2a)^ 34

9267. (Professor Hanumanta Rau, M.A.) — Given the base and tho
vertical angle of a triangle, prove that the envelope of the nine-points
circle is itself a circle .' 120

P\?7I. (Irofessor De Wachter.) — A straight rod is divided at ran-
dom into four parts ; prove that it is an even chance that these parts may
be the sides of any quadrilateral 24

9272. (Professor Ignacio Beyens.) — R^soudre en nombres entiers et
positif 8 1* equation x^—yz ± a' = 22

9277. (Rev. T. 0. Simmons, M.A.)— Prove that the Taylor-circle of
a triangle is always greater than its cosine circle, and that in an equi-
lateral triangle the respective areas are in the ratio of 21 to 16 98

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V

XVIU CONTENTS.

9293. (Elizaheth Blackwood.)— Find the nuraher of permutations of
n letters, taken k together, repetition being allowed, but no three con-
secutive letters being the same ; and prove that, if this number be denoted

by P*, P*.i- Pt = (n2-«) ^L^,

a — p

where a, $ are the roots of the equation x*— (n — 1) a?— (n— 1) * 0. ... 46

9301. (Professor Sylvester, F.R.S.) — Prove that the points in which a
pair of circles are cut by any trausversal will be the collinear foci of a
system of hyper-cartesians having double contact with one another at
two points 37

9303. (Professor Neuberg.) — Sur les c6te8 du triangle ABC, on con-
struit trois trianerles semblables BCD, CAE, ABF; demontrer que la
Bomme (DE)2+(EF)2 + (FD;^ est minimum, lorsque les points D, E, F
sont les sommets du premier triangle de Brocard 38

9304. (Professor Schoute.) — Of a triangle ABC there is given the
vertex A, the angle A, and the line of which BC is a part ; find the loci
of the remarkable points of the triangle ABC 49

9307. (Professor Genese, M.A.) — In the ordinary conical projection
of one given plane on another from a given vertex, prove that there is a
point in space, other than the vertex, at which every line and its projection
subtend equal angles 21

9314. (Professor Beni Madhav Sarkar, B. A.) — Solve the equations

ar + yz = a=384, y + «jr =- i = 237, « + a:y « c = 192. ... 120

9315. (Professor Mukhopadhyay, M.A., F.R.S.E.)— Prove that (1)
the locus of the mid-points of the chords of curvature of the conic
b^x^+a^l/^ = aH^ is the sextic 2, = ^-2x2+^ 2^2 = {a'^-x^-b-^i/^)i pass-
inc^ through the origin ; (2) the area of 2i is half the area (A) of the
ellipse ; (3) the envelope of the chords of curvature of the same conic is
the sextic ^^ = {a-^x^ + b-^t/^-4)^+27 (a- x^-b-'^y^Y^ -= ; (4) the area
of 22 = ^A ; (5) trace the locus 2i and the envelope 22, and show that
they touch each other and the conic at the ends of the major and the
minor axes. 56

9316. (Professor Wolstenholme, M.A., Sc.D.)— In any curve OM = jr,
MP = 1/ are coordinates of a point P, MQ is drawn perpendicular to the
tangent at P and bisected by it ; prove that the arc <r of the locus of Q is
given by the equation

- = ± ( 2y- ^- V where '^^ = tan 6 ; and that
do \ "^ del dx

(1) when a;2 + y2 _. ^s^ the whole arc of the locus of Q = 12a ;

(2) when y^ = 4«a:, the arc from the vertex = a;+ 2fllog (I -^xja) ;

(3) when -^ + ^ = 1 (a >*), the whole arc = 4a f 1 + ?— ^^ log ^~\ ;

a^ Ir \ e l — ej

(4) =(«<*), =4*{(l-^)* + 2/esin-»^};

(5) when x = a {2<p + sin 2<^), y = a (1 + cos 2<p)f rr = 2.r ;

{6) whon X — a (2<^ + sin 2</)), y = a (I - cos 2<^\ the locus of Q is a cycloid
of half the linear dimensions and having the same tangent at the vertices;

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CONTENTS. XIX

(7) when the curve is such that the radius of curvature is n times the
normal at P terminated by the axis of a;, the arc = ± («— 2)/w .a:, n
being any constant number 28

931 9. (Professor Bhattacharyya.)— (9319.) Show that
(2m+l)(2ffl-«-3) ...(2w + 2r-l) (2w-i- l)(2w-t- 3) ... (2 m-t-2r~ 3) 2rt-l

r! ir-l)l * 1
(2m+l)(2>«-f3)... (2m + 2r~6) (2»i-l)(2n-t- 1)
(r-2)! • 2!
^ (m + n + r-l)l ^^ ^g

(m + »-l)!r!

9320. (Isabel Maddison.) — Four lines, p, g^ r, «, in a plane are cut
by a line a. Prove that the point a [(pg) {{^8 .rg)(ar ,8p)}'\ is un-
changed when any of the letters p, g, r, s are interchanged. [In the
above complex symbol the combination bf two line symbols represents a
point, and the combination of two point symbols represents a line.]... 23

9324. (Rev. T. C. Simmons, M.A.)— Prove that

Jo {a^ + b^ tan- x)» ^ U^ * (a^ - b^)^~' 16a« * {a^ - 6^)» '

when n=2f 3 ; and deduce, if possible, a general formula for this type of
definite integitd 26

9325. (S. Tebay, B.A.)— A, B, C are the dihedral angles at the base
of a tetrahedron ; X, Y, Z the respective opposites ; show that, if

Ti = (1 -cos2 B -cos2 C-cos^X- 2 cos B cos C cos X)*,
■with similar expressions (denoted by Tg, T3, T4) for the other solid angles,

T2T3 cos X + T3T1 cos Y + TjTj cos Z =. 1 - cos^ A- cos^ B - cos* C
— cos B cos cos X —cos G cos A cos Y— cos AcosB cosZ + cos Xcos YcosZ.

79

9327. (F. R. J. Hervey.)— The point; O is fixed, P describes a
straight line A ; OP and a line T passing through P rotate UDiformly (in
the same or in contrary senses) with angular velocities as 1 : 3, and be-
come simultaneously perpendicular (or, in the limiting position, parallel)
to A. Show that the envelope of T is a cardioid 64

9337. (W. J. C. Sharp, M.A.) — If S,. denote 1^ + 2'. ..+n% prove
that(l)rS.-i + ^(^)s,-2-H ^^"~^^^^^-^) s.-3-H...+So^(n.Hl)«'--l;
(2) deduce therefrom Febmat^s Theorem ; also (3) show that

^ 'lr + 1 (r-1;! r (r-2)! r-1 j'
where («)('') stands for n (w— 1) ... (w— r + 1) 48

9338. (A. Russell, B.A.) — Show that the solution of the partial
differential equation

dx* dJt.^ dx^ dx dy^ dy

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XX CONTENTS*

9340. (R. Knowles, B.A.)— In Question 9149, if BD and AC intersect
in O, and CA meet KH in M ; prove that the lines GM, GA, GO, GB and
LC, LO, LA, LH form harmonic pencils 60

9350. (Professor De Wachter.)— A point being taken within a tri-
angle, prove that the chance that its distances from the sides (a), {b), (r),
may form any possible triangle will be 2abc/{ (b + e) (c + a) (a + b)\ . 87

9352. (Professor Hudson, M.A.)— -Prove that

(tan7r + tan37i° + tan67i°) (tan 22i° + tan 52i° + tan82n = 17 + 8^/3.

52

9353. (Professor Asutosh MukhopadhySy, M.A., F.R.S.E.)— Points
D, E are taken in the sides AB, BC of any triangle ABC, such that
BD = w . DA, BE = n . EC. If O be the intersection of AE, DC, prove

that C0^,»+2 ^d ^-?i±i 65

9354. (Professor Mahendra Nath Ray, M.A., LL.B.)— A pencil of
four rays radiates from the middle point of the base of a triangle, and is
terminated by the sides. If the segments of the rays measured from the
origin be a;,, i/u x.^^ y^y a-j, ^3, and 2:4, ^4, show that the identical relation
connecting these lengths is

124

^:\

'i\ ^3-'.

'I'

JT^

yi\ y^S

yi'

(*I.Vi)-

'. (a-ivj)-', (^syj)-'

, (*4y4)-'

1,

1. 1,

1

9369. (J. O' Byrne Croke, M.A.)— Prove that the area of the simple
Cartesian oval formed by guiding a pencil by a thread having one end
attached to the tracing point and brought once tensely round a fixed pin
of negligible section, the other being fastened to a second pin at a dis-
tance a from the former, and the whole length of the thread being 2ay is
5a-(2ir-3'/3) 60

9360. (R. Curtis, M.A.) — A tetrahedron ABCD is circumscribed to
an ellipsoid, and straight lines are drawn through the centre from the
comers to the opposite sides meeting them in X, Y, Z, W ; nhow that

OX^OY^OZ 0W_
XA^YB^ZC'^WD""^ ^^

9361. (F. R. J. Hervey.) — A line A bisects at right angles the radius
OM of a circle (centre 0) ; three lines U, V, W, passing through M,
rotate uniformly with angular velocities as 1 : — 1 : — 2, and cut re-
spectively A in P, and the circle in Q, R ; V and W passing through O
at the instant that U becomes a tangent. Prove that P, Q, R are always
collinear, and PQ.QR constant 23

9364. (W. J. Greenstreet, M.A.) — If q is any positive integer,

prove that JL = l+^^M ^. .? (f -!)(?- 2)te- 3)^ 78

5'+ 1 2 ! 4 !

9365. (W. J. Barton, M.A.) — In the expansion of (l-3ir + 3a;2)-i
show that the coefficient of a:^'»-i is zero 110

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CONTENTS. XXI

9367. (F. Morley, B.A.)— In the sides AB, AC of a triangle ABC,
find points D, E, such that BD - DE « EC 61

9369. (W. J. C. Sharp, M. A.)— Prove, from the theory of com-
binations, (1) that 1 + y . ^^ + i . 2 nfTT mini

must he true ; and (2) deduce that, if (m) be a prime greater than (n),

Twi-i-wV-m ! « ! and ^^'*"^^' are respective multiples of (m^), (w). ... 74
^ ' m!

9371. (J. Brill, M. A.)— Prove that in any triangle, n being a positive
integer, «* cos wB + b*^ cos «A

- c* - nabc^''^ cos (A- B) + ^-^|^ «2*«^-< cos 2 (A-B)

__ n(/i-4)(n-5) «3i3^n.6cos3 (A-B)
3!

^n(n-5)(n.-6Hn-7)^^4^„_8^a4(^,B)-&c 78

9376. (A. E. Thomas.) — Solve the equations

a;4 + 3y2z2= a^+2^(y3 + a3) (l),

i^ + dzh:^ ^ b* + 2t/ (^ + x^)j «* + 3icV = <^ + 2«(*' + S^) (2.3).

88

9378. (Rev. J. J. Milne, M.A.) — PSQ is a focal chord of a conic.
The normal at P (aJi, y,) and the tangent at Q intersect in R. Show that
the coordinates of R and the locus of R are respectively

/ 2^2- *2 \ «2 ^ b^t/^ , .-

(,"''>' — ^^^V' 'i^^W^^'^ ^

9380. (Sarah Marks, B.Sc.) — Tangents are drawn to a parabola
from a point T ; a third tangent meets these in MN ; prove that the polar
of the mid-point of MN and the diameter through T meet on the
parabola 77

9381. (Professor Sylvester, F.R.S.) — If (^ and r being prime num-
bers) 1 +p-j-p^ + ... J?*""' is divisible by q, show that, unless r divides j' — 1,
it must be equal to q and divide i?— 1 54

9384. (Professor Bordage.)--Show that the roots of the equation
(x + 2y + 2{x + 2)^x-2x^SVx-ie = are 9, 4, ^ {-l3db3(-3*)}.

77

9386. (Professor Neuberg.) — Si suivant les perpendiculaires abaiss^es
du centre O du cercle circonscrit k un triangle ABC, sur les c6tes de ce
triangle, on applique, dans un sens ou dans P autre, trois forces 6gales,
la r^ultante passera par le centre de Pun des cercles tangents aux
trois cotes 65

9389. (Professor Hanumanta Rau, M.A.)— Prove (1) that sin 6° is a
root of the equation lea^-i-Sx^— IGor^— 8a: + 1 =» ;
and (2) express the remaining roots in terms of trigonometrical functions.

IK

C

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XXU CONTENTS.

9390. {N'Importe.)--In any triangle ABC, prove that

aco8 2Aco8(B-C) + &c. =-?^ = -?^ 110

9391. (Professor Satis Chandra Ray, M.A.) — If the diagonals of a
cyclic quadrilateral A BCD intersect in O ; and if AB « a, BC = b,
CD = <?, DA « rf, Z ADD = ADB ; prove that

{be + ad){edi-ab)/(ae + bd) ^ a* 76

9392. (Professor Genese, M.A.) — If the tangent at any point P of a

folium of Descartes meet the tangents at the node in X, Y, and the curve

11 3
again at Q, then prove that r— +^^- -— 72

9401. (J. Brill, M.A.) — Prove that, if n and r he positive integers,

{a-\-l){a^2) ...{a + n) __ [b+l) {b + 2) ... (b + n) {c+\) {c+2) ... jc + n)
nl («-l}I (11-2; I 2!

(«-3)!3! V / »

where fl = fir, i«(it-l)r-l, <? = (n-2)r-2, rf=(n— 3)r-3, &c.

100

9403. (Rusticus.)— Bahy Tom of hahy Hugh

The nephew is and uncle too.

In how many ways can this he true P 114

9406. (W. J. Barton, M.A.) — Show that, if R = 2r, the triangle is
equilateral, without employing the expression for the distance between
the centres 70

9407. (W.J. Greenstreet, M.A.) — From a point outside a circle
centre C, APQ is drawn cutting it in P and Q ; AT is a tangent at T :
show that it is always possible to draw such a line that AP shall equal
PQ, as long as AC < 3CT ; and that then 3 cos TAC = 2 a/2 cos PAC.

109

9410. (A. E. Thomas) — If n and rare positive integers, and n>r,
then {e being the Naperian base)

l^!L±i^l.(>',tJ)(^2)l(n^lJ(n^2)(n^3) ^^^^^
r-ri 2! (r+l)(r+2) 3 ! (r+ l)(r+ 2)(r + 3) "^

^4l^>Ll.%l(!L-^)(^-^^+i(5rZ)(>L-J' -l)(>»->-2) 4. etc.)
I r+l 2! (ri-lj(r + 2) 3! (r + l)(r + 2Xr + 3) "• j

112

9412. (A. R. Johnson, M.A.)— Show that, if 1, 2, 3, 4, 6, 6 be six
points on a conic, then = 2 (023) (031) (012) (466),

2 denoting summation with respect to all terms obtained from the one
presented by cyclic interchanges ; O denoting any point in the plane of
the conic, and (456), etc. the areas of the triangles 466, etc., described in
the ordernamed 123

9413. (J. O'Byme Croke, M.A.) — If D be the distance between the
centre of the ciicumcircle and the point of intersection of the perpendicu-
lars of a triangle, prove that 2D/(1 — 8 cos A cos B cos C)' = «/ sin A.

93

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CONTENTS. XXUl

9414. (R. W. D. Christie.) — If 2p— 1 is a prime, show that^ is also
prime. [Better thus : — What prime p will make 2^— 1 a prime ?] ... 75

9416. (J. O'Byrne Croke, M.A. Suggested by Question 9360.)—
The sides of a polyhedron are of areas inversely as the perpendiculars on
them from a point O, and 00' meets them in Pj, Pg, P3 ... Pn, respec-
tively; prove that J'|i + ^^.OT,^ .^aT„_ ,,,

9418. (Professor Sylvester, F.R.S.)— If p, iyj are each prime num-
bers, and \-\-p-\-p^-\- ...+p^-'^ = (f ^ prove that j is a divisor of q — i.
Example: 1 + 3 + 32+33 + 3* = 11'^, and 2 is a divisor of 11-6 69

9423. (Professor Neuberg.) — On casse, au hasard, deux barres de
longueurs a et ^, chacune en deux morceaux. Quelle est la probabilite
qu' un morceau de la premiere barre et un morceau de la seconde, 6tant
juxtaposes, donnent une longueur moindre que c ? 69

9425. (Prof( ssor Hanumanta Rau, B.A.) — Prove that the sum of the
products of the first n natural numbers taken three at a time is

,V«2 («+ 1)2 (»- l)(w-2) 109

9427. (Professor Genese, M.A.) — If A, B, C, D be points in a plane,

,, . HC .AD CA . BD AB . CD

prove that = = ,

^ sin (BAC - BDC) sin (CBA - CD A) sin (ACB - ADB)'

where any angle BAC means the angle through which AC must bo
turned in the positive seme to coincide with AB 76

9430. (Professor "Wolstenholrae, M.A., Sc.D.) — In a tetrahedron
OABC, the plane angles of the triangular faces are denoted by a, /S, or 7 ;
all angles opposite to OA or BC bt ing a, those opposite OB or CA are
i3, and those opposite OC or AB are 7 ; the angles at O have the
suffix I, those at B, C, D the suffixes 2, 3, 4 respectively ; prove that, if
«i + iSi + 7i = «2 + i82 + 7j = '» then

7i + ai-^i = 74"*-a4-^4; «i + ^i-7i= «3 + ^3-78

72 + 02-^2 = 74 + 03-^; 02 + ^2-72 = 044^4—74 88

9433. (G. Heppel, MA.)— If, within a triangle ABC, O be a point
where the sides subtend equal angles ; then, putting OA = jo, OB = ^,
OC = r, show that the equation to the ellipse with locus O, touching the
sides in D, E, F, is in (1) rectangular coordinates, with O as origin and
OA as axis of y, and (2) triliuear coordinates, ABC triangle of reference,

{x^-^y^)^^\{pq + qr-^rp)-^\{pr-^pq-%qr)y-p{q-r)x^'6-^^pqr']...(\),

a^p-fi:^^-b'^q^0^-k-<?r^-2bcqrfiy''2carpya-2.ibpqa^ « 2).

90

9436. (W. Gallatly, M.A.) — AB is a miiTor swinging on a hinge
at A. At C is a candle flame, and at D an observer ; the line ACD being
perpendicular to the axis of the mirror. Find geometrically the position
of the mirror, when the observer at D sees the image of the flamu on the
point of disappearing 73

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XXIV CONTENTS.

9437. (H. Fortey, M. A.)— Show that, if a, jS, &c. are the p roots
(excluding unity ) oi xP*^~-mx^ + m—\ = 0, the number of ways in which
m letters can be arranged « in a row, repetitions being allowed but not
more than p consecutive letters being the same, is

_!5_S— (i^J-'""^!- 94

(W— 1)'^ OP*^ — (j9+l)o+i?

9439. (A. Kahn, M. A.)— Show, by a general solution, that the roots

of 4a^ + 4a:3+i3a;S + 6a: + 8 = are i{-l±(-7)H, i|-l±(-3)*}.

94

9440. (Rev. T. C. Simmons, M.A.) — Vtove geometrically that the per-
pendicular from the Lomoine -point of an harmonic polygon on the Lemoine-
line is the harmonic mean ot the perpendiculars drawn on the same line
from the vertices of the polygon, [A proof by trigonometrical series is
given in Lotid, Math. Soc. Froceedings^ Vol. xviii., p. 293.] 16

9444. (R. W. D. Christie.)— Solve (1) in integers x^-¥X'y^-^y^=ah;
and (2) note the result when a — b 175

9449. (Professor Sylvester, F.R.S.) — If there exist any perfect
number divisible by a prime number p of the form 2»+ 1, show that it
must be divisible by another prime number of the form j^a: ± 1 85

9459. (Professor Genese, M.A.) — If p, B be the polar coordinates of
a point whose coordinates referred to axes inclined at any angle w are
Xy y, then xjp, yjp may be denoted by C (6), S (6). Prove that
S(6-(/,) = S(0).C(4>)-C(O).S(4,),

C(a + 4>) = C(e).C(</))-S(0).S(<^) 107

9462. (The Editor.) — If the radius of the in-circle of an isosceles
triangle is one-w*** of the radius of the ex-circle to the base ; prove that
the ratio of the base to each of the equal sides is 2 (/*— I) : w + 1. ... 86

9468. (R. \V. D . Christie, M.A.) —Show that the tenth perfect number
is Pio = 2'«>(2«-l) = 2,417,851,639,228,158,837,784,676.

9469. (W. J. C. Sharp, M.A.) — lip be a prime number and r<p - 1,
prove that (1) r! (jt? — r — 1)! + ( — 1)*" is a multiple of jp ; and hence (2), if
p=2g-l, {((?-l)!}2+ (-!,«-! is a multiple of 2^-1 110

9477. (Swift P. Johnson, M.A.) — A, B, C and a, h, c are two triads
of points on a sphere ; show that, if the circumcircles of the triangles
Abcy B(?fl, Onb meet in a point, then the circumcircles of the triangles
aBC, iCA, fAB will also meet in a point 107

9478. (Rev. J. J. Milne, M.A.) — If p be the sum of the abscissae, q
the sum of the ordinates of two points P, Q of an ellipse ; prove that (1)
the equation of PQ is ll^px + la'^qy = b^j>^ + cfiq^ ; and hence (2) if either
(a) /> or 5- be constant, or (&) if p and q be connected by the relation
(p + w^' = 1, the envelope of the line is a parabola 94

9479. (A. Kahn, M.A.)— Solve the equations xyz = 24,

ar(y-z)2 + y(s-a-)2 + z(a;-y)3 - 18, x^ (y-z)-\.y^(z-x)-^z^x-y) ^ -2,

117

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CONTENTS. XXV

9481. (W. S. McCay, M.A.) — AB is the diameter of a semicircle;
show how to draw a chord XY in a given direction, so that the area of
the quadrilateral AXYB may he a maximum 106

9482. (S. Tehay, B.A.)— AB, AC, AD are edges of a tetrahedron ;
BE, CF, DG perpendiculars on the opposite faces ; P, Q, R their areas ;
j», q, r the areas CED, DFB, BGC; and S the area of the hase BCD ;
prove that Pj9 + Q j + Rr - S^ 112

9499. (Professor Ath Bijah Bhut.) — Prove that the orthocentre of a
triangle is the centroid of three weights, proportional to tan A, tan B,
tan C, placed at the corners A, B, C 112

9503. (Professor Bordage.) — Show that the roots of the equation

22«+2 + 4i-«==, 17 are a:= ±1 Ill

9505. (Professor Wolstenholme, M.A., Sc.D.) — Prove, witliout
evolution, or the use of tahles, that 3x2* — 2* Hes hetween 3*6022831...
and 3*602282... ; the latter heing nearer to the exact value 101

9506. (Professor Hudson, M.A.) — Prove that (1) the parahola
y^ = 2l{x + 1) can be described by a force to the origin which varies as
r/{x + 2lj^'f and find (2) what ambiguity there is in the case of this law
of force 102

9511. (E. B. EUiott, M.A.)— Of inhabitants of towns p per cent,
have votes, and of country people q per cent. Also of voters r per cent,
live in towns, and of non-voters s per cent. Find the proportion of the
whole population who have votes ; and show that^, g, r, s are connected
by the one relation 100 {qr^ps) t= (p + 8) q7'—{q + r)p8 113

9516. (D. Biddle.) — Prove or disprove that (1) a circle B is not
properly drawn at random within a given circle A, unless its centre be
first taken at random on the surface of A, and its radius be subsequently
taken at random within the limits allowed by the position of its centre ;
(2) putting unity for the radius of A, r for the radius of B, and x for the
distance between the two centres, there are two things requisite in order
that B may include the centre of A, namely, that x be less than i, and
that r be between X and l—x; (3) from a favourably placed centre, the
chance of the radius of B being such as to make it include the centre of
A is (1 — 2jr)/(l — :») ; (4) the chance is identical for 2irx.dx positions,,
which form the circumference of a circle of radius x, around the centre of
A ; (5) the probability that a circle B, drawn at random in a given circlet
A, shall include the centre of A, is not correctly found by the formula

.i A-x A A-x

P = 2ir xdx^di/-i-2ir\ xdxdy = \,

since this assumes that the number of circles capable of being drawn
from any centre is proportioned to the upper limit of the radius ; leaves,
out of account that one centre, one radius, one circle B, are taken each
time ; and gives a result which actually does not fall short of the chance-
that the centre alone shall be favourably placed ; (6) the probability in
the case referred to is correctly found as follows : —

P = *2,rj x[^-^\dx-i-2ir[x,dx^\\'¥2\o^,\

+ 2-61370564 =- 011370564, or less than ^- 10^

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XXVI CONTENTS.

9621. (R. W. D. Christie.) — Prove that (p*m,ir*)l5 is an integer
where JO is a/iy perfect number and ir any prime number except 5. ... 176

9624. (Rev. J. J. Milne, M.A.) — If yi, p^y y^ are the ordinates of
three points P, Q, R on the parabola y^ = Aax^ such that the circle on
PQ as diameter touches the parabola at R, prove that

yi + ys- 2^3, yi^yj- 8a 119

9661. (W. J. C. Sharp, M.A.)— If (1.2), (2.3), &c. denote the
edges of a tetrahedron, and Dj, Dg* I^a *^6 shortest distances, and ^„ 9^, 0^
the angles between (2 . 3) and (1.4), (3.1) and (2.4), and (1.2) and
(3.4), respectively ; prove that

^^^ "*" *' " 2 (2 .SKI. 4) ^^^ • ^^' ■" ^^ • *^'~ ^^ ■ *^~ ^^ • ^^'^' *•'■' *''■•
and (2) the square of the volume

« ?4 {* (2 • 3)' (1 . 4)2- [(1 . 2)2 + (3 . 4)2- (2 . 4)2- (1 . 3)2]2} - &c., &c.
^■** 138

9608. (Septimus Tebay, B.A.) — Find the least heptagonal number
which when increased by a given square shall be a square number... 176

9629. (Professor Gorondal.) — Partager 90° en deux parties ar, y
telles que la tangente de I'une soit le quadruple de la tangente de l*autre.
et prouver que tan^.i; = 2 sinlb® 176

9643. (R. W. D. Christie.) — If 2r4 = I'' +2'' + 3'-... n*", prove that
2w is exactly divisible by 2i when r is odd 177

9643. (R. W. D. Christie.)— If 2n - l'* + 2»- + 3'*... «»•; prove that
2j is divisible by 2l» 178

9668. (Professor Vuibert.) — Si Ton d^signe d'une mani^re generale

5ar S,„ la somme dos puissances de degre m des n premiers nombres entiers,
emontrer qu'on a (3S6 + 2ISi'*)/5S4 « 83/83 177

9683. (R. W. D. Christie.) — If 2r = l*' + 2'* ... +«»*, prove that
72(, + 624 = 125323 178

9767. (R. W. D. Christie.) — Prove that n^ is the sum of n con-
secutive odd numbers 178

9876. (R. W. D. Christie.)— Prove that

2 tan-14 ± tan-i ,, ^\ ;; - Jir,

b d^ + 2ab~a" * '

whore a is the coefficient of x*^ and J of a;"* ^ in the expansion of -.

l + 2j;-a;-

180

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CONTENTS. XXVU

APPENDIX I.
Solutions of some Unsolved Questions, by W. J. Curran Sharp, M.A. 125

APPENDIX II.
New Questions, by W. J. Curran Sharp, M.A 141

APPENDIX III.
Unsolved Questions 151

APPENDIX IV.
Notes, Solutions, and Questions, by R. W. D. Christie —

(A.) Diophantine Analysis 159

(B.) Besolution of Squares ; 162

(C.) Resolution of Cubes 170

(D.) Solutions of Old Questions 180

(E.) New Questions 186

CORRIGENDUM.
Page 64, line 9 from bottom, /or 8737 read 8337.

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MATHEMATICS

VBOM

THE EDUCATIONAL TIMES.
WITH ADDITIONAL PApSeS AND SOLUTIONS.

2352. (Prof. Sylvester, F.R.S.)— We may use P»Q to denote the
third point in which the right line PQ meets a given cubic ; P«Q«B to
denote the third point in which the line joining the one last named and
B meets the cubic, and so on. Thus P«P will denote the tangential or
point in which the tangent at P meets ^q given cubic, and[P#P]#rP#P]
will denote the second tangential, i.e., the tangential to the tangential at P.

1. Prove that [P<>P]^[P#PJ « I<|P<j[P<>P]<>P^I, where I is any point
of inflexion in the given curve.

2. Obtain a function of P, I which shall express the point in which the
curve is cut by a conic having five-point contact with it at P.

Solution by Professor Nash, M.A.

(1) This theorem may be stated as follows : — If T„ T, denote the first
and second taugentials of a point P on a cubic, I a point of infiexion, and
if IP meet the curve in Q, QT, meet the curve R, RP meet the curve in
8, then SI will pass through Tj.

IP and ETj are coresiduAl, U being residual to both pairs of points.
But the tangents at I and P may be considered as a conic through the
six points I, I, I, P, P, Tj ; therefore the four points I, I, P, Tj are residual
to IP, and therefore also to RTj ; therefore I, I, T,, Tj, P, R lie upon
a conic, and every conic through the four points I, I, Ti, Tj will meet
the cubic again in two points, the line joining which will pass through
the coresidual of the four points, t.tf., S. But the tangents at I and T|
form such a conic, and the two points are I, T3 ; therefore I, T,, S are
coUinear.

(2) The required point is the intersection of PTj with the cubic
(Salmon's Higher Plane Curves, Art. 155), and this may be expressed in
Prof. Sylvester's notation as P»{(P»P)»(P#P)} or (P»P)#(P#P)<>P, and
therefore by (1) the same point is represented by I«P«(P«P)«P«I«P.

9307. (Professor Gbnbsb, M.A.) — ^In the ordinary conical projection
of one given plane on another from a given vertex, prove that there is a

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22

point in space, other than the vertex, at which every line and its projection
subtend equal angles.

Solution by the Proposer.

Draw a plane j9 through the vertex V and the line of intersection / of
the given planes ; through I draw the plane k which is harmonically con-
jugate to p with respect to the other planes; from V draw VO perpen-
dicular to k ; then is the point in question. Let any straight line
through V meet the given planes in P, P' and A; in L, then (VPKLP') is
harmonic, and RVOL is a right angle, therefore OP, OP' are equally in-
clined to OL, and they are in a plane normal to k. Similarly for a second
line VQQ ; whence, by symmetry y |ngle POQ = angle FOQ'.

927^. (Professor Ionacio Beyens.) — Resoudre en nombres entiers
et positifs r^uation x^^yz db o* = 0.

Solution by R. W. D. Christie, M.A. ; E. Ruttbr ; and others.
"We have a?^— ya =» db a^ ■» (x^yz/n)^ say, whence we get

n^ — 2nx + ys = ; therefore x^—yz = ±a^ = ±{n—x)^;
therefore x ^ ±{n—a); and hence, easily, y — n and « = « — 2a, where n
and a may be any integers, regard being had to the signs.

^ (Isabel Maddison.) — Four lines, ;?, y, r, «, in a plane are cut

by a line a. Prove that the point a \_{pq) {(aw.r^) far . «jo)|] is un-
changed when any of the letters p, q^ r, « are interchanged. [In the
above complex symbol the combination of two line symbols represents a
point, and the combination of two point symbols represents a line.]

Solution by Prince de Polignac ; F. R. J. Hervby ; and others.

The equations of the lines being a = 0, &c., assume p = a + lr + ms,
q = a+l'r + m's. The equations to as .rq and ar . sp are respectively
a + m'« s and a + /r = 0. To find the line joining their intersection to
pq^ assume the forms a + /r — X (a + m's) = 0, j» — ju^ = 0, and equate ratios
of coefficients ; we find fi = Im/ (^m'), and the line is

{lW-lm)a + ll'{m'-m)r + mm'{l'-l}s = (1).

To find the result of interchanging r, s or ^, ^, we either interchange
/, m or displace the accents. These changes leave the line (1) unaltered ;
hence the points ^qr, {as . rq) (ar . sp), {ar . sq) {as . rp) are collinear. Thus
the permutations arrange themselves in six groups, to each of which cor-
responds a single line passing through one of the intersections of p, q, r, s.

Interchange «, q ; the equations to aq ,rs and ar . qp are evidently
/V + *n'« » and («'— m) a+(/m' — rm)r =* ; the corresponding line
through*/? is lf{m'^m)a + ll\m'-m)r + mm'{l'-l)s := (2).

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23

It follows that the lines through any two points, such a;Bpq tndpSf having
a common, line j», intersect on a ; which proves the theorem. The equa-
tions of the lines through pr^ qvy and qs are derived from (2) by inter-
changes. [The point on a is, by Brianchon's theorem, the point of
tangency with the conic that touches the five lines a, py ^, r, <, &c.]

9361. (^' R' J« Hervey.) — A line A bisects at right angles the radius
OM of a circle (centre 0) ; three lines IJ, V, W, passing through M,
rotate uniformly with angular velocities as 1 : -- 1 : — 2, and cut re-
spectively A in P, and the circle in Q, R ; V and W passing through O
at the instant that U becomes a tangent. Prove that P, Q, R are always
collinear, and PQ . QR constant.

Solution by B.. F. Davis, M.A. ;
D. BiDDLE ; and others.

Let TJ meet the circle in S and QR
in P. Since the angles QMN, QMR,
SMT are (by hvpothesis) equal, so also
are the arcs QN, QR, 8M. Hence OP
bisects at right angles the parallels
QM, RS; and therefore the angles
POM, PMO are equal, being the com-
plements of equal angles, and P lies
on A.

Since Z QOR = 2QMR

= OPM = OPR,

QO touches the circumcircle of OPR and

Q02 = PQ . QR.

7178. (W. J. C. Sharp, M.A.)— If three concyclic foci of a bicircular
quartic, or circular cubic, be given, and also a tangent and its point of
contact, determine the curve.

Solution by Professors Matz, M.A. ; Nash, M.A. ; and others.

Let A, B, C be the three given points, P the point of contact of the
given tangent. The quartic (or cubic) is the envelope of a circle whose
centre moves on a certain conic through A, B, 0, and which cuts ortho-

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24

gonally the circle ABC ; the curve will therefore be completely deter-
mined if the conic can be determined.

Take P' the inverse of P with respect to the circle ABC, then P' is also
on the curve, and the line which bisects PP' at right angles touchen the
conic at the centre of the variable circle whieh touches the curve at P and
P'. This tangent is therefore known, and its point of contact is its in-
tersection with a perpendicular to the given tangent at P. Hence four
points are given on the conic, and the tangent at one of them, so that the
conic is completely determined.

0271. (Professor Db Wachtbr.) — A straight rod is divided at ran-
dom into four parts ; prove that it is an even chance that these parts may
be the sides of any quadrilateral.

Solution hy Artemas Martin, LL.D.

Denote the parts by Xy y, «, and a— a:— y— z, a being the leng^ of the
rod. The following conditions must be satisfied, viz., ^<jt^, V<i<'»
«< Ja, a; + y + «>a— ar— y-«.

The required probability will be

p^ [[[dxdydz I uAdxdydz,

In N, X may have any value from to ^a ; y may have any value from
to ^a ; z may have any value from \a — x—y to ^a when y is less than
ia—x, and any value from to a-x-^y when y is greater than \a — x.
In D, ic may have any value from to a ; y may have any value from
to a— » ; z may have any value from to a^x-y. Hence

Jo LJo Jja-«-y Jia-«Jo J * JoJo Jo

-mr\:..-.'''-<..r '''■]'■

[If we take a regular tetrahedron, the altitude of which is the length (/)
of the rod, and from any interior point draw perpendiculars to the faces,
then the sum of these four perpendiculars will be = L Any interior point
represents (by those perpendiculars) a distinct chance of division of the
rod ; and the favourable points are situated so as to have each of their
distances < \U If, therefore, planes be drawn parallel to the faces, and
equi-distant from each vertex and the opposite face, it is easy to see
that the favourable points are included between those four planes and the
faces of the tetrahedron, and they form a regular octahedron. Hence
the probability will be the ratio of the Octahedron to the Tetrahedron,
that is to say, |.]

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25

8132. (W. J. Johnston, M. A.)— Prove that, if the section of a
quadiic by a plane is given, and also a straight line in that plane ; then,
if through this line a plane can be drawn to cut the quadric in a circular
section whose radius is also given, the locus of the centre of this circular
section is a circle in a plane perpendicular to the given plane.

Solution by G. G. Sto&r, M.A. ; A. Gordon ; and others.

Let the given line be chosen as axis of y, the given plane as plane of
xyy and a normal to it as axis of z. If

ax^ + V + <^«' + 2«'y« + ^Ih'zx + 2c'xy + 2a" x + 2b" y + 2o"z + rf -
a b t^ «" b"

is the quadric, then ---, --, --, ---•, — - are known, since the section

a a a a a
s = is known.

Let the plana 2 — revolve about y, through an angle till it forms

the section required, so that j? » | cos 0~ ^sin 0, i » { sin + ^cos 0.

Substituting after putting ^ » 0, we have

I* (acos'a + tf sin' $ + 2*' sin« cos e) +b^ + 2^ (a' bid •¥ c^ cob $) •¥ d

+ 2| (a" cos « + e" sin $) + 2y*" * 0.

In order that this may be a circle, it is necessary and sufficient that

a cos' + (; sin' + 2^ sin cos » b, a'ainB-^e' cobB » ;

the coordinates of the centre are given by

*" /o «^,.af««f N t ^' cos + </' sin
yi = - y (a constant), {i ^ ,

and (radius)' ^^djb-^ |i' + y^,

therefore || is a constant. But |i » rrj cos + 2i sin » x suppose, there-
fore a?i* + zf =a x', yj — constant, is the locus required.

[The conditions amongst the variable coefficients <?, a', b\ </', in order that
the circular section can be obtained of given radius, are that the equation
(a-*)cos«0+(tf-*) sin' + 2*' sin cos = 0,

must give real roots for tan 0, or *''>(a— *)(tf— *) (1),

and tan « — c^ja' must satisfy the above, or

(a^b) «''+ (e^b) a'^-2b'ae = (2),

also the condition that (radius)' + (f/6-(*'7i)'>0 (3).]

i

0324. (Rev. T. C. Simmons, M.A.)— Prove that

dx _^ 2a3~3a«3 + ^ ir 8a»--16a<^-f 10a'^-3y

(a' + ft'tan'a;)'»*"4«' • (a3-*2)' ' 16a» * {a^-ti'f *

when n»2, 3 ; and deduce, if possible, a general formula for this type of
definite integntl.

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26

Solution iy Prof. Wolstbnholmb, Sc.D. ; J. W. Sharps, M.A. ; and others.
Writing a^^p, b^^q, let U» - f*';^ ^ , . then

Un -T (XJ„-i) ' T-i(^«-2)i "^d 80 on.

JoJ» + ytan«« Jo(1+«')CP + ^2*) i'-^Jo \1+«' P + q^J

^^~g\'\p} ^^\P + Vpgl " V^\V>~v^MVi)'
HenceU, = iir^-^-^-^^^-^^ -J-^^irJ^^Hf ^ )

ir 1.3.5... 2 (»-3)
2a^^'^d 2.4.6...2(»-2)

(-1)" / 1 ±Y'' ( 1 \.

6...2(»-2) \a rfa/ \a + */ '

"*"2^ 2.4.6...2(»-2)
where a^ ^ are positive. Thus

* 4 o^A 4*(a + ft)«* 4 a5(a + A)2 4 a» (a2-*8)» '

IT =?5±-JL W 1 I ^ ^ = ir /g aM3g + 3) \

* 16a***16A a U2(a + A)« a (a + 3)3/ 16a*A \ (a + 3)» /

" 16 a* (a + *)3 16 a» (a»- ^)»

In the same way the value of the definite integral
f>' tan^^a; ^
Jo (j» + (7tan2a:)« '
where m, n, and n—m are positive integers (or zero), will be found to be

2 «! \dp) \dql \p+Vpq)'

(Obviously, iip^ qhe of opposite sign, the integral will be infinite.)
Another method of evaluating Un is : — Put Vq tan x a Vp tan z, then
p + qteai^x^ paet^z, VqBec^xdx ^ Vpsec'^zdz,
and the limits 0, ^w are unchanged, so that

'**I>«V? Jo (1+i?/? tan2«)(l + tan2«)"-i j^** Jq j^sin^z + jcos^* '

or XJ^^ V^^y !*'( P*" p*-(p-q)»cos^z \ ^^

P**(i'— ?)** Jo V^ sin"'' « + y cos' « p^{p^q) coai^z J

— i»" ''(/>—?)' cos* «—... to « terms j dz

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2 l>'»tp-?rlv/i^? VJ' y; 2 ^ ^^ ^^ 2.3

— . . . to » terms J ;

-i-i ra2'-2 + Ja2n.4 (a8_i2) + lii ^2„.8 (a2-J2)2+ ... to » terms'])

(a2-^2)»

The result in this form may always be reduced by the factors («—*)*»,
which must be a factor of the numerator, otherwise the integral would be
infinite when a ^ b. Putting n » 2, n = 3, we get the results.

Equating the two values for I r— — , we get an expression for

Jo (i^ + jtan^a;)"

fli f _L_ \ ^d since this = J- m (J -J— ] ^. also

fl(pn -1 \p + ^pq I ^q dp"*-^ \y/p ^p + ^qj

get a finite series for r ( | .

The integral 1 * , ^^ "'^ v ^ (n>m) may be evaluated in the
Jo (a^ + A^tan^a;)" ^ ' ^

Jo (a^ + A^tanSa:)'
same way, being equal to

s/pq f >* sin^"* z cos2*»-2w z <fg ,
^m^»-mj^ j?sin2a + jcos^s

^^22 T'sinZ^Z J" ^^^I^!! «"-m-l-«n-m-Jc082f

^n-m^m (^ — y)»»-»» J^ (^^ sin2« + q COS" «

— j3**-"*-"co8*«— ...to (»— m) terms | rf«,

in which the value of each term may be written down at once, with the
exception of the first ; and

(*" sin^"* z dz _ 1 r f4» ( jp — y) »» sin^*" z±.q^ a m 9"* JL\
J j^sin^aj + ycos'z {P — q)^Xjo {p—q)8m^z + q Vpq 2 )'

which gives the result as a series of m terms.

Thus p''_tan2£ ^ ^ jr^ 1

Jo (p + qt^ri'xy 4 Vpq(Vp+Vqf

fi' tan^a- ^^ ^ ir f 1 2 •>

Jo (/>rytan2a^)3 " 16 i -/p^(V>+ V^)' Vpg2(v>+ v'j)S)

but I think the former method preferable.

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9316. (Professor Wolstbnholme, Sc.D.) — In any cnrre OM « ar,
MP = y are coordinates of a point P, MQ is drawn perpendicular to the
tangent at P and bisected bj it ; prove that the arc a of the locos of Q is
given by the equation

~= ± f2y-^V where ^-tan«; and that
d9 \ ' de I dx

(1) when a;2 + y' =■ /»», the whole arc of the locus of Q - 12a ;

(2) when y' — 4aa;, the arc from the vertex = of + 2a log (I + xfa) ;

(3) when ^ + ^ - 1 (a >*), the whole arc - 4a ( 1 + ?— ^log ]-^\ ;

(4) = (a<b), =4*{(l-<»»)* + 2/<rsin-i#};

(5) when x = a (2^ + sin 2^), y « a (1 + cos 2^), o* « 2jf ;

(6) when x^ a (2^ -i- sin 2^), y ^ a (1 -cos 2^), the locus of Q is a cycloid
of half the linear dimensions and having the same tangent at the vertices;

(7) when the curve is such that the radius of curvature is n times the
normal at P terminated by the axis of x, the arc = sb (n— 2)/ft.«, it
being any constant number.

Solution by J. W. Shaapb, M.A.
Let 1, 1} be the coordinates of Q ; then

I « ic— y sin 2a, ij « y (1 + cos 2a), and dy ^ dx . tan $,
therefore d^ = {dx - 2y dd) cos 2a, dfi= {dx- 2y de) sin 2$,

therefore da^ ±{dx—2yd$),

(1) «8 + y2-a2; therefore tan $ = ± — , and rfa ^— .,

y (a»-ir2)*

therefore o- « 4 T 3<2r ; therefore o- -> 12a.

(2) y«=4a« ; therefore tan a « ~; and rfa - - ./^f'', ;

therefore <r = f'^-^+lW = « + 2alog(l+ —J.

(3) £. + ^ « 1. Put a;»aco8^, y^^sin^; then
a* 0*

*,_ eH± ; therefore^- ( '*'"°'» + Bin*^<to:

therefore f - 24» f »' [__££|±_—- +8m^<*«.|

4a Jo (.«»—(«»—*') cos* ^ >

$ ° 1-tf

^Ua Jo la> + (6a-a«)co88^^ ^ ^j

^ tan->— ^+1;

^(1 -<?«)* (l-<^)*

tr 2

therefore li '^ "" "^"^ * ^ (!-«')*•

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(5) X = a{2<p + sin 2^), y = a (1 + cos 2<^) ; than tan e = —tan <p ;

therefore $ + ^ « ir ; therefore ~ » I 4 (1 + cos 2^) «f^ — 2x.

a Jo

(6) (B = a (24> + sin 2^), y = a (1 — cos 2<p) ; then tan 9 «> tan ^ ;
therefore B =^<^\ therefore | « ia (4<^ + sin 4<p), ij « ^« (I — cos 4^).

(7) "*", ^ = »»y sec 9, where i> = tan e ; therefore ^ =^ ny;
dp jdx d$

therefore a- =» T (» — 2) y rffl «= T ^^-^^ dx : therefore <r = ± -^^ x,
Jo Jo » »

8954. (W. J. C. Sharp, M. A.) — If seven tang^ents to a cuspidal cubic
(or tricuspidal quartic) be given, and a conic bo described to touch any four
of those, the conic which touches the other three given tangents and the
two remaining common tangents of the first conic and the curve, will
always touch a fixed tangent to the curve.

Solution by Professors Nash, M.A. ; Sarkar, M.A. ; and others.

A cuspidal cubic (or bicuspidal quartic) being of the third class, its re-
ciprocal is a cubic, and the reciprocal theorem maybe stated as follows : —
Given seven pointia A, B, C, D, E, F, 0- on a cubic, if through four of
them. A, B, C, D, a conic be desciibed meeting the cubic again in P, Q,
the conic which passes through P, Q, and the other three points E, F, Q-,
will pass through another fixed point H in the cubic. This follows at
once from the well*known theorem that PQ passes through a fixed point
R in the cubic, the coresidual of the four points A, B, C, D, and also of
the four points E, F, G, H. Therefore, &c.

9128. (^' F- *f' Mann, M.A.) — Find the sum of all numbers less than
n and prime to it is divisible by n.

Solution by the Proposer.

If a is a prime to n, n— a is also prime to n ; hence, therefore, all the
numbers less than n and prime to it, may be arranged in pairs, the sum of
each pair being n.

9227, (W. J. C. Sharp, M.A.)~Show that (1) 1 . 2. 3...n^ is divi-
sible by (w) to the power of («**- 1) / («— 1) ; and (2) when (») is a prime
this is the highest power of {n) which will measure it.

VOL. XLIX.

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Solution by ProfesBor Ignacio Betkks.

Si notiB &u0on8fi^s> N, la plus grande puissance d*iin nombre pre-
mier CQntenae dans le prodoit 1 . 2 . 3.. .N est

a a a
N' 6taiit => — , et aiasi de suite ; mais si (a) n'est pas nombre premier alors

a a a '"

sera I'exposant d'une puissance de (a) qui divisera 1.2. 3...N. Cela
pos^, conmie N = n**,

n n \n I n-1

sera le degr6 d'une puissance de (w) qui sera f&cteur de 1 . 2 . 3...«'*; et si
(«) est premier, (n^— i) / («-i) sera le nombre plus grand de fois que
(1.2. S.-.n**) contiendra au facteur (w).

8742. (R. Knowles, B.A. Suggested by Quest. 8521.)— The circle
of curvature is drawn at a point P of a parabola, PQ is the common
chord ; if 0, 0' be the poles of chords of the parabola, normal to the
parabola at P and Q respectively, and if M, N, R, T be the mid- points of
00', OQ, O'P, PQ respectively, prove (1) that the lines MT, NR intersect
at their mid-points in the directrix, (2) that OP, O'Q are bisected by the
directrix.

Solution by Rev. T. R. Terry, M.A. ; Professor Nash, M.A. ; and others.

Let the coordinates of P be ap^, 2ap; then equation to PQ is
x-ap'-+p{y^2ap) = 0; therefore coordinates of Q are (9ap^, -6ap) ;
normal at P is y^2ap+p {x-ap^) =. ; therefore coordinates of 0, O',
M, N, R, T are '

(-2a-«p2, -2«;>-i), (-2a-9fljp2, fap-i), (-2a-5aj»2, -|ap-i),
{-a + iap^, -ap'i-Sap), (^a^iap^, ^ap-^ + ap), (6ap2, -2op) ;

therefore coordinates of middle points of MT and NR are
(-a, -^ap-^-ap),

whence (1). Also abscissa of middle points of OP and O'Q is -«,

whence (2).

8463. (J. C. Stewart, M.A.) — Solve completely the equations
a; + 2y-^^2+ ^/S{l'2xy-y^) = y + 2x-x^y + {2-^ ^Z){l-2xy''a}^ = 0;
and show that one system of values is a; = ± ^ a/S, y = 1 and ^/3-2.

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Solution by Professor Sarkar, M.A. ; Bblle Easton ; and others.
The first equation may be put into the form

or mir + fir = taii-*ip + 2tan-*y.

Similarly, from the second equation,

fiir + -^ir = 2tan->a: + tan-*y,
therefore tan-* a: ■» (2n— w)^ir + Jr, and tan-*y = (2m— «) Jir + Jir.

8095. (H. G. Dawson, B.A.)— If «, *, e be the axes of a quadric
having the tetrahedron of reference for a self-conjugate tetrahedron,
(6 ^> C ^) ^^^ tetrahedral coordinates of the centre of the quadric, and
(^i» Mi» viy wi), (Xj, /U2, vg, ir^ii (^3» A«3» "a* 's) *^® tangential coordinates of
its principal planes ; prove that (I)

and hence (2), if a tetrahedron be self-conjugate with respect to a sphere
of radius 31 and centre 0, show

- R2 (ABCD) « A« (OBCD) + m= (OCDA) + v^ (ODAB) + ifl (OABC),

where A, B, C, D are the vertices of the tetrahedron, X, /a, v, » the per-
pendiculars from A, B, C, D on any plane through O, and (ABCDj, &c.
are the volumes of the tetrahedra.

Solution by the Proposer ; A. Gordon ; and others,
Let (^i, yi, h)> (^2> y%t H)> fe ^3* «3)> (^4» ^4. «4) ^Q the four vertices of

the tetrahedron, and -t> + -^ + -- = 1 the quadric. Then we have
«2 0^ c2

£i£3+yiy3+«i^- 1 £i£5+yi.y3+5j^« 1 ^1^4 ,yiy4 ,gig4- 1

Hence, if

(A1B3C3D4) =

Ai B, C, Dj

Aa Bj Ca Dg

A3 B3 Cs Dg

A4 B4 C4 D4

-^ (a?2 y» «4) = (1 y3 ^4)
-^ (-^2 yj «4) » (^2 1 «4) !
-^ (^2 ya 2-4) = (j-2 ys 1)

.(1).

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We ihall have three other groups of equationB of a siimlar character, viz.,

•^(^1 yi u) = (I yi ^O") ^(^\ t/i «4) - (1 yj «4)'

■^(^1 yi O - (^i 1 V, I , -|-(^i yi «4) - (^1 1 «4) > -(2, 3),

-^(^i yi «4) = (-^1 ys 1) j ^ (^1 yj «4) » (^i y» i) ^

|^(^iyj*'3> = (i y,»3);

^(^iyj*'») = (^i 1 «,)y (4).

-^ (^1 yj ^i) = (^1 yj 1)

Multiplying the first equations of groups (1), (2), (3), (4), by 9^ — ar,,
^i> ~^4> A^d adding, we obtain

•^ {^1* (^syj«4)-^' (^iya24) +'a' (^iyj24)-a'48 (xiyj*,)} = (a?,yj«,l).
Now ^«> (^ 8y^ ^4) „:»j£ii^aM. ^ = -j£iy2^, e.i£:ii^i^.

* (^ly^^i)' ('lyj^^ji)' ^ (^lyi^ai) (^lya^i)

Hence, as Xi ss xi, fii = «,, vi = xs, »i « iP4>

Similarly the other equations are established. The remainder follows
easily.

9215. (8. Tebay, B. a.) — The growth at any point of a blade of grass
varies directly as its distance from the root. The respective heights of
grass in three meadows, of 2, 3, and 5 acres, are 3, 3^, and 4 inches. The
grass in the first and second meadows is cut in 32 and 30 days, respectively.
If 12 oxen consume the produce of the first meadow in 66 days, and 16
oxen consume the produce of the second meadow in 63 days, find when
the grass in the third meadow must be cut so that 1 8 oxen may consume
the produce in 80 days.

Solution by the Proposer.

If h be the height of the grass at first, and m the rate of growth at
unity, the rate at the height h-\-x is tw (A + a:) = dxfdt. Hence, if x
vanishes with ^, we have h-vx = he^^y which is the height of the grass at
time t. Hence the consumptions in the three cases are 6^'*'", lO^c*^,
20c"»'. Now 12, 16, 18 oxen consume 6tf82», lOJer*"*, 20«»* in 66, 63, 80
days ; therefore ttJ^^** = ^s^***" = jt^*' From the first equation we
have ^ = (J)* ; therefore ^^ = (J)»6 = ^^e^* = V (*)**>

i^ = 16 - l^RLltzi^ » 13-13375, and t » 262676 days,
log 7 -log 6

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8577. (B. Hanumanta Rau, M.A.) — Prove that the arc of the pedal
of a circle, of radius a, is equal to the arc of an ellipse (0 » f ), the origin
being at a distance ^a from the centre of the circle.

Solution by Professor Mathews, M.A. ; Sabah Marks, B.Sc. ; and others.

Let SP = r, Z POA -= ^ ; then
QQ' or rfS = rrf^, ultimately.
Now we have

r2 = SO2 + 0P2 + 280 . OP cos4>,
or, if SO « \a, OP == a,

'•'*(« + icos<|>)a2
= (ff-8inn<^)a2,
Hence, if ^ = 28, we shall have

where « = |, therefore, &c.

8855. (Profe8sorMuKHOPiDHYiY,M.A.,F.R.A.S.)— Prove that (l)the

solution of the system -^ . -^^ = a, -^ . "'"^, = i^ is given by
a; l + y2 x^ \+y^

h y^ ar* 1 + y"

where X satisfies ( ""^ | = — ^^ — ; and obtain (2) all the solutions
by the transformation A + X"^ = /*.

Solution by W. J. Barton, M.A. ; R. F. Davis, M.A. ; and others.

Putting y = \x, we have A ^ = a, whence a;2 __ __ . _ZLfL »

1+A^;c2 A aA-1

and therefore y' = A ~ ; hence, by substituting in the second equation,
aA— 1

... (^-a3)(x3--l)-3a(*3-a) (a^- ^ ) +3a(aft3.i) (x-1) =0;

whence A — A"' = 0, or A = ± 1, which give a:2+ 1 = 0, y^^ j _ q, or,
putting A + A-i = /x, (^-a3)(^2_i)_3^(^3_^)^ + 3a(a^3_i) . q. Let
hij Ms ^^ roots of this quadratic ; then

a2-/*iA+1=0, or a2-/U2A+1 = 0,

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the roots of which may ho denoted hy Aj, — ; Aj, — . Suhstituting in

Aj Aj

yalues of «*, y' ahoye, we get ar = ±a?i, ± — ; y — ±yi> ± — •

^1 yi

9140. (Emile Vioari^.)— Si B, Rj, R, designent respectivement lea
rayons du cercle circonscrit du premier cercle de Lemoine {triplieaU ratio
circle) et le deuxi^me cercle de Lemoine {cosine)^ d^montrer la relation

E« = 4Ri»-R52.

Solution by Professors Ionacio Bbyens ; Bordaob ; and others.

D'apr^ les valeurs de B| et R} qui sont dej^ connus (voyez Libbek, Uber
die Oegenmittellinie und den Grebis^ehen Funkt) on a :

P B,{b^c^ + a^c^ + a-b^^ r, abe

**" a2 + *2 + c2 ' ^^a^ + b^ + c^*

•*• ^^'^^ = {a^^b^^c^r •

Mais, d^signant A la surfEice du triangle, on a A » -— , et on d6duira :

4R

^^ " 1^' "* ^^ ^^ ' (^T^T^jui^

_ n^&V (4aV -t- 4&»g3 -I- 4fl8&g ~ 1 6 A») _ a^b^^e^ {2a^c^ + Wc^ -h la^l^ + q^ + M + g^)
"* 16A2(a2 + i2 + ^j2 " 16^2 ■ \^tk) "

9264. (Professor Hudson, M.A.) — Prove that y = 'v/2 (a?— 4a) is hoth
a tangent and a normal to 11 ay^ «» 4 (a;— 2a)3.

Solution by R. F. Davis, M.A. ; R. W. D. Christie, M.A. ; and others.
The abscisssB of the points of intersection of the straight line and curve
are given hy the equation

27a(a?-4a)2=2(a:-2a)«,
or 2a:3-39flw;2+240a%-448a3 « 0, or (:c-8a)2(2ar-7«) = 0.

By examining the value of dy/dx at those points it will he found that the
straight*line is a tangent at the point {Sa, ia V2) and a normal at the
point (7a/2— a/>/2).

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[We may write the normal to 27ay* = 4 (x- 2a)3 as

3a 2a

and the tangent (being the normal to y* = iax) as y = «M;-2ffm— am^ •
and if these lines coincide, we have the equation w'— Sm^ + 2 = 0, whereof
the only real roots are m = ± ^/2. ]

9338. (A. Russell, B.A.)— Show that the solution of the partial
differential equation

Bar* d^ dx^ dx dy2 dy

Solution by J. W. Sharpb, M.A.

Substitute C^\ogx\ then ^^^a^l^^a\ c; therefore the solution
is the sum of the solutions of

^ = a(i-a^., and ^^^.a[l^a\z

Take the first, and put z = Ae^y*^v^ and ^ = D^;

then h - Dya ; therefore « = «?«>'. eW«DY(a:),
/being arbitrary J therefore

^^ff ^-«'/(ar + 2w*/i^W«.

ox^
Let «2 « — , then we obtain for z the value

de

The other solutions are obtained by changing the sign of a under the
integral sign.

8752. (Professor Gbnesb, M.A.) — If AL, BM, CN be perpendiculars
from the vertices of a triangle ABC upon any straight line in its plane,
then, three letters denoting an area, and signs being regarded, prove that
AMN + BNL + CLM = ABC.

Solution by the Proposer.

Let py Qt r be the perpendiculars from A, B, C on the line, then its
equation in perpendicular coordinates is apa-i-bqfi + cry = 0. The line

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is therefore the line of action of the resultant R of forces ap, bq^ cr along
BO, CA, AB. Taking moments about A, Rp = ap . 2a / a^ or R » 2a.
Let R make angles 0, <pf i|^ with the sides. Resolving along the line,
ap COS e + bq COB <l> + cr COB }}/ = R or /7.MN + j.NL-t-r.LM = 2A,
whence the theorem. The line may be drawn through A and the theorem
verified by Euc. i. 37, then easily extended.

9146. (R- Lachlan, M.A.)— If two circles (radii p, /) intersect in
A and B, and any straight line cut them in the points (P, Q), (R, S) re-
spectively, show that

(AP . BP . AQ . BQ)/p2 == (AR.BR. AS . BS) /p",
(AP . BP . AS . BS) / SP« = (AQ . BQ . AR . BR) / QR2.

Solution by Professors Ionacio Bbtens, Matz ; and others.

(1) Soient PH, QH' les hauteurs des triangles PAB, QAB, de mtoe
Boient ST, RT' les hauteurs de SAB, RAB : nous aurons :

PH .2p = AP.PB, QH'. 2p = AQ. QB,
d'oti PH . QH'=- AP . PB . AQ . QB / 4p2.

De la m^me mani^re dans T autre circonf^rence on a

ST . RT'« AR . BR . AS .BS/ 4p'3 ;
mais les triangles semblables KRT', KPH (K 6tant le point de rencontre
de PS, AB) et KH'Q, KST nous donnent :

(a) ?^-?? QH'^KQ ot ,,ar Quito ^^-Q^' -^ ^^-^^ -l-
^^ RT' KR' ST KS* *^ RT'. ST KR . KS

done AP.BP.AQ.BQ/p2=AR.BR.AS.BS/p'2 (1).

(2) Des relations AP . BP = PH . 2p, AS . BS = 2p'. TS, on deduit :
AP . BP . AS . BS - 4pp' . PH . TS, et d'une mani6re analogue on a :
AQ . BQ . AR . BR = ipp'. RT'. QH', et par suite

AP.BP.AS.BS ^ PH . TS .
AQ . BQ . AR . BR RT'. QE' *
mais des relations (a) on a :

PH . ST ^ KP KS ^ KPg ^ KS^ ^ PS'
Kiy. QH' KR • KQ *" KR2 KQ2 RQ^'
cardeKP.KQ = KR . KS on obtiendra :

KP^KS PS KPg ^ Kffl ^ PS»

KR KQ ■ QR ' ^ KR- KQ2 QR '
et par suite la seconde rotation est demoutree.

[The results may be otherwise deduced from the following theorem : —
(o) If two circles, centres H, K, intersect in A and B, and if OP be the
tangent from any point O on the former, drawn to the latter, then
OA.OB = 0P2.0H/HK.

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To prove this, let OA cut the other circle in A', and let E, F be the
mid -points of OA, A A' ; and lot O'
be opposite extremity of the diameter
OH. Then the angles OO'B, EHK
are equal, and therefore
EF ^ OB
HK 00"

""'- ^-ii

whence the result follows. P

[The theorem (a) is a particular case of the following general theorem : —
(;8) If through a fixed point O, a variable circle, with radius R and centre
H, be drawn to intersect a circular curve of (n + m)^ order, having n double
foci F„ F2, ..., in the points P„ P, ... ; then the product of the distances
0P„ OPj, . . ., say (OPj, varies as R*'/(HF). This general theorem is easily
proved, and may be regarded as an extension of Carnot's theorem. The
similar extension to the case of a circle cutting a non -circular curve has
been given by Laoubrrb, Comptes Retidus^ Vol. lx., pp. 71 — 73.]

9229, 9259, ft 9301. (Professor Sylvester, F.R.S.)- (9229). Prove
that the points of intersection of any given bi<ircular quartic by a trans-
versal, will be foci of a hyper -cartesian capable of being drawn through
four concyclic foci of the given quartic.

(9269). Prove that, if one set of four coUinear points are the foci of
a hyper-cartesian drawn through a second eet of the same, the second set
will be the collinear foci of a hyper- cartesian that can be drawn through
the first set.

(9301). Prove that the points in which a pair of circles are cut by any
transversal will be the collinear foci of a system of hyper-cartesians
having double contact with one another at two points.

Solution by Professor Nash, M.A.

A hyper-cartesian is the inverse of a bicircular quartic with respect to
a point on one of the focal circles. Hence the first two theorems can be
at once derived from the general theorem that if F, G, H, K be concyclic
points on a bicircular quartic of which A, B, 0, D are concyclic foci, then a
bicircular quartic may be described through A, B, 0, D having F, G, H, K
as foci. This may be proved as follows : —

Let (a;— o)2+(f/-;8)2 = p2 and x^-k-y^—p'^ be the equations of the
circles ABCD, FGHK, and ax^ + 2hxy + by^ + 2ffx + 2fy + c = that of tkm
focal conic corresponding to ABCD ; then the equation of the quartic is
CS2-4S{G(x-o)+F(y-^)}

+ 4A(a;-a)2 + 8H (a:-a)(y-i3) +4B (y-^)2 =- 0,
where S « a:8 + y«-a'-i3« + p« =- x^ + y^-t^,

.and A = bc^f*, B =» ea—y^ &c.

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At the intenectioDS of the curve with the circle FGHK, S = p'^-^^t'^
suppose ; hence, if these intersections lie on the conic,
4A (a;-.o)» + 8H (a;-a)(y-i3) + 4B (y-)3)«

-.4^'3 G (ar-a) -4^ F (y-iS) + C^* = 0.
Foimrng the reciprocal coefficients, the equation of the bi circular quartic
having this conic as focal conic and FGHK as focal circle is seen to be

<yS'« + 2S'<^ {gx +/y) + H^ (aj;« + Ihxy + hy^ « 0,
where S' = (a:-a)2+ (y-i3)«-p2 + ^'2.

hence at intersections with the circle ABCD, S' ^ ^, and these lie upon
the conic oar^ + 2A4;y + &c. « 0, i.«., the quartic passes through ABCD.

(9301). Prof. Casey has shown that two circles may be considered as
a particular case of a bicircular quartic when the focal circle and conic
have double contact, i.e.^ when AB coincide and also CD. Therefore by
what has already been proved, if F, G, H, K be the int^'rsections of the pair
of circles with a straight line or circle, a bicircular quartic can be
described having FGHK as foci, and touching the focal conic at A and C.

9303. (Professor Neubeeg.) — Sur les c6t^s du triangle ABC, on con-
struit trois triangles semblables BCD, CAE, ABF ; d^montrer que la
somme (DE)2+(EF)' + (FD)2 est minimum, lorsque les points D, E, F
sent les sommets du premier triangle de Brocard.

Solution by the Proposer ; R. F. Davis, M. A. ; and others,

Soient X, /i, v les angles en B, C, D du triangle BCD que nous suppo-
sons toumi vers Tint^rieur de ABC. On a :

BD « asin/i/siny, BF « csinx/sini^;
d'oti, dans le triangle BDF :

(FD)' = {a2sin2/i + c2sin2x-2<wsin/isinXcos(B-x— ^)} cosec'v,

<rs(DE)2+(EF)2+(FD)«
= {(a^ + lil^ + e^){Bm^ ti + an^ \) + 2BinfABm\2aecoB(B + y)] coaed' y.
Or, si V est Tangle de Brocard de ABC, et S l*aire ABC, on a :

a2+^ + c2 = 4ScotV, sin2;i + sin2x-2sin^sinXcoftir = siu'k^
21 ac cos(B + 1^) = cos y 2 2ac cos B— sin u 2 2ae sin B
=-C08ir(a2 + *2^.^_12Ssinv;
done <r «= 48 cot V + 128 sin X sin/ii cos (V + f) cosec* v cosec V,

<r - a2 + *« + r2+ i2Scosec V . DB . DCco&(V + v)/a2.

iConstruisons une droite CD'rencontrant BD sous Tangle BD'C=90° - V;
le triangle BCD' donnera DD'= CD cos (V + 1^) / cos V. Done

ff » a3 + *« + c«+ 128 cot V .DB . DD'/a'.

Ainsi la difference <r— (a^ + i^ + c^) est proportionelle k la puissance du

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point D par rapport au cercle du segment capable de Tangle 90°— V con-
Btruit sur BC. Le centre de ce cercle est le sommet Aj da premier tri-
angle de Bkocakd. La puissance DB . DD' a sa plus grande valour
negative lorsque D coincide avec Aj ; alors <r passe par un minimum.

Scolie. — Le lieu du point D tel que <r a une valeur constante est une
circonference ayant pour centre A,. En particolier, lorsque Tangle
BCD = 90°- V, on a : <r = a^-\-i^-¥cK

8868. (Professor Schoute.)— If ABC and A'B'C are two positions of
the same triangle in space ; if A", B", 0" are the centres of the segments
A A', BB', CC, and if the planes through A", B", C" respectively perpen-
dicular to AA', BB', CC, intersect in P, the tetrahedrons PABC and
PA'B'C are not congruent, but symmetrical.

Solution by Professor Q. J. Leobbeke.

Displacing first the triangle A'B'C parallel to itself into the position
AB(,Cq, a being the vertex of the triangle ABC corresponding to A', we
may afterwards turn ABqCq round an axis AG until it coincides with the
triangle ABC. This axis AO is perpendicular to the lines BoB and CqC.

The point P considered as vertex of the tetrahedron PA'B'C will share
the movement of the base A'B'C and first describe the line PPq equal in
lengfth and direction to A'A. If now the two tetrahedrons are congruent
the rotational displacement of ABqCq will bring Pq to coincide with P, the
vertex of the tetrahedron PABC ; then the axis AO is perpendicular to
the line P^P. The line AO, being perpendicular to PqP and consequently
to AA', BBoand CCq, is also perpendicular to the three lines AA', BB', CC,
or, what is the same, the displacements of the vertices of the triangle ABO
are parallel to the same plane. In this case, however, the planes which bisect
and areperpendiculartothelines AA', BB', and CC, donotmeet in one point.
When, therefore, those three planes meet in one point, the tetrahedrons
cannot be congruent, but must be symmetrical. When the displacements
A A', BB', CC are parallel to the same plane BB'B(,or CCCq, the axis OA,
being perpendicular to BqB and CqC, will in general be perpendicular to
that plane. Kow it is not difficult to prove that in this case the planes,
which bisect and are perpendicular to the displacements, meet in one line.
Therefore we project the figure on a plane parallel to the displacements
or perpendicular to AO. Then, of course, the projections of the three
positions of the triangle are congruent. Now the lines bisecting perpen-
dicularly the lines joining the corresponding vertices of the projections
of ABC and A'B'C' will go through the same point, and therefore the
planes in question will meet in one line.

In particular, when the lines BBq and CCq are parallel, the axis AO
need not be perpendicular to the parallel pianos BB'Bg and CCCq. In
this case we find, by projecting the figure on a plane parallel to the dis-
placements, that the bisecting planes meet in three parallel lines ; for the
projections of the triangles ABqCq and A'B'C are congruent, and that of
ABC is symmetrical with that of ABoC©.

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2437. (The lato Rov. J. Blissard, MA.)— Prove that
1 ^ 1 ^ 1 ir . » j:

12»aj2 32_a;i 62-ic2 - 4^ 2

Solution by Gbobob Goldthorpb Sto&r, M.A.
Wehave ooe, = (l-l^') (l-^^) ... .

hence logcosy - log ^^1^% log ?l^i^ + &c.

DiflFerentiating, we have tan y -» „ ^^ -i- ^, ^ -^^ o +&c.,

and, putting y « Jtx, we obtain the result given in the question.

8818. (Professor MuKHOPADHY Ay, B. A., F.RJS.EO—Show that, (1)
the equation of the directrix of the conic which is described having the
origin for focus and osculates *%2 ^ ^CyS _ ^2^ ^t the point ip, is

(a-2-*-2) (aa?cos'^-*y sin'0) = I;
(2) the envelope of this for different values of ^ is the quartic
*2a;-2 + ^2y.2 = (»A-i-*a-i)2,

which curve is also the reciprocal polar of the evolute of the conic
o%2 ^ ^^2 _ ^2^ with respect to a circle whose radius is a mean pro-
portional between the axes of the ellipse.

Solution by Professor "Wolstenholmb, M.A., Sc.D.

1. The equation of any conic having its focus at the origin is
aj2 + y2 j_ ^^\x + By + C)2 ; and, if this osculate the conic x-ja^ + y^lh'^= 1 at
the point {a cos 0, b sin <^), and we denote (a^cos* ^ + ^2 ^\j^ ^ji y^y ^^ t^g
equation r = A« cos <^ + B3 sin ^ + C must have three roots <^ ; or we may
di£ferentiate it twice with respect to <p. This operation gives for A, B

the two equations {a'—b-) sin ^ cos ^ . r-^ = Aa sin ^— B* cos <p,
(rt2- ^(a2 cos* 0-^2 gia4 <^) r-5 « Aa sin + B* cos ^ ;
whence Aal{a^ — b-)

= sin <p sin <^ cos ^ . r"^ + cos ^ {a^ cos^ 0— ^ sin* <(>) r-' = a' cos^ ^ . r-* ;
and similarly B*/(^-a2) = ^ sin' <^ . r-^,

whence C = r (a2 _ ^2) (^^2 cos* (p-b^ sin* «^) r - *

S {{a^ C082 + ^2 gui2 <^)2_ (a2-*2)(a2 COS* <^-*2 81^4 ^)} y-8 = a2i2^-3^

Hence A : B : C = «cos'0 : -*sin30 : aH^I{a^-b^i

and the equation of the directrix (Ax + By + C = 0) is as stated.

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2. In the envelope of the directrix, we have, on differentiating,

or ■ B — 2: — » x — ^ ^~ — = / (am ib cos 4>\

sin^ — cos^ sin ^ cos ^ (COS? 4) + sin^ ^) b^—a^l '*

ah b . ^ ah a

or COS <b = -- — i — , sin ^ » — — - — ,

whence the equation of the envelope is

(x^ y^ \ ab I U ,a / ;

The reciprocal polar of the evolute of the conic a^/a'^ + y^lb*^ ^ 1, with
respect to a circle a?* + y' =» A:* is the locus of the pole of the normal

— — -^ « a'2_ ^'j ^ith respect to this circle ; and if (XY) be its pole,
cos® sin A N / r »

k'2 * (a'2-A'2)co8 0' ;fc» " (Z»'^~rt'=»;sin4)'

and these equations coincide with those for the point of contact of the
directrix with its envelope if

^V a&» m' aH

a"^-b'* " b^-a^' b'^-a'^ "" a--*2*

whence oa' = 4^ = k^. Thus the envelope of the directrix is the reciprocal
polar of the evolute of the conic x^/a^^ + p^/b^ ^ I wiih respect to the
circle x^ + y^ = k^ provided aa* ■■ bb' = k^^ and one case is when k^ = ab,
a' = b, y ^ a,

[If the osculating conic in this Question osculate in P and cut the
ellipse again Q, the equation of PQ will be

{a2 sin' 0(1+ sin^ 0) + b^ cos^ 0}

{a^ sin< (^ + ^ cos* ^ (1 + cos- 0) } = a* sin* 0- ** cos' 0,

acos0

y_

dsin^ '

the envelope of which it would be interesting to find. Tiie equation of
the conic having its focus at the origin and osculating the ellipse
x^/a^ + y^/b- = 1 at the point (« cos tf, b sin 0) is given as Question 1218 in
Wolstenholmb's £ook 0/ Frobktns.]

2396, 6931 & 8935. (W. 8. B.Woolhouse, F.R.A.S.)— LetABCD be
any convex quadrilateral, having the diagonals AC, BD intersecting in E ;
and let p, p' denote the ratios 2 AE. EC : AC, 2BE.ED : BD- respectively.
Then, if five points be taken at random on the surface of the quadrilateral,
prove that the probabilities (1) that the five random points will be the
apices of a convex pentagon, will bo ^^^ (I I + 5p/) ; (2) that the pentagon
will have one, and one only, point reentrant, will be J ; (3) that it will
have two reentrant points, will be -^ (1 — p/).

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Solution by the Proposer.

This question was designed as an exercise on my general theorem for
all convex surfaces, an inTostigation of which is given as a solution to
Quest. 2471 (Vol. vin., p. 100), and of which theorem the following brief
extract contams all that relates to the question ahout to be discussed.

Theorem. — Let a given plane surface having a convex houndary of
any form whatever be referred to its centre of gravity and the principal
axes of rotation situated in its plane ; and, corresponding to an abscissa
a?, let y, y' be the respective distances of the boundary above and below
the uxis ; also let h, k denote the radii of gyration round the axes, M
the total area, and

Then, if five points be taken at random on the surface, the probabilities

of a convex pentagon = 1 — — ^ + —jj- '

c * 4. • X IOC 20A2A;2( ,.,

of one reentrant point = ^^7:7 TFr-> (I)'

of two reentrant points <

M

One object in giving this extract from the general theorem is to effect an
improvement in the formula for determining the value of the subsidiary
quantity C, and to conveniently adapt the same to polar coordinates.
We have, according to the above,

MC - i Jx2y3fl?x + 3ja:yrf^Jy«rf2:,

in which the integration is to be carried round the entire boundary. For
the purpose of modifying this formula, we are of course at liberty either
to retain or reject any function which vanishes between limits. The
function jxydx is one of this kind, and therefore between limits we have

= (jicydx\ (Jy^dx^ = jxydx jy'^dx+ {y^dxjxydz.

Subtracting three times this, the expression for MG becomes
i J x'y^dx — ZJy^dxjxy dx.

Deducting = p^y^ = f J {x^dx + ar^y^ ^y) = i J a^V^ + i J ^ {^P^ «V,
the expression for MC is reduced to

--Zjy^dx^xydx-^jdixy^x^y,
Lastly, we have =^ xy^jxydx = j xhj^dx + J tf {xy'^) J xydx^
by the addition of which we deduce

MC = \x^y^dx-Z\y-dx\ xydx-^jd{xy^) x^ + j d {xy^) j xy dx
= J {y^dx-^d {xy^)} {x^y^z\xydx)
* t J (y*^^ -^y<^y) J {(C^dy—xy dx)
^ I [y^d^{ ci^d^:^ I [u^dB Bine {WdBiio^e (2),

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the algebraic sign being necessarily reversed in the last expression for
polar coordinates because x decreases when $ increases. The result Inst
obtained is both elegant and symmetrical. Moreover, by chang^g into
6 -I- a it might now be easily shown that the value will be precisely the
same for aU rectangular axes of coordinates passing through the centre of
gravity of the surface. This fact presents us with a most important
advantage, as we are thereby relieved from all difficulty as to the former
necessity of using the principal axes.

We now proceed to apply the last formula (2) to the
case of the quadrilateral. In the annexed diagram G is
the centre of gravity of the quadrilateral, and is taken
as the origin of coordinates; GAB « A^, GBO = A^,
&c. , are the component triangles, P is a variable point
xy in the side AB of the boundary, the integration being
taken from A ; GP « R, the angle AGP = a, m = the
sectorial area AGP, which is a triangle having its centre
x^ at n, (hi ^ r, the angle a(ht = J, and m = the
sectorial area aGn described by n. Then an is parallel
to AP, making Ga — |GA ; also w = f m. The coordi-
nates of the centre of gravity of the sectorial element
dtn » iB?d0 being |R cos $ and |B sin 0, we have

mx = ^JB^de COB e, f«y -ijR'rfesine;
hence the first part of
f MO ^ jd (3i»y)(3mf ) = 9 J (m rfy +^dm) mx ^ 9 j m^x efy + 9 J {mdm) xy
= 9 Jm^^r efy + fm'fy-f J m^ {xdy-^y dx)
- f J»|2 (xdy-gdx) +fm2xy = f J mV^ <^+ im^xy
= 9 Jm2dw + fw2iy =. 2jw2«f»f + fm2;py « |m» + fm'i^ (3).

When P has reached the comer B and the first component triangle is

completed, then m — Ai, ^ = i (j?i + a?,), y — i (yi + y^j ; and if we make

Xi « Ai {Xi + a?3), Xg = Aa (a?^ + x^^ &c.,

Yi = Ai (yi + yj), Yj = ih^Vi-^yi}* &c.,

the first part of |MG will be f Ax> + iXiY, ; and we shall have

Xi + X3 + X8 + X4 = 0, Yi + Yj + Y, + Y4 = 0.

In ascertaining the integration with respect to the second component

triangle when P passes from B to C, it should be observed that in the

formula |MC = J R^ rfd sin « J R3 (fd cos = J rf (3my)(3mf ),

the second factor Zmx represents the integral f R'(^0 cos 0, and that it is
indispensable that this integ^l shall always be counted from the same
epoch. From B, its value having there become Xj, it must now be re-
garded as ^tnx + Xi ; and in the third and fourth parts of the integration
it must, in like manner, be taken as ^mx + X3 + Xj and Zmx + X3 -1- Xj + Xj
respectively, the quantities Xj, Xj, X3, X4 operating as constants. Thus
the second, third, and fourth portions of f3ie integral are found to be

iA/+(iX2 + X,)Y^ |A8» + (iX3 + X, + X,)Y3 = iV-(JX3 + X4)Y3,

IA4H (iX4 + X3 + X2 + X,) Y4 = i A4'»- JX4Y4.

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By collecting the four sections of the integral we obtain, as its complete
value, fMC = §2(a') + JX,Y, + (^X^ + XO Y,-(iX8 + XJ Ya-iX.Y^.
Adding the equality

0--i(Xi-^X2)(Y, + Y5) + i(X, + X4)(Y3 + YJ,
it reduces to |MC - |2 (A^) + J (XiYa-XjYO + i (XjY^-X^YJ

- t2(A») + 4(AiAj.A('2 + A,A4. A^'O (i),

in which A 1^2 denotes the area of a triangle formed by the centre Q- and
the middle points of the sides AB, BC ; and A3' 4 the same with respect
to the other two sides CD, DA.

We have yet to determine, from the geometry of the quadrilatet-al, the
values of the several triangular areas involved in the last formula.
Furthermore, we have afterwards to ascertain the value oih^k^ where A
and k specially relate to the principal axes.

Let ABCD he the quadrilateral with the
diagonals AC, BD intersecting in E ; m the
middle point of AC and m' that of BD ; g the
centre of gravity of the triangle CDA, and g'
that of ABC. Put Am = y, wE — r, Bm' — ^',
w'E = v', and the angle AEB = E. Then, with
AC and BD as oblique axes of coordinates, the
line through g^ g' is j? — —ft?. Similarly, the
line through the centres of gravity of the
triangles DAB, BCD is y = - |r'. The centre
of gravity (G) of the quadrilateral is therefore
a; = — |v, y = — Jt>'. With this centre as the origin, the coordinates of
the four comers A, B, C, D are respectively arj ■■ — (^ + it;), y^ = \v' ;

a?2 = K y2 = -(y'+K); aJs-y-K ya-K; ^k^\^^ y^^f^-W-

The axes of coordinates are now respectively parallel to the diagonals AC,
BD. To change the axis of y into rectang^ular axes, for x put ar + y cosE,
and for y put y sin E. Thus we get, for the rectangular coordinates of
the four points, the following values : —

^i = - (? + i*') + \^' cos E, yi « \v' sin E ;
H - f «'- (/ + K) cos E, yg « - {q' + iO sinE ;
^8 = (? - \'o) + \^' cos E, yj = ft;' sin E ;
a?4= ft; + (2'-it;'jcosE, y^^ (/-K) sinE.
The values of 2Ai, 2A2, 2A8, 2A4 are to be had from the fxpreesions

*iy2-^2yi» ^^vz-HVi^

hence found to be

^zV^-x^yzi ^^yi-^iVk respectively, and are

2Ai = {(^ + it;)(^' + JvO-Jt?t;'} sinE,
2A3- {(^-it;)(/ + Jt;')+4t;t?'} sinE,

2 A3 = {{A'-\o){(i'-¥)-¥A «"^ ^»
2A4 = {(2 + Jt>)(/-Ji;') + Jt^t;'} sinE.
These values ako give

A, + A2 = j(g' + Jt/)sinE, A3 + A4- j'(j'-Jt/)sinE,
Ai + A3 =: {qq' - \ov') sin E, Ag + A4 = {qq' + Jt;t>') sin E ;
M = A1 + A2 + A3+ A4 = 2j/sinE

.(6).

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Again, the coordinates of the mid-points of AB, BC, CD, DA are

< i{(!?-if) + (!?'-K)«>8E}, y'{ i(?'-K)8inE;

<- i{(? + i'')-(*'-*'0co8E}, y'^ K?'-if08inE;

'>t= 4{(? + i«) + (?' + 4»0co8E}, y'^~ JCs' + Jt-OsinEj
<' — i{(?-i'')-(*'+K)oo8E}. y:'= J(j'+l*')8inE.
From these, we find

4A;;,-2«y^'-<:yn and 4^-, = 2 (.:-y--<V30
^K[% = ? (?'- JO sin E = A3 f A4,

4<4 = ^(^' + W8inE=Ai + A2 (6).

Hence, by substitution in (4), we find

|MC = 12 (A3) + Ai As (A3 + A4) + A3A4 (Ai + A,)

= t2(A3) + AiA3(A2 + A4) + A2A4(Ax + A3) (7).

We have also 82 (A') = 8 (A^a + A2' + A33 + A43)

= {(y + Jt')(?' + JO-4«'»'}'8ui'E-|-{(^-Ji;)(^' + Jt;')+4vv'}3sin»E
+ {(^-•j2')Y-i<'')-4vv'p8in3E+ {(^ + Jy;(j'-Jv') +*«'«''}•'' sin^E

4A1A3 (As + A4) + 4 A2A4 (A J + A3)

+ {(?- J«')(?' + K) +4*^'} {(? + J*') (^-K) +*«'«''} {n'-¥'^') sin'E
- 2^^' {(?^"4t^(?'2_^^'2)_ j^e^V2} sin'E,
whence |M0 = \qq' {(f + Jt>') (^'2 + j,/2) g{n3 g

+ i?y' {(?^-4t;2)(j'3_j^/2)« 16^2^/2} sinSE

>i-i('-f) (•)■

It now only remains to determine the radii of gyration round the
principal axes of the quadrilateral. Suppose one of these axes to make
an angle ^ with the axis of x which has been taken parallel to the diagonal
AC The ordinate of any point xy when referred to this principal axis
will have a value equal to y cos^— a;sin ^ ; and accordingly the new
ordinates of the four comers will now be the following : —
Vi = i«^sin(E-<^) + (j + Jv)sin<^, yg — ft/sin (E-<^)-(2'-Jr)sin^,
yi = - (j' + JvO sin (E~0) - \v sin <^, y^ = (y'- J*') sin (E-«^) — f y sin <^\

t/i + y^ = fv'sin(E-0)+tf;sin0,

ya + y4 =-fv'sin(E-<^)--ft;8in<^;

and hence, by the theorem given by me in Quest. 8922, we get

6A2 =-(y, + y3>(y3 + y4)-yiy3-y2y4
= {q^ + }v') sin2 <t> + (q'^ + }t/' ) sin^ (E - <^) + pv' sin <^ sin (E - 0) .

VOL XLIX. F

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To abbreviate, let a » ^+ Jr", 3 - ^'s + Jt/', 7 = Jn/ ; then

12A«- o(l-cos2^)+/B {l-co8(2E-.20)} +7 {co8(E-2^)-co8E}
— a + /9— 7CoeE-(a + /9co82E— 7COS E) cos 2^
- (/B sin 2E - 7 sin E) Bin 2^.
To farther abbreviate, denote this by

12*« - U- V cos 24»- Wein 24» ;
then for the other principal axis, changing 2^ into x + 2^, we get

12A?« = U + Vcos20 + W8in2^.
Therefore 6 (h^+k^ = U, 6 (A'-Ar^) = -V cos 2<^- W sin 2<p,

Now, when these relations refer to the principal axes of the quadrilateral,
the value of h^—k^ must be either a maximum or a minimum as regards
the variable angle 2^. Hence, by differentiation,

= V8in2<^-Wcos2<^.
Adding together the squares of the last two equations, we get

36(^2-^-2)2 = V2 + W«,
which being subtracted from the square of 6 (A' + A^ := XT, we obtain
144A2>t2 = U2- (V2+ W*) = (4ai3-72) ginSE

- {4 {q^^\vi)(q^-^\v'^-^vh'^} 8in2E
= % {4j2/=^-(V-f^(^*-v'2)} 8in2E.

Therefore h^k^ ^ "^ll ^t' ^' . i^\ «-^(l-ppO (9).

108 \ 2^2 2j'2 / 108^ '^'^' ^ '

Hence, by the general theorem premised at the beginning of this solution,
if five points be taken at random on the surface of the quadrilateral, the
probability of their being the comers of a convex pentagon

. I^IOC 5A«^2 ^ i_20 /^^pp^x J_ ^ 11 ^W .

M« M2 27 V 4/108'^ ^*^^ 36

And the probability of

X X . X IOC 20A2A:2 20 /, pp'\ 5 „ « 6
ooereentrantpoint-— — ^^--(1-^)--(1-PpO-3»

two reentrant points » = — (I— pp').

[That the probability of one reentrant point should be for all quadri-
laterals the simple fraction {, is most remarkable.]

9293. (Elizabeth Blackwood.)— Find the number of permutations
of n letters, taken k together, repetition being allowed, but no three con-
secutive letters being the same ; and prove that, if this number be denoted

a-iS
where a, jB are the roots of the equation «2— (n - 1) a;— (*i— 1) « 0.

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Solution by Professor R. Swaminatha Aitar, B.A.

Of the Pft permutatious taken k together, let f^ose that do not begin
with any specified letter, say a, be represented hjpt in number ; of these
Pk permutations those that begin with a single b are evidently jt?*_i in
number, and those that begin with two i*s are Pk-2- We thus have

'Pk^Pk+Pk-\+Pk-2, and Pk^{n-l){pk.i+Pk-2)*
Observing that j»j =n — 1, we see thatj^i, p^j p^.-.^ie the successive
coefficieDts in the development of

-, therefore pk^i = — ,

l-(«-l)a:-(«-l)ar2

and

Pa.i-P*-

jj* + 2_g**2

i-a*-i

o-iS

This seems to be the correct result.

-i3

9378. (Rev. J. J. Milne, M.A.) — PSQ is a focal chord of a conic.
The normal at P (a?,, yi) and the tangent at Q intersect in R. Show that
the coordinates of R and the locus of R are respectively

(-^if -

b'^

yi).

*V

a2 (2a«-*2)2

1.

Solution by C. E. Williams, M.A. ; R. Knowlbs, B.A. ; and others.
Let the normal PR meet the axes in
G, fff and the diameter conjugate to CP
meet PSQ in E ; then PE = CA, and
PE/7 is a right angle. Again, R is the
centre of the escribed circle of the tri-
angle PS'Q, whose perimeter = 4CA;
hence, if RK be drawn perpendicular to
PQ and parallel to E^, we have

PK = semi-perimeter = 2CA ;
therefore PE = EK, Ty = ^R,
and CN«CM;

also RM:PN = GM:GN

= CN + CG:CN-CG
= l+<^: l-«2
« 2a^-b^ : *2.
[The equations to PSQ and the other focal chord PS'Q', are
yix^-ae) ^yi{x-ae), yi^f^ + ot) ^ Pi {x + ae) ]
hence, combining these with the equation to the curve, we ought to get
the tangent at P, and the chord QQ', so that

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therefore QQ' is, by comparing coefficients,

and the pole of QQ' is therefore (— x„ ""--7^"~yi)» ^^t;h can be
shown to Ue on the normal at P, and is therefore the point R required.]

9337. (W.J. C. Sharp, M.A.)— If S^ denote K + 2*' . . . + W, prove
that (l)rS..i + '^(p^yS,_2+ ''^''7\l^V^^ S.-3>...+So - (ii + l)--l ;
(2) deduce therefrom Fbbmat^s Theorem ; also (3) show that

^ '(.r + 1 (r-1;! r (r-'l): r-1 y

where («)('") stands for « («-l) ... («— r+ 1).

Solution by R. Enowlbs, B.A. ; Professor Matz, M.A. ; and others.
It is known that Sr = ^ + irS,..i- ^^^^^ Sr-2+&c.,

and hence we obtain S© ■« «, Sj — i« (n + 1), &c.

In the series in the Question, putting r = 1, 2, 3, &c., we have

2Si + So-if«+2»-(l + w)»-l, 3&, + 3Si+8o = (I + »)»--l;
therefore rSr-i + ^-~^ ^r-2 + ... So = (1 + «)» - 1,

therefore r|Sr.i+ t32Is^.2 + &c.| - (1 +#i) {(1 +»)'->- 1} ;

hence, when r and 1 + n are prime to each other, (1 + m)**-*— 1 is divisible
by r, which is Fermat's Theorem.

(3) In De Moaoan's Calculus, p. 257, the formula

18 given, its rth term being — .— . — ^ = — ■ ,
® r!r + lr+l

and with this substitution we have formula (3).

[We may prove (I) thus, without assuming the value of Sr : —

Sr + rS,.i + ?l^:iils._2 + ... +So = |^«'' +r,n''-i + ?1^^

+ f (n-.l)»' + r (n- 1)*--! + ^^"""^^ (w- ly-^ + ... + 1| + &c.
= (n+I)'' + »''+(»-l)'' +... + 2'' = (M + ijr + Sr-l**;

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and hence (1) holds. The slightly more general formula

where S^sar + (x + l)''+ (« + 2)»'+ ... + (x + f»- l)^

may be proved in the same way.

Thus, if r be a prime number, (»• + «)*•—««*•—« is dinsible by r.]

2448. (J. S.BBKRIM AN, M.A.)— Let AEB, CED be two lines of railway,
whereof AB is perfectly straight, and CD curved as fur as F, the
remainder being straight ; then, if FE be 25 feet long* and the curve CF
have a radius of 3000 feet, and the angle BED = 25 26' ; show that the
distance from B to E, so that a curve BC may be struck with 1000 feet
radius is 342*765 feet.

Solution by D. Biddlb.

Let O be the centre of the curve CF, of radius
3000 ft. Draw the arc IK with the same centre, O,
and radius 2U00 ft., and, taking the point G, 1000 ft.
from the line AB (produced), draw GH parallel to
AB, cutting the arc IK in L. Then L is the centre
of the required curve. In order to find the length
of EB, produce OF to cut AB in P ; then

OP = 3000 + 26 tan 25° 26'= 3011-8887025.
Again, MP = 1000 sec 25° 26'= 1107*3147,
and OM = OP -MP = 1904*5740025.
Moreover, OL : sinLMO = OM : sin MLO ;
whence Z MLO = 24° 8' 24*5",

and LOM = 1° 17' 35-6" ; also LM = 105*1.
Now, BP = 1000 tan 25° 26'-LM « 370*4481,
and EB = BP - 25 sec 25° 26' = 34276523 ft.,
the required distance.

9304. (Professor Schotjtb.)— Of a triangle ABC there is given the
vertex A^ tiie angle A, and the line of which BC is a part ; find the loci
of the remarkable points of the triangle ABC.

Solution by R. F. Davis, M.A.

The locus of the orthocentre H is the perpendicular AD on BC, and
that of the centroid G is a line parallel to BO trisecting AD. If OM be
the perpendicular from the circumcentre on BC, OM = R cos A ; hence
the locus of O is a hyperbola having A for focus, BC for directrix, and
eccentricity = sec A. Similarly the loci of I, J^, J 2, J3 the in- and ex-

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centres are two hyperbolas having the same focus and directrix as the
preceding one, and eccentricities cosec JA and sec ^A respectively. If O'
be the image of O with respect to BC, and N the mid-point of AO' ; then
N is the Nine-points centre and describes a curve similar to the locos of
O', or of O, and is also a hyperbola.

9359. (J- 0*Byrnb Croke, M.A.)— Prove that the area of the simple
Cartesian oval formed by guiding a pencil by a thread having one end
attached to the tracing point and brought once tensely round a fixed pin
of negligible section, the other being fastened to a second pin at a dis-
tance a from the former, and the whole length of the thread being 2a, is
4^2 (2»- 3^/3).

Solution by H. Forte y, M.A. ; D. Biddle ; and others.
Let P be a point on the curve,
AP = r, and zPAB==fl,
and area of curve ■= A ; then

AB = a, 2BP = 2a-r,
and 4BP2= 4(AB2 + AP2- 2AB . AP cosfl),

(2a-r)2 - 4 (a2 + r2-2arcostf), ^,

or 3r = 4a(2cosfl— 1) ;

therefore A = jfrVe

= fa2 f (4cos2tf-4cose+l)ifl

= i«' I (2 cos2e-4cos 0+ 3) d0 = {a^ (sin 20-4 sin fl + 3tf)

= 4a2(2»-3<v/3).

[The curve as algebraically represented is a Cartesian oval ; but we
are concerned only with the inner loop, which is the only part of the curve
that can be generated in the manner described.]

9250. (Major-General P. O'Connell.) — If s = the length of an arc
of a circle, v = the versed sine of half the angle subtended by the arc,
e = the chord of the arc ; required a series for the value of a in terms
of V and c.

/Solution by K.W. D. Christie, M.A. ; Sarah Marks, B.Sc. ; and others.
We have v « (1 — cos \0)y c = 2r sin |fl, « = rfl ;

therefore /- « ^^" ^^ « tan i0 ;

2vr 1-cosifl

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hence, by Grboory*s series, we get

4r * 2vr

-*(4y-*(2^r-*-

9367. (F. MoRLEY, B. A.)— In the sides AB, AC of a triangle ABO,
find points D, E, such that BD = DE - EC.

Solution by Prof. "W. P. Casbt, M.A. ; D. Biddlb ; and others.

Make AH = AC, and divide BC in P in the ratio
of ATI : HC, and let the arc PO be the locus of
that ratio. Join BO and draw OK parallel to AC
Then KH : HO = BO : OC « AK : 00, and
therefore BO = AK. Make BD « BO and AE = KO,
hence BO = BD = DE « EC.

[Take I the in - centre of the triangle — a
figure is readily drawn or imagined — and P on
BA, so that BP =. CA ; draw PQR parallel to AI, cutting the circle
BIO in Q, R ; then if BQ, CQ meet AC, AB in E, D, these will be points
required. Another solution will be obtained by using P in place of Q,
unless Z A = 60°. Similarly, if the lines are drawn in different direc-
tions AB, CA, or BA, AC. We thus have a construction for a triangle
when they are given a + A, a + c, and LA. The problem has been long
since solved in our columns, in its present form, and a generalized form
of it is solved, under part (2) of the Editor's Question 7675, on p. 64 of
our Vol. xLii.]

8331. (H. G. Dawson, B.A.)— Show that the solution of

depends on the solution of

a(p_fl)«-i + j(p-i)»-i + c(p-c)»»-ic«0 (4).

Solutions by Professor Aiyar, B.A. ; the Proposer ; and others.
Divide (1) by a:", (2) by y*, (3) by «~, and add; then

-^+-^ + -^=0 (6).

^-1 yn-\ 5;»-l > *

^- f5-:-(]4)/(i-i).

whence we have — : — : — — (p--«) '> (p— ^) * (p--^)>
X y z

where p is a quantity that has to be determined. Substituting these
ratios in (6), we have (4).

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[Otherwise: (l)-(2) gives (a?-y)(a)-» +y-» + «-») « ajr-3y (6).

Treating (2), (3), and (3), (1) similarly, we have

ax—btf bu — cz ez—ax

« ^ — <r say,

a:— y y—z z—x

therefore (<*— p) * =* (*~p)y ™ (^~p) « ^ <r,

therefore g — — ^, y — --^, « — -^ ;

substituting these expressions in (I), (2), and eliminating a^, we get (4).
From (6), we see that <r» « (a-p)»» + (*-p)* + (<f-p)".]

8315. (Professor Booth, M.A.)— If

tan"* (Jir + ii|<) = tan" (Jir + \<p).

show that

Solution by FrofeBSors Mahbnoba Natk Kat, LL.B., a»<2 Aitab, B.A.
tan-Ciir + i,^) - {tan2(iir + i,^)}»« - ( J-±^y"'.
Therefore log tan"» ( J» + i t|^)

= im . 2 (sin ^r- J sin^ if^ + i sin* ^-\ sin^^,...) = mi tan-» [ M_t\ .

Similarly, log tan« {^ + \e) = in tan-» ^ i^^ \ .

9352. (Professor Hudson, M.A.) — Prove that
(tan 7i° + tan 37^° + tan 67i°) (tan 22i° + tan 62^° + tan 82^ = 17 + 8v/3.

Solution by K. "W. D. Christie, M.A. ; G. G. Storr, M.A. ; and others.
We have tan 7^ = ( -v/3 - -v/2) ( -v/2 - 1) ;

tan37i** = (^/3-^/2)('v/2 + l); tan67i°= ^/2 + l.
Also tan22r - 'v/2-1 ; tan52i** = (^/3+ ^/2) (a/2-1) ;

tan 82^ = (-v/3 + -v/2) ( a/2 + 1) ;
therefore product -(2^/6+ a/2- 3)(2v/6 + -v/2 + 3) = 17 + 8 v/3.
[By easy reductions, the product can he brought to

(5 + 6co876*')(6-6c08 76°)/(2v/2cos76°) « the given result.]

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9195. (Sir James Cockle, F.RS.)— Integrate

d^u

dt^ (aH*3^'»)

-, when m = 1 or when m «= 2.

Solution by the Pboposer.
1. Letm= 1; put 4^=A2^2_c2; then ^+i '

X*

: is flatiflfied by

«= Co(l + Vx)e"^'' + C, ( -^ + ^^) (»~''*^^ + C, ( J+ ^*) «"^*^*'

where a and fi are the unreal cube roots of xinity.
2. Let m = 2 ; put ^ — n sin and 4%' ._ _ ^2 . tj^e^

Icostfe^fl/ " a3(cose)3' e^flUosd^/eJ " "* a^ (cos «)* *
whence ^.3tan«^.3 {(tan*).^}^ = ^ ; and.if »' = f"

' ^0 aS

rf»'

rf3f/

.<^«'

rfw'

^+3tanfll^ + 3{2(tanfl)« + |}^ + 3{2tana[l + (tan6)2]}./-^,.

Now put u « (cos e) y ; then -^ + -r 5-^ = 0, whereof the coeffici-

d^ dd a*

ents are constant and wherein a » ~6 V^— 1. The solution of case (1) is
obtained by an analogous process.

8743. (C. BiCKBBDiKE.) — Prove that (1) the length of a focal
chord of the parabola is I cosec' (p ; (2) when the chord is one of quickest
descent, cos (p ~ (|)' ; and (3) the time of quickest descent down the
chord then is a/(3^/)/^, where I is the latus-rectum, and ^ the angle
made by the chord with the axis.

Solution by Geobob Goldthorpb Stork, M.A.

1. SP + SF:

2a

2a

1 — C08<^ 1+COS^

_ ia ^ I
sin* <t> sin* <l>'

2. Here t = {2Z/(8in*<^.ycos<^)}* ... (a) ;
and, as this is a minimum, sin* <^ cos <^ =x— a:'
(where x ^ cos (p) must be a maximum, which
is the case when Goa<f>sx = (^)^,

Substituting this value in (a), we find the time of quickest descent
to be as stated in the Question.

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9381. (Professor Sylvester, F.R.B.) — If {g and r being prime
numbers) 1 +j0+p'+ ...p'-* is divisible by g, show that, unless r divides
q— 1, it must be equal to q and divide j0— 1.

Solution hy Professor Gbnesb, M.A.

Let N( denote a number which is expressed in the scale of ^7 by 111 ...
to t digits. Now, q being a prime,

^-/,-M(^), !.<?., j0(/?-l)N,.i -M(^);
and, by the question, N^ = M (q) ; thus N, . i, N^ have a common measure
q, unless q divide (js— 1), for it clearly does not divide i?.

1. The arithmetical process shows at once that N,_i, N,. caimot have a
common factor unless of the form N^ where t divides q^\ and r ; but r is
prime, therefore r divides q^\ (which is not prime).

2. If i> — i»^ + 1, we find Nr « M (j) + (1 + 1 -i- ... to r terms), whence
q^r.

2353. (The late Professor De Moboan.)— The
late Dr. Milnek, President of Queens' College, Cam-
bridge, constructedalamp which General Pbreonet
Thompson remembered to have seen. It is a thin
cylindrical bowl, revolving about an axis at P, and
the curve ABCD is such that, whatever quantity
of oil ABC may be in the bowl, the position of
equilibrium is such that the oil just weto the wick
at A. Find the curve ABCD.

Solution by D. Biddlb.

Let ACi, ACj, AC3 represent
the surface of the oU at three
different times ; then in PBi,
PB2, PB3, perpendicular to these
respectively, will lie the centre
of gravity of the oil at those
particular times. Consequently,
the level of the oil, always
flush witii A, and the line join-
ing the centre of gravity with
P, describe equal angles in a
given time.

Taking a wedge-shap^ed por-
tion of oil, of infinitesimal
depth, with its apex at A, and
its base at C|, its particular
centre of gravity will be at
fAOi from A, say at N, ; and,
if Ki be the gentre of gravity
of the mass of oil when at the level AC^, then N|E| wiU be^^tangential to
the locus of the centre of gravity of the mass.

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Let :ri, y^ be the coordinates of the centroid K|, x, y, the coor-
dinates of the centroid E (that is, of the oil when the vessel is full).

Let M a the mass having the centroid E), and a/, y* be the coordinates
of the centroid of AC^D (= 1 — M) now empty ; also let Z DAC| » a.

Then PKisinaM = (x'-ar) (1-M),

and (PKiCoso-PK) M = (y-y') (1-M),

whence TK^ ^ ^^"^'^^^ ^^^^, and PK - cota(«-a;l)-(yl-y)•
co8a sino V v wi

If ABCB were a semi-circle, and P its centre, these conditions would
be fulfilled, but as the oil sank in the vessel there would be no bias to
cause rotation. For the purposes specified in the question, it is essential
that, as the oil is consumed and its level sinks, the layer taken oS. shall
have been unequally divided by the perpendicular from P, the lesser
portion being always on the side next A, so that what remains is heavier
on that side, until by rotation equilibrium is restored.

A very near approach to the curve required is given in the accom."
pAnying fig^e, where r — cos* fl, dr/d$ = sin0/2r, and the area of the
side of the vessel = JAD*. With this curve, if, as may be supposed,
equal quantities of oil are consimied in equal times, an indicator parallel
to AD and projecting from the vessel as if drawn from P, will nse at a
uniform rate, because the sine of the angle DAO lengthens at a uniform
rate. Such being the case, wo have at once not only a convenient light,
but also a time-keeper scarcely inferior to Alfred the Great's graduated
candle.

The positions of E and P, though easy to find by practical methods, do
not yield readily to the integral calculus.

But AL==}[ cos*fl(?0 + J cosflifl,

LE == I 008^0 sin d<^d-i- Q0%9dB.

9386. (Professor Neuberg.) — Si suivant les perpendiculaires abaiss^es
du centre O du cercle circonscnt h, un triangle ABC, sur les cStes de ce
triangle, on applique, dans un sens ou dans 1' autre, trois forces ^gales,
la i^sultante passera par le centre de Tun des cercles tangents aux
trois cotes.

Solution by Professors Genese, M. A, ; Betenb ; and others.
Let L, M, N be the mid-points of the sides, P, Q, R the points of con-
tact of any one of the four circles touching the sides. Then we have

PL- J(PB + PC), &c.;
and if, following Laguebre's principles, we define, for this case, the
positive direction of the sides as that leaving the circle in the positive
sense of rotation, we have

PC + QC = 0,&c., .-. PL + QM + RN = 0.

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Whence, for Boitable direcHonB of the foroee in qneeiion, the sum of their
moments about the contra of the circle is seen to be zero.

[Professor Gsnesb adds that this problem also occurred to, and was set
by, him at Aberystwyth, in 1886 ; and that, if the equal forces be repre-
sented by radii of the drcum-cirde, the lines representing the resultants
terminal at the centres in question.]

9316. (Professor MukhopIdhtAt, M.A., F.R.S.E.)— Prove that (1)
' the locus of the mid-points of the chords of curvature of the conic
4V + fl^2 « a'*2 ig the sextic SiSa-^^c^ + i-ayS = (a-«a;'-*-2y«)l pass-
ing through the origin ; (2) the area of 2i is half the area (A) of the
ellipse ; (3) the envelope of the chords of curvature of the same conic is
the sextic ^2^(a-^x^-^b-^t/^-4)^+27 (a- x2-*-2y2)2 - ; (4) the area
of 2] = f A ; (6) trace the locus 5i and the envelope Sa, and show that
they touch each other and the conic at the ends of the major and the
minor axes.

Solution by Professor R. Swaminatha Aiyab, B.A.

1. The line
a-^a;co8 0— i"V8in0=co820...(i.)
passes through the point (<^) on the
ellipse, and makes the same angle
with the axis that the tangent at
the point does : it is therefore the
chord of curvature at <p. The
diameter conjugate to it is the line
a-»ir8in^ + *-^yco8 0= 0...(ii.);
and, eliminating <p between (i.)
and (ii.), we have 2 as the
equation of the required locus.

2. From (i.) and (ii.), x = Ja (cos + cos 30), y = J* (sin <^— sin 30),

I ydx « \ah J (sin 30 - sin 0) (3 sin 30 + sin 0) dip « \ab f F ^0.
The required area ^ ab I ' Frf0 = ^irab = JA.

3. Differentiating (i.), we have a-ia:sin0 + *-iy cos0=2 sin20...(iii.).
From (i.) and (iii,), a? = Ja (3 cos - cos 30), y « i* (3 sin + sin 30) ;
and eliminating 0, we have 2^.

4. f ydx =« ^ab J (sin 30 + 3 sin 0) (sin 30 — sin 0) (^ » | a& f F (£0 ;
hence the required area of Sj is \ab I Fe^0 === firad = f A.

6. As increases from to 2ir, the locus 2i is traced in the order
PQKSTUVW ; and the envelope Xj in the order ACBDA'EB'FA. The
points (± (1^2, ± bV2) are cusps in the latter curve, the equi-conjugate
diameters of the ellipse being the cusp -tangents. In the first curve the
origin is what might be called a double tacnode.

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8989. (ProfeoBor Wolstbnholmb, M.A., Bo.D.)— In a tetrahedron
OABO, OA - fl, OB = 4, OC - fl ; BC - x, CA - y, AB - 2, and the
dihedral angles opposite to these edges are respectively A, B, C ; X, T, Z.
Having given tiie equations i — y = J (a + a?), «— ««ia— a?, B » Y,
C + Z - 180°, prove that B - Y = 60°, 0-A » Z-X - 30° ; and find
the relations between a, h, e.

Solution by Sbpthcus Tbbay, B.A.
We have x » 2d— a, y ^ by « » « + 26— 2a ; and therefore
BO + CO = AB + AO, and BC-BO - AC-AO ; or (Quest. 8606)

A + Z = X + C, andA-Y = B-X; andsinceB^Y, andC+Z= 180°,

therefore A = B + C-90°, X-B-C + 90°,

and sinAsinX — 1— cos'B— cos^C.

Let the areas of the faces BOO, COA, AOB, ABC be denoted by Ai,
A3, As, A4 ; then

sinB ^ sin BOO ^ ^ Aj sin B ^ sinAOC ^h^ Aj .
sinA ^ sinACO x ' A,' sinX " sinBOO " a * Ai *

.-. sinAsinX = ^.8in2B- 1 -cos^B-cos'C, .•. cosC= ~^sinB.

Also sinB ^ sin ABC _ b A4 ^ sinBCO ^ Aj

sinC sinOBO" « * Ai "" sinACB A4*

Therefore sin' C = ^ . sin* B. These equations give

BinB-— A-=-^, sinC=-*:=^5 ±, (1,2).

b + c—a c + a o + c—a c + a

putting 6 - fl = a.

Now the general relation among the dihedral angles of a tetrahedron is

2 (sin* A Bin2 X) - 22 (cos X cos Y cos Z) - 22 (cos B cos C cos Y cos Z) « 2 ;

which in the present case reduces to

8co82Bco82C-4cos2B-4cos2C + l = (3).

From (1, 2, 3), wo have

48in2B = 2-sec 2C = 3+ ^^-^^ = r^«-

(c + o)2— 2o^ (c-j-o)'

This equation reduces to

a* + ^(c-b)a^-2{7(^+ebc-'b^a^+i{Zc^-¥7bc^ + b^c + bi^)a') ...

= 3A4-4^^(j+106V+12*c3 + 3<H > ^ ^'

which is the general relation among a, *, e.
Thus the least value of B is 60°, which makes

C = Z = 90°, a = b =x = yy and 2a = c V3.
In the other cases we must have

gin B = — > -8660254 < 1.

b-k-c—a

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We have assumed e > b > a. Now b/{b + e^a) is less than el{2e—a) ;
and since e is the greatest value of b, B will be a maximum when
sinB == e/{2c-a). liet b^e^l ; then, from (4), a*-24a« + 48a-24 = 0;
from which we nnd a » '8618704. These values of a, b, e make

B = Y - 60°34'22"-76, and C - 82°36'13"-11.

If a be small in comparison with c, so that c? and higher powers may
be neglected, we find

2b^e^/Z (l+ -+^). and sinB « i^/3 f 1 + j^V

which is not sensibly affected if a = or < 'OOl. Hence any solution
depending upon a small value of a makes A+X =» 120°, nearly; and
since C + Z = 180°, therefore C-A + Z-X = 60°, or C-A=Z-X=30%
sinceC-A = Z-X.
For the maximum value of B we have

a - -8618704

a? = 1-1481296

B = 60°34'22"-76

b= 1

y-1

C«82°36'13"-ll

<; - 1

z = 1-2962692

A - 63°9'36"-87, X = 67°69'9"-66.

[There being 6 equations gfiven, apparently independent, it would seem
that the shape of the tetrahedron most be fixed, but there is certainly
more than one solution. One obvious solution is when a = x — b ^ y^
when it will be found that C'v/3 = 2a=Z, C=Z=-90°, A = X=B=Y = 60°.
The tetrahedron in which

Ca^ 4*8023, x = 4-8044 I satisfies the conditions; and
lb = 4-80336, y = 4-80336 A = 69°69'20"-96 = C-60°,

(c = 6-64538, z = 6-64738 X - 60° 0'39"-04 = Z-60°,

B = Y = 60°].

8701. (A. Russell, B.A.) — Resolve into quadratic factors
(a2_ bcY (b + e)^ {b-e) {fl2 + 2a(b + c) + be}
+ (b^-eay {c + ay {c-a) {*« + 2b{c + a) + ea}
+ (c^-ab)'i{a + bY{a-b){(^-{-2c{a + b) + ab}.

Solution by R. F. Davis, M.A. j Professor Betens ; and others^

Let A = {a^-bc)(b^e), B= ..., C = ... ; so that A + B + C -0.

Then, since B -0 =- (* - c) {a2 + 2a (* + c) + *<?},

the given expression may be written A* (B —C) + ... + ... , which can easily
be reduced to the form -ABC (B-C) (C-A) (A-B). The given
expression (which is homogeneous and of the 18th degree) is therefore
equal to the product (with its sign changed) of nine quadratic factors ;
three of the form b^—c^, three of the form a^—bCf and three of the form
a" + 2a (i + c) + be.

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7759. (Ptofeasor Hanumanta Rau, M.A.) — From one end A of
the diameter AB (a 2a) of a semicircle, a straight line APMN is drawn
meeting the circumference at N, and a given straight line through B at
M, at an angle a ; show that the locus of a point F, such that AP, AM,
AN are proportionals, is the cubic through A,

r as 2a sin^a sec 6 cosec^ (a - 6), or 2a sin^ a (x^ + y^ « (a; sin a— y cos o)',
which, when a = ^ir, iir, becomes

2a2(a;3 + y2) ^ ^^ 2a^{x^ + y^ ^ x^x-^yf.

Solution by G. G. Storr, M.A. ; Rev. T. Gallibrs, M.A. ; and others.
The polar equations of the line and
the circle are respectively

r == 2a sin a cosec (a — 6), r = 2a cos d.

But AP.A:N'=AM2; hence the
locus of P is given by the equations
stated in the Question.

8852. (J. Gbifpiths, M.A.)— If a, jS, 7, 8 be the roots of the quartic

ax^-^^ba^-^Qcx^-^^dx-^e - 0, and if ^ - ^^^^ + ^^l3i • show that

a— 5 i8-5

(2-^)2(1-2^)2(1+^)2 108J2'

where I « ae- Ud +Zc^, J = ad^ + eb^ + (^- ace- 2bed.

Solution by D. Edwardbs ; G. G. Storb, M.A, ; and others.

Let the quartic be linearly transformed into a' (l—nui^) {l—nx'), and,
Bfi in the Fundamenta Nova, let

U-7V = A(l+w^, U-5Va:B(l-f»^, U-aV-C(l+«*;c),
U— /8V *» D (1— n*ic). Putting x « — w-i, +n-i successively in these,

Butwehave {m^^n'^^Umnf ^V_

(m + nf (Z^mn^m^-n^)^ 27J3
(Caylby's Elliptie Functions, Arts. 413 — 14) ; hence substituting for m/n
in terms of q, we have the stated result.

8850. (W. J. Grbbnstrbbt, B.A.) — Prove that the sum of all the
harmonic means which can be inserted between all the pairs of numbers
whose sum is «, is ^ (w*— i).

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Solution by A. W. Cavb, M.A. ; W. J. Barton, M.A. ; and others,
8am -2r!t^ + 2J«-22^3ii^^ -I
L *• *• ** J

-2 [1 + 2+ ... +(n-l)]-2>[l« + 22+ ... +(»i-l)*]
«n(«-l)-i(n-l)(2n-.l)-i(n«-l).

9340. (R. Knowles, B.A.)— In Question 9149, if BD and AC intersect
in O, and CA meet KH in M ; prove that the lineB GM, GA, GO, GB and
LC, LO, LA, LH form harmonic pencils.

Solution by G. G.
MoR&iCB, M.A., M.B.

G{M.A.O.B}

-G{M.A.O.C}

-H{M.A.O.C},

which is an harmonic
pencil, by the known
property of a complete 5
quadrilateral; similarly

L{C.O.A.H}

-H{C.O.A.M}.

8300. (Professor Hanumanta Rau, M.A.) — Prom any point P on the
circle described about an equilateral triangle ABC, straight lines PM, PN,
PR are drawn respectively parallel to BC, CA,
and AB, and meeting the sides CA, AB, BO H

at M, N, and R. Prove that the points
M, N, R are collinear. _d/

Solution by D. O. S. Davibs, B.A. ;
R. Knowlbs, B.A. ; and others.
Since ABC is an equilateral triangle, evi-
dently ANP, PMO, PRO are angles of equi-
lateral triangle. Hence N, A, M, P and
P, M, R, are conoyclio. Join PA, PC,
NM, and MR ; then

ZPMN-PAN = P0B;

hence N, M, R are collinear.

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4251. (Colonel Clahkt?, C.B., F.R.S.)— If A, B, Cbe three circles, B
being within A, and C within B ; prove that the chance that the centre

of A is within C is -.

7

Solution by the Pkoposer.

In the accompanying diagram,
let a, b, c be the centres of A and
of B, C, two circles lulfllling the
condition of which the probability-
is required. It will be convenient
to denote such cirrles by accented
letters B'C. Let (C')a?y represent
the number of circles C within
a B' of radius ijS = y and for
which ab = x\ while (C) the
entire number when B' has taken
all magnitudes and positions.
Then, lastly, if (C) denote the
entire number of C*s, the chance
required is (C')/(C).

It is easy to see that the centre e
of any C cannot fall outside the
ellipse whose foci are a, b and major axis y = radius of B', for this ellipse
is the locus of the centre of those circles which, passing through a, touch
B'. Let c be on an interior confocal ellipse whose minor and major axes
are z and v — {x^-\- «^)* ; then the number of the circles C with centre c is
c^—ca^ y-v. And so, if c be any point in the elementary area \ird {vz)
contained between two consecutive confocal ellipses, the number of circles
C' whose centres are in that area is iir (.v — v) d {vz). This integrated from
« = to « = (y2— a;2)* gives us (C%y. Now we have

[(y-'V)d {vz) = (y — v) vz + J vzdv^
which between the limits, and since vdv =^ zdz, becomes simply \z^ ; that is,

"We may include all the circles B', corresponding to x and y, by multiplying
this by the area contained between the circle whose centre is a with radius
X and the consecutive concentric circle whose area =rf(ira;'). Hence,

between the proper limits, (C) = Jir^ ^^ (y^ -x^)^xdx dy.

Taking unity as the radius of A, x < \, and y > do and < 1— a?, that is,

{^')^is'^^^^^' (y^-x^^xdxdy (o),

the y integration is to be first effected. After some reduction and a
substitution 1 — 2x=u^, we get, after the second integration, (C) =ir2/7-l80.
It is, of course, inmiaterial which integration is performed first. If we
commence with the x integration, the integral (a) is transformed into

VOL. XLIX. H

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C2

The first doable integral embraces all circlos B' for wbicb the radins is
less than i ; the second, those having a radius greater than 4 ; the separate
values are easily found to be

180 2« 180 \ 7 2« y

180'

180

But (a) is most easily integrated by taking two new variables such that
1-^2 «a? + y, ij-v/2 « y— x. Thus we have, instead of (o),

(CO =*«»£'""' J* ({!,»- {«,4)^{rf, = t-r=£'''^*(f-«i''Q.

giving immediately the result as before, viz., n^ 17 'ISO,
Finally the number of C*s within a B of radius y is

2ir J r rfr (y — r) = Jiry* ; thus we have

Jo Jo Je Jo

This gives the required chance = — .

[Assuming that in each case
the position of the centre is first
taken, and then the radins, within
the proper limits, Mr. Biddlb
proceeds to find the required proba-
bility as follows : —

** Let O, Q, V be the centres of
A, B, C respectively ; let OD = 1,
OQ«ar, QT(-QE)-y,
QS(=QF)««.
Then we readily discover the follow-
ing limits for success : —

Bero<a;<J, a;<y<(l— a?),
zero<«< J (x + y).
Make OS = ST = EF - y-«,
and join QV (« QS) ; also join OV, and let Z OQV = ^. Then we have

the further limits, zero< *< SQO ( - cos-' ^'^^-^^''^^ ^

\ 2xz J

and OV < « (s rad. of C) < (y — «), which may be rendered

(x« + «'- 2a« cos 4»)* < w < (y - «).**

By this process he arrives at an integral very troublesome to evaluate.]

7986. (J- Brill, B.A.)— ABCD is a quadrilateral, AB and DO when
produced meet in E, and AD and BO when produced meet in F ; prove
that AB . CE . DF cos (ABD + CEF + OAF)

+ AD . OF . BE cos ( ADB + CFE + OAE)
- BO . AF . DE cos (OFE + ADB + DCA)
-OD.AE.BFcos(OEF + ABD + BCA) « AO.BD.EF.

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Solution by A. Gobdon ; Professor Nash, M.A. ; and othert.

Let t, y, k be unit vectors along 0-c, Oy, Oz, and a, /3, y, J, &c. unit
yeotoTS in the plane of ary, such that

a a t cos a +j sin a (inclined at a to Oj;, &c.), j3 » t cos fi +J sin fi, &o. ;
then 0/3 = cos o - i3) + A: sin (o— jS), &c.,

alfi.jl^ « cos(o + 7-i8 — 5) r A; sin (0 + 7-^-8) — a/5. 7/^8;
hence, if

oi = /i/8 + ^7 + y + ..., a,r=i»ia + »«i7 + «»88 + ..., OJ«f•li3 + «^7 + '»^'^ •»
and 0, ^, 4^, ai, a,, 09, jS, 7, 8 ... are all vectors in the plane of ^y, we
have aje . 02/4» . ag/if^ = «i/<^ . «4/e . oj/if^ - &c. - ^Jli;" iS/fl . y/<p . 8/if^,

where 2 constitutes a summation of terms each formed of the product
of 3 quaternions, such as li$l$ . fn^yl<t> . nj^l^ (one from each vector),
(the three numerators being the same in none of the products).
Let large letters denote vectorSf small letters lengths, so that
AB » — BA, but ab ^ ba; then
, _ AF -AE AF AC_AE AO
EF " EF * AO EF * AO
AF.AC-hAF.AE-AE.AF-AE.AC
EF.AC
_ AE.CF~AF.CE AE /BF-BC\ AF /CD + DEX
** EF.AO ""acV EF j AOV EF /

AE.BF.CD-hCE.DF.BA-i-FO.BE.AD + FA.DE.BO . ,
" AC.EF.BD *

, ^ BA CE DF AD BE FC DE FA BO AE CD BF^
* BD * EF * AO "^ BD * AC * EF ^ AO ' BD * EF BD ' EF ' AO '
Also, by Ptolemy's theorem, CD . BE . AF = CE . BA . DF,
and BO . AE . DF = FO . BE . AD ;

therefore 1 - ^ . ff . ^' {cos (ABD + OEF + FAC)

^ ""-^ ""^ +Asin(ABD + CEF + FAO)}+....
Hence bd . ef. ae^^ba.ec. df. cos (ABD + OEF + FAC),

the result required,
and =« J^a.^^J.^C/'sinCABD + OEF + FAC).

8771. (W. J. G&BENSTBEET, B.A.) — Provc that the Beriae
U g sina l^i -I- J^sin^a + J^sin^a + ^]'^^'^^^ Bin«a + ...| » tan^a.

Solution by Prof. Ionacio Bey ens ; B. Enowles, B.A ; and others,

ff «cosa[i+ L3 8in»«+ 144 sin^«+...]
da C 2.^ 2.4.6 >

= -r-~ { (1 - sin^ a)-* - 1} = J sec' ia ; therefore m = tan |a.

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9327. (F. R. J- Hbkvby.) — The point O is fixed, P describes a
straight line A ; OP and a line T passing through P rotate uniformly (in
the same or in contrary senses) with angular velocities as 1 : 3, and be-
come simultaneously perpendicular (or, in the limiting position, parallel)
to A. Show that the envelope of T is a cardioid.

Solution bf/B^F. Davis, M.A. ; G. G. Sto&b, M.A.^ and others.

Let the circle (centre R)
roll upon an equal circle
(centre C), U being the
point of contact and CURV
the common diameter
through U. Then, if Q be
an invariable point on the
moving circle, QV and QU
are the tangent and normal
respectively to the cardioid
described by Q, whose
cusp is at E, the point
passed over by Q when in
contact with the fixed circle.
Since arc UE » arc UQ,
angle

UCE = UEQ = 2UVQ ;
so that, if CP bisect the angle UCE, CP = PV. Draw PN, PM perpen-
dicular to CE, CR respectively, and take CO = CV = constant. Then the
angles OPN, CPN, CPM, VPM are all equal, and VP may be conceived
as having revolved from PN through an angle three times as great as that
revolved through by OP from PN in the opposite direction.

8737. (Professor Mukopadhyat, M.A., F.R.8.E. — Extension of
Qut^stion 8107.) — If 0, 4>, ^ be the angles of inclination of any two tan-
gents to a conic, and of their chord of contact, to a directrix, show that,
if « be the eccentricity of the conic,

^^ A-^ sing + M-^ sin » .,_1-A».1-m2

A-*COS6 + fi-*CO8 0'

e« =

sin' sin'

Solution by R. Enowles, B.A. ; Professor Matz, M.A. ; and others.
Let (x^y ^i), (A, k) be the coordinates of the ends of the chord and its

pole, then tan ^ = ^,
b*h

therefore

sin'tf*

coi9 *» +

a'y;

*«x,

COt0 B

fl2#/2

b'^ifi^-e^x^)^'

and, since ^ =

1-A«
sin'fl'

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therefore \ = , and u = -.

and A-Uine + A*-'8in» _ a«(y, + y,) _aaAr_^^.

9353. (Professor Asutosh MukhopAdhyAy, M.A., F.R.S.E.)— Points
D, E are taken in the sides AB, BC of any triangle ABC, such that
BD = m . DA, BE = #1 . EC. If O be the intersection of AE, DC, prove

that C0^,n+1 ^^ AO^^l

OD « OE m

Solution by R. F. Davis, M.A. ; Professor W. P. Casby, M.A. ; and others.

The point O is (by hypothesis) the centroid of masses m, 1, n placed at
A, B, C respectively. Whence, &c.

9089. (Emile Vioari^.) — Par leg sommets A, B, C d'un triangle on
m^ne des parall^les aux c6te8 opposes qui rencontrent le cercle circonscrit
en A', B', C Les droites A'B', A'C, C'B' rencontrent respectivement
AB, AC, BC en a, jB, y. Demontrer que Torthocentre du triangle atiy
est le centre du cercle ABC.

Solution by R. Enowles, B.A. ; Professors Matz, M.A. ; and others.

The equations to AA', BB', CC and the coordinates of A'B'C are
respectively by + ez^O, ax + ez='0, ax + by»0; 1/a, [c^ — b'^jaHy
(b^-c^)la^e; {e^-n^/ab^, l/b, {a^''C^)l{b^c); {f^-a^)lae\ {a^-^b^/b^c, l/c
(omitting 2a in all the coordinates) ; hence we find the equations to A'B',
A'C, B'C, and also the coordinates of a/87 • —

{a^'-<^)la{a^^b^, {(^--b^) I b {a^^b^), 0;

(^-«-)/a(c3-fl2), 0, {<P-b-)lc{e^-a^); 0, {l^-^a^) j b {l^-<?),

(a^-c^)lc{b^-'<P);

therefore the equations to afi and the perpendiculars from 7, fi on ai9, ay are

a {c^-b^) X + b (c^-a^) y + c (a^^b^) z = 0,

a [(^2 + ^2) cos A-*<?] a; +* (aS-^j cos Ay + <? (a2-^ cos A? = ....(1),

a(b^-c^)coaBx + b [(a^-^c^ cos B- ac]y + c{b'^'' a^) cob Bz ^ (2),

and (1), (2) are each satisfied by RcosA, RcosB, RcosC; hence the
orthocentre of the triangle a/87 is the centre of the circle ABC.

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8667. (N'Importb.) — Two equal perfectly elastic balls, moving in
directions at right angles to each other, impinge, their common normal
at the instant of impact being inclined at any angle to the directions of
motion : show that, after impact, the dii^ctions of motion will still be at
right angles.

Sohttim by F. R. J. Hbbvbt ; Rev, T, Gallibrs, M.A. ; and others.

Let OA, OB represent the velocities, either before or after impact, and
C be mid-point of AB. Then OC, velocity of mass centre, is invariable ;
and so (elasticity being perfect) is magnitude of relative velocity, or length
AB. But, if AOB be a right angle, length AB « 2 x length OG ; and
the converse.

9360. (B. Curtis, M.A.) — A tetrahedron ABCD is circumscribed to
an ellipsoid, and straight lines are drawn through the centre from the
cornel's to the opposite sides meeting them in X, Y, Z, W ; show that

ox oy oz ow^
xa"*"yb zo"*^wd

Solution by J. O'Btrnb Cb.okb, M.A.

Let Pj, jp„ Pa, P2i P3, Pit P4, Pa* «i, «2» *i> *4 ^ t^© parallel perpen-
diculars from the angles of the tetrahedron, and the point O, and the areas
of the sides on which they respectively fall ; then

Pi'i +P^3 +i'3*3 +-^4*4 = 3 times vol. of tetrahedron:* Pi»i = P2«2= P8«3= P4«4.

Therefore '&- + $ + ^+'&==l> whence the result.

Fi Fa Fa Fi

8270. (^' Edwardes.) — Let ABC be an acute-angled triangle, and
L, M, N the points where the angle bisectors meet BC, CA, and AB re-
spectively. Prove that (1) the circles ALB, ALC cut one another at an
angle A, the circles ALC, ANC at an angle ±| (0— A), and the circles
ALC, BN<^) at an angle 90°- JB ; (2) the centres of the pair of circles which
pass through L are equidistant from the centre of the circle ABC, and
similarly for the other two pairs ; (3) if p^, p'^ be the radii and Sj^ the
distance between the centres of the circles wMch pass through L, and
similarly for p^, p'^, &c., p^ />m Ps = p'l Pu p'n ** 'l^m^n *» (^) ^ ^1 ^® ^®
distance of the circle ALB (or ALC) from the centre of the circle ABC
(radius R), and similarly for d^, rf„ R^ - R {d^d^ + d^^ + d^i) — 2d^d^2=^ J
(5) if the base BC and the circum-circle BAC be given, the envelope of
the line joining the centres of the circles ALB, ALC is a parabola whose
focus is at the centre of the given circle and latus rectum 4Rsin3 ^A.

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Solution by Professor Swaminatha Aiyar, B.A.

(1.) Let 0, P, Q be the circum-centres of ABC,
ALC, ALB ; then OQ, OP are at right angles to AB,
AC respectively, therefore QOP and BAG are
supplementary, also

l AQO = ALC ; and I APO = ALB.
Therefore QOP and QAP are supplementary and
a circle passes through Q, A, P, 0. Also
Z QAP = BAG. Again, AP : AQ = AC : AB,
and aQAP is similar to ABAC, and the straight
line AO bisects the angle QAP.

B

(2) Therefore QO, OP, the chords of the circle QAPO, are equal ; that
is, P and Q are equidistant from 0.

(3)

Ag.AP^^.^j^^j^ P^^Pj,^

AB

AC BC

Similarly,

and ?N^?N^?N.

Therefore

a e a c

(4) AO.QP«AQ.PO + AP.QO = (AQ + AP)//i.

R

a

Therefore

Similarly A = ^^f and -3^ =
R3 b-k-e e + a a + b

did^d^

b + e
a

b
e + a
b
R
^1

Therefore
a + b
e

a b e rf, ' '

^ ' dz

(5) If R be the middle point of PQ, OR is at right angles to PQ.
The distance of R from BC = } the distance of P and Q

- i (LCcotJA + LBcotiAJ _ j^^^^j^

The locus of R is thus a straight line parallel to BC. Therefore PQ
touches a parabola whose focus is O and whose tangent at the vertex is
the locus of R. •

Its latus rectum = 4, distance of O from the locus of R

= 4(JacotJA^JacotA) = ataniA=: 4R8in«iA.

8540. (Rev. T. R. Terry, M.A.)— Show that the series
1 ,ffl g , fn(w+l) q{q + r) ^ w (fn + l) (w-t-2) y (g + r) (y ->• 2r ) ^
p 1.2 p{p + r) 1.2.3 j»(P + r)(i? + 2r)

is convergent if jo > q + mr.

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Solution by the Proposer ; Professor Nash, M.A, ; and othert.
Denoting the hyper-geometric series

«jB^^ ai[«4j]3(^+J)^
^1.7 1.2.7(7-»-l)

by F {a, i8, 7, a;}, it is well known that F {o, jB, 7, 1} is convergent if

7>o + i8. Now the given series —FJtn, -^, =^, 1 y, and is there-
fore convergent if i? > g + mr.

7244. (D. Edwardes.) — The circles of curvature at three points
of an ellipse meet in a point P on the curve. Prove that (1) the normals
at these three points meet on the normal drawn at the other extremity of
the diameter through P ; and (2) the locus of their point of intersection
for different positions of P is 4 {a^x^-¥bhf^ = {a^^b^f.

Solution by the Rev. T. C. Simmons, M.A.
Let a be the eccentric angle of P', the other extremity of the diameter,
that of P being of course ir + a, and let i3, 7, 8 be the eccentric angles of
the points of contact of the circles of curvature, then

a;/<icos^(ir + a + i8)+y/*8ini(ir + a4iB) = cos^ (ir + o+ jB),

or ir/a8inJ(a + /S)-y/*cosJ(o + /3) =* 8inJ(a-i8) (1),

the chord joining P with i8, and xja cos ^ + y/i sin jB = 1 (2) ;

the tangent at i9, must make equal angles with the axes, therefore

sin J (o + /3) sin /3— cos i (a + jB) cos jB « 0,
or o + 3iB - (2m+ l)ir.

Similarly, o + 87 =» (2« + 1) ir, a + 35 = (2^? + 1) » ;

therefore a + i3 + 748i8an odd multiple of ir, or the normals at a, jB, 7, 8
are concurrent.

Again, if a, jB, 7, 8 occur in this order, since their differences must be
even multiples of ^ir, it is evident that, when they are unequal, iB — 7,
7—5 each = fir ; in other words, jB, 7, 8 are at the vertices of a maximum
triangle in the ellipse. Consider now the normals at jB, 7 : they are

2<w;Bini8-2*ycosi9= {a^-h^%m2fi (3),

and Ux sin (/3 + fir) - 2by cos (/8 + Jir) - [a^ -lP)%m (2/3 + |ir),

which reduces to 2ax ( \/3 cos /3 — sin 0) + 2by ( >v/3 sin /8 + cos 0)

« _(fl2_i2)(8in2/8+ >v/3 cos2iB) (4),

whence, by (3) + (4), 2aircosiB + 2iysini9 «-(a2-i2)cos2/8 (6).

Squaring and adding (3) and (5), we obtain for the locus of the inter-
section of the normals 4 (aV + V^y^) = (a^^h^^

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9418. (Professor Sylvestek, P.R S.)— If /?, i^j are each prime num-
bers, and 1 +j»+j»'+ ... +y-* = ^, prove that j is a divisor of q — i.
Example : 1 + 3 + 32+ 3=» + 3* = 11*, and 2 is a divisor of 11-6.

Solution bt/W.8. Foster.

therefore $>-' +^'-» + ... +y+ 1 « -^— (1 +i?+i^+... +i?*-2),

a.ndp is a prime number ; hence, as in Question 9381, y divides p—l, or
J = p and divides q~l. In the first case, p " AJ-^-l; therefore

1 +(A;+ 1) + {Aj + 1)«+ ... + (A; + l)-i - q^ ;
therefore i + By = qj ; therefore B;' = q'—q + q—iy

and, sincey is a prime number, q^—q is divisible, by j\ therefore q — iis
divisible byy. In the second case, we should have

l+i^+Z^+.-.-CQ^ + l)"- 1+C,i?« + C,i^+...,
which is impossible if jn is a prime number.

9423. (Professor Neubeho). — On casse, au hasard, deux barres de
longueurs a et ^, chacune en deux morceaux. Quelle eat la probabilite
qu'un morceau de la premiere barre et un morceau de la seconde, 4tant
juxtaposes, donnent une longueur moindre que c ?

Solutions by (1) Professor De Wachteh, (2) Professor Schoutb.

1. Assume two rectangular axes, OX,
OY ; take OA = a, and OB = b, and
describe the rectangle OAMB. From
any interior point if perpendiculars be
drawn to OA and OB, they may repre-
sent, in one of the possible combinations,
the parts of a and b to be added
together.

The amount of possible chances will
be represented by the area OAMB and
measured by ab. On OX and OY, re-
spectively, make OC = OC = given
length c, and join CC. The sirni of the
distances of any point in CC from the axes is = c. First of all, we must
have a + b > c^ if CC is to cut o£E a part from OAMB. The required
probability P is the ratio of the area limited by the axes to the rectangle
OAMB. Draw AA.' and BB' parallel to CC Assuming a + b > c and
b > a, three cases are possible.

(1) b > a > c. CC falls between AA' and 0, and P — c"l2ab.

{2) b > a. CC lies between AA' and BB', and P = {2c -a' 12b.

{^) b > a, CC falls between M and BB', and P = 1 - (a + A - cY[2ab.

TOL. XLIX. I

c

\

8

'\ M

\. N

C

S X

\

A

\^^ ^\

^N

, \

a \

C

\^ '^^

^\

\ \. '

V

\ \

X V

C A

C\ C

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2. Otherwise : — Si-' Von repr^ente les longueurs des deux morceaox
qu*on reunit par x et y, on a lea conditions generales < 2^; < a et
a < 2y < b, tandis que les cas favorables sont soumis a la troisieme con-
dition x-^y < c. Ea considerant x Qi y comme les coordonnees rect-
angulaires d*un point dans le plan, on trouve pour la probabilite en
question d^aprds les figures suivantes :

k.

pour c <\b (Fig. 1) P = 4r/2fl*,

pour \b<e< Ja(Fig. 2) P = [4c2-(2c-*)2]/2a*,

pour \a< c< J(a + *)(Fig.3) P = [4c2-(2(?-ft)=i-(2c-a)2]/2a*,

pour i(«+*) < ^^CFig- 4) P = 1.

On a suppose a > b\ quand a = i, le cas de Fig. 2 disparaJt.

9406. (W. J. Barton, M.A.)-Show that, if R = 2r, the triangle is
equilateral, without employing the expression for the distance between
the centres.

Solution by Professor Emmerich, Ph.D.
If R « 2r, the in circle is equal to the nine-point circle. After
Feuerbach*s theorem, the incircle of each triangle is touched by the nine-
point circle, and it may easily be seen that the incircle lies entirely inside
the nine-point circle ; for, if the bisector AD of the angle A meets the
circumcircle at E, the incentre T lies between A and E ; therefore the
projection of the point T upon BC lies between the projections of the
j)oint8 A, E ; that is to say, the first projection, which is a point of the
incircle, lies inside the nine-pcint circle. Therefore, if R = 2r, the two
circles coincide. Hence the sides are touched in their mid-points by the
incircle, etc.

1898 & 4043. (Hugh MacColl, B. A.)— Find the number
situation of the real roots, giving a near approximation to each, of
a^ + 4-37162a:3-24-S64235876U2+ 34-129226840859882^

-14-63442007818570452204 = 0.

and

Solution by D. Biddle.

The ordinary 7 -figure Tables of Logarithms do not permit of a near

approach to accuracy in this investigation. But we can proceed thus : —

Rendering the equation x^-^-ax^—bx'^-^cx — d'^ (I),

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form another equation thus : {x^ + yx— s){a^—uz + r) =- (2),

in which y— « = a, uy-i-z—v « i, uz-\-vy= <?, and vz = rf, whence v = iilz^
y = {az^-¥cz)l{^-\-d), and m = (cz—ad)/{z^ + d). Then find any one real
root of X from the original equation. A very good method of effecting
this, in equations like the present, is to separately record the portions of
X as they are found, and amend the coefficients «, A, c, rf, so as to form an
equation similar in kind, hut of which the remainder of x is the unknown.
Thus, let ArtS the last portion of x found, consisting hy preference
of only one figiire, and rf„+i s the error resulting from it; also let A =
the addition necessary to raise (Ai + Aj + ... A„) to x. Then

A* + a.Ai-*„Ai + rnA,-rf„ = -e/„,i (3),

(A„ + A)4 + a„(An + A)3-*„(A^ + >i)2 + ^(AH + A)-rf„«0 (4);

expanding (4) and suhtracting from it (3), we have
A*+ (a„ + 4A,.) A3-(*,.-6Ai-3A«fl„) A«

+ (c + 4A?, + 3A?.«„-2A„*„) A-^„*, = (5),

which may he rendered A^ + fl'n+iA'— *»^i A2 + e?,»+i A — (f„4i = (6).

This is a condensed form of the old method of extracting roots of the

fourth power, and is scarcely to he surpassed for equations such as the
present untU full logarithmic tahles to 24 places of decimals are provided.

A a b

1-000,000,0 4-371,620 24-964,235,876,1

-100,000,0 8-371,620 6-849,375,876,1

-100,000,0 8-771,620 3-277,889,876,1

-010,000,0 9-171,620 0-586,403,876,1

•006,000,0 9-211,620 0-310,655,276,1

•001,000,0 9-235,620 0-144,630,116,1

•000,900,0 9^239,620 0^116,929,256,1

•000,080,0 9-243,220 0-091,981,472,1

-000,001,0 9-243,540 0089,763,060,9

-000,000,9 9-243,544 0-089,735,338,4

1-217,981,9 = oTj

d

34-129,226,840,859,882 14-634,420,078,185,704,522,04

1-315,615,088,659,882 0-097,809,113,425,822,522,04

0-400,888,513,439,882 0016,269,743,320,834,322,04

0-012,459,138,219,882 0-000,088,170,737,846,122,04

0-003,486,546,697,882 0-000,013,038,123,257,302,04

0-001,617,538,344,682 0-000,001,311,427,089,610,04

0-001,355,988,972,482 0-000,001,130,953,285,928,04

0^001, 147,764,455, 162 0^000,000,093,452,567,496,24

O^OOl, 133,244,891,498 0^000,000,002,215,359,935,12

0-001,133,065,393,107 0-000,000,001,082,204,797,44

The final remainder = 0-000,000,000,062,445,936,90

Having thus found that Xi =« 1*2179819, we revert to (2), whence we oh-

tain x^ + yx-z ^ 0, and x^-^tix + v ^0 (7,8).

Expressing (7) in the terms given under (2), we have

z^-(ax + x^ z'^''{cX''d)Z'^dx^ =^ (9),

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or s*-6-8080339445656l23-26*93434047497581s

-21-70986819346878 « (10),

which has two real roots, namely «i « 9*7874906, zj = - 1-4846695.

iVo«(7),(8). .-=.f..}("^)'j'-i^fi^) (.1).

"-b('-^:y-~M'-^) (->•

Taking «i, we obtain, from (11), ^i = 1-2179819 and r, = -8-0352616,
and from (12) the two imaginary roots l-2227875±(- -0000075)'.
Taking z^, we obtain from (II) two more imaginary roots,

l-2l84609db(--0000225)*,
and from (12), X2 — -803.32616 and ^3 « 1-2266901, which is quite dis-
tinct from Xi.

For the service of those who may wish to carry the investigation to a
great degree of nicety, the logarithms of a, ^, <;, d are here given to 24
placet of decimals, by aid of Peter Gray's admirable method : —
log a - 0-640612401175296337879050,
log* - 1-397318277386039907465102,
loge; - 1-533126449938424905356617,
logef - 1- 1653755 17215043330021209.

9392. (Professor Gbnese, M.A.) — If the tangent at any point P of a
folium of Descartes meet the tang nts at the node in X, Y, and the curve

11 3

again at Q, then prove that p^ + p y * pH-

Solution by Professor Wolstenholme, M.A., Sc.D.

The equation of the curve being a^* + y* « oay, any point P may be

taken, x « — -, y « - — - ; and if any transversal px + gt/=^a meet this
1 + ^ 1 + ^

in three points, the values of t at the three points will be given by

pt-^qfi ^ 1 + t^i if t^t^^i be the three ^1^3^ = — 1. Hence, if ^, ^ be the

values at P, U, iH' ■» — 1 ; also the equation of the tangent at P will be

X {2i -'t*)-¥y (2^'- 1) — aflf and if be the origin, and PR the harmonic

mean between PX, PY, the equation of OR will be y = — tx. Now, the

coordinates of a point dividing PQ in the ratio m : / will be in the ratio

m mn , It mt' It* mt^ , U mt* ,

l + <a 1 + <'8'1-H^3 i + t'i* ' i + ^'*"^«_i • 1 + ^3 ^_i'

which = t {/(^-l) + m} : /(^»-l)-»t^»; if m = 2/', this ratio is -t : 1.
Hence PR = ,PQ, and j^ . ^i^ = X . ^^.

The pedal of the parabola y^ ^ iux with respect to the point ( — 3ff, 0) is

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an orthogonal projection of the folium, and hence the property will be
true for this pedal, hut I find on trial it is not true for any other pedal of
the parabola. In this pedal, the circular points are inflexions ; as, in the
folium, the three points at infinity are inflexions.

8503. (N*Importe.)— A rod of length flr\/2 rests in equilibrium
in a vertical plane within a rough sphere of radius a, one extremity
of the rod being at the lowest point of the sphere ; show that the
coefficient of friction is >v/2 — l.

Solution by Gboboe Goldtuokpb
Storr, M.A.

Let AB be the rod, O the centre of the
sphere, /* the coefficient of friction, and W
the weight of the rod ; then, resolving ver-
tically and horizontally, and taking moments
about B, we have

/iRa + Ila=Wja,
whence /*=» v'2— 1 = -4141 nearly.

9436. (W. Gallatly, M.A.) — AB is a mirror swinging on a hinge
at A . At C is a candle flame, and at D an observer ; the line ACD being
perpendicular to the axis of the mirror. Find geometrically the position
of the mirror, when the observer at D sees the image of the flame on the
point of disappearing.

--.£

Solution by Professor Schoutb.

Let the ray emitted by C, that would
be observed at D when the mirror in the
arbitrary position AB was long enough, fall
on the produced mirror in E. Then AE is
the external bisector of angle E of triangle
CDE. Therefore EC/ED = AC/AD. This
proves that the locus of E is the circum-
circle of the triangle AEF, EF being the
internal bisector of angle E. So the two
limiting positions of the mirror are given by
the joins of A with either of the points P, Q
common to this circle and the circle with A as centre and AB as radius.

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6911. (W. R. Westropp Roberts, M.A.)— Let H and H' be the
Hessians of two binary cubics respectively, B their intermediate co-
yariant ; then, using the notation ot Salmon, proye that
9e'-36HH'-6PJ + H(6J).

Solution by D. Edwardbs.

"With the notation of Salmon's Higher Algebra^ Art. 216, the sources
of'H, H', e, H (6 J) are respectively,

flkJ-62, aV-*'2, ac'-\-ca'-2hb', Z&ia^^-^a^),
Now 4 (ac-l^) (aV-*'2)-(a<?' + aV-2W)2

=« 4 {be') {ab')^{eay « 4 K-JP) Oo-W ;
hence, for the corresponding covariants, we have the stated result.

9369. (W. J. C. Sharp, M.A.) — Prove, from the theory of com-
binations, (1) that i+i?L.iL + '!Li!?iz:Il.!Ll«=ii) + ...=.fi^^
^ ' 11 1.2 1.2 mini

must be true ; and (2) deduce that, if (m) be a prime greater than («),
(m + «) ! — w! nl and ^ — ^ are respective multiples of (w^), (i/i).

Solution by Professor Iqnacio Beyens.

1. Designant par Cm le nombre des combinaisons de {m) lettres prises (»)

^ («), on aura d'abord ^^±^' == C«+» et aussi,
ml nl

Cm + » = Cm . C + Cm • C» + Cm . C/* + . . . + Cm CJi,

mais Cr* = Cl, Cr* - Cn,

done 1+ ^ . ^ + ^i!^i^.^i!!:i±) + ... «fei±^!,

111.2 1.2 ml nl

2. De cette Equation on deduira

(m n . m(m—l) «(w— 1) . \ . i /_ . m t i
-~ . — + —^ — -—!■ . — ^ — p-^+ ... ) m! ft! = (m + «)! — *n! «!,
1 1 1.^ 1 , Z J

et si (m) est un nombre premier plus grand que («) est Evident que le
premier membre est toujours multiple de m ,m «= w^, done

(m + «) !-m ! n ! « M . (m^) ;
et de la mSme relation on a

,uifl + ^ M_ »«(m--l) «_(n-l) \ (m + w)2
■ V 1*1 1.2*1.2 •*; nl '

done -^ — J — * est aussi multiple de m.

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9149. (Charlotte A. Scott, B.Sc.) — If ABCD be a quadrilateral,
in which the sides BA, CD meet towards A and D in H, and the sides
BC, AD meet towards C and D in K ; and if from a point L in HK,
LAG, LFC be drawn meeting BC in G and AD in F, respectively ; show
that BF and GD meet in HK.

Solution by D. 0. S. Davies, M.A. ;
G. G. MoRKiCE, M.A. ; and others.

The triangles ABG, CDF are co-
planar. Hence the diagonals of the
quadrilateral ABCD, AGCF pass
through same point. Therefore the
triaugles AGD, CFB are copolar, and
therefore coplanar. Therefore GD,
BF intersect in HK.

9414. (R- W. D. Christie.) — If 2^—1 is a prime, show that p is also
prime. [Better thus : — What prime p will make 2'' — 1 a prime ?]

Solution by Professor Ionacio Be yens ; D. Watson ;

Si ( p) ne f usse pas nombre premier, supposons p = np'
2P-1 = 2"»»'-l « multiple de (2»»-l) = multiple (2?'-
serait pas de nombre premier, ce qui est contraire k V
Be YEN 8 adds the following generalization : — * * Si (« + 1 )p —
premier, Texposant {p) est aussi un nombre premier, parce
pas de nombre premier, soit p = mp' ; alors (» + !)''—«'' =
serait un multiple de (m + 1j»^-«p') et de (w+ l)"*— «"».'*
remarks that the solution shows that p is a prime is a
tion, but not that it is a sufficient one, which is the real
feet numbers.]

and others.

; alors on aurait
1), et 2P-1 ne
enonce. [Prof.
nP est un nombre
que si {p) n'etait
: (w+l)'"P'-w»«P'
Mr. Christie
necessary condi-
problem in per-

9164. (Professor Nilkantha Sarkar, M.A.)— Prove that

— c« co«* Bin (e sin x) sin nx dx « — -.
» Jo « !

Let

Solution by D. Edwardes.
Sf = 1 + c cos a: + — cos 2a: + —^ cos 2x + &c.,

c^ -:„

S, = csina;+ -— sin2a; + &c., and^* — — 1.

Then So+^'S, = €'^(co8*+> .inx) „ ^ccos* |cos (c sina:) +ysin {c %\nx)} ;
therefore S, = c*" «<>•' sin {c sin x) .

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Hence, substituting the series for €*«»•' sin (tfsin ^), the integral reduces
to — . — ^ sin' nxdx ^ — -.

9427. (Professor Genbsb, M.A.)—If A, B, C, D be points in a plane,

prove that HC.AD ^ CA /BD ^ AB.CD

^ 8m(BAC-BD0) 8in(CBA-CDA) 8in(ACB-ADB)'

where any angle BAC means the angle through which AC must be
turned in the positive sense to coincide with AB.

Solution by Professors W. P. Casbt, Matz ; and others.

Describe a circle about ABD, produce ^^ "2:^^

AC to H, and join DH, BH.

Then Z HDC - I BAC - I BDC,
and Z CBH « Z ACB - Z ADB,

but CD/CH = sin DHC/sin HDC,
and CH/BC =« sin CBH / sin CHB ;
thus

CD/BC « sin DHC sin CBH/sin HDC . sin CHB,
but sin DHC/sin CHB - AD/ AB,
therefore CD . AB / BC . AD - sin CBH/sin HDC

= sin (ACB- ADB)/sin (BAC -BDC).
Again, Jiake Z CDR=» Z CBA, and therefore Z ADR= Z CBA- Z CDA.
Produce BC, DC to V and S. Join VS, SB, and AR, and from the
similar triangles of this figure, after a little reduction, we get

BC . AD/AC . BD - AD . CD/CH . AR = sin HDC /sin ADR,
or BC . AD/sin (BAC-BDC) = AC . BD/ sin (CBA - CDA).

9391. (Professor Satis Chandra Ray, M.A.) — If the diagonals of a
cyclic quadrilateral ABCD intersect in ; and if AB = «, BC — = *,
CD = <?, DA « rf, Z AOD = ADB ; prove that

{be + arf) {ed + ab)/ (ac + bd) ^ a*.

Solution by J. Young, M.A. ; G. G. Storr, M.A. ; and others^

For all cyclic quadrilaterals — - — =» ■ « -— ,

AC Du AO

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and in this case d^AO; hence (^<^+f^{f+'^^) « ,

AC . BD
and AC.BD - ac + bd.

9380. (Sarah Marks, B.Sc.) — Tangents are drawn to a parabola
from a point T ; a third tangent meets these in MN ; prove that the polar
of the mid -point of MN and the diameter through T meet on the
parabola.

Solution by C. E. Williams, M.A. ;
R. Knowlbs, B.A. ; and others.

The diameter through T is TQV, QO
the tangent at Q, meeting MN at ; Mm,
L/, Oo, Nn diameters; then w, «, o, V,
are the mid-points of P/, P7, rV, PFj
therefore o, are the mid-points of #w«,
MN ; hence the polar of O, the mid-point
of MN, is QL.

8826. (Professor Sircom, M.A. Suggested by Question 2845.)-

Show that l+-|.a;2+|^

6 3.6.7

x{l-x-)^

Solution by the Proposer ; Professor Chakravabti, M. A. ; and others.
The differential equation satisfied by the given series is found by the

usual methods to be (jp— a:^) dy/dx + ( 1 - 2x^) y « 1 ,

of which the solution is ar (1 — a:^)* y = sin-^ a; + C,

which is satisfied by the given series if C =» 0.

9384, (Professor Bordaob.)— Show that the roots of the equation
(a; + 2)«+2(x + 2)A/aj-2a;-3>v/ar-46 = 0are9, 4, | {-.13±3(-3^)}.

Solution by Eleanor Bobinson ; G. G. Storr, M.A. ; and others.
This is (a; + a;* + 2)2- 3 (a: + a;* + 2) « 40, whence a; + a;* + 2 — 8 or —5,
a;* =-3, +2, i{-l ±3 (-3;*}, and ;p « 9, 4, J{-13 =F 3 (-3)*}.

VOL. XLIX. K

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9371. (J. Brill, M. A.)— Prove that in any triangle, n being a positive
integer, a* cos «B + b** cos « A

« c» - nabc^ -2 cos (A- B) + li|zi) aSJV -* cos 2 (A - B)

-^(^-^H^-^)a3y^-6co83(A~B)
3!

H.^(^-^)^^7^'(^"^)a4^V-Scos4(A-B)-&c.

Solution by Professor Sircom, M.A. ; H. Fortey, M.A. ; and others.
We have a cos B + * cos A = <?, a sin B— i sin A « 0, whence

Now the sum of the «*^ powers of the roots of x^^px + q = (Tod-
hunter's Theory of Equations, p. 182) is

and fle'B, be-'^ are roots of a?2—ca: + ai^-*(A^-®) = 0, a^-*^, ic*^ are roots
of ar^— fa? + a3e^(^-B) «= 0, whence substituting for p and y from each of
these equations, and adding, we obtain the required result.

[The same method applies to Quest. 8290, which is otherwise solved on
p. 96 of Vol. XLV.]

9319 ft 9364. (Professor Bhattacharyya.)— (9319.) Show that
(2w+l)(2m-t-3) ...(2m-t-2r-l) (2m + l)(2m-t-3) ... (2f» + 2r~3) 2/1-1
r\ {r-'\)\ ' \

(2m+l)(2m4-3)... (2m + 2r-6) (2«~l)(2n + l)
(r-2)! • 2!

^ (m + « + r-2}_! 2r.
(w + «— l)!r!

(9364.) (W. J. Grebnstreet, B.A.) — If q is any positive integer,
prove that -JL ^ i ^^^ to-J) ^^^ (g"l)(y-2)(y-3)^ ^
^ q-¥l ^ 2! ^ 4!

Solution by R. F. Davis, M.A. ; W. J. Barton, M.A. ; and others,

(9319.) This identity follows from equating the coefficients of a;** in
the expansion of (1— a;) to the power — (m + «) and in the product of the
expansions of the same binomial to the powers — \ (2m > 1) and
-i(2«-l).

(9364.) This identity follows from the fact that the sum of the odd
coefficients in the expansion of (1 + x) to the power (g + 1) = i . 2« ♦> = 2'.

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9325. (S. Tbbay, B.A.)— a, B, C are the dihedral angles at the base
of a tetrahedron ; X, Y, Z the respective oppofijtes ; show that, if

Tj « (1 - C082 B - cos^ C - C082 X - 2 COS B cos C cos X)*,
with similar expressions (denoted by Tj, T3, T4) for the other solid angles,

TaTaCosX + TaTiCosY + TjTjCosZ = 1 -cos^A-cos'B -cos^C
— cos B cos C cos X — cos C cos A cos Y— cos AcosB cosZ + cos Xcos YcosZ.

Solution by D. Edwardes ; Prince de Polignac ; and others.
Denoting the areas of the faces by P, Q, R, S, we have by projection

the equations P cos A + Q cos B + R cos C — S — 0,

-P + QcosZ + RcosY + ScosA = 0, PcosZ— Q + RcosX + ScosB«0,
PcosY + QcosX-R + ScosC = 0,

therefore

cos A, cos B, cos C
cosZ, —1, cosX
cosY, cosX, —1

= S

Therefore, squaring, P^ | &c. | ' =
metrical determinant

— 1, cos A, cosB, cosC
cos A, —1, cosZ, cosY
cos B, cos Z, — I , cos X
cosO, cosY, cosX, —1

— 1, cos B, cosC
cosB, —1, cosX
cos C, cos X, — 1

P2

. S2 I &c.

1, cosB, cosC
- cos B, — 1 , cos X
— cosC, cosX, —1
I 2. But we have the sym-

cosA, cosB, cosC '
cosZ, —1, cosX
cos Y, cos X, — 1

therefore

— 1, cosZ, cosY
cosZ, —1, cosX
cos Y, cos X, — 1

S2

— 1, C9sZ, cos Y
cos Z, — 1 , cos X
cos Y, cos X, — 1

1, cosB, cosC
-cosB, —1, cosX

- cos C, cos X, — 1

t.(f., P/Ti«S/T, where T-l-co82X-cos2Y-co8-Z-2co8Xco8YcosZ;
therefore P/Tj = Q/ Tj = R/T3 = S/T.

Now, from above, PQ cos Z + PR cos Y « P^ - SP cos A,
and two similar equations. Adding these,

2(PQcosZ + QRcosX + RPcosY)« P«+Q2 + R2-SS

and substituting for P, Q, R, S the quantities Tj, T2, &c., to which they
are proportional, and reducing on right side, we have the required result.

[If we form similar relations for the other three faces, and add all four
together, we obtain

T2T3 cos X + T3T1 cos Y + TjTj cos Z + T1T4 cos A + T2T4 cos B + T3T4 cos
« 2-cos"A— cos^B-cosSC— cos^X— cos^Y-cos^Z-cosB cosC cosX

— cosC cos A cos Y— cos A cosB cosZ—cosX cosY cosZ.]

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9200. (Professor Nbubbbo.) — On casse, au hasard, une barre, de
longueur Sa, en trois morceaux. Demontrer que la probabilite que le
produit des longueurs de deux quelconques des morceaux soit moindre
que a* est :

I lofiTe tt (3 + ^6)] + 2 - -v/6 = 0123 (k tr^s-peu pr^).

Solution by Professor F. X. De Wachtbr.

Dans le triangle Equilateral de hauteur 3a, deter-
minons le lieu des points dont les distances u deux
des c6tes ont pour produit a'. Ce lieu se com-
pose de trois arcs byperboliques ayant le centre du
triangle pour sommet commun et les deux c6tes /

correspondants pour asymptotes. II est aise de /
voir que I'espace favorable h Pev^nement consid^re /^,/^
se compose de 3 quadrilat^res mixtilignes egaux, L.i..
situ^ dans les coins du triangle. Done la pro-
babilite cberch^e a la valeur donn^.

5440. (^' Rawson.) — ^Prove that the general solution of the equation

dx^ i Xi \dx J dx^]dxdx^ x^ \dx )
t xi \dx I dx^ ) dx Xi \dxj

where o, )8, Xi are given functions of x, and

i^dx dx )

M = ^3 — ' f — €'i*(*)-^*(*) . <t> {ay^*^ip' (a)
dx i^dx

-^ €'.♦(-

dx

Solution by the Proposer.

Let o, $f Xi be any functions of ar, and u = 1 €*»*<*) }l/{z)dz + e (2) ;

then, differentiating (2) with respect to (x)y (Todhuntbr's Int. Calc.,^. 198)

dx j^ ax ax ax

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or

S = S^t *"*^'^ • ^ W ^ W ^ + N (3),

where N =» ^ €*.♦(•) 4,(0)- ^c'l^W.V'CiS) (4).

Differentiating (3) with respect to x, we have

where M - | . £ [^.« (•). + (,)]- g . £ [.-.♦'») . .^ {0)]

(6)

= gi{g •'■♦'•' ^ («) + (»)- g .»■♦ W. ^(J8) ♦(j8) j (6).

Ffom (3) and {6) we obtain
rfN

.(7).

Integrating the part affected hy the integral sign in (7) by parts in the
usual way, we have

r.'.*w . ^ (»)'+(«) <fe - r t^iM I [,-.♦(.)] &
if }$ 'f* («) *«

. L- r*-^^ |2^(.) ♦ (.)+ 1^^ - f('y%if)fmdz

Xidxildx dx Xi x^Jfi ^'(2) C 0' W j

(8);

where L - JL(^"^^'^» f (f f H - ^"^^^^ • f (f ^ (^) ] (9).

Substitute the value of (8) in equation (7) ; then

dh* I q dxi dx^ I dxi \ du \

dx^ \ x^dx dx^l dxJdx j

^(2^1-^/f^ON^^ + ^ + M+f^VL I (10).

^ \Xidx dx^ldx) dx dx \dx I f ^ '

x,\dx)}/ 1>'(z)V^^ 9'(z) i )

In equation (10) take

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then we have ^(«) = i^j e -<*»♦('>. 0(z)^>0'(») (12).

Obsenring the conditional equation (11), equation (10) becomes

dx^ \ x^ dx dx^l dxidx x^\dx I ^

\ Xi dx dx^ I dx I dx dx Xi\dx I \dx I J

(13).

The integ^l of equation (13) leads therefore, obserying the equation
(12), to the results in the Question, the constants M, N being as there
given, and

L = £a. {€*>*(•) -<^>W .0 (o)^« **-€'»♦('») -'^'♦('') . 0()8)^t+2j.

[By assuming the particular values, arj = — a*a?*», <? = 0, <?2 = 1,
nc^ = - 1, e^n — — «— 1, ^ {z) = z-**, a == oc , jS = a*:c*, we obtain
herefrom a solution of Question 5100 ; see Vol. xxviu, p. 76].

8020. (Asparagus.) — ^A conic circumscribes a given triangle ABC
and one focus lies on BC ; prove that the envelop of the corresponding
directrix is a conic with respect to which A is the pole of BC ; and, if A
be a right angle, the envelop is the parabola whose focus is A and direc-
trix BC. [If (0, 0), (a, 6), («, — <j) are the coordinates of A, B, C, the
equation of the envelop will be

Abc{be-a^ x^^^a (* + r) {be-a^) xy ■¥a'^[^a^-¥ {b^cy]y^

+ a2(i + <j)2(2aa;-a2) =0.]

Solution by K. Lachlan, M.A.

1. Let px + qy + rz = 0, be the equation in areal coordinates of the
directrix of a conic which peisses through the angular points of the triangle
of reference ABC. If S be the corresponding focus, and e the eccen-
tricity, we have at once SA « ep, SB — eq, SC = er.

2. If S lie on a circle, we have /. SA^ + w.Sfis + w .SC^ « k.

. And, if/+m + » «• 0, then S lies on a straight line ; thus we shall have
Ip^ + tnq^ + nr^^ kje^.

3.

Again, S, A

,B,C

being

0,

1, 1,

1,

1

1,

0, SA^,

SB',

SC^

1,

SA', 0,

AB',

AC-

1,

SB', AB2,

0,

BC

1,

SC", ACS

BC,

-0;

0,

l/»», 1,

1,

l/«»,

0, P^

*».

1,

t^, 0,

«',

1,

«', 0'.

0,

1,

r\ i^

«',

1

*'

= 0.

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4. If then the locus of S be /.SA^ + m. SBHti .SC^ = ;t,
the tangential equation of the envelop of the corresponding directrix is
0, Ip' + mq^ + nr^, k, A;, k =0;

k, fi, 0, <^, *2

k, q\ e\ 0, aa

k, r2, ^, aS,
which may also be written

0,
k^c^m — h'^ny

k^m-a-m,
where

k-i^m^bhi, k-i^l-a^n,

k-bn^ahn

= 0,

0,

0, P\ q\

i^, 0, c\

q\ cs, 0,

r», ^, «^

Thus, if the locus of a focus of a system of conies circumscribing a
triangle be a circle or a straight line, the envelop of the corresponduig
directrix is a curve of the fourth class.

5. If the locus of the focus be the straight line BC, we have
a.SA2-*co8C.SB2-<jcosB.SC2« a*<?cosA,
and the envelop of the directrix is

= 0;

-8^

\ 0,

0,

0.

0,

P\

(l\

r2

8a2

i"',

0,

^y

ft2

0,

«',

^,

0,

a2

0,

r^.

h\

««,

which reduces to { -p'^a^ + qU^ + rV}a - 4iV^V2 cos^ A = .

Thus, if the focus of a conic circumscribing the triangle ABO lie on
BC, the envelop of the corresponding directrix consists of the two curves
-j»2a2 + ^2^2 + y2c2 + 2*<?cosAgr=0, -j!?V + j2*2 + ^c2-2*ccosA^r = (1,2).

If D, E, F be the mid-points of ABC, (1) is clearly touched by DE and
DF, and (2) by EF and ijie line at oo . Thus (2) is a parabola.

Again, it is clear that A is the pole of BC with respect to (1) and (2).

If A be a right angle, Jl) and (2 J coincide, and the envelop is clearly
the parabola whose focus is A and directrix BC.

6. If the focus lies on the circum-circle of ABC, we shall have
;.SA2 + m.SB» + «.SC«-A:,
where m(^-\-nl^ = ky Ic^-^na^ = k, Ib^ + ma^ = k ;
hence the envelope of the directrix is

= 0; or pa ±.qb ■;^rc = Q,

0,

I^,

«',

r"

p'.

0,

0%

«2

?'.

«',

0,

a"

r^,

*'.

«^

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Thus, if the focus of a system of circum-conics lie on the circum-circle,
the corresponding directrix passes through one of the centres of the
circles which touch the sides of the triangle.

7. More generally, if the locus of S be given by an equation of the
form Wn + «»-!+ ... +Wo = ®»

where w,» is a homogeneous function of SA, SB, SC of the «th degree,
then, substituting SA ^ ep, &c., we shall have

«- ./« (Pqr) + e^-Vn-i {pqr) + ... = 0.
And if <; be eliminated from this and the equation in § 3, we obtain the
tangential equation of the envelop of the directrix corresponding to 8.

8. If the locus of S be given by an equation of the form

/(SA, SB, SC) - 0,
where /is homogeneous of the nth degree, then clearly the envelop of the
directrix is / (/?, j, r) » 0, a curve of the wth class, and conversely.

7949. (B. Knowlbs, B.A.) — Prove that the sum of the series

-3-i^log (^-/t.^')% 3-i^(tan->g4fl^.cot-i3i].

Solution hy Rev. T. C. Simmons, M. A. ; J. O'Keoan ; and others.
Putting X ^ y^, and dividing by y, the left-hand side becomes
11^-1!^ + iy«-...sS.

the constant being added in order to make S and y vanish together.
Hence the stated sum of the original series follows.

8668. (Alpha.) — The ellipse whose eccentricity is i 'v/2 is referred to
the triangle formed by joining a focus to the extremities of the latus
rectum tlurough the other focus : prove that its equation is
'y^+9($y + ya + afi)=^ 0.

Solution by A. Gordon ; Rev. T. Galliehs, M.A. ; and others.
If x^Ja^ + y^/b^— 1 = be the ellipse referred to its axes, we have

-x + y2^/2-'^p ^a, -x-y2V2-Zp ^ fi, x-Zp ^ y (1,2,3),

and cosa - -J, p = a/3V2, x^ + 2y^ « 18^?* (4).

Eliminating x^ y, p between (1), (2), (3), and (4), we obtain the result.

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9449. (Professor Sylvester, F.R.S.) — If there exist any perfect
Dumber divisible by a prime number p of the form 2'*+ 1, show that it
must be divisible by another prime number of the form^^ ±, 1.

Solution by W. S. Foster.

Let the number N ^p^ ,q^ .r^...; then, since N is a perfect number,
we must have one of the Victors (say, ^*) such that ^* ♦ * — 1 is divisible
by j7, therefore ^*** « M (^) + 1 ; and, since p and q are prime numbers,
ji»-i = M(j7) + 1, therefore d + 1 is a divisor of 2'» » 2* suppose. Let
q — xp±hj then k^—l = M(i?) ; hence h must be some power of the
remainder after dividing (2»»)***~* by 2'* + 1 ; therefore h must equal 1,
and q ^ arp±\f which is a prime divisor of N.

9468. (R. W. D. Christie, M.A.)— Show that the tenth perfect
number is P^o = 2« (2«- 1) = 2,417,851,639,228,168,837,784,676.

Solution by Professor Ionacio Bbtens.

Le dixi^me nombre parfait est donne par M. Carvallo dans Touvrage
ThSorie dea Nombrea parfaits.

[Every divisor of 2f — 1 is of form 2px + 1 when piaa, prime ; but 2** — 1
is indivisible by ii2x + 1 ; hence 2^ {2*^— 1) = &c. is a perfect number.]

3419. (Artbmas Martin.) — The point Ai is taken at random in the
side £C of a triangle ABC, B^ in CA, and C^ in AB ; the point Aj is taken
at random in the side Bfii of the triangle A^BjCi, B^ in CjA^, and C^ in
AjBi, and so on ; find the average area of the triangle AnB^C^.

Solution by D. Biddle.

It is difficult to understand why this question has remained unanswered
so long ; but the reason may be any one of three : — (1) its extreme sim-
plicity, ^2) the fear of some concealed pitfall, or (3) a mere disinclination
to consitter the matter. Unless (2) be well grounded, there can be no
doubt that the correct answer is (J) "ABC. For, let Ai, A], ... A« repre-
sent the successive triangles drawn at random upon the given ABC, in the
way described. Then, the average area of An will be Ja»-i ; of An-i,
iAn.2; aiid so on, until by retrogression we arrive at Aj, which on the
average is JABC. The fact, in regard to each pair taken separately, is
well known. But, if A^ be on the average J of Aj, which on the average
is i ABC, it seems impossible to escape from the conclusion that on the
average Aj = (i)* ABC. In being J, on the average, of any Aj, on which
it may be drawn, A2 is on the average i of ^ ABC. It is a case of multiple
integ^tion in which, as each variable is eliminated, the additional factor
^ is yielded to the result.

VOL. XLIX. L

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9402. (The Editok.) — If the radios of the in-circle of an isosceles
triangle is one-»*** of the radius of the ex-circle to the base ; prove that
the ratio of the base to each of the equal sides is 2 (n- 1) : n + 1.

Solution by Professors Emmerich, Ph.D. ; Ionacio Bbybns; aftd others.

Draw perpendiculars TD, T«D« from the centres T, Ta of the in-circle
and the ex-circle to the base on AB. From similar triangles, we have
AD : AD« = 1 : w. But AD = J (2b- a), AD. « J (2* + a), a denoting
the base, and b the other sides ; Uieref ore 2b + a:2d — a=nll; hence

2a : 4ft — «— 1 : «+l, and a : ft — 2 (n-l) : « + 1.

9440. (Rev. T. C. Simmons, M.A.) — Prove geometrically that the per-
pendicular from the Lcmoine-point of an harmonic polygon on the Lemoine-
line is the harmonic mean of the perpendiculars drawn on^he same line
from the vertices of the polygon. [A proof by tiigonometrical series is
given in Lond, Math^ Soe. Proceedingsy Vol. xviii., p. 293.]

Solution by R. F. Davis, M.A.

If a polygon ABC ... L (n sides) inscribed in an ellipse is such that each
side subtends the same angle (2ir/n) at the focus S ; then, projecting
orthogonally, we get a harmonic polygon abe ...I {n sides) inscribed in a
circle whose Lemoine-point a is the projection of S, and whose Lemoine-
line yy' is the projection of the S-directrix YY'. This property is the
basis of Mr. Simmons' theory of harmonic polygons, as set forth in the
above paper.

The properties of the polygon ABC ... L may be derived in turn by
reciprocating with respect to any point S a regular polygon of n sides
circumscribing a circle. Since (by a well-known theorem) the sum of
the perpendiculars of 8 upon the sides of the latter ^lygon = n (radius),
we have 2 (k^/SA) = w (Ar^/SD), where SD is the semi latus-rectum of the
ellipse. Hence, if SX, AA', BB' ... be perpendiculars upon Y Y', SX is
the harmonic mean of A A', BB' .. . ; and this relation is unaltered by pro-
jection.

(W. J. C. Sharp, M.A.)— If (x^, yj, z^, m>i), (a^s* ^i, «2> «^t)»
(^3» !/zi «3» ^p) he any three points, and A, fi, v the areal coordinates of any
point in their plane referred to the triangle of which they are vertices ;
show that the equation to the section of any surface U = by the plane
will be obtained by substituting for x, y, », ta from the equations

(\ + ;*+y) Jf « KZi + fiX^ + yX^, (a + ^+i.) y » Ay j + ftyj + •'^j*

(A + fi -I- v) « = ATj + ^j + yz^f {\ + fi + y)w*^ AtTj + fiw^ + viTj.

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Solution by D. Edwaudes.

Let A be the area of the triangle formed by the three points. Then
any point in the plane of the triaogle may be expressed

l + m + n * l-tm + n

Also k — ' A, &c., therefore Ax = XXi + fix^ + yx^f Ay — &c. &c. ;

l + m + n

and A = \+fi + v, therefore, &c.

fThat the point ( ^^l+>»^^-^>>^3 If/i^my^-^ny, ^ \
L \ l&m + n l + m + n J

is a point in the plane follows at once, because if this point be (or, y ...) it

I point ^

satisfies the equation

Xi

X2,

= 0, which is the equation to
the plane of the triangle.]

9350. (Professor De Wachtbb.) — A point being taken within a tri-
angle, prove that the chance that its distances from the sides (a), (6), (0),
may form any possible triangle will be 2abc/ {{b + e) {c + a) [a + b)}.

Solution by Professor Ionacio Beyens.

Soient A], Bj, Cj les pieds des bissectrices
du triangle ABC ; il est ais^ k d^montrer que
la droite Bfii est le lieu g^om^trique des
points tels que leur distance au c6t6 BC est
egale k la somme des distances aux autres
deux cdtes AB, AG, et que par suite pour
tout autre point situ6 dans Pint^ieur
du triangle AB^Ci la distance k BC est plus
grande que la somme des autres distances k
AB, AC, et que pour un point du quadrilat^re BCBjCi la distance k BC est
plus petite que la somme des distances k AB, AC. La m^me chose
arrivera aux droites BjA^, AjCi, et par suite tout point interieur k AiB^Cj
aura la propriety que Pune quelconque de ses distances aux cAtes AB, AC,
BC, sera plus petite que la somme des deux autres, et par consequence la
probabilite demand^e sera AA^B^Ci : A ABC — 2abc : {b + e) (e + a) (a + b).

. 8344. (K. Knowles, B. a.)— ad, be, CF are drawn from the angular

points of a triangle ABC, so that the angles BAD, EBC, ACF are each

equal to the Brocard-angle of the triangle ; show that their equations are

bcy—jah = 0, b'^x—acz = 0, abx-e^ « 0.

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Solution by Gbobob (Joutthorpb Storb, M.A.
It to "be the Brocard angle of the triangle, we have
cot « s cot A + cot B -I- cot G.
Now the equation to AD is

Jl_ . «^(A-a,) ^ rinA cot— COS A - -J$^^ = f', or bcy-aH^ 0.
« sin ft» sin B sin C *c

Similarly for the equations to BE and CF.

9376. (A. E. Thomas.) — Solve the equations

x4+3y22i^a*+2d;(y* + a^ (1),

y< + 3«%««4< + 2y(2' + :c^, 2*+3a;V = «* + 2« (a' + y') (2,3).

Solution by Professor Sebastian Sircom, M. A.
Adding, ^fl+y^+s^^xy—yz^zx^ (x + 9ty + uh){x + t^ + att)

-(«^ + ^ + <r*)» (4),

(2)x«, (3)x«i8 give (a; + y + «)(a: + «*y + «2) = (o^ + wi^ + ci'i?*)* (5),

(x + y + »)(ar + «y-l-«%) = («< + ^^^ + »<?♦)* (6).

(5)x(6) gives ^ + y + . = (a^-^<^^^-^c^^)* (a* ^ a>^b* + ^)\
(4) *" ' (a* + ^ + <?*)*

with similar expressions for x + »y+§^f &c. ; adding, we ohtain a; in a
form that can easily he rationalized, and then the valnes of y, z can he
written down.

9430. (Professor Wolstbnholmb, M.A., Sc.D.) — In a tetrahedron
OABC, the plane angles of the triangular faces are denoted hy a, jB, or 7 ;
all angles opposite to OA or BG heing a, those opposite OB or GA are
i3, and those opposite OG or AB are 7; the angles at have the
suffix 1, those at B, G, D the suffixes 2, 3, 4 respectively ; prove that, if
«i + i^i + 71 = as + iSs -h 7s s T, then

71 + 01-^1-74+04-^4; Oi + ^i-7i- 08 + ^8-78

7s+aj-^- 74 + 09-^5; oj + ^-72 » 04 + ^4-74-

Solution by Professor Swaminatha Aitab, B.A.

^i> ^2> ^3 fti'o >^7 three points not in the same straight line. is the
middle point of Gi, Gj, A of Gi, G3, and B any point in the plane equi-
distant from Gs and G3. Now the tetrahedron of which the &ices are the
triangles GAB, OAGi, OBG3, ABGjis of the sort described in the question,
and, naming the angles as directed, we see at once from the figure,
since OA is parallel to G2G3, that

ai + 7i-^i= 04 + 74-^4; 02 + 32-72 = 04 + ^4-74-
A similar proof is easily seen to hold for the other part.

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[Professor Wolstbnholmb remarks that he had not attempted a de-
ductive proof of this theorem ; the lengths of the edges of the tetrahe-
dron, by means of which he noticed the property, are

DA - 7069273, BC = 713973,
DB - 7-376329, CA - 64644,
DC - 7-316126, AB « 8* 13924 ;

the half angles at D are 29*»4'40"-45, the half angles at A are 28*»36'42"-86,
21**38'39"-03 28**46' 4"-73

34**16'40"-62 32°39'12^'-42

(Ti = 90° 0' 0" ffi - 90°

those at B are 26° 68' 14"- 76, those at C are 30° 42' 8"-66 ;
24° 43' 23"-96 30° 40' 19"-78

30° 14' 59"-77 36° 40' 63"- 19

0' 0"

<ra«81°d6'38"-48

<r4 « 98° 3' 21"-52

the dihedral angle opposite DA is 66° 1' 3"-76^
BC is 66° 31' 60"-4 i

opposite DB - 65° 68' 29"-34 7
„ CA = 69°ll'49"-62j»

.yj + ai-i3i«73"26'23"-88 7
74 + «4- i84 = 73° 26' 23"-92 j

«, + fi^^y^ = 42° 63' l7"-92 7
«3 + /38-78-*2°63'l7"-96j

opposite DC iB 79° 42' 30"-7 7
„ ABis86°46'22"-96j»

7i+ai-i52 = 64°69'41"-08 7
78+«B-iBs - 64°59'41"-12 5'

09 + ^^-7% = 49°23'10"-32 7
' a4 + i34-74 - 49°23'10"-28)'

the difference in each case being "'04 ; which is as near as can be expected
with 7 figure logarithms.]

9006. (H. L. Orchard, B.Sc., M.A.) — ^Inside a hemisphere (of radius
p) a luminous point is placed, in the radius which is perpendicular to the
base, at a distance from the base » ip \^3 ; show that the illumination of
the surface (excluding the base) is = 3irC.

Solution by Rev. T. Galliers, M.A.

Let be the luminous point ; ABD the vertical section
of the hemisphere (its centre being at C) through CO ;
Z CPO = ; Z FOB = e ; radius of hemisphere = a ;
CO=e,a^eVZ\ also let OP » r.

Then the illumination of a band generated by the
revolution of the elementary arc at P about CB

Now

= 2irC (r sin tf . cos 4) <fo)/r2 « ^ (say).

cosd) =s^-J^ ^ "^ . also a^ = r2 + c2+ 2(T cose ;
^ 2ar

therefore

(r + c cos 6)dr ss er sin . rfd,

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■^^ ^-?l±££2B», therefore *--?-;

dr tfBine dr eam$

thus ^.I^J-W.lJ..;

therefore illumination of hemi-spherical sur&ce

- — f^"'*"*^ i^-^ + 1| rfr - 3»0, the result given.

ap a^bq,fi^er,y, .-. ^ = *-2, ...M«N^^^^

9433. (G^. Hbppbl, M.A.)— If, within a triangle ABC, be a point
where the sides subtend equal angles ;. then, putting OA >« p, OB » q,
00 « r, show that the equation to the ellipse with focus O, touching the
sides in D,- E, F, is in (1) rectangular coordinates, with O as orig^ and
OA as axis of ^, and (2) trilinear coordinates, ABO triangle of reference,

{x^+y^^'=^\(pq + qr-¥rp)'^[{pr+pq'2qr)if-p(q-.r)xy/Z + Zpqr]...{l)^

aYa^ + b^q^0^+A^-^2bcqr$y^2earpya-2abpqafi - '.(2).

Solution by W. S. Foster.

Since is the focus of the ellipse touching the sides of the triangle, the
angle AOE = AOF, and AOB = AOC, therefore BOF = OOE, therefore
BOD » COD, therefore OD bisects the angle BOC, therefore AOD is a
straight Une. Let L<r» + M/32+N7«-2LMa/5-2LNo7-2MNi87 = be
the equation to the ellipse ; then the line AD is MjS— N7 = 0, and since
this passes through O, whose coordinates are given by the equations

^^Ji

N er* ' ' bq cr "^^^ ap'

Substituting these values of L, M, N in the equation to the ellipse, we
have the equation given in the question for trilinear coordinates.
; Let hjp a 1 + « cos be the equation to the ellipse, referred to the major
axis as initial line, and let this make an angle Bi with OA ; then, since
AE touches the ellipse at E,

k

— = tfCOs0i + co8AOE = tfcos^i + i,
P

and — «tfCos(-^ + aij+i««(-icos0i — sin^jj+i,

|.-fco8(^+a,)+i=i,^-4cos0i+^sin0i)+i,

therefore

k^—-ML -, .cosO,= ^g^-^-i^g , .sine,»^/^^(^'^) :

^{pq-^qr-^rp) 2{pq + qr-tpry 2(pq + qr+pr)

therefore the equation to the ellipse referred to a line perpendicular to

OAis A;/p - l+« sin (01 + 0),

.-. (a?' + y2)*= i{pq + qr'^pr)-^{(pr+pq-2qr)t/-^/3p{q^r)x + Zpqr}.

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8461. (F. R. J. Hervey.) — Find in how many ways n lines of
verse can be rhymed, supposing that (I) no line be left unrhymed, and
(2) the restriction as to unrhymed lines be removed ; and show that, in
the case of the sonnet, the respective- numbers of ways are 24011157 and
190899322.

Solution by the Pkoposeb.

Let the numbers of ways be denoted, (1) by 0», (2) by/« ; and first
consider /n. Let line-endings which rhyme together be considered the
samei and denoted by the same letter. Let the first euding be denoted by
fl, the next distinct ending by b, the third by c, and so on. Two lines
give two cases, aa and ab. Three lines give five, for aa may be followed
by a or 6, and ab by a, b, or e. Genersdly, an arrangement of this sort
containing only the first r letters of the alphabet may be followed by any
one of the first r + l letters ; and the complete sets of arrangements for
successive values of n may be exhibited in a table such as the following.
Each letter gives rise to a ffrovp in the next line ; a g^oup of r letters,
wherever found, gives invariably r—l groups of r followed by a g^oup of
r + l; and the number of letters in the n^ line is fn. I shall show that
fn a A**fl ; this is contained in the two following propositions.

b

d

ab abc abc abc abed abe abe abed abc a\be\a\ bed abed a \ bed \ a \ bed \ a \ bede

I. Suppose the above table continued indefinitely , and form a new table
thus. Suppress the first line {a) ; suppress a of tbe second line, and all
that it gives rise to in the following lines. Call this the first derived
table. From this derive another table, and so on continually, the rule
being : — To form the (m + 1)^ derived table, suppress the first line of the
m^ ; then suppress every a of the second line, and all that they give rise
to afterwards. The numbers of letters in the successive lines of any
derived table are the differences of the numbers in the successive lines of
the preceding table.

Let A denote either the original, or anv derived table ; B the next,
and S the table consisting of all that part of A, with the exception of its
first line, which is suppressed to form B. To each letter {k) in the first
line of A corresponds a letter {a) in the first line of S ; A; gives a group in
the second line of A ; the corresponding a (the first of this group) gives
a similar group in the second line of S ; and these two groups, being
similar, g^ve similar sets of groups in the next lines of A and S respec-
tively ; and so on continually. In fact, A and S are, with the exception
of their first lines, identical, letter for letter. But the(j3+l)**» line of
A is made up of the^?*^ lines of S and B ; whence the proposition.

II. The first line of the m**^ derived table consists of the same number
of letters, arranged in the same numerical groups, as the m**» line of the
original table. Let the first lines of the m^ and {m -i- 1)*** derived tables
be denoted by P and Q respectively ; and assume the truth of the pro-
position for P. Consider a group of r letters in P ; this is a group of
r+l with its first letter suppressed ; it gives rise to r— 1 groups ^f >• -t- 1,

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n

12 3 4 5

T

1 2 5 15 62

A

1 3 10 37

A2

2 7 27

A'

6 20

A*

15

92

and a group of r + 2 ; and these, with their first letters suppressed, are
found in Q. Hence a group of r in P gives r - 1 groups of r, and a
g^oup of r + 1 in Q. But this is just what arises in tie (m + 1)"» line of
the original tahle from a group of r in the m^ line. In fact, the table
consistmg of the first lines of the series of derived tables only is an exact
copy of the original table, with this exception, that the letters a, 3, ^ . . .
of the latter are everywhere replaced by *, <?, <f ... respectively, in the
former.

From this property fn can be calculated (see
example). The oblique rows are formed in suc-
cession ; each value of /obtained is transferred to
the left-hand column, giving a new oblique row,
ending in a new value of/; and so on to any
extent.

Another process, and an expression for /», are
thus obtained.

If (r, n) denote the number of groups of r in the n**» line of the original
table, we have evidently (r+1, « + l)=(r, «)+r . (r+1, n). Forma
table in which the r*** number of n*^ column is (r + 1, « + 1) . The first row
is 1, 1, 1, ..., the first column 1, 0, 0, ..., the above relation gives all the
rest, and the sum of the n^ column is fn. But the above conditions are
precisely those for the formation of a table in which the r*^ number of
«ti» colunm shall be A*'0"/r!; (compare the above relation with the
following — aW 0" = A^**"*)©"-* +r . A^*^ 0"-^ in which A^*") 0*» stands for
A*" 0»/r !) Hence (r + 1, « + 1) « A(*"^ 0", and

fn = A0'»+A20»/2 + A30»/3! + ... + A»»0»*/«t;
in which the r^ term, being (r + 1, « + 1), is evidently the number of cases
with just r distinct endings.

From this expression an algebraic proof may be given of the theorem
already derived from elementary considerations; namely, that the
«**» term of the 8eries/i,/2,/3 ... is the same as the »*^ difference of its
first term. We have, A and D referring to x and y respectively,

D'». A*"a:i' = A»» .D^arv = A»» .arv (a;-l)" ;

whence D'».A"*a;<*— A*"(jr— 1)*». Now, the above expression may be
written without error as an infinite series, and the general term A** 0^/r I
is also A»-i !»•-'/ (r-1) ! Hence

D» . A»*OVr! = D" . A»-i lo/(r-l) ! « A'-iO«/(r-l) ! ;
and the theorem follows at once.

It remains to determine the relation between ^ and /. A line which
rhymes with no other may be called an odd line. I shall show that the
number of arrangements of n lines containing at least one odd line is
exactly equal to the number of arrangements of « + 1 lines which contain
no odd line. Take any one of the first set, and add an odd line at the
end ; then replace the odd lines by as many lines rhyming all together,
but not with any other line. The result is one of the second set. Con-
versely, take any one of the second set, and replace the last line and those
which rhyme with it by as many odd lines ; then erase the last line.
The result is one of the first set. Hence the two sets correspond
definitely in pairs, and the number in each is the same; that is,
/n — ^ =s 4) (« + 1), whenife the values of 4) are derived in succession from
thoseof/. Also <^(« + l) =/»-/(«- 1) +/(fi-2)-...dr /I.

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Of the two relations, fn = A"/l and/«— ^« = ^ (« + 1), either may be
made a coosequence of the other, as follows. It is clear that the number
of arrangements of n lines containing k odd lines is, in general,
^{jfi—h) .» («— 1)...(«— A;+l)/A;! \k ^n gives one; k ^ n—l gives none,
but <pl = 0. Hence, if /O = 1, we have

/«= 1+«4>1+J«(;j — l)«^2 + ...+«4>(n-l) + «^w,
true forn = 0, 1 , 2, . . . J which givjes <f>n = A**/0, and </>» + 4) (w + 1)' « A*»/l .

The question of finding /« may be approached somewhat differently, as
follows. Of the complete permutations (r" in number) which can be
formed from r given letters taken n (not < r) at a time, let the number
of those in which all appear be deiv>ted by Ur ; let P be one of them ; and
let ab .,., hk ...f represent any two of the r ! simple permutations of the r
letters. If then the result of writing h for every a, k for every b, &c., in
P, be called Q ; it is evident that P and Q represent exactly the Fame
arrangement of line-endings. (The selection of one out of each such set
of r ! equivalent permutations may be made by requiring that the r letters,
so far as the^r«^ entrt/ of each is concerned, shall present one invariable
permutation; compare the notation already used.) Hence (»'+l, w+1)
= nr/r !, and «r = A** 0". But the last result may be obtained indepen-
dently ; (it is in fact essentially the same as that contained in the first
part of Question 8390 [Vol. xlv., p. ] ; as may be seen by putting letters
for persons in that question, and regarding iheplaceSf 1st, 2nd, 3rd, &c.,
occupied by the letters as the things distributed among them) ; the formula
for fn will then follow, by the above reasoning.

The values of/ and 0, as f ar as « = 14, are given below.

n

/

n

/

1>

n

/

1>

3

6

1

7

877

162

11

678670

98253

4

16

4

8

4140

715

12

4213597

680317

6

52

11

9

21147

3425

13

27644437

3633280

6

203

41

10

115975

17722

14

190899322

24011157

9413. (J' O'Byrnb Croke, M. a.) — If D be the distance between the
centre of the cii'cumcircle and the point of intersection of the perpendicu-
lars of a triangle, prove that 2D/(1 — 8 cos A cos B cos C)* = «/ sin A.

Solution by D. Thomas, M.A. ; R. Knowles, B.A. ; and others.

Let o, )8, 7 be the vectors of A, B, 0, cooriginating from the circum-
centre, so that To = T)8 = T7 = R, S$y = -R=2 cos 2 A, &c. The vector
of the orthocentre = cot B cot C.0 + ... + ..., therefore

-D2 = -R2 {5 cot2 B cot2 + 2 cot^ A cot B cot cos 2A + ... + ...},
.-. D» = R2 {(5cot B cot 0)2-2 cot2 A cot Boot (1- cos 2A) + ...}
•= R2 .[l — 2 cot A cot B cot 2 sin 2A} =» the result required*
[By Question 8872, we have D = R (1 - 8 cos A cos B cos C)*,
hence 2D / (I - 8 cos A cos B cos O) = 2R = a/ain A, &c.]

VOL. XLIX. M

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9478. (Hev. J. J. Milne, M.A.) — If p be the sum of the abscissae, q
the sum of the ordinates of two points P, Q of an ellipse ; prove that (1)
the equation of PQ is 2J^x-¥2a^y = i^ + a^q* ; and hence (2) if either
(a) p or qhe constant, or (/3) if p and q be connected by the relation
Ip-i-mq ^ If the envelope of the line is a parabola.

Solution by R. Knowles, B.A. ; Prof. A. W. Scott, M.A. ; and others.
If hk be the pole of PQ, its equation is

l^hx-'ta^ky = a»*3 (1),

h ^piffiJe^-^b^h^l^a^lfif k =r ^(a2;t« + A2A8)/2a2^,

^ + aV = 4a**V {a^ + ^2^2) .

making these substitutions in (1), we obtain 2ll^px + 2a^y = b'^p'^+a^q^,

(a) If p is constant the equation to the envelope is the parabola
aV = ^P (p-2a:) ; if q, b^x^ = a-q {q-2y),
(/3) If Ip + mq = 1, the envelope is

(aHy-bhnx)^ + 2a^b^lx + 2a^b^y^d^b^ - 0.

9439. (A. Kahn, M.A.)— Show, by a general solution, that the roots
of 4x* + 4txi+lZx^ + 6x + S = are i{-l±(-7)*}, i{-l±(-3)*}.

Solution by Professor Cochez ; R. W. D. Christie ; and others,
L' equation du 4« degr6 x* + ax^ + bx^ + ex + d = pent se mettre sous la
forme {x^ + ^ax)^ + {b - ^a^ {x^ + exl{b-ia^} +d = 0,

et pourra etre r^olue par les m^thodes du second degr^ dans le cas oh

L' Equation propos^e ix* + 4^ + 1 3:r2 + 6j; + 8 = est dans ce cas ; on peut
I'ecrire (x^ -i- ^x)^ +Z{x^ + ^x) + 2 = 0.

Posant x^ + ^x = y, on a k r^soudre y^ + Zy + 2 =0, dont les racines
Bont — 1 et — 2. Par suite les quatre valours de x seront donn6es par
les equations x^ + ^x ^—l et a;' + Ja: = — 2.

[It p^—ipq + Sr = 0, any biquadratic x^ + psfi + qx^ + rx + s « can be
immediately reduced to quadratics ; hence the given equation

= {x^ + tnx + 2){x^ + nx+l) = 0, ^(xi + \x + 2){x^ + ^x+l) = 0.]

9437. (H. FoRTEY, M.A.) — Show that, if a, /3, &c. are the p roots
(excluding unity ) of a^*^—mxP + m—l = 0, the number of ways in which
m letters can be arranged « in a row, repetitions being allowed but not
more than p consecutive letters being the same, is

m 2 (a-l)^a* *-^P

{m-iy aP''^-{p+l)a+p'

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Solution by Professor Swamin atha Aitar, B.A.

Referring to my solution of Question 9293 (Vol. xlix., p. 26), let Q^
stand for the required number of ways ; of these Q,, ways let those that
do not begin with the letter a be 9» in number. Then we have

Q* - ?»» + ?«-i ...••- 7h-f and qn = (»»— l)(^»-I + ?i.-8...+^M-.p);

therefore Qm = ——^ ?»• And y„ is the coefficient of x^ in the expansion
tn — 1

of {l-(m-l)(x + a;«...+a:^)}-»;

therefore Qn^-gL-^ [-^)'-'''' .

8177. (Professor Hanumanta Rau, M. A.) — The images of the circum-
centre of a triangle ABO with respect to the sides are A', B', C ; prove that
the triangles A'B'Cand ABC are (1) equal, (2) have the same nine-point
circle ; also find (3) the equation of the i^kcum-circle of A'B'C and the
angle at which the two circum-circles cut each other.

Solution by A. Gordon ; R. Knowlbs, B.A. ; and othsrs,

1. OA' = AI and is parallel to it, /^
OB' = BI and is parallel to it,

therefore A'B' is parallel and equal AB,
&c. ; therefore the triangle A'B'O' is
equal triangle ABO.

2. I is the circum-centre of A'B'O'
(for lA'isparallelandequalOA^Rs &c.)^
also I A' « A'B (each = R), therefore
B» = «I ; therefore the lines AI, BE, &c.
are bisected by B'C, A'O', &c., at m, «,

&c., and these points are on the nine-point circle of ABO.

But n is also the middle point of A'O' ffor In is a perpendicular from
centre I on the chord A'O'), and is thereiore a point on the nine-point
circle of A'B'O'. These two circles have therefore three points m, «, &c.
common, and are therefore coincident.

3. If the mid-point of 01 is taken for origin of coordinates, and any
rectangular axis for reference, and if x cos a + y sin a—Pi is the equation
of BO, &c., then the circle about A'B'O' is

5 sin {x cos o + y sin a +j?j) (x cos jS + y sin +pi) = 0.
The equation can also be written in trilinears —

R2 (a sin A + fi)^ = 5 (iSy -7i9')«- 25 cos A (70'- 07') (ajS'- a'jS),
where o', jS', 7' are the coordinates of I the orthocentre.
The angle at which the two circum-circles intersect is given by
(01)2 = R2_2Rr = 2R8-2R2co8d>, or cosrf> = 2»-fR

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9217. (Major-General P. O'Connell.) — In using either the French
or English Arithmometer, any two nnmhers each containing less that
nine figures can he multiplied together, and the sum of a series each term
of which is the product of two sudi numhers, whether positive or
negative, can he ohtained without writing down any figures. It is
required to find a formula for the product true to, say, thirteen figures on
two numhers each of sixteen figures, so that the result may he ohtained
by the use of the Arithmometer alone, i.e., without intermediate record.

Solution by the Proposer.

Let A and B he two large numbers, and let their digits, counting from
left to right, he indicated by numbers written under A and B respectively.
Let Ai_8 mean Uie first eight highest digits of A, and B9-12 the 9th, 10th,
1 1th, and Tith digits of B ; then, if A and B each contain sixteen figures,
the following formula will give a result true to thirteen or fourteen
figures.

AxB = Ai.8 xBi_g + Ai.4xB9_i2 + Bi_4X A9_i2 + Ai_2 x Bi3_h
+ Bi-2xAi3-14+A6-lcxB9-10 + B5.«xA9-l0

+ A, XB15 + B1XA16 + A3XB13 + B3X Aia + AsxBa + Bjx An

+ A7XB9 + B7XA9 (I).

If A =» B, we have

A2= (Ai.8)^ + 2{Ai.4xA9.i2 + Ai.2xAlS-14 + A5-6xA9-10

+ Aix A15 + A3X Ai84 A5X Aii + Ajx A9} (2).

In using the second formula, first sum the series under the vinculum,
add the result to itself, and finally add (Ai g)^ J ^7 this means A^ will be
obtained true to 13 or 14 figures.

The following formula, to be worked with pen or pencil and paper, will,
when its total is added to the above, give a result true to all but the last
figure, which may be looked upon as approximative. The dots are to be
understood as decimal points.

Remainder = 2 x {A, + -Au + Aj x -Ais-w + A3 x -Au-ie + A4 x 'Aw- is

+ A5X •Ai2_i4 + A6X An-is + Ayx •A10-12 + A8X A9_ii} (3).

By formula (2), if A= 3 99999 99999 99999,
A2 = 15-99999 99999 9867.
Byformula (3), remainder = 132, giving A' =. 15-99999 99999 9999 for
the corrected value.

If A = 3-16227 76601 68379 33 - >v/10,

by formula (2), A2 =. 9-99999 99999 9972,

by formula (3), remainder = 27-418, giving

A2 = 9-99999 99999 999948 instead of 10.

8333. (Professor Hanumanta Eau, M.A.) — Prove that the equations
x^ 4 19a?- 140 = 0, and 7^'- 12.i;^ + 46*-2+ 12.i;+ 7 = 0, have two common
roots.

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Solution by Profs. Aitab, B.A. ; Sulcom, M.A. ; and others.
If a, /3 be two roots of the second equation, the other two roots are
evidently and . Therefore the second equation is reducible to

the form (s^—px + q) 1^^+ ^x+ ]»0, and comparing the co-
efficients we have p » 2 and ^ — 7 ; and 3^—2x-k-l is a factor of
«« + 19iB-140.

9018. 0^* '^' Grbbnstbbet, B.A.) — If the Earth and Jupiter are
in heliocentric conjunction at the same time as Jupiter and one of his
satellites, show that the times when the satellite will appear to an
observer to be stationary are the roots of the equation

£. + il + £ + ^ (J + ,)co82, ( i-J-V- -('' + «)co.2, (I-IV
a b c be \ b e I ae^ ' \a e I

-J(a + *)co.2,(i--J-).-0;

where e^ jj s are radii of the orbits of the Earth, Jupiter, and the satellite,
a, b, e their periodic times, the orbits circular and in one plane.

Solution by W, J. Gbbbnstreet, B.A. ; Sabah Marks, B.Sc. ; and others.

St

J
E

8

St'

N' N N"

Consider motions in two directions perpen-
dicular to each other. After a time t^ draw
E'N, J'N', S^N" perpendicular to &r.

llien SN = eBmwt.

The position of satellite relative to Earth
is given by

y sin 0)1^ + « sin fl»2^— ^sin »n

ycos«i^ + «cos»2^— ^cosaO
Kelative velocities in same directions are

wij cos «i ^ + t0^s cos wj^— we cos »^, — «iy sin a»| ^— a»2' sin «3< + we sin co t.
M^en the satellite is stationary,

jsmwit + ssmaf^—eBJnwt w^j cos oa^t 4- <a<^ cos w^t^we cos w t _ ^ .
jcoswit-¥SGOBw.2t—eQOBwt wiJ^nwit-^w^&iiLw^t—oieBUiwt '
therefore, simplifying, and putting wi » l/b, w^ =» l/e, » « l/a,
WiJ^ + w^ + <at^ + (»! + ft»2)y» cos (aj—aj) <—«(» + «2) cos(«— wj) t
— (« + wi)je cos (« — aj < = 0,

or ^ + £ + i.' + f' (j + ,)co82x(i— LV--(* + «)co82»(l— L V
a b c be \b e I ea \ e a I

- ^ (rt + *) cos 2ir ( w = 0, an equation for t,

ab \ a b I

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9277. (Rev. T. C. Simmons, M.A.)— Prove that the Taylor-circle of
a triangle is always greater than its cosine circle, and that in an equi-
lateral triangle the respective areas are in the ratio of 21 to 16.

Solution bp R. F. Davis, M.A.
If R be the circum-radias and » the Brocard-angle, we require to show
that R{sin«A8in«Bsin'C + cos»Aco8>Bcos20}*>Rtan».

If tan <f> a — tan A tan B tan C, since

cot A + cot B-i- cot C— cot A cot Boot C « cosec A cosec B cosec G
— a positive quantity,
cot» + cot^ is always positive, sin(^ + «) is always positive, and
^ + » < 180^. The expression for the Taylor-circle may be written

T> cosec <t> -n sin « .

it — . — s t\ ;

cot « + cot ^ Bin (^ + w)

hence the above inequality reduces to R . ^'^^ — is always > tan w«

' 8in(«^ + ft») ^

or cos a > sin(a + ^), [« > and < 30°], cos^a > sin'(« + <^).
We may square, as both are positive ; therefore cos <j> . cos {<j> + 2w) > 0.

For an ocM^-angled triangle cot <f> is negative, and ^ lies between ^t
and ir, and, since ^ + 2« < v + w{< 210°), both cosines are negative, and
the inequality holds. When the triangle is o*^«*«^-angled, cos <t> is positive,
and (p lies between and ^n ; hence cos (p . cos (^ + 2(a) > 0.

But when the triangle is right-angled <t> » }ir, and a has any value
from up to cot-^2. The radius of Taylor-circle « radius of cosine
circle in this case. Therefore it would appear, for a triangle having one
angle a little greater than a right angle and the other two angles nearly
equal, the above theorem is not true.

When the triaugle is equilateral, a — 30°, cot « -■ -v/3 ;

cot^— (-L)' = -3-L^ooBec»^=l^l-|.

1 8

cot w + cot <b = 'v/3 — - — — »« - — — ;

and ratios of areas - — ^^^^ ^ ,, : tan'« = 21 : 16.
(cot « + cot <t>)^

8781, (Professor Hanumanta Rau, M.A.) — If S be the sun, and A
dnd B two planets that appear stationary to one another, show that
tan SB A : tsm SAB = periodic time of A : periodic time of B.

Saiuiion by W. J. Greenstreet, B.A. ; C. Bickebdike ; and others.
By a well-known formula, we have

tan SBA : tan SAB = \rr^V I (-7^)* » a* : **
=- (by Kepler* 8 Law) periodic time of A : periodic time of B.

I

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9044. (S. Tebay, B.A.)— If A be the area of one of the faces of a
tetrahedron ; X, Y, Z the dihedral angles over A ; and

M = (l-co8«X-co8'Y-cos2Z-2cosXcosYco8Z)*;

show that A/M has the same value for all the solid angles.

• Solution by the Proposer; Prof, Ignacio Betens; and others.

Let Aj, Aj, A3, A4 be the areas of the faces ; «, J, c conterminous
edges ; and a, /3, 7 the angles contained by bcy ca, ab. Then, from the
polar triangle, we have cos X + cos Y cos Z •= sin Y sin Z cos a.

Squaring and reducing, we find M s sin Y sin Z sin a.

Now V - I . ^^ sinY = f . ^^ sinZ ;

b

therefore V^ = ± . ^^^^^ sin Y sinZ

be

^ VA5A5 sinYsinZsina = 4.-^l^;^M.
® ^csmo ® 2Ai

Therefore ^ = 4 • ^'^'f^\

which is the same for all the solid angles.

9122. (Professor Hudson, M.A.)— Prove that the locus of the feet of
I)erpendiculars from the vertex of y^ — ^ax on chords that subtend an
angle of 46° at the vertex is r^^2^ar cos e+lSa^ cos 2fl =• 0.

Solution by R. Knowles, B.A. ; Rev. T Galliers, M.A. ; and others.

Let PQ be the chord, A the vertex, M the foot of the perpendicular,
and Xiy-iy x^^, hk the coordinates of PQ and its pole respectively ; the

equations to AP, AQ are 2^1^ » yi^;, x^^y^ (1> 2).

By the condition, yiX^—x^2 — ^1^2 + y 1^2 ^^ ^ (k^—4tah) = (4a + A)2...(3).

The equations to AM, PQ are 2ay + kx = 0, ky = 2a {x-^-h) (4, 6).

From (4) and (5), k '^—^ayjx ^-'2aiSi:D.e, h ^^{a^ + y^)lx ^r^—r.Beoe,
and substituting these values in (3) we have the result in the question.

(D. Edwardes.) — Prove that (1) the squares of the lengths of
the normals drawn from a point xy to the ellipse b^x^ + a*y^ « a^^, are
given by the equation {p^-^-iV +pW ■\'9q*)f^ + JJY}^

= 4{r*-(2V + 3i?2)r2 + 3U + V2}{(^4-3^)r*-(2p2U-3^4V)r2 + U2},
where U = b^x^ + a^y^ - a^^, V= x^ + y^-a^-b^, p'^a^ + b^, an'd ^=a«*2;
and (2) if on the normal at P, a length PQ be measured inwards, equal
to the semi -conjugate diameter, the squares of the lengths of the other

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three normals drawn from Q are given by the equation

+ {4(a-*)2PQ2(2aS+2*2 + a*)-4a«*2(2a2 + 2*2-7a*)} r'

-4 {(«-*)2pQ2-a2*2}2 = 0.

Solution by Professor Sebastian SmcoM, M.A.

1. To eliminate k from the equations

A;2"(A; + a2)2 (A- + ^2» (^k + a^y (k + 1^)^ ^
Eliminating a^ and y2 alternately, we obtain

k* + 2*2^:3 + ^.2 {^4 + (^2 _ ^) y2_ ^2^:} _ 2>fca2^2^- a2i4y2 =, 0,
k* + 2a2^ + ^2 1^4 + (^2 _ ^2) a;2 - ^2^2 J _ 2ka'^l^^ a*l^r^ = 0,
whence, introducing IJ, V, &c., we have

2A:S + X:2(H«-V)-(?4r2 = 0, k^ +k {V-ph'^-2g^r^ ^ ...{I, 2),
the eliminant of which is the required result.

2. The coordinates of Q will be given by

whence U = («-*)« PQ2«a2i2, V = -2«*, then (I) may be written

2X;2(A; + ad) + r2(A;2-a2^) = 0, 2;t2 + ^2;t-a^ = 0,
and (2) becomes k^ + k {{a'-b)^FQ^ -a^^ ^{a* + H^^} ^2a^h^ = 0,
and the required result at once follows by elimination.

9401. (J. Brill, M.A.)— Prove that, if « and r be positive integers,
{a+l){a + 2)...{a + n) _ {b+l) {b + 2) ... (b + n) (e+^) {c+2) ... (e + n)
«i (»-!)! («-2)!2!

wherea=»«r, *-(n-l)r-l, tf = (»-2)r-2, rf=: («-3)r-3, &c.

Solution by W. S. Foster ; Sarah Marks, B.Sc. ; and others.
Let (a + 1) (a + 2)...(a + ») = a'»+i?ia'»-i+ ...+jt7„;

then the given expression
= 5|2- J a»-«-.n . ^n-,+ i^ziL) ^.,_ ....| from y - to 17 = n,

-^^J^^^'^-^'n[a-{r^l)y-^^ ^-^ [«-2(r+l)]n «-...jp,.
and »'^-n[a-(r + l)]'^+^^il^[«-2(r+l)]*-...

=s A! . coefficient of a;* in ^«—7j^<»-(»- ♦!)]«+ ^i!iziile[a-2(r+i)]«_

1.2 *"'

f.<?.,in^[l-«j-(r+i)x-]n^ =0,if Ais less than «, and =»:w!(r+l)«,ifA=«;
therefore the g^ven expression — (r + 1)».

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9505. (Professor Wolstbnholmb, M.A., Sc.D.) — Prove, without
evolution, or the use of tables, that 3x2* -2* lies between 3-5022831...
and 3-502282... ; the latter being nearer to the exact value.

Solution by D. Biddle.

Let a = 2*. Then Za^-a=^ x\ also a* = 2, and 3fl»— a* « ax, whence
a2 « 6 - aa; ; thus 3 (6-*r) -a = a-, and a {Zx + 1) = 18-a?. Cubing both

sides, we obtain a^+ l%x « 106 (1).

"We can now find x approximately by a series of trials, correcting accord-
iDg to the successively reduced errors. Let A = the portion of x already
found, say 3, which is easily seen to be the first figure, and let A + e = a;.

Then we have (A + «)'+ 18 (A + z) = 106, A3+18A= 106-A; (2, 3).

Subtracting (3) from (2), we further have

3A«z + 3Az2 + z3+18s = A:, 2 = A;/(3A'+ 18 + 3Az + ««) (4,5),

or roughly, especially as z diminishes, z — A;/(3A'+ 18) (6),

A, A3, 18A, k, 3A2+18.

3- 27- 64- 25- 45-

3-5 42-875 63- 0125 64-75

3-50228 42-95784 63-04104 0-00112 64-7979

3-5022821, 42-9589211, 63-0410778, 0-0000011, 54-79794
3-50228213 nearly = x.

[Otherwise : If x be the value, and y = 2a:- 7, we shall then have
y+7 = 6x2t-2^ (y + 7)3 = 216 x 4 -16-18 y 4 (y + 7),
or(y + 7>3 + 72(y + 7) = 848sy3^.2iy2+219y + 847, or y3 + 21y2 + 219y«l.
This cubic has only one real root which is positive, and

which are the given limits. Since 21y ^y^\A nearly = -^, it is clear that
the value of x is nearer to the inferior Hmit.]

" Something or Nothing ? " By Charles L. Dodqson, M.A.

In the years 1885, 1886, there appeared in regard to a Solution of Quest.
7695 (see Vol. xliii., p. 86, and xliv., p. 24) a discussion about a
difficulty in the Theory of Chances, of which the following question was
treated as a typical example : — *' A random point being taken on a given
line, what is the chance of its coinciding with a previously assigned
point ? '* On one side it was maintained that the chance is absolute zero :
on the other side it was maintained," by myself and others, that it is some
sort of infinitesimal, and not absolute zero. The arguments on both sides
were fully stated, and my only excuse, for re-opening the discussion, is

VOL. XLIX. N

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that I have a n^w view of the dif&culty to offer to the supporters of the
** absolute zero ** theory.

I assume that both sides accept the following axioms: — (1) that no
aggregate, however infinitely numerous, of absolute zeroes can constitute
a magnitudey however infinitely small ; (2) (an example of the preceding)
that no aggregate, however infinitely numerous, of points can constitute
any portion, however infinitely short, of a line ; and hence (3) that, if
the chance of a random point on a line coinciding with a single selected
point be absolute zero, so also is its chance of coinciding with one or other
of a selected aggregate of points ^ however infinitely numerous.

I now propose two questions : —

I. *' A random ^int being taken on a g^ven line, what is the chance of its
dividing the line into two commensurable parts ? " It seems clear that we
are here dealing with a selected aggregate of points ^ since it is impossible to
mark off any portion of the lincy and to say " Wherever, in this portion,
the random point shall fall, it will divide the whole line into two com-
mensurable parts." I assume, then, that my opponents would answer
** It is absolute zero,''*

II. ** And what is its chance of dividing the line into two inewnmen-
surable parts P " Here again they must answer " It is absolute zero,^*

And yet one or other of these two events must happen ! Hence, the sum
of the two chances must be mathematically represented by unity ; that is,
one or other (though we cannot say which) must be — not only
** something y^^ not only a certain infinitesimal y of some inconceivably
high order — but must actually reach, if not exceed, the finite value
of one- half !

9506. (Professor Hudson, M.A.) — Prove that (1) the parabola
y3 = 2/ (a; + /) can be described by a force to the origin which varies as
r/(« + 2/)3; and find (2) what ambiguity there is in the case of this law
of force.

Solution by Professor Wolstenholmb, M.A., Sc.D.
The polar equation is sin^ — 2— cos 6 + 2— ,
or 2/i# « -cos + (1 + sin' 0)*, (« = 1/r),

\ rf02/ ^ <i« V(l + sin»0)*/

= (l+sin20)» + -^5l26L._8in2^co8^e= 2 ^

(l+sin-e)* (l+sin'e)* (l+sin^e)*
therefore the central force

The centre of force is the centre of curvature at the vertex of the para-
bola, and, when the moving point reaches the vertex, it can describe, with-
out any extraneous pressure, either the circle of curvature or the other

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half of the parahola, apparently at its own choice. Of course the Ques-
tion is not one of physics at aU.

[Otherwise .—Let A be the vertex, S the focus, C the centre of force,
AS = SO « il; draw SY, CZ
perpendiculars on the tangent
at P, and SR perpendicular on
CZ: then AY is perpendicular to
AS, and the triangles CHS, ASY
are similar ;

hence CR : CS = AS : SY,
or CR.SY = AS2,

and CZ.SY = CR.SY + SY2 = AS' + AS.SP = AS(AS + SG)

= AS (AC + CM + MG) = AS (aj + 21),
where PM is the ordinate and PG the normal at P. But we have
central force : normal force = CP : CZ,

and normal force = ..^^^'^^y'' Z ■ ^^

radius of curvature CZ2 2SY^

central force = ^^'^1'r.tF = ^'- CP/ 2AS {x + 21)^ = ^/"^ ^, .
2SY3CZ» ' ^ ' l{x + 2f)^

It would be easy to make this strictly geometrical, or rather to
frame a geometrical proof on these lines: an expression involving
A* cannot be made geometrical, but it could readily be proved that the
central force must vary directly as CP, and inversely as the cube on AC]

9087. (H. FoRTKY, M.A.) — Show that, when the cards are dealt out
at whist, the probability that each player holds two or more cards of each
suit is -2062806, &c. ; or the odds are about 4 to 1 against the event.

Solution by the Proposer.
If one suit be distributed anyhow amongst 4 persons o, jS, y, 8, the num-
ber of possible distributions is the sum of the coefficients in the expansion
of (a-i-/3 + 7 + 8)**. But, if each person must have at least 2 cards, the
number of distributions is the sum of the coefficients of the terms divisible
by a^BrY^ ; that is, the sum of the coefficients in

13 ! o2i8V82( _M_ + — ^^ — + &c. \ ,

Call this expression M ; then, if all the 4 suits are distributed subject to
the same restriction, the number of distributions will be the sum of the
coefficients in the expansion of M'*. But at the game of whist 13 cards
are dealt to each person. Therefore, with the above restriction, the
number of possible deals is the coefficient of a^^0^^^d^^ in M*, or of

2tt^ Sa^a Sa^fl^ 2a3j3y

•(2!)3 6!3!(2!)2 6!4!(2!}2 6!(3I)22!

{i3!r{^

^ Sa^fl^y ^ ^o?Byd

(4 I;2 3 ! 2 » 4 ! (3

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or (reducing) in

86804 {18 (6) + 42 (41) +63 (32) + 84 (311) + 105 (221) + 140 (2111)}*,
where for simplicity for 5a*, 2a*j3, &c., I write (6), (41), &c.

Now, for 18, 42, &c. ... 140 write a, b, .../, and let a (5) = A,
*(41) « B, .../(2111) = F. Then (omitting for the present the factor
8580'*}, we have to find the coefficient of (5555) in the expansion of

(A + B + C + D + E + F)^
or in 5A* + 45A3B + 65A-B2 + 125A2BC + 242ABCD.

As a preliminary step I give the squares of the quantities A, B, &c., and
the product of every two in terms of the symmetnc functions ; but as we
want ultimately the coefficient of (5555), 1 omit from these values all
functions involving an index greater than 5.

A2 = a2.2(55),

B2 = d* {2 (55) + 2 (541) +2 (442) +4 (4411)},

C« = c2 {2 (55) + 2 (532) + 2 (433) + 4 (3322) } ,

D2 = rf2 {2 (442) + 4 (4411) + 2 (4321) + 2 (3322)},

E2 =r tfS {(442) + 2 (4411) + 2 (433) + 2 (4321) + 6 (4222) + 4 (3322)},

F2-=/2{(4222) + 2(3322)},
AB =a*(54l), AC=a<;(532), AD = eK;(5311), AE»«^(5221), AF=0,
BC = be {2 (442) + 2 (433) + (4321)},
BD = bd {(541) + (532) + 2 (5311) + 2 (4411) + (4321)},
BE = be {(532) + (5221) + (4321) + 3 (4222)},
BF = */{(5311) + 2(5221)},

CD = ed {(541) + (5311) + 2 (433) + (4321) +6 (3331)},
CE = «J {(541) + (532) + 2 (5221) + 2 (442) + (4321) + 3 (4222) + 2 (3322)},
CF =c/{ (5311) + 2 (4411) + (4321)},

DE^de {(532) + 2(5311) + 2 (5221) + (433) + (4321) + 3 (3331) +4 (3322)},
DF = df {(5221) + (4321) + 3 (4222) } ,
EF = <?/{(4321) + 3(333l) + 2(3322)}.

We shall now have no difficulty in finding the coefficient of (5555) in
every term of the expansion of (A + B -r C + &c.)*. Take, for instance,
the term 12D2EF ; then, omitting for the present the factor 12 which
multiplies all terms of that form, we have
D2EF « DE . DF = d'-t/ {{5'S2) + 2 (5311) + 2 (5221) + (433) + (4321)

+ 3 (3331) + 4 (3322)} x {(5221) + (4321) + 3 (4222)}.
But (5555) can arise only from the products of the complementary func-
tions (5221) X (433), (4321);4321), 3 (4222) x 3 (3331) (1, 2, 3),

and the coefficient is 12 in (1), 24 in (2), and 9x4 in (3) ; therefore the
coefficient of (5555) in D^EF = (t^efiVI + 24 + 36) = 72d^ef.
It is convenient for the present to divide each coefficient by 24, which
factor can be re-introduced after summation, and this being understood
the coefficient of (5555) in D'^'EF is Zd'^ef, Determining the other co-

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efficients in like manner, and collecting those corresponding to 2 A"*,
2A-^B, &c., we get

Group of
terms.
2A^
2A3B

2A2B2
2A2BC

2ABCD

Coefficient of (6556).

a* + 9{b* + e* + d* + e*) +f* = P, suppose,
^(2a + 6(i+/) + c3(2a + 6tf)+rf3(a + 6* + 3c + 6tf + 2/)
+ e'^{a + Sb'^6e+l2d+7/)+P^ = Q,

+ er^(8^+/2) + 2«2/2 = R,
i^{ad+ Zed + 2ce + Scf+ 2de)
+ c^{ae+2bd+ Zbe + 2bf+ Sde + df+ 2ef)
+ rf- {2bc + 2be + b/+ Zee + 5cf+ Zef)

+ e^iadi-2be+ibd + ib/+\^ed + y-i-Zdf)+P(ed + ee-^^)'S,
abe (d + <?) + aede + be (ode + 2df+ef) + def{b + c) = T.

Substituting for a, by e, &c., the numerical values 18, 42, 63, &c., and
reducing, it will be found that

P = 2,090,086,733, Q = 4,366,141,668, R = 1,676,987,172,
S = 3,888,459,288, T = 366,476,292.
Therefore the coefficient of (5555) in (A + B + C -r &c.)'*, or in

^A-* + 42A^B + &c. = P + 4Q + 6R + 12S + 24T = 85,079,518,906.

Now, remembering that every coefficient Was divided by 24, and that
we have omitted the factor 8580^, we see that the nimiber of ways of
dealing out the cards so that each player may hold 2 or more of each suit,

= 24 X 8580-* X 85,079,518,906 = H suppose.
Then log H = 28*0439855.

Let K = number of ways of dealing out 4 hands, without any restric-
tion. Then K = 52 ! -^ (13 I)-* and logK = 28 7295271, therefore

logH/K = log H -log K = -3144584 = log -2062806,
therefore H/K = -2062806 ... = the required chance,

and the odds are about 4 to 1 against the event.

9481. (W. S. McCay, M.A.) — AB is the diameter of a semicircle;
show how to draw a chord XY in a given direction, so that the area of
the quadrilateral AXYB may be a maximum.

Solutions by [\) the Proposer ; (2) D. Biddlb and Prof. MacMahon.

1. Drawing the chord XX' perpendicular to XY, the quadrilateral is
equal to the triangle XX'B, and if the chord BC be drawn parallel to
XX' the problem is reduced to construct the maximum rectangle standing
on BC and having its vertices (XX') on the circle, the solution of which
is well known ; X is the middle point of the intercept on the tangent at X
between BC and its perpendicular bisector (Townsend, Vol. i., p. 47).

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2. Let be the mid-point of AB, and centre of the semicircle. Draw
AC parallel to the j?iven direction, and OD perpendicular to it, cutting
AC in E. Bisect AE in F, through which, with centre A, draw the arc
GFH. Also draw GI perpendicular to AB, and make IK = AG. Draw
KL at right angles to AB, and make AM = AL. Bisect FM in V, and
make EP = VF. Finally draw PX parallel to OD, and XY parallel to
AC. The quadrilateral AXYB, being completed, is that required.

For the conditions are fulfilled when, with infinitesimal bases tangential
to the semicircle at X, Y, the pairs of triangles whose apices are respec-
tively at A, Y, and at B, X, counterbalance each other, that is, when
YT-YS = XR-XQ. Let AB = 1, then YS = XR = J(l-cosXOY) ;
also XQ = i (1-cos AOX), and YT = J (1-cosBOY) ;

whence i (cos AOX + cos BOY) = cos XO Y = 1-2 sin« DOX.
But it is easy to see, by reference to p, q, r in the diagram, that

i (cos AOX + cos BOY) : sin DOX = AE : AG.
"We therefore have a quadratic, whence

sin DOX = {(iAE)2+i}*-iAE.
The construction follows this formula.

[ Otherwise: — For the maximum area, the squares of the variable sides mus*
be m Arithmetical Progression. For, if we take X'Y' a consecutive position
of XY, then it is evident that the areas XX'Y + YY'X differ only by an
infinitesimal of the second order from the areas AXX' + BYY' ; hence, in
the limit. (AX)2 + (BY)2 = 2 (XY)^. Now take, further, Z the diametral
j}uiiit ut X ; draw AZ, YZ ; draw the diameter GM parallel to the given
direction and hence bisecting YZ at right angles, and draw AP' perpen-
dicular to OF. Then AY^ + AZ2= 2YZ2= SM'Y^ ; hence AM'^ = 3M'Y2,
AO^ + Oil's + 2PG. 9M' = 3(OY2-OM'2;, (P'G + 2GM0 (20M') = 20A2;
and tbiiB the quadrilateral may be constructed by finding Q', so that
l>(i\ i Hi'= 20A2, and through M', the mid-point of OQ', drawing M^
perpt'iidicular to GM' ; this determines the vertices Y and X.]

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9459. (Professor Genese, M.A.) — If py 9 he the polar coordinates of
a point whose coordinates referred to axes inclined at any angle w are
«, y, then a:/p, y/p may be denoted by C (0), S {$), Prove that

S(e-<^) = S(e).C(<^)-C(e).S(<^),
c (e+<^) = c (e) . c (^)-s (e) . s (<f>).

Solution by Profs. MacMahon, B. A., Ignacio Beyens ; and others.

The following is a general proof by the method of projections. As cos B
is the orthogonal projecting factor, so C (0) may be called the o^-gonal
(om6gonal) projecting factor. Take any two angles e and <^, positive or
negative, the initial line of ^ being the terminal line of ; then a unit
distance on the terminal line of <b resolves into C (^) on its initial line and
into S {<!>) in the w-gonal direction which makes angle « + d with a;-axis.
Take the w-gonal projection of all three on this axis, then (whatever
a.ud<p) C{0+<f>) =C{<t>) .C(0)+S(<p).C (« + e); but it is evident from a
figure that C (co + 0) = -S {0), therefore

C(0 + <t>)^C{0),C{<i>)-8{e),8(<p) (1).

Now change into (a— 6), and observe that

s («-e) = c (0), c («- 0) = s (e),

therefore S{0-<t>) « 8 {0) .C {<!>)- C{0) ,S{<f>) (2);

hence also 8(0 + <t>) ^ 8(0) .C (-<l>) + C (e) . S (<l>) (3),

C{0-<l>) = C(0),C{^<l>) + 8{0).8{<p) (4).

It is worthy of remark that the orthogonal formulas for sin (0 + <f>) and
cos {0—<l>) are not so general in their nature as the other two formulas,
since they require C (— ^) = C (<^), which is true only when tp = nir [n an
integer). It may also be noted that, when <p ^ n ,^ir, (n an odd integer),
C(-<^)«-C(^).

9477. (Swift P. Johnson, M.A.) — A, B, C and a, h, c are two triads
of points on a sphere ; show that, if the circumcircles of the triangles
KbCf "Bca, Oab meet in a point, then the circumcircles of the triangles
aBC, bCA, cAB will also meet in a point.

Solution by Professor Schoute.

A stereographic transformation and a transformation by reciprocal radii
vectores, the centrum of which is the point common to the circles Abe,
Bcttf Cabf lead to the following self-evident problem. When on the
sides b'e\ c'a', a'V the points A', B', C are given, the circles a'B'C,
h'Q'A!y c'A'B' meet in a point.

9516. (I^- Biddle.) — Prove or disprove that (1) a circle B is not
properly drawn at random within a given circle A, unless its centre be
first taken at random on the surface of A, and its radius be subsequently

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taken at random within the limits allowed hy the position of its centre ;
(2) putting unity for the radius of A, r for the radius of B, and x for the
aistance hetween the two centres, there are two things requisite in order
that B may include the centre of A, namely, that x he less than |, and
that r he hetween x and 1 — ir ; (3) from a favourahly placed centre, the
chance of the radius of B heing such as to make it include the centre of
A is (\-2x)l{\—x)\ (4) the chance is identical for 2ifX . dx positions,
which form the circumference of a circle of radius x, around the centre of
A ; (5) the prohahility that a circle B, drawn at random in a given circle
A, shall include the centre of A, is not correctly foimd hy the formula

14 rl-ar t\ r\-x

xdxdy -f- 2ir xdxdy = i,

ix Jo Jo

since this assumes that the numher of circles capahle of heing drawn
from anv centre is proportioned to the upper limit of the radius ; leaves
out of account that one centre, (me radius, one circle B, are taken each
time ; and gives a result which actually does not fall short of the chance
that the centre alone shall he favourahly placed ; (6) the prohahility in
the case referred to is correctly found as follows : —

P = 2ir f*a: (^,^) dx-^ 2Tr T x , dx ^ H + 2lDgei

+ 2-61370564 = 0-11370664, or less than |..

Solution hy W. S. B. Woolhouse, F.R.A.S.

By " drawn at random,** as stated at the heginning of this question, it
should he understood that there is not to he any conditiort whatever affect-
ing the circle B, excepting that it must he wholly included within the
circle A. In order that the circle B may be properly regarded as so drawn at
random, the correct mode of procedure is to discuss all the cases that can
arise from an indiscriminate admission of every position of the centre, and
also, at the same time, of every possible magnitude of the radius consistent
with the foregoing condition alone. This important principle is strictly
carried out in (5), the first working stated in the question by which I con-
sider the probability P = J to he correctly obtained.

The probability otherwise deduced in (6) would be correct if the circle
B were drawn at random in a very restricted sense, that is, assuming (I)
that its centre is first taken at random within the circle A, and (2) that
onlij one radius is allowable. This is undoubtedly special, and the circle
B is not drawn at random in the free sense of the term, the cases taken
into consideration being but an infinitesimal portion of the total cases.

To further show how misleading partial considerations are in questions
of this nature, suppose that the circle B is first taken at random from an
infinite set of circles having radii from to 1, and then placed only once
on the circle A, radius = 1 . Now, if the radius p of B fall between and |,
it may or may not include the centre of A ; and if it should fall between
\ and 1, it will, when placed on A, be certain to include the centre. The
resulting probability according to this arrangement will therefore exceed
\. Thus, when p = ... J, the probability will be

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109

and, when p «= ^ ... 1, it will be unity. The resulting probability is
therefore f * -^— dp+Cdp^2(l- log, 2) = -61 21 .

Jo (1-pr h

If the gfeneral problem be considered absolutely, so as to include all
possible cases of the circle B and its positions, then, when the radius p
ranges from to \, the successful positions of the centre = 2irp2 ; and when
p ranges from J to 1, the number of positions = 2ir (I— p)^, all of which
are successful. Whence the total successful positions

-2Tj'pVp + 2Tj\l-p)2^p=2ir(^ + V,) -iT.

Also for every value of p the number of positions = 2ir (1 - p)', so that
the total number of positions - 2ir [ {l-p^dp = Jjt. And therefore by
division P « i, as before, which is the true result.

9407. (W. J. Greenstreet, B.A.) — From a point outside a circle
centre C, APQ is drawn cutting it in P and Q ; AT is a tangent at T :
show that it is always possible to draw such a line that AP shall equal
PQ, as long as AC < 3CT ; and that then 3 cos TAG =» 2 V2 cos PAC.

Solution by R. Knowlbs, B.A. ; Sarah Marks, B.Sc. ; and others,

PQ cannot be greater than the diameter of the circle ; it may be equal
to it, and in that case AC » 3 times the radius.

CosTAC = AT/AC, cos PAC - AN/AC (N the mid-point of PQ),
therefore cos TAC / cos PAC = AT/ AN . AN = JAP ;

and (Euc. iii. 36) AT =v/2 AP, therefore 3 cos TAC = 2^2 cos PAC.

9425. (Professor Hanumanta Rau, B.A.) — Prove that the sum of the
products of the first n natural numbers taken three at a time is

Solution by Prof. Aiyau, B.A. ; Rosa Whapham ; and others.
Let „P3 stand for the sum of the products taken three at a time. Then

nP3= n-lPa + ♦»«-H 2 = n-lP3 + ♦»•!] ^ -f ^^ q J

Therefore „P, - 2«„.,P, = 16 ^Ji±llf' + 2o'^^^6 (5±i)!i'

6 ! ! 4

^i^n^{n^\Y{n--\){n-2),

VOL. XLIX. O

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Similarly,

.p. - :S«->P, = 105 M!!V210 (?Lt |J!!! + 130 (l±l)!l' + 24 (l±l)^'.

8 ! 7 ! o ! o !

And, calling the coefficients (106, 210 Ac), «, *, c, d, we have
.P. = 9« (^^Jl^Vs (a.*) (!i±lll!l. 7 (*..) ^^'

and 80 on for nT^, ^Py, &c. ; the first coefficient in J^r being
1.3.6... (2r-l).

9390. (N'Importb.) — In any triangle ABC, prove that
a cos 2A cos (B-C) + &c. = - ?? - - ^.

Solution by G. G. Storr, M.A. ; J. Young, M.A. ; and others,
a cos 2 A cos (B — C) = 2R sin A cos 2A cos (B - C)

= R (sin 3A- sin A) cos (B- C) ;
2 8in3Acos(B-C) =0,
2 sin A cos (B - C) = sin 2A + sin 2B + sin 2C =» 2A/R* ; hence, &c.

9469. ("W". J. C. Sharp, M.A.") — If ji? be a prime number and r<p^l,
prove that (1) r! (ji?— r — 1)! +(— l)** is a multiple otp; and hence (2), if
p^ 2^-1, {(^-l)!}»+(-l)«-^ is a multiple of 2^-1.

Solution byW.S. Foster.

1. Since (jp-r- 1) ! = M (;?) + {^l)P-r-i (r + l)(r+ 2) ... (p-l)f
therefore r! (jj-r-l)! ^^{p) + {-l)P-r-i (p-l)];

and, since j9 is a prime number, we have

(i)-l)! + l =-M(j3), therefore r!(p-r-l)! = M(j5)-(-.l)i»-»-»;
and p must be greater than 2, or r would be nothing, therefore p — l
must be even, therefore (- 1)**"*""^ = (— 1)%
therefore r! {p-r- 1)1 + (-ly = M (p).

2. Let p ^ 2^-1, and r = ^-1, therefore {(^-l)!}2+(-l)«-i is a
multiple of 2g— 1.

9365. (W. J. Barton, M.A.)— In the expansion of (1-3« + 3j;3)-»
ow that the coefficient of x*""^ is zero.

show

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Solution by H. Fortbt, M.A. ; C. E. Williams, M. A. ; and other*.
Let
(l-3d? + 3a:2)->-.(l-«a:)(l-/B») = (1 +aar + a2x« + &c. + a«— ^x*— l + &c.)

X (1 + i3x + ^2a;2 + &c. + i8^- 1 a*»- 1 + &c.).
Here the coefficient of ««"-J = a)'*'-* + a**-* /S + &c. + a/S^"-* + )3*"-^

= (a'^-ZS^-j/Ca-iS).
Now a + i8 = 3, a3 = 3; therefore a -i[3 + (-3)]*, 3 = i [3-(-3)]*,
.«'-t[l + (-3)]*-3«, i8««i[l-(-3)]» = 3a>«,
where « is a cube root of unity, therefore a* = 3^ « pfi, and a'"— /S*" = 0.
[We know that, when ^ < 4atf, the coefficient of x^ in the expansion of
(«r» + ^a? + <r)-iiB r;^ sin (n-h 1)7 ^

c sin 7

where rco8 7 = — */2<?, and rsiny- (4ac— *')*/(2(j) ; hence, with the
given values, we get for the coefficient of sfi'^- ^

r\ (6«-i) sin 6«7 ^ »i(«« - 1) sin 180£ ^ ^ ,
sin 7 sin 30° *-'

The readiest way of obtaining the development is perhaps by Hornek's
method of Synthetic Division ; and by observing the law of the series, we
see that if A, D be two consecutive coefficients, these and the following
coefficients will be A, 0, -3A, -9A, -18A, -27 A, -27A, 0, &c. ;
hence, if occur in any term, it will occur every 6th term. But it occurs
in the 6th term, and therefore occurs in the 6«th term, that is to say, as
the coefficient of x^ *-i.

9503. (Professor Bordage.) — Show that the roots of the equation
22x+2 + 4i-«^ 17 are a? = ±1.

Solution by A. M. Williams, M.A. ; A. H. Lewis ; and others.
The equation becomes 2^* + -^ — -»/-.
Solving in the usual way, we get 2^ • 2', or 2*'.

9389. (Professor Hanumanta Rau, M.A.) — Prove (1) that sin 6° is a
root of the equation IQjc* ^^x^-l^x^-^x-^-l =» ;
and (2) express the remaining roots in terms of trigonometrical functions.

Solution by J. Young, M.A. ; Professor Nash ; and others.
The equation 2 sin 6a— 1 = has » = 6° for one solution, and 9 = 30**
for another ; expanding and removing the factor 2 sin d— 1, we have the
equation given in the question, on writing x for sin 0. The remaining
roots will be found to be sin 78°, sin 222°, sin 246° ; or, in terms of functions
of acute angles, the four roots are sin 6°, cos 12°, -cos 24^, — C08 4b°.

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9410. (A. E. Thomas.) — If n and r are positive integers, and n>r,
then {e being the Naperian base)

l+!i±l^.i . (!i±^Ji!L±2) ^l(n^ l)(n ^ 2){n ^ 3) ^^^ ^
ri-l 2! (r+l)(r+2) 3! (r+l)(r+2)(r + 3) ''

1 h**-^t ^ {n-r)(n-r-l) 1 (n-r)(n~r- l)(n-r-2) . •)

r + 1 2! (r^l)(r + 2) 3! (>•+ l)(r + 2Xr+ 3) '*• *j

'[

Solution by R. F. Davis, M.A.
This identity follows from equating the coefficients of «♦•-•• in the de-
velopment of (l+a:)'»^+* = ^{(l+a;)»»^}, and dividing each side by
,»Cr ; and the necessity for f» >r is easily seen.

9499. (Professor Ath Bijau Bhut.) — Prove that the orthocentre of a
triangle is the centroid of three weights, proportional to tan A, tan B
tan C, placed at the comers A, B, C.

Solution by W. J. Greenstreet, B.A. ; Col. H. W. L.Hihb ; and others.

With trilinear notation, o. /8, 7 being the centroid of masses oaj bfi, ey
or a sin A, /B sin B, 7 sin C, at the angular points, we have

a = aabBinc / {aa + bfi + cy) ■■ o . 2 A/2a = o.
Similarly, 3 = 3 and 7 = 7.

Now, the orthocentre is sec A, sec B, sec C, therefore it is the centroid
of masses sin A, sec A, etc., or tan A, tan B, tan C.

[The in-centre is 1, 1, 1, therefore is centroid of masses sin A,
sin B, sin C ; the circum-centre is cos A, cos B, cos C, therefore is cen-
troid of masses sin 2 A, sin 2B, sin 2C ; the ex-centre (A) is —1, 1, 1,
therefore is centroid of masses —sin A, sin B, sin C; the nine -point centre
is cos (B — C), etc., therefore is centroid of masses sin A cos (B— C), etc.,
i.e.y of masses sin 2B + sin 2C, etc. ; the symmedian-point is sin A, sin B,
sin C, therefore is centroid of masses sin^ A, sin' B, sin- C ; the cen-
troid of the triangle is cosec A, etc., therefore is centroid of masses 1,1,1.]

9482. (S. Tebay, B.A.)— AB, AC, AD are edges of a tetrahedron ;
BE, CF, DG perpendiculars on the opposite faces ; P, Q, R their areas ;
jp, q, r the areas CED, Di B, BGC ; and S the area of the base BCD ;
prove that Pj» + Q^ + iir = S^.

Solution by W. S. Foster; Professor Beyens ; and others.

Let vectors AB, AC, AD be o, P, 7 respectively ; and let xYafi be the
vector DG ; then, since G is in the plane ABC,

SajS (7 + xVa/8) = 0, therefore x (Vo/3)- = - Sa/8y,

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113

Now r : R = tetrahedron DBCG : tetrahedron DBCA

= S(a-7)(i3-7)^Vo/8 : S {a-y){&^y)(-y)
= xS{a^-ay-yfi)Yafi I —80/87,

and R- = - i {Ya6)^

therefore Rr : - i ( Voi8)2 = ( Vai3) - 2 S (Va& + V70 + V/37) Va/3 ;

therefore Rr = - JS ( Vaj8 +- VJ87 + V7a) VajS.

Similarly, Q^ = - iS ( VajS + V/87 + V70) Yya,

and P^ « - iS ( Va/3 + Vi87 + ¥70) Y$y,

therefore Pi> + Q^ + Rr = - ^ ( VajS + VJ87 -i- Yya)^

= i ( ^AH^ )^ ^^ ^^ ^® *^® perpendicular on BCD
6 vol. ABCD ^

1;

-('

2AH

I \2 / 3 vol. ABCD \2 ,, . . ^_,^. ,
■ ] = ^ -^ ) = (triangle BCD)'.

[The theorem has heen otherwise proved by ordinary methods.]

8941. (W. J. C. Sharp, M. A.)— Prove that the cond-tions that th©
binai y quaaiic (a, b, c . . . ^ .t-, y)" bhould be reducible to a binomial form,

ff, b, Cf
b, c, d,

=■ 0.

[This is a generalisation of the catalecticant of the quartic ; those of
quantics of higher order admit of similar extension.]

Solution by D. Edwabdes.

Let the factors of the binomial form be given by Aj-' ^ giry + Cy^. Then,
substituting differentiating symbols for the variables and operatiug on
the quantic, the result must vanish identically. We thus get

Ac — Bi + Ca = 0^ or a, b, c, d ... =-0.

M-^c-^Cb =
Ae-Bd^Cc =

A/-B«+C<? =
&c.

a,

i.

0,

*,

0,

d,

1 c,

d,

«,

/.

9511. (E. B. Elliott, M.A.)— Of inhabitants of towns p per cent,
have votes, and of country people q per cent. Also of voters r per cent,
live in towns, and of non-voters * per cent. Find the proportion of the
whole pojjulation who have votes ; and show that jo, ^, r, « arc connected
by the one relation 100 {qr—pb) = {p + f) qr— {q + r) ps.

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Solution by E. F. Elton, M.A. ; Rev. J. L. Kitchin, M.A. ; and others.

Let V and N = total numbers of voters and non-voters ; then

100 — « : p = town non-voters : town voters = * N : -^ V,

100 100 *

100 — tf : q = country non- voters ; country voters = - ^^ N ; V,

* •" -^ 100 100 '

therefore ^-^^^ . ^ = iL = L?^^ . 12?J1-'-

p 8 \ q 100-*

H^nce lOO(^r-/?») = (j» + »)^r — (^ + r)i?*,

«» V

xind = • — number of voters : whole population.

lOOr-^r+i?* K + V ^ ^

9403. (RusTicus.) — Baby Tom of baby Hugh the nephew is and
uncle too. In how many ways can this be true ?

Solutions by {\) Professor Macfaklane, LL.D. ; (2) D. Biddle.

1 , Using the method of the analysis of relationships described by me
in Vol. XXXVI., p. 78, let T denote Tom, H Hugh, m male, / female, e
child, c* parent; then the data are

T = mccc-^inR, T = mcc-^c-^ mH (1, 2).

By combining the two, we obtain ^ — mc c c-^ mc c c-^- mH (3) .

Now, if the sex symbol before the third and fifth symbols of relationship
ftxo the same, the equation reduces to

H = mcccc-^mH, i.e., c-^ mH = c c c^ niE. (4).

Now the sex of c-^ on each side cannot be the same, for then cc would be
1 ; thkit is, a person could bt» identical with his or her own grandchild.
Nor can the sex be different, for then one parent would be the grandchild
of the other parent, which is against the law of marriage. Hence the
Bpx isymbol in the third and fifth places cannot be the same.

Kor can the sex symbol in the second and sixth places be the same,

for (3J is equivalent to c-^niK = e c-^ mc c c~^ niK (5);

and, Tinder the hypothesis, the sex of the parent of H is the same ; then
cc-^mcc =ly or c~^mcc=c-^ (6).

Now the sex of the first and third symbols cannot be the same, for
then c = e-^, i.e., the child of a person could be the parent of the person ;
nor ean the symbols be different, for then the consort of the child of a
person could be a parent of the person, which is against the law of
mairiJigo. The only cases left are

mcincmc~^mcfefc-^m = 1, mcfcfc^mcmcmc-^m = 1 ...(7, 8),
f)icmcfc-^mcfemc-^m= 1, mcfcme-^mcmcfc-^m = l...(9, 10).
JTt>w (7) means that Hugh's father and Tom's mother have married each
thtj Eippropriate parent of the other; (8) that Hugh's mother and Tom's
fiif,h(jr have married each the appropriate parent of the other ; {9) that
the fiithors of the two babies have married each the mother of the other ;
And ( I Oj that the mothers of the babies have married each the father of

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the other. Hence there are four, and only four, possihle ways in which
the phenomenon may he true.

2. Otherwise: — In relation to an uncle, the species "nephew" com-
prises the following varieties : (1) a brother's son, (2) a sister's sen, (3; a
half-hrother's son, (4) a half-sister's son, (5) a wife's brother's son, (6) a
wife's sister's son, (7) a wife's half-brother's son, (8) a wife's half-
sister's son. This list comprehends all the varieties.

Of these eight varieties, the four last are excluded from our present
consideration by the epithet **baby," attached to both individuals,
which puts the existence of a wife of either out of the question.

Again, although we may suppose a grandfather to marry a grand-
daughter, and that she and her mother bears sons about the same time,
one named Tom, and the other Hugh ; and, although we may suppose,
what is still more preposterous, that a grandmother marries her grandson,
and bears a son about the same time his mother does ; such marriages are
contrary to law, so that we must exclude also varieties (1) and (2), and
restrict our attention to (3\ and (4).

Let A, B represent husbands, and A', B' their roRpective wives,
whilst figures at the foot signify first or second. We then obtain the
following eight cases : —

Case 1.
A'l ^ A = A'j

B'=pBi

T

Hugh Tom
Case 3.
A'i=pA = A'2
J _ ^_

^2 T B' T Bi

I

A'2=^A

Tom Hugh
Case 5.

AiyA'rsAs

A2=FA'

■gh Tom

Hui

Case 7.
Ai =F A' = A,

AajA'
Tom Hugh

Case 2.
A'l =r A = A'a

B'2 nr B "T B'l

A'2=f A

I
Hugh Tom

Case 4.

A',=T=A = A'2

B'2 =F B =j= B'l

A'2=r A

Tom Hugh

Case 6.
Ai=f=A' = Aj

A2=fA'

Hugh Tom
Case 8.
Aj q= A' = Aj

I — — ,
B 2 "T~ B -J- B'l

A2=r A'

Tom Hugh

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It is quite possible, in these cases, that no disparity of age shall exist be-
tween A and A'j, or between A' and Ag.

Note on a Rectangular Hyperbola. By R. Tucker, M.A.

The equations to the hyperbola with respect to which the triangle of
reference ABC is self- conjugate, and to a tangent (or polar) thereto, are

4, = a2(/>2-c2)+/3'(er2_a^)+y(«2-^S) = (1),

aa'{b'2--(^) + fifi'(c^-a-)+rY'{a^-b^ = (2).

From (2) we see that the polar of either of a pair of inverse points,
i.e., foci of an in-conic, passes through the other point. The same property
holds for the related points

a=* ^ C * 6' + ^ C' + a^ a2 + A2 ^ ^*

From (1) we see that the curve passes through the circumcentre fO)*,
the Symmedian p')int (K), and the in- and ex-ceutres. Its centre, which,
of course, lies upon the circumcirele, is

a(*--c2)/a = fi{c-^-a^)lb = y{a^-b^)lc (4),

the ** inverse'* of which is the **isotomic" of the Steiner point

The tangent at the in -centre is a (i'-c^) ■»-... + ...= 0, a line which
passes through {a\l^yC% {b^-\-e\ c^-\-a\ a^-k-l^),

(cot A, cot B, cot C), and (l/*4-<?, 1/c + a, \la-\-b) (5).

The tangent at K is oa (**— c^) + ... + ... =0, which passes through the
centroid (1/a (A + c), ...), (cot A/rt, cot B/*, cot C/c),

the last point in (5) and other points through which Kw passes (6).

(See Note on ** Symmedian-point Axis, &c.'* Q. J., Vol. xx.. No. 78.)

The tangent at O is a cos A (*2_^) + .,. + ... = q (7),

which passes through the orthocentre and centroid.

The polar of the centroid is o (^ — c^) /« + ... + ...= 0,.
which is the circum-Brocardal axis.

The polar of the Steiner point is aja + fijb + yje = 0, that is, the axis of
porspective of ABC and of the triangle formed by tangents at A, B, C to
the circumcirele. The "isotomic" of is the point (1/a-cosA, ..., ...),
i.(?., the point P of my paper on " Isoscelians '* ?cf. Proc. L. Math. Soc.y
Vol. XIX., No. 316), hence the inverse of P is (a^ cos A, ..., ...), the polar

of which is a^ocos A (^2-c2) + ... + ... = (8),

i.e., the line Pit of the above cited Note (Q. J., Vol. xx.).

The polar of the centre of the Brocard ellipse is

aa(*4-c*)f... + ... = (9),

[♦ This follows also from the fact that O is the orthocentre of the
ox-centric triangle.]

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which passes through the centroid and the inverse of Kiepert's point

Similarly we can ohtain results from the polars of the two Brocard
points.

The asymptotes are given hy P^j + ^— c^. <?2— a', a'— i» (aa + .. . + .. .)*=0.
where P « a^^c^ (1-8 cos A cosB cos C).

9479. (A. Kahn, M. a.)— Solve the equations xyz = 24,

«(y-2)2 + y(«-dr)2 + «(:f-j^)2 - 18, x^ {y-z) ^y^{z-x) -k- z^ {x -y) --2.

Solution by R. Knowles, B.A. ; H. L. Orchard, M. A., B.Sc. ; andolhes.
Let y — mxy z ^ nx\ then

ars- 80/(m + m2« + «2) =. 82/(n f iw«« + m^) - 2^1 mn.
Eliminating m from these equations, there results

12«»-41n2 + 40/i-12 = 0, or («-2) (4«-3) (3«-2) = 0,
whence » =« 2, f , J, and m = f , 4, J ; and, since x^ = 2i/mn, therefore,
when m = 5, II — 2, a: = 2, y = 3, « «- 4.

[By inspection we have iP = 2,y=»3,2 = 4; and the symmetry of the
equations shows that congruent values are

a: = 3, y = 4, 8 = 2 ; a? == 4, y = 2, « = 3.]

9183. (A. R. Johnson, M. A.) — Investigate the induced magnetisation
of an elHpsoidal shell composed of any number of strata hounded by con-
focal surfaces.

Solution by the Proposer.

Let there be m strata bounded by confocal surfaces which may be con-
veniently indicated by the suflfixes 0, 1, 2, ... m, counting outwaids.

Let fin% V„ be the magnetic inductive capacity and the total potential
in the stratum between the (w — 1)*^ and n^^ surfaces. Then, if the in-
ducing potential be ES, the proper assumption is Vn = (A,»E + BhF) S.
The conditions at the «*^ surface give

An>iEn + B„+iFn = A„E,» + B„F„,

whence
where

/t„^i(A„,iE,. + B„,iP,.) = fin (A„E,.+ B„Fn) ;
fl„A»+i = -<7«Art — CrtB,,, a„B,,*i = inA„+/»B,,

, E„ Eh a — / '*'» — 1 \ ^'*^»* /. _ Jf!L __ 1

„ ^ — — i On — \ 1 I —t Cn — 1

^n F„ ^M#»*l 'F..F„ /*«♦!

F„

f iln E„ E„

/t« + l *M Frt

M» En E„

gn =

VOL. XLIX.

M«.l F„

P

(1).

(2).

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From(l) ?^A„j- lii -"-^Xa^^i^ (^-^ -^^ A„ = (3).

The internal potential Vq cannot contain F. Therefore Bq = 0, and
therefore, from (1), ^qAi = — ^o-^*

From (3) and from the relation just obtained, there results, after some
reductions,

*A,. . 1 ^ ( fln(Hn.l-fl>^ [" /*/.- !-/*» 1 M»(^»-^»»-l)-Mti-l(^»-^n-l) 1
A„ 1 Mm t 1 (/*/» - /*«-l) L H-n^l — ^n On J

fin- fin- 1 On )/ C M« (/*«-! — /^n-2)

M'l-M'i-l 0«_l J /t»-I-/*'»-2 0M-l3'

(M3(Mi-Mo) L'^i-Mi «i J Mi-Mooii'

f Ml A) — Mo3i) 7 /^\

I oo i

F

wlierp /8„ = ^^', /S; =^; On = fin (fin- fi'„).

Hunoe An,\IA^ « product of continued fraction (4) and those derived
frriiii it by writing? «-l, w-2, etc., instead of n; the last being
{;ti$ii ^^Bo)lao' That is to say, A„ + i/Ao = N„ + i, where N„+i is the
numerator of the continued fraction (4) expressed as a proper fraction by
the method of converarents.

Now the potential due to the induced magnetisation can have no term
in E in external space ; i.e., A,„ + i = 1. Therefore A^ = l/N„,+i, and

A„ = N„/N,„u (5).

Wy Diny in the same way find B„ from the equation

0«+l \ On On Ml \ On Onhnl

or proceed as follows.

Frrjin (1) whenM;;f>w, B,. = ^JL^-X-^9n^n jgj^

a,n a'»
so thnt (Mm+i-M»») B"»*i = ^^^-^0/» + M»n + l )8'«— MwiSw* (7).

[* When the fraction is written in the customary way, the last term
within each each pair of parentheses is to be placed over all that follows

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In the case of a composite spherical shell, E = c*, F = r-'-', and '

». = >•;'*>, 0:, = -:-^ '^», «» = ^4^ '-^r •,

* " » + 1 '* t + 1 "

SO that (4) becomes

( fln{flu*l-fln) r /*nM-Mn-l _ {>'/*n4l + (* + 1). Mnj { 1 ~ (^n-l/^-'"' '} "[

_ /i,..i(M».i-M.O ('Viiy'^^^/etc. / ipi+{i + l)fio
Mi.(m«— /u«-i) \ r„ / )/ '. / (2i + l)/ao *

Fri^m the comp'ete potential subtract the inducing potential, and the
result is the potential of the induced magnetisation.

8853. (A. Russell, B. A.)— Prove that

V =
Jo

J -^ V 4a%'^' U-p^' ic^q^J ^ ^

is a solution of the differential equation

Solution by the Proposer.
Since

r r f / (^- _£L, t^ -^„-„, ^- — ^ e-i'^*'P'^'i')dHdpdq,
Jo Jo Jo \ 4«%2' 4^2^2' 4^^.; ^ ^»

therefore 3^ = 3 ( f \ f{...)e-i'^'-P'^9')dndpdq,

at JQ Jo Jo

also

therefore
a'

e^^^ Jo Jo Jo V 4a2;*2' 4Z»V 4(:V/

Thus we see v satisfies the given differential equation. This solution of
the equation of the motion of heat in an eolotropic solid is suitable for the
case of a time-periodic source.

9524-. (Rev. J. J. Milne, MA.)— If yi, J/a* 2^3 ^^^ *^® ordinates of
three points P, Q, R on the paiabola y^ = 4«.r, such that the circle ou
PQ as diameter touches the parabola at R, prove that

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Solution hy A. E. Thomas, M.A. ; Rev. T. Gallibrs, M.A. ; and others,

(1) is derived from the well-known theorem, *' The algebraic sum of the
ordinates of the points of intersection of a circle and parabola is zero," by
supposing two of the points to coincide.

(2) The points where the circle on mj, mj, as diameter meets the para-
bola again are given by

(am*— awi') [am^—amj) -k- {2am i — 2am) {^am^^2am) = 0,
or, discarding the factor a' (m— /Mj) {in — m^jj

nfi + (wi + iwj) m t »»! *»3 + 4 =0,
which has equal roots if iwi <*» mj = 4, i,t. if yi '*' ya = 8«.

9267. (Professor Han VMANTA Rau, M.A.)— Given the base and the
vertical angle of a triangle, prove that the envelope of the nine-points
circle is itself a circle.

Solution by R. F. Davis, M.A. ; D. O. S. Davies, M.A. ; and others.

Since the circura -circle is fixed, the radius of the nine-points circle is
also fixed. The nine-points circle therefore touches, at the other extremity
of its diameter through the mid -point of the base, a circle having this
latter point lor centre and radius equal to the circum-radius.

9314. (Professor Beni Madhav Sarkar, B. A.) — Solve the equation
4- + yz =a a = 384, y + «x»»* = 237, « + a:y = <? = 192.

Solution by R. F. Davis, M.A. ; D. Watson, M.A. ; and others.

Since x — a-yz = (b^y)/z - {c—z)/y, we find y = (az— ft)/ (a^— 1)

and X = (fta-«)/ (22-1) ; hence («-c) («»- l)^ + (az-b) (bz-a) = 0.

There are, therefore, five solutions as each value of z determines a single
value for both x and y. In the given example, a; = 10, y = 17, « = 22 by
trial. The equation determining z is the subjoined quiutic,

2*- 1922^-2s3 + 9l392s2_ 203624s + 90816 « 0,

which is reducible to the following quartic by rejecting the root 22,

24-170.»-37422« + 9068s-4128 = 0, or

(22-852 + 40-728)2-=11048-46z2-15992z + 5786-77=- (10511z-7607)2&c.,

whence z = 189-6, 1-62, -63, -21-9 roughly; so that (-11*7, -18-1,
-21-9), (--8,238, 1-62), (1*24, 2-02, 189-6), (380, '5, -63; are approxi-
mately the other four solutions.

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9092. (A. E. JoLLiFFE, M.A.) — Prove that

Ml (2n-l)! i2n-2)l ... to (« + 1) terms = 1.

nlnl l!(«-l)!(n-l)! 2!(«-2)!(«-2)! ^ ^

Solution by K. E. Thomas, M.A. ; Prof. Matz ; and others.

3^ (1 +a:)- ^ (I ArxY [(1 +a:) -l]*-

= (l+ar)'»*'--r(l+a:)«**-i+ ^-^^ (1 +ar)'»+'-2-...

1 < ^
Equating coefficients of sf on both sides we get

1 « (jL±r)l ., ('Ll^jdll + Li!^^ {n.r-^y.

r\n\ r\ («-l)! 2! r! («-2)! ^ ^ ^ '^

r!«! l!(r-l)! («-l)! 2! (r-2)! («-2)! * »*" ^ -^ ^ ^
If r =» «, we have
, {Iny. ('2M-1)! ^ (2M-2)! . , ...

8782. (A. Russell, H.A.) — Prove that, if

a3 (* + <y) + ^3 (c + a) + c» (a + *) = lahc (/? + * + <?), then

<'>(*T^'--)/(.*-)-r-r=-«)/(.^-*)

(3) (*2-(;2)fa-^V{3a' + a(i + c) + 3r}
+ (c2-a«)(*--^y {352 + *((. + «)+,.«}
+ («3-*2)fc--?^)^{3c2 + c(a+^) + a*} =0.

Solution by the Pkoposer ; Professor Sarkau, M.A. ; and others.
1, 2. The given rehition may be written
{a^—bc)(b-{-c)a t ... + ... = 0, and since {a^-bc){b-\-c) + ... + ... = 0,

therefore (a^-hc) ^ = {b'^-ca)^-^^ {iP-ab) 'i^\ (a).

b — c c—a a — b

c.. ., ^ ab-i- ac—2'>c bc + ba-2ca ca + cb — 2ub ...

S""'l«rfy -iJZ? ^-i ^^ZiT- w.

a{b-c) b(c-a) c{a-b) ^^^'

From {a) we deduce ('i), and from (b) and {d) (1) follows.

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3. The given relation may also T)e written

a— ^■b +c =- ;

-r c c -t a a -t o

and putting a: = a — - — , y = &c., z = &c., x + i/ + z = 0,
■*■ c

and therefore . a* (y - r) + v** (z - a-) + r^ (x - »/) ^ .
Substituting for x, v, &c. their values, after a little reduction, the re-
quired result follows.

9416. (J 0*Byrne Croke, M.A. Suggested by Question 9360.)—
The sides of a polyhedron are of areas inversely as the pnrpendiculars on
them from a po'nt O, and 00' mtiels them in Pj, P^, I'a ... Pm, respec-

tively ; prove that ^^^4 ^^-^ . — ^ ... ^ -^,-; = n.

Solution by Professors Curtis, M.A. ; Beyens ; and others.

T\\^^ volumes of the pyramids which are subtended by the sides of the

Eolyhudron at O are equal each to n-'V, where V = vol. of polyhedron;
unrt.', ciiUiug the vols, subtended by these sides at O', t'j, r.^, t'a, &c., we have

OPi OP2 •• ' w-iy u-^Y

SBSG. (E. ViGARiE.)— Dans un triangle ABC si (a) est le pied snr BC
do lu jivmediane issue du sommet A, et si (a') est le point conjugue
hanniHiiquo de (a) ; deniontrer que Aa' est egale au rayon du cercle
d'AiKiilonius coirespondant a BC.

Solution by Professor Ignacio Beyens.

Be k propriete de la symediaue bien connuj — = — on deduit

aC b^

c/B _ oB ^ c^ J, . ,-D _ gg" ,Q _ f^^*^
a'C aC " Z»2' * - ^.2_^2' * ^' - ^2_^ '

Ertaia m [m) et (;«') sont les pieds des bissectrices qui partent du sommet

A, on Liura mC = - — et Qm' = ,, d'ou le rayon du cercle d'Apollonius

o-\-c c — b

*^ ^ «^'"' niC-\-Q)n' abe ...

ftera = =» = n (1).

Jlais, par le theorerae de Slewart, dans le triangle ABa' nous aurons
Ka'- . BC + AB^ . Ca' = B/ (AC^ + BC . C/},

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et rempla9ant Ca', Bo', AB, AC, BC par leurs valeurs on aura
rayon du cercle d'Apullonius d'apr^s (1).

9412. (A. B. Johnson, M.A.)— Show that, if 1, 2, 3, 4, 5, 6 be six
points on a conic, then = 2 (023) (031) (012) (156),
2 denoting summation with respect to all terms obtained from the one
presented by cyclic interchanges ; O denoting any point in the plane of
the conic, and (456), etc. the areas of the triangles 456, etc., described in
the order named.

Solution by Professor Curtis, M.A. ; G. G. Storr, M.A. ; and others.

If the six points (j'i, yi), (j^o, ya)* &c., all lie on the conic
(flf, ^ ^/, ^» ^'2^, y, 1) = 0,
we must have six linear equations in rr, ^, c, /, ^, ^, the condition for
whose being simultaneous is the determinant

= 0, or 2

^1% ^L'/i* i/i"

X

^4« y4» 1

^5» Vb. 1

^6» y6» 1

« 0.

V> ^eye* ye'* arg, yg, 1
Now, the first of the two determinants here written down is equivalent to

- (^2^/3 - ^3^2^ (^3^1 -^\yz) (^1^2- ^2^1) »

or to eight times the product of the triangles (023), (031), (012), and the
other determinant is double the triangle (456). Hence the theorem
stated.

If we make O coincide with the point (jTg, yg), we see that the condition
that six points may lie on one conic may be written

2 (023) (031) (012) X (045) = 0,

there being ten such products to be taken each with its proper sign. Of
course, each when developed would express the property of Pascal's
Hexagram.

8766. (S. Tebay, B.A.)— If AX, BY, CZ be opposite dihedral angles
of a tetrahedron, show how to construct the solid in order that

{tan i (B - Y) " tan i (C - Z) } tan i ( A + X)

+ {tan i (C-Z)-tan \ (A-X)} tanHB + Y)

+ {tun i (A -X) - tani (B- Y)}tan i (C + Z) =0.

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Solution by the Phoposer.

Let a, 3, <; be conterminous edges ; x^y^z the oppositcs, and A}, A], A3, A4
the areas of the faces bex^ eay^ abz, xyz ; then

volume = I ^^^sin A = J ^^sin X ;
a X

therefore ^i^:^ - i*!"? A.» = *^ A,» = ^^^ Aj'.

A4 ar sm A y sin B z sm G

Tj sin A sin B sin O ,
If . ^ = -i—^r — -r— Ti» wo have
sm X sm Y sm Z

tan \ (A + X) = Mtan ^ (A-X), tan i (B + Y) = Mtan \ (B- Y),

tan|(C+Z) =Mtani(C-Z);

M being a common factor. Whence the proposed relation. We also have

xyz
These conditions furnish the required construction.

9354. (Professor Mahendra Nath Ray, M.A., LL.B.) — A pencil of
four rays radiates from the middle point of the base of a triangle, and is
terminated by the sides. If the segments of the rays measured from the
prigin be x^^ yi, aig* V^y ^3» Vzt and 3-4, y^^ show that the identical relation
connectiog these lengths is

= 0.

*:'.

'i''

*.-'.

*;'

y:^

yi\

yi•^

y;'

('.V,)-'

, (^-jyj)"'

. (^ay.)-'

, (^4^4)-'

1,

1,

1.

1

Solution by J. O'Byrnb Crokb, M.A.
Obviously the relation which is to be established must be independent
of the position of the origin. Let » = vertical angle of the A ; and let
c, e' be the parts into which it is divided by a line of length k drawn to
the origin of rays. Then, we easily find that

1/A;2 8in2 » = a- -2 sin^ € + y{^ sin' €'-2 (j-i^i) "* sin € sin c' cos « ;
which may be written

«i ^r H 'C2 yf ^ + 'ca (j-iy 1) - * + ^4 = 0.
And, eliminating the constants between this and the three other similar

equations, we have

r-2

x-2,

^4"'» ^4"'* (^4y4)-^

a relation identical with that in the Question.

y,-^ (^iyi)-S I
y-2, (x2v^-\ 1

2. (T.yA-K 1

= 0;

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APPENDIX I.

SOLUTIONS OF SOME UNSOLVED QUESTIONS.
By W. J. CuRRAN Sharp, M.A.

2144. (Professor Wolstenholme, Sc.D.) — If from the highest point
of a sphere an infinite number of chords be drawn to points uniformly-
distributed over the surface, and heavy particles be let fall down these
chords simultaneously, their centre of inertia will descend with accele-
ration Iff,

Solution,

Let A be the highest point, AB the vertical diameter of the sphere,
and let P bo a point such that BAP = ; the acceleration on a particle
falling down AP is ff cos 0, and its depth at time t is ^ff cos^ . ^.

From symmetry it is evident that the centre of inertia lies in AB.
Now the area of a belt bounded by horizontal small circles such that
the lines to them from A makes angles and + b0 ^ 27rrsin20 x 2r86,
and the depth of the centre of inertia is therefore

f *' ^Sf cos2 0fix 4ir>'2 sin 20 d0 f *' cos^ sin d0
«Ji>__- «|^^Jj- = i^^,

4irr2 sin 20^0 cos sin 0^0

Jo Jo

and the centre of inertia descends AB, as a particle would which started
from rest at A, under the action of a uniform acceleration \g.

I have, of course, assumed that the particles were equal, as the data
would be insufficient without some rule as to their mass, and this simple
supposition gives the correct result. All question might be avoided by
inserting the word equal — thus, ** and equal heavy particles be let fall."

The result may be confirmed by the fact that all the particles reach the
sphere at the same time, viz., that in which the one which falls along AB
reaches B, and that then the centre of inertia is at the centre, as it
should be.

2146. (Professor Nash, M.A.) — D, E, F are the points where the
bisectors of the angles of the triangle ABC meet the opposite sides. If
a:, y, z are the perpendiculars drawn from A, B, C respectively to the

VOL. XLIX.

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opposite sides of the triangle DEF ; p^^ p^, p^ those drawn from A, B, C
respectively to the opposite sides of ABC : prove that

^ + ^' ^.^-ll + SsinAsin-S sin-^.
x^ y* ^ 2 2 2

Solution,

AE - -*^- , AF - — - (EucKd, vi. 3) ;
a+e a+b

therefore the triangle AFE - i -— sin A = lil^^

and FE2 - *«(j2 i-A- + -J_ - . -? ^^ cos a)

= , ^r— 7^ {2a2+2aft + 2a<? + A2 + <52-2(a2 + a* + fl<j + *c)cosA|

(a + r)2(a + i)5 «^ > ' j

A2-2

— — V {2a2(l-cosA) + 2a(*-<JC08A) + 2a(tf-*cosA) + an

(a f cy {a -k- b)^ ^ ^ ' "*

= ^fjf^ {j-cosA + cosB + cosC} ;

,.. ,, ^'^'f-^' ,, {i~cosA-hcosB-hcosC}«i- ^^^l__«in«A;

therefore ki^^a^ {{-cosA + cosB + cosC} - Ji^cSginS^ ^ iPi^a^;
therefore 3 - 2 cos A + 2 cos B + 2 cos C — Pi^/x'.

Similarly, 3 + 2 cos A - 2 cos B + 2 cos C « P2^li/"f

3 + 2cosA + 2co8B-2cosC =p^lz^\

therefore ^ + -^ + -^ = 9 + 2(cosA + cosB + cosC)

«= 11 + 8 sin JA sin |B sin \G,

2173. (Professor Wolstenholmb, Sc.D.) — The qnadric
aa:2 + V + C22« 1

is turned ahout its centre until it touches a'x^ + b'y^ + c'z^ = 1 along a
plane section. Find the equation to this plane section referred to the
axes of either of the quadrics, and show that its area is

Tr{a-^b^-c-a'-b'-c')^l{abc-a'b'c^)^.

Solution.

Let kx^ + By2 + Cz^ + 2Fys + 2Gzx + 2Yixy = 1 he the equation to the
quadric, the onginal equation of which was ax'^ + by^-¥cz^ «» 0, after it has
heen turned about its centre. Then, by question,

Ar2 + By- + 0^2 + 2Yyz + 2Qzx + 2Ha:y s a'x^ + h'y^ + cfz^

' +R(a:co8o + ycos/3 + «cos7)2,

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therefore A = a' + R cos'o, B -= *' + R cos^ /3, C « (?* + R cos^^,
F = Rco8/3co87, G = Rco8 7C08o, H — RcosaoosiS.
Now, by a Paper in the Froe. Zond. Math, Soc, Vol. xin., pp. 193 — 94,
A + B + C = « + * + c,
AB + BC + CA-F*-G2-Hi»a* + ^ + M,
ABC + 2FGH-AF2-BG2-CH2 = abc;
therefore a + b + e^ a' + b' + e' + R, because co8'o + co8?/3 + cos'7 = 1,
ab + bc + ca => a'V + b'c' + /a' + R {/»' (cos' /3 + cos' 7)

+ V (cos' 7 + cos^ o) + c' (cos' a + cos' /3) }
s= «'*' + b'e' + <^a' + R {rt' + *' + c'- a' cos' a-b' cos' /3 - c' C08'7},
abc = a'^<^ + R {^V cos' a + tfV cos' /3 + «'A' cos' 7} ;
therefore a^b^c^a' -V -^ _ 1

aAc — of b'e' b'(f cos' a + cfa' cos' 3 + a'b' cos* 7'

which is equal to the product of the squares of the axes of the section of

a'«' + yy' + tf'z' =s 1 by arco8o + yco8i8 + «cos7 =» 0,
(Salmon's Oewn, of Three Dim., Art. 97).

R and cos' a, cos' /3, cos' 7 are easily found from the above equations
which are linear in those quantities, and so the equation to the plane
j;cosa + yco8/3 + 2COS7 » 0.

4721. (Professor Sylvester.)— Prove that every point in the plane
carried round by the connecting-rod in Watta' or any olher kiud what-
ever of three-bar motion has in general three nodes, and that its inverse in
respect to each of them is a unicircular quartic.

Solution,

Let AB, BC, CD be the three bars of a three-bar system, where
AB = a, BC « bf CD = c, DA = d, and P be a point rigidly connected
with BC, and h and A;, respectively, the perpendiculars from P upon BO
and the portion of BC intercepted between B and this perpendicular.
Then, if BAD = <p, and be the angle which BC makes with AD, taking
A as origin of rectangular coordinates, and AD as axis of ^, the coor-
dinates of B are a cos <p and a sin ^, and those of C,

a cos <p+b cos 0, and a sin ^ + ^ sin ;
therefore c' = (a cos <^ + * cos d — rf)' + {a sin <^ + 3 sin 6)',
or 2acos<^ (ftcosa— rf) + 2asin<^.isine = c^ + 2bdcoB0—tfi^b^—{P,., (1).
And, if (iP, y) denote the point P,

X ^ a cos (p-k-k cos 6 — A sin 6, and y = a sin <p + k sin 6 + A cos 0... (2),
therefore 2(Arj? + Ay) cos0 + 2 (%-Ad;) sinO = ir2 + y'+ A' + ^-'-a' ... (3),
and from (1) and (2),

2(*a; + rfA?-W)cosd4-2(^y-rfA)sin0 = c2 + 2i^ + 2dii;-a'-A2-<i'... (4);

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from (3) and (4), 4 {{bi/^dh) {kx + hi/)-(bx + dk-bd) (ky^hx)Y
= {(*y-rfA)(a:2 + y2 + A2 + ^s_a2)-(Ary-Ax)(c2 + 2W- + 2d:r-a2-^-rf2)}«
+ {{bx + dk-bd)(x^ + y^ + h^ + ki-a')

^{kx + hfj){c^ + 2bk + 2dx-a^^b'-d^y (5),

which is the equation to the locus of P. This evidently has double

points, where

by-dh ^b -^ dk-bd ^ <?-^2bk-^2dx--a^-b^-'d^ ^ sunnose •
icif-hx kx-^hy a;2+ya + A2 + A-2_«3 ^' PP ♦

therefore p = dhx^dy{h-k)^h(c^^2bk-^a^-i;^^d^)

__ by — dh __bx + dk-^bd ,^y ,

" ky-hx kx-ithy ^ ^'

and, eliminating x and y from the equations (6), a cubic is obtained to
determine /?, and each value of jp g^ves one node, and there are therefore
three.
The equation (5) to the locus of P is easily reduced to the form

^{bh{x^^-y'^-bdhx-d{h'^^y^-bk)yY

-= {*2(;r5 + y2)-2WAy-2W(*-Ar)-^ + rf2(A2 + *2 + ;t2_2*Ar)}

X {a;2 + y2+A2 + A-2_aa} \f^2bk^-2dX''a^-V^^d^'\

which meets any circle at its intersections with a cubic, and therefore six
times at infinity, i.e.^ it is a tricircular sextic.

And, if the origin be at one of the nodes (and one must be real, since
the cubic in p must have a real root), the equation will be

Uo (a:2 + y2)3 + XJi (x2 + y2)2 + XJ3 (a;2 + t/VU3 + V2 = 0,
where Uq, Ui, Uo, U3 are homogeneous functions of x and y, of order
0, 1, 2, 3, respectively, and Vj one of order 2 ; therefore the equation to
the inverse will be Uq + Ui + Uj + U3 + Vj {x^ + y^ = 0,
a circular quartic.

4828. (The Editor.) — If the comer of a page of breadth a is turned
down in every possible way, so as just to reach the opposite side;
(1) show that the mean value of the lengths of the crease is

j{7A/2 + log(l + A/2)}fl,

and (2) the mean area of the part turned down is \\c^.

Solution ,

1 . Let ABHK be the page, and ADE the comer tumed down into th©
position DCE, meeting the edge BH in C, and let ar, y be the lengths
interctptud upon the edges AB, AK from the corjier to the crease ; then

\

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we have AB = a, AE =» ar, AD = y, and (C being the point where the •
comer A rests after folding) if AC cut DE in F, the angles at F are
right angles ; therefore

a2 = AB3 := AC3- BC2 « 4AF2- CE^ + BE^

therefore = 42:V- 2aa; (x* + y^),
and 2a;y2 = a(a;2 + y3),

and the crease :

'<-*»=)'- (-^j'
-•(^.)'

Now ar may have any value from x ^ iatox — a;
hence the required mean value

^ 1 p / 2^\ efa; = i- r(c2 + «2)Ww, if c« = «,and2a;-a = w2,
\ajia ylx—al a Jo

= f^{7V2 + log(l-hV2)} = |- {7V2 + log(l+V2)}.
2. Again, the area turned down = Ja-y, hence the required mean value

iajja \2ar-a/ 4>v/a Jo Vz ' ' 16

6391. (J- J* Walker, M.A.) — If O, A, B, C, D are any five points in
space, prove that lines drawn from the middle points of BC, CA, AB
respectively parallel to the connectors of D with the middle points of OA,
OB; OC, meet in one point E, such that DE passes through, and is bisected
by, the e^ntroid of the tetrahedron OABC. [^Quest. 6220 is a special case,
in two dimensions, of the foregoing theorem in three dimensions.]

Solution,

This may be generalised into — If ABCD... be a simplicissimum in
space of n dimensions, and Bi, Cj, D, ... the mid-points of the lines drawn
from any point P to B, C, D ... (all the vertices but A), and if parallels to
AB„ ACi, AD, ... be drawn through the centroids of CDE ..., BDE...,
BCE..., &c., (the faces of BCDE ...) respectively ; these will all meet in
the same point Q, AQ will pass through the centroid of PBCD ..., where
it will be divided in the ratio of «— 1 : 2.

Any straight line through (A.', /*', v ...) may be represented by

a

where a + b-\-c.

b e

And the parallel through (a.", /ia", v" ...) is

\-x"

_/t-M__^

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If, then, ABCD ... be the simplicissiiiium of reference, and P (\', /*', i^ . . .),
Bi,Ci,Di... are {K, i(M' + V), */...} {K, i/*', i(»'' + V) ...}, &c.,
and the equations to ABj, AC], ... are
X— y __ ji ¥ jr __ A.— V ji^ V __ jr_

&c. &c. &c.,

and those to the parallels through the centroids of (C, D, E ...)
(B, D, E...), *.<?., through

are

X « M ^ p-y/jn- 1) ^ ir~V/(«-l) ^
X'-2V"/i' + V 1^' "* ir'

;i~V/(w-l) ^ _p_ ^ y~V/(n~l) ^

X'-2V /*' iZ + V ^^

which all meet where

^ _ M~V/(n-l) _y~V/(n-l)^ ^ / ^\

x'-2v / »^ "* V«-ir

80 that, if (x'', /a", p" ...) be the point of intersection Q,

X" = - — 2--, /'«-'•-+ -, w = +- -, &c.,

fi— 1 n-r «— 1 «-l n— 1 n-1

X — V it V

and the equations to AQ are -T/ -rr "--=#/-=*••• »

X —V fJL" V

x-V ._. M. p _

x'-(» + l)V /+V / + V **•'

and the Hne passes through (-^, '^^t-Y, ^;^j •..) or (X,^,7...),

the centroid of (PBCD ...) ; and, since

(«+l)X = (fi-l)x" + 2V, (» + 1)m= («-1)/*" + 2x0,

(« + l)ir« («-l)v" + 2x0, &c.,

this centroid divides AB in the ratio ft— 1 : 2.

[Solutions by Professors Minchin and Gbnbsb are given in Vol. xxxiv.,
p. 40.]

7131. (W". J. C. Shaup, M.A.) — Prove that the vector equations to
the centrodes of a three-bar motion, which are easily derived from one
another by a linear substitution, are of the third degree in the vectors, and
reduce to the second where the algebraical perimeter of the figure is zero.

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Solution.

If AB, BC, CD be the bars, a, b, e their lengths, and AD = d the
distance between the fixed ends, the instantaneous centre is O, the inter-
section of the lines AB and CD. (Clifford, Di/namie, p. 166.)

It then follows at once that, if AO = r and BO = r',

2 cos AOD = tf±r^ = ^f)!±irl-_£)!=:^.

®' rr" (r''a){r'^c) '

which is the yector equation to the fixed centrode, and is, in general, of
the third order. If, however, r + r'±<?=s(r + r'— a— <?)±*, i.e.,
a+c ^ ±(b ±d), a factor will divide out, and the equation is of the
second order.

If BO =» R, and CO = R' the relation between R and R' will determine
the movable centrode, and it is easily seen that this is derived from the
above by putting r = R + a, and r' « R' + c ;

and the equation to this centrode is of the pame order as that to the other,

(R + R^ -g-g)^-rf3 ^ (R.|.R72_^2

,^** (R-a)(R'-<j) R.R' '

which may therefore be deduced from that to the fixed centrode by writing

b for d and vice verad.

8177. (B. Hanumanta Rau, B.A.) — The images of the circum-
centre of a triangle ABC with respect to the sides are A', B', C ; prove
(1 ) that the triangles A'B'C and ABC are equal ; (2) that they have the
same nine-point circle. Find the equation of the circumcircle of A'B'C
and the angle at which the two circumcircles cut each other.

Solution,

If a, b, c be the middle points of the sides, evidently B'C is parallel to
and double of be (Euclid, vi., 2), and therefore parallel and equal to BC,
and so for the other sides of A'B'C, and this triangle is equal to ABC in
all respects. Also O, the circumcentre of ABC, is the orthocentre of
A'B'C, since A'O, B'O, and CO are perpendicular to BC, CA, and AB,
and therefore to B'C, C'A', and A'B', respectively ; and a, i, e are the
middle points of the portions of the perpendiculars intercepted between
the orthocentre and the vertices, therefore the circle through abcy the
nine-point circle of ABC, is also the nine-point circle of A'B'C. The
relation of the triangles AliC and A'B'C is mutual, for, since
BC s= BO = BA', the perpendicular from B on AC or A'C bisects A'C,
and the orthocentre of ABC is the circumcentre of A'B'C.

If 0' be this point, the angle at which circumcircles cut is

where R is the circumradius «= cos-^ i (1 - 8 cos A cos B cos C).

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132

If {jr\ /, r') be the trilinear coordinates of O, the points A', B', C are
-x', p,— y*, ;»3-s'; p^-x', — y', ^-r'; Pi-r', ;>j-y', - c' respectively,
where Pi, p^ p^ are the perpendiculars of the triimgle ABC, and by
sabstituting these in

yz sin A + ex sin B + jry sin C — (jr sin A<f y sin B + z sin C) (Ax + Ary + fa) = 0,
hj k, and / are easily determined, and the equation to the circumcircle of
A'B'C The line hx + Ay + fc = 0, the radical axis of the circumcircles is
the line bisecting 00' at right angles.

The equality of the triangles ABC and A'B'C may be otherwise demon-
strated in a way which explains the geometrical meaning of the equation
to a conic circumscribed to a triangle, and demonstrates an interesting
property of such figures.

If P (y, y', z') be any point, A", B", C" its projections on the sides of
the triangle of reference, and A', B', C points on PA", PB", PC", such that
PA' = /. PA" = /x', PB' = y . PB" = y/, and PC = A . PC" = hz' ;
the area of the triangle A'B'C

= i{PB'.PC'sinA + PC.PA'sinB + PA'.PB'8inC}

= i {yA sinAy'c' + A/sinB z'x'+^sinCx'y'} (1).

Now, if/ = y=A = 2, and x', y', / be the circumcentre, the points
A', B', C are those in the Question, and

A A'B'C = 2 (// sin A + c'x' sin B + x'/ sin C)

= 4 (A*Or+ AeOa+ AaOb) = aABC.
Now, returning to equation (1). If P lie upon the circumscribed conic
yA sin Ayz + A/ sin B zx +fy sin C xy = 0,
the triangle A'B'C vanishes, i.^.. A', B', C lie on a straight line. Or,
in other words, if the perpendiculars upon the sides of the triangle
of reference from any point on the circumscribed conic

Ays + fizx + vxy =
be produced to points A', B', C, such that

PA'=/.PA", PB' = y.PB", PC = A. PC,
the points A', B', and C will lie in a line, if

/: y : A :: sin A/ a. : sinB//i : sinC/v.
The well-known property of the Simpson lines is a particular case
of this.

8592. (Professor Mathews, M.A.) — Through a point P are drawn
three planes, each parallel to a pair of opposite edges of a tetrahedron
ABCD. Prove that the 12 finite intersections of these planes with the
edges of the tetrahedron lie on the same quadric surface ; and that, if
BC'« + AD2= CA2 + BD2 = AB2 + CD2 {i.e., if each edge of the tetra-
hedron is perpendicular to the opposite edge), there is one position of P
for which the quadric surface is a sphere.

Solution,

If Ai, fij, vi, TTi be the tetrahedral coordinates of P referred to the
tetrahedron as tetrahedron of reference, the equations to the three planes

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133

through P parallel to opposite edges of the tetrahedron are

\-\-fi __ y + ir \ + y __ /i-t-ir , X-t-ir __ /i-t-y

Now, if Aii\2 + Aj2fi2+ ... +2Ai2\fJL + 2A^X»+ ... -

be any quadric, this meets x=:/a=0 at points at which

A85i'2 + 2A34i/ir + A44*^=0 (1);

but, by Question, these points lie on

A- + y ^ M + ir ^n^i A + ir ^ t^ + p ^

and therefore the equation (1) is equivalent to

(-1 ^U-!^ =^)-0;

vAj + i'i Mi + ^'l' ^Ml + ''l A.i + iri/
therefore Ajs : — 2A34 : A44

.. 1 . 1 , 1 . 1

" (^1 + "1) (Ml + yi) ' (Ml + iri) (Ml + "1) Ui + "1) (^1 + iTi) ' (Xi + iTi) (/ii + Ti)'
and so on. And the quadric

X2 f^

(^1 + Mi)i(^i + ''i) (^1 + »i) (Ml + "1) (Ml + »i) (Ml + A^i)

-^^^( I X 1 ]...«0

(^1 + Ml) l(>^i + ''i) (Ml + ^i) (Ml + ''i) (^1 + »i) 3

meets the edges at their finite intersections with the three planes.

If Aii\2 + Aj2/i'+ ... +2Ai2\fi-" = be a sphere,

•°-ii •** -^22 ~ 2-^12 _ Aji 4 Aa3 -- 2Aj3 _ A32 -j- A33 — 2 Ag __ «
(1.2)2 (i.3)a (2.3)8 •'

where (1 . 2), &c. are the edges of the tetrahedron of reference {Proceedings
o/Lond, Math. Soe,, Vol. xvm., p. 341).

And {ui + iTi) (Ai + Ml + 2vi) (\, + Ml + 2iri) « r (1 . 2)«, &c.,

or (^i + iTi) (O— (w 1 - I'l)'} = r (1 . 2)2, &c., where X + M + v + ir=0,

and therefore, if Aj = /*i "■ •'1 = 'i> ♦•*•> if P be the centroid,

(,., + iri)C-=r (1.2)2, (^j + »j)C2=r(1.3)2, (mi + i'OC^ = r (1 .4)',
(^i + Mi)C« = r(3.4)«, (Ai + v,)C'=r(2.4)3, (a^ + ^,) C* = r (2 . 3)»,
and (1 . 2)2 + (3 . 4)« = (1 . 3)2+ (2 . 4)2 = (1 . 4)2+ (2 . 3)2 ;

and the tetrahedron is rectangular (see Solution of Question 3228,
Appendix III., Vol. xlviii., p. 168).

8940. (W. J. C. Sharp, M.A.)— If
S = a;r2 + by^ + cz^ + dw' + 2/ys + 2mzx + 2nxt/ + 2pxw + 2gyto + 2rzw,
and Fi. 2 ^(1X1X2 + by i^j + WjZj + <?m?i«72 + ^ (yiSj + yjjc J + &c. ;

VOL. XLIX. R

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134

show that Si88S,+ 2Pi.8P2.3P8.i-SiPj.3-S2l^.l-S3P? 2

yi» i^2» ^3 M «1» «2> «3 W'l, ^fiy ^Z

«i» «2i «3 + ... + 2L t/;i, tt'a* "'s h^i> «2> ^3

t^l, «'2» ^^3 I I ^1» ^2» ^3 I I yi» t/%» VZ

where A, m are the first minors of the discriminant of S.

+ &c.y

^ Solution,

If X, y, Zf w be eliminated from S by means of the relations
l\ + fi + y)x = XXi + fix^ + px^,
(\ + /i + >/)y= Xyi + zuya + yyai
(A. + jU + v) « = A«i + /*22 + •'«3»
(A. + fl + v) M> « Aw?! + ^"^2 + ''"^3>

the result is

,,^^^ ,2 {^'Sl + M'S2 + I.SS3+2\MPl2 + 2/il'P23 + 2l'XP3i}.

Now, as I have shown in a paper read before the London Mathematical
Society in December, 1883, and in a note, JReprint,Yol. xliii., p. 47, A, fx, v
are areal coordiDates of any point (a:, y, «, w) on the plane through three
points (:r„ yj, «i, «?i), (a-g, yg, Zj' «'2»)» (^3» S^s* ^» *^3) referred to the tri-
angle of which those points are the vertices. And the plane equation to
the section of the quadric by the plane is

A2Si + /i2S2 + I^Ss + 2X/iPi2 + 2/U>/P23 + 2«'AP31 =0 (1).

If the plane satisfy the tangential equation to the surface, the section
must have a double point. Now the equation to the plane is

= (2),

^1

y,

2,

W

^1»

yi»

«ii

iTi

a-2,

y2i

«2,

w^

Hy

^3,

53,

W3

and therefore SiS883 + 2P23P8iPi2-S.P28-SaP3i-S3Pj2 = 0,
the condition that (1) should have a double point, and

yii

!^2» y^

«i»

Hy «3

m;„

m;2, w^

. + 2L I

M?i, iTa, t^al +...= 0,

^i> ^2» H

y\y y%y yz I

«1> «2> ^3
I ^11 ^2» ^3

the condition that (2) should touch the surface, hold simultaneously.
The sinisters are both of the same order in the coefficients and variables,
BO they can only differ by a factor which is easily found to be unity.

[Mr. Edwardes sends the following solution: — ^As in the solution
by 8970, wg have

^1
^3

rfSj

^1,

y\y

«i»

w^

« 16

Si,

P12.

Y,3

^2»

y<i^

«2»

w^

P12.

S2,

P23

Hi

yai

Hy

w^

Pl3,

P23,

S3

1

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135

and writing S = aj!^ + by* + ez* + dtt^ + 2fi^+2ffMX'\-2hxi/-¥2lxia

+ 2myw + 2mw,
the same is 2 dSj ^ dOi

dxi difi dzi

rfSg dS« rfSg

dx^* di/2 dz^

rfSa rfSa rfSa

dx^ dy^ dz^

2

«,

A, ^, M

^i» yij

A, ft, /, m

^2, y^y

9y /, ^ »

^3» ya.

= 2JD

iPi, yi> «i

yi,

x^ y^y «2

+ L

y2»

^.

*3» y3» H

ysi

«2>
^3,

«'2
W?3

yi,
y2»

^8. ysi

+ N

= result J

t^8»

2i»

«2»
«3»

^3»

^2 +M 22> «^2i ^2

^^3 I I «3» ^'Sl ^3

yiK |^i» yi»

y2 ? ^ U2» y2,

ysK Us, ys,

8969. (W. J. C. Sharp, M. A.) — ^If the ternary n-ic be written

1 1 • ^

and <m; + ft|y + b^ be written for a,

hix + Cjy + Cjg be written for ftj,
b^ + c^ + e^ be written for ftg, and so on,

in any invariant or covariant ; the result will be a covariant of the

(m + l)-ic <m;»»*i + "-ii (bji/ + ftjz) x** + &c.

For any covariant of the (n + l)-ic, the operation y — - must be equivalent

dx

to

"T.^2*.J-.*,^ + 3..|^ + 2..^^,3^^,&c.

^ft|

Writing the (« + l)-ic.

aV» + Y (^'i y + ^'a *) ^**"* + &C'> where a' =^ax + biy + ftj?, &c.,

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Then

136

I in question, we have

y ^ • y f (n + 1) <w» + n + 1 . n (*,y + ij«) «*-*
dx (^

+ ftj {Mf^-^yz + ff«t"- V + « . n— 1 y« z*-*)

— (»+ 1) aa^y + « + 1 . fi (*iy + *j«) a^-^

+ ^L+LllliLli (ciys + 2<?2y« + <y«) a:"- V + &0.

— yj(» + l)ajc* + n + l.fi (*,y + b^) a?"-*

+ ^:tiili±li(cjy8 + 2<?5y8 + c3a;3)a^-2 + &c.| = y^ from above.

Similarly, « -— is equivalent to
aar

and therefore, &c.

[Any invariant or covariant of a quantic maybe expressed as a function
of the differential coefficients of the quantic, and this same function of its
differential coefficients will be a concomitant of any other quantic.
Now, if Un and «n +1 denote the gfiven quantic and the one derived from it by
making the proposed substitutions, dP*<i*' Un^ij dsiP , dy^ , d^ is the result
of making the same substitutions in df^t*^ Uh/ dxP . dy^ , dsi^, and the
symbolical form of the transformed concomitant is the same as that of the
original one ; therefore, &c. This may, of course, be extended to A;-ary
quantics, and to substitutions of a higher order in the variables.]

8970. (W. J. C. Shaep, M.A.)— If X, Y...U denote the deter-

minants

^l> t/lf «i» ^u ««i

^2» t/3y H* ^2> «2

^8, t/3y Hy «^8» «3

^4» y4» «4» ^iy ««4

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137

and Vi, Vj, Vj, V4 be the values of the quinary quadratic V when
(^1, y\y 2i» «^ii ««i), (^2»'yj. «2i w's* «*2)> &«. are put for {x, y, «, «;, «), and

S1.2 , &c. stand for J ( a>i -- + yi — + ... J Vj, &c.,
\ 00:3 ay2 /

«AX» + BY« + &c.,

v„

8, .a,

8,.,,

S,.4

s,.„

v„

8,.,,

ai.4

Si. J.

Sa.j,

v„

8, .4

81. 4>

S,.4,

8,.«,

V4

where A, B, &c. are the first minors of the discriminant of V.

Solution.

This question is a generalisation of 8960, which is itself one of the
property (otherwise) proved in Salmon's Conica^ Ed. 6, Art. 294, and
which is the analytical ground of the theory of reciprocal ^olars.

The proof in this case is the same as that in S940 ; for, if or, y, s, w, u
be eliminated from V = 0, by means of the equations

(\ + t^ + y + ir)x — \xi + fir^+yx^ + nx^

(x + fi + i' + ir)y « Ay, + fiy2 + i/y8 + Ty4,
&c. &c. &c.

the result will be the equation to the section of V — by the linear
locus through (ari, yi, «i, «;„ w,), (a^a, yg, ar„ t(?2, «,), (arj...), (3:4...),
and, if this locus have a double point, the determinant in the question, the
discriminant of the locus of section, will vanish, and also the result of
substituting X, Y, ...U, in the reciprocal equation; and those two
quantities being each of the same order in the coefficients of Y, and in
X, .V, ... &c., can only differ by a factor which will be found to be unity.
The property may easily be extended to space of any dimensions for
A;-ary quadratics) and proves that the theory of reciprocal polars holds for
space of all dimensions.

[Mr. Edwardbs sends the following solution : —

Since Wi —-* +yi -—1 +&c. - 2Vi, we have

dVj dVj rfVj
dtoi du^

dw2 ^^

dj^^

dwJ

dxi

rfVj
dx^'

dVj
dx^

dx,
dyi

dVj

dyi
dy^

dzi
dz2

dV^
dzA '

2Vi,
2Si2,
2Si3,
2Su,

2S21, 2S3,,

2V2, 2S30,

2523, 2V3,

2524, 2S34,

3

rfV4

dwl

2S41

2S4,
2S43
2V4

or,,
^2>
^3>
a^4»

yii
y2>
yzy

^4.

*2»
«3»
«4,

du^
dV,
du^

and obviously S12 = S21, &c., therefore
this determinant becomes

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16

138

V2, 823, S24

Sii> Sjj, Vj, 834

Sl4» S34, S34, V4

But expanding in another way, we have
cfV, rfVj ^i dVj

tfo-j dt/2

rfV4

and this = 162

<^3

&c.

^4» y4>

IT,

«, A, ff, I

If frt, w, <f

a?i, yit «!, wi

^2> yj» «2> <^2

^3» y3> ^3' <^3

a?4» y4> «4> «^3

viz., 16XA.X^, the quadratic being written

(abedefghlmnpqrs IS^xyzwu)^,^

9561. (W. J. C. Sharp, M.A.)— If (1.2), (2.3), &c. denote the
edges of a tetrahedron, and D}, D^, D^ the shortest distances, and &i, ^ji h
the angles between (2 . 3) and (1 . 4), (3 . 1) and (2 . 4), and (1 . 2) and
(3 . 4), respectively ; prove that

^^^ ''''' ^' ^ 2(2.3K1.4) ^^^ • ^^'^ ^^ • '^^'"^^ • ^^"^^ • ^^'^' *''" ^''•'
and (2) the square of the volume

= ^{4(2-3)Ml-4)'-[(1.2)» + (3.4)»-(2.4)»-(1.3)»]»} = &c.,&c.

Solution.

Let ABC!D be a tetrahedron. From D draw
DE parallel and equal to BC ; join BE and AE.

BCDE is a parallelogram and its diagonals
bisect each other (in F).

Now 61 = ADE ;

therefore cos ^i =» cos ADE

= ^ (AD» + DE2-AE2>

2AD.de ^ ^

1
'2AD.BC

{AD2 + BC2-AE2};

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139

but AB2 + AE2 = 2BF2 + 2AF»,

D A8 + AC? - 2DF2 + 2AF2, DB^ + BC? = 2BF2 + 2DF2,

therefore DA^ + DB^ + AC^ + BO^ = 2BF2 + 2AF2 + 4DF2

= AB2 + AE2 + CD2,
therefore ADs + BCa-AE^ = AB^ + CDS-DBS-AC^,

and therefore cosei = _^ {AB^ + CD^-DB^-ACaJ

2AD.B0

1
'2(1. 4) (2. 3)

{(1.2)2+(3.4)2-(2.4)2-(1.3)2},

From these values it follows that the opposite edges are at right angles,
if (1 . 2)» + (3 . 4)2 - (1 . 3)2+ (2 . 4)2 « (1 . 4)2 + (2 . 3)2,

and conversely. If HK be the shortest distance Dj, it is at right angles
to AD and BC, and therefore to AD and DE, and so to the plane ADE ;
and since BC is parallel to that plane, it is equal to the perpendicular
from C upon the plane ADE.

Now Vs tetrahedron ABCD = tetrahedron ACDE

— sin 01 X perpendicular from C to ADE

6
AD. BO

sin 01 . Di ;

therefore V2 = 42!.i^ D^a (i « co82 0{)

36

== ~ {4 (1 . 4)2 (2 . 3)2- [(1 . 2)2+ (3 . 4)2- (2 . 4)2- (1 . 3)2]2} « &c.

From the above (2.3) (1 . 4) Di sin dj = (3 . 1) (2 . 4) Dj sin 63

-(1.2)(3.4)D3sin03.

[Mr. Edwardes sends the following solution : — Denote the sides of the
tetrahedron (1 . 4), (3 . 4), (2 . 4), (2 . 3), (1 . 2), (1 . 3), by «, *, e, d, e,f,
respectively. Through the point (c, d, d), on the plane of abe, draw a
parallel to a, and let angle (a, d) — 0i, &c., also, let angles (^, c), (e, a),
(c, a), (c, d), {ed«y abe) be X, /ia, if, o, <j), respectively. Then we have

—cos 61 = cos a cos 1^- sin a sini^ cos d), and cos d> = cos /^~<*08^ cosy

sin A sini' '

therefore -cosOi - cosy (cosa + ?^5L?L22i^\ -^Iffcos/*
\ sinA / smX

2ac \ 2dc d lee I d 2ae

2ad 2ad 2ad *

hence we have cos 0, = —^—P

2ad

" 2(2.3K1.4) ^^^ • ^^'^ (3 .4)2- (2 . 4)2-(l . 3)2}, &c., &c.

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140

Again, Dj « — — ; therefore, V = -J arfsin e, ;

odBmBi 6

therefore, V»^gLWBin»(>,»gl%«^[l-( ^^^^"f-/ Mn
36 * 36 i V 2a^ / j

= ^{4(2.3)M1.4)2-[(1.2)«+(3.4)»-(2.4)»-.(1.3)2p} = &c.,&c.]

7384. (Professor S. RfiAus.) — Etant donn^e la s^rie ilKmit^ 7, 1 3, 25,
43, 67, 97, 133, 137, ..., dont le tenne general, celui qui en a « avant lui,
est An — 3 (f»^ + f»j + 7 : demontrer les propositions suivantes : — (1) sur cinq
termes consecutiis, pris k volonte dans la serie, un terme est divisible par
6 ; (2) sur sept termes coDs6cutifs, deux sent divisibles par 7 ; (3) but
treize termes consecutifs, deux sont divisibles par 13 ; (4) aucun terme
de la s^rie n'est egal k nn cube; (5) une infinite de termes, tels que
Aj = 26, Aj* = 4225, etc., sont aes carr^s divisibles par 25 ; (6) la
deuxi^me et la troisi^me proposition sont comprises, comme cas particu-
liers, dans la suivante : si N est un nombre premier, de la forme 6m + 1,
sur N termes consecutifs de la serie, deux sont divisibles par N ; (7) on
pent affirmer aussi que, k I'exccption de 5, aucun nombre premier de la
forme 6in— 1 no peut diviser aucun terme de la s6rie.

Solution,

Out of any five consecutive numbers one must be of the form 5p + 2,
and the corresponding term of the series will be

3(5i? + 2)(5p + 3) + 7= 25(3p2 + 3p + l),
which proves (I) — indeed, that one term in five is divisible by S^— and
(5), since an infinite series of squares of the form Zp^ + Sp-k-l can be
found [see Solution of Question 4535, vol. xxiii., pp. 30, 991. Similarly,
out of seven consecutive values of n two are of the forms ip and 7p + 6,
and in each case 3 (w^ + n) + 7 is divisible by 7 (2) ; and out of thirteen,
two are of the forms 13;?+ 1 and 13p+ 11, which each of them makes
3 (w^ + fj) + 7 divisible by 13 (3). No term is an exact cube, for if

3 (n« + «) +7 = 3 (n« + « + 2) + 1
be a cube at all, it is the cube of a number of the form Zp + l, and
n''^ + n + 2 must be divisible by 3, which it never is (4).

Again, 4A„ = 3 (4«2 + 4» + 1) + 25 s 3j»2 + 25

if p — 2n + lf and if 6w + I be a prime, 3/>2 + 25 = (6m + l)y has real
solutions. If 6m— 1 be a prime, Sp^ + 25 = (6m- 1) y has not, except in
the special case m = 1 (7) ; and if, when n — q, 3 (w^ + «) + 7 is divisible
by 6m + 1, itis 80 whenu = 6m — ^ or (6m + l)p + q ot {6m + l)p + 6m-qf
which proves (6).

It may be interesting to point out that

2oA«ar« = (7~8a:-7rr2)(l-ar)-3.

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APPENDIX i^l.

NEW QUESTIONS. By "W. J. Curran Sharp, M.A.

9776. If perpendiculars jp, q, r he drawn from the vertices of a
triangle upon any tangent to the circumcircle, these are connected by

the relation

9, ^\

c2, *2

-0.

r, l^y a\

Prove this, and show that a similar relation holds in space of n
dimensions.

9777. If /A + W/A + WV+... « be the equation to a linear locus in
space of n dimensions, in terms of the simplicissimum content coordinates
(areal, tetrahedral, &c.), (see Question 8242) ; show that /, m, «, &c. are
proportional to the perpendiculars drawn from the vertices of the simpli-
cissimum of reference upon the locus.

9778. If the variables a, jS, 7, 8 be removed from the tangential
equation to a surface, by substitution from

(X + M + >')a = Xai + Ma2 + i'a3, (A + /i + 1') jS = AjSi + /i/Sj + v^j,
(X + /i + i^)7= A7i + /i78 + i0'3, (A. + M + >')8=^ ASi + fiSj + irJa;

show that the resulting equation in (x, /*, y) is a tangential equation to

the tangent cone whose vertex is at the intersection of the three planes
(«!» ^1, 7i» 81), (o,, /32, 72, 83), and (oj, ^3, 73, 83).

Hence determine the number of tangent lines of different classes

which can be drawn to the surface from any point.

9779. Prove the following identities

0, 1, 1, 1, ... 1

1, 0, iPl + ^8> ^1+^a ••• *l + ^n + l
1, jri + a?3, 0, X^-¥X^ ... X^^-Xn^l
1, ^1 + ^8) fl's + 'Si ••• ^8 + ^»» + l

1, Xi+Xn^li *i + *n + li a's + ^n*l ...

^^{^2Yx^X^,.,Xn^\\^ + — +... + —),
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(1)

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U2

^1 + ^8, 0, r^ + x,

a-j + a-j, a-j + a-,,

^1 + <F«*

Xz + Xn^l

l[(«^l-

0, 1, 1 ... to n columns

1, 0, 1 ...
1, 1, ... „

(3)

= {-!)-' (»-I),

the last to be proved independently, and then shown to be consistent
with the first two.

9780. In ppace of n dimensions, the ppherical loci (hyper-spheres)
described ahout the n + 2 simplicissima (see Question 8242), eadi oi which
is bounded by m + 1 out of n + 2 given linear loci, all pass through the
same point, when n is even ; but not, in general, when n is odd. In
space of two dimensions this is Micquel's Theorem.

978 1. If a.v + bt/ + cz -^ dw = 0, and a'x + b\i/ + c'z + d'w = 0, where
fl, i, c^ df a', b'y (?', d' are functions of a parameter 6, be the equations
to a line, this line will generate a ruled surface of order m + «, where m
and n are the orders of the two given equations as functions of e.
Especially examine the case when m = « = I, and show that a second set
of lines exists in this case.

9782. If Pnr denote the coefficient of a:*" in the expansion of
(1 +iF)'*, &c., C„r denote the number of combinations of n things taken r
together; form the equations of differences which determine P„r, and
Cnn and hence show that these are equal.

9783. If ABC be a triangle, in which AC > AB, AD the perpen-
dicular from A upon BC, and if DC be taken upon CD produced, so that
DC = CD ; the circle through A, B, C will also pass through the ortho-
centre of ABC, and will be equal to the circumcircle of tiiat triangle,
which will pass through the orthocentre of ABC.

9784. If the sums of the squares of the opposite edges of a tetra-
hedron be equal to one another, show that the nine-point circles
inscribed in the triangular faces are all sections of the same sphere ; show
also that this is the condition that the perpendiculars from the vertices on
the opposite faces should meet in a point. Also show in space of n
dimensions, that if (r«) denote the edge joining the Hh and «th vertices
of a simplicissimum (Question 8242), and (r.«)2 = Ay + A, (where
Ai, A2 ... An .1 are w + 1 arc^l magnitudes) for all values of r and «, the
perpendiculars from the vertices upon the opposite faces will all meet in
a point, and the nine-point circles of all the triangles formed by joining
the vertices are sections of the same spheric (hyper- sphere).

9785. li m *^ m', there are in general (m— 1)2»» («+ 1) points which
have the same linear polar with respect to each of two loci, of orders m
and m', in space of n dimensions. Hence deduce the conditions that the
loci may touch.

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9786. If a circle cut the sides of the triangle of reference at the feet
of concurrent lines from the vertices, the line joining the isogonal con-
jugates of the points of intersection passes through the centroid.
Enunciate the corresponding proposition in the Geometi^ of Higher
Space.

9787. If the sums of the opposite edges of a tetrahedron he equal to
one another, show that the circles inscribed in the triangular faces are
all sections of the same sphere. Also show, in space of n dimensions, that
if for all values of r and s [{r.s) denoting the edge joining the rth and «th
vertices of a simplicissimum (Question 8242)], {r ,s) ^ dr + dg, where
<?,» d.2f &c., dn^i are « + 1 linear magnitudes, the circles inscribed in the
triangles formed by joining the vertices are all sectioijs of the same spheric
(hyper-sphere) .

9788. In space of n dimensions, the Jacobian of n + 1 quadratic loci
(which is a locus of the n + l^^ order) is the locus of the points which are
conjugate with respect to each of the loci, or the locus of the points whuse
first polars with respect to all the quadratic loci meet in a point.

9789. Show that

(fi-y) (?-«) (»-iB) (a;-a)2+(8-y) (?-«) (a- 8) (x-^fi)^

+ («-«) (a-iS) (i3-8) (;r-7)«+ (i3-a) (7-/8) («-.7) (:p-«)«
vanishes identically, and hence deduce that the sextic covariant J of the
binary quantic {x — o) (aj - 0) {x -y){x- 5)

= {{a''fi){x^y){x^n)^{y-B){x-a){x^0)}{(a-y){x--fi)[x-9)
-(8-i8)(a:-a)(2:-7)}x{(«-8)(a.-/3)(a:-7)-(i8-7)(2:-a)(a;-8)}
s {(a-iS) (^-7) (^-8) 4- (7-8) (a:-a) (a;- i3)} {(«-7) (.^-i8) (^-8)
+ (8-i8) (2:-a) (a;-7)} X {(a-8) (a:-i8) (a;-7) + (/3-7) (a;-«) (a;-8)} ;
and confirm this by showing that

(a-/3) (ar-7) (ir-8)-(7-8) (a:-a) (a^-iS)

= (a-7) (^-/3) (^-8) + (8-/3) (a?-a) {x^y).
Also explain the geometrical relation between the points determined by
the roots of the covariant, and those determined by the roots of the
quantic.

9790. If u be a rational and integral symmetrical function of

Xi, X2 ... Xny show that x? -^'- x^ — and x^ ^ — x^i ^
dXf. dxg dxr dxg

are divisible by Xr — Xg for all positive integral values of p, r, and »,

9791. Show that the secondary form of Maclaurin's Theorem, given
in Boole's Finite Liffereticea^ p. 23, Ed. 1., viz.,

0W = ^(O)^0(|)o.^ + ^«(|)o«.j^+&c..
leads at once to the equation ---^ = log (1 + A) ^ ^. ,^

€r — \ A

wheye the A*8 only act on the zeros. Also from this form deduce
(^~1)»= 2 (A". 0''») — -.

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9792. Show that, if (A) «*-i»i «•-»■»•/?, «•-«...)"•

and that, if m - 2, ^^ - PrPo+Pr^i -Pi + Pr-i -i^ ... +J?oPr.

Also deduce the expansion of ^ (/r) in powers of x, where (^ (z) is an
integral and rational function of x, and

^ 9793. If XJ =» be the equation to an tn-ic locus in space of n dimen-
sions, referred to simplicissimum content coordinates, and

Ssx>i(1.2)« + /ii'(2.3)« + ...,
where (1.2), &c. are the edges of the simplicissimum of reference, the
equations to the normal to U at the point (X/i ...) are

dhf d\* dfi''^ dfi '

dV dV -0,

dk' dti

1, 1

where xV, &c. are the current coordinates. Hence show that m" normals
can, in general, be drawn from any point to U «*0.

9794. If (1 . 2), &c. denote the edges of the tetrahedron of reference,
and X, /A, Iff » be tetrahedml coordioatos, show that the circle at intinity
is represented by the equations

X+/i + v + xs=0, and X/i(1.2)2 + /ij*(2.3)2+ ... - 0;

and that in space of n dimensions, if (1 . 2), &c. be the edges of the
simplicissimum of reference, and X, /i, v.., content coordinates (see
Question 8242), the Lypersplicre at infinity (in space of n dimensions) is
represented by X + /i + y+ ... « 0, and X/a(1 . 2)* + fir (2 .3)*+ ... — 0.

HencQ determine the conditions that an equation of the second degree in
tetrahedral or higher content coordinates may represent a sphere or
hypersphere.

9795. (Suggested by Question 6420.) If S,,^ denote the coeflScient
of tj in the developed product of (1 + /) (I + 2<) ... (1 +tO> 8how that

Si.i« Si.l., + iS.-8j-l+»(»-l)S<_3.y-2+...

+ i(»-l) ... (»-y+ 1) Si-j-i.o.

and that the product itself is the coefficient of x**^ in the expansion of
(l-tx)- 1/< multipUed by (» + 1) !

9796. Show that the transformation from rectangular to areal coor-
dinates, or vice versd, may be eflfected by substitution from the equations

{\ + fi + y)x = \Zi + fix.2 + yX3, (X + /it + v)y == Kt/i + fiy^+^Jsy

where (j*i, yj), (a-j, yj), {x^, y,) are the vertices of the triangle of reference.

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And similarly, that the rectangular and tetrahedral coordinates of a point
in space of throe dimensions are connected hy the equations

(X + /i + K-i-»)a; = ^Xl + /UTj + yx^-\-wx^f

(X + /4 + y + x)y-Xyi + /*y2 + yy8 + iry4,
and (X + /i + y + x) « =• A«i + fiZ^ + W3 + irr^,

or more gf nerally that, in space of n dimeneions, the connection between
orthogonal and simplicissimum content coordinates (see Question 8242) is
given by the equations

(A + /* + !'... +t) z « Xi*i + ^2+ ••• +Tir„+i,

{\ + /i + y... f T)y « Xyi + /iiy2+ ••• +Ty«,i,

&c. &c.

9797. If P ^ + 2Q ^ -h % = X, where P, Q, R; X are functions

UX' ax d / V \ PR
of X only, and are subject to the condition — ( — ] + — ^ 1=0,

show that y - f-J^'^*^ [[ J- cK'^ dxK

9798. A quadratic locus in space of n dimensions, has n principal
axes {i.e.f axes which are at right angles to the linear loci which bisect
chords parallel to the axis).

9799. Deduce the solution of ^ — a^ -t « from the expansion of

dx' dy^ *^

^ (x, y) in ascending powers of x and y.

9800. If «, A, A be given in a spherical triangle, deduce the con-
ditions that the triangle should be impossible, unique, or ambiguous, from
the discussion of the equation

cos a =: cos ^ cos e + sin * sin ^ cos A,
where there are two triangles ; show that, c and (/ being the third sides,

tan i (c -I- c') ^ tan b cos A,
and confirm this by the case when the radius of the sphere is infinite.

9801. If Pr denote the Legendre's coefficient of the r**» order of
J(A; + 1/A:), showthat

j:

{(1-2:2) (1-^2^2;}* '3 '6 2r + l

9802. If there be two series of functions of x, P©, Pi, Pj..., and
Qo, Qi> Qa •••» and one of operations, Rq, Ri, Rj, &c., each of which gives
a result independent of x : then, if K,» . Pn . Qp = 0, whenever t/i, n,
and JO are not all equal, but not when they are, any function /{x) may
be developed in the forms 2 A»P„, or 2 BnQn- Apply this to some known
expansions.

9803. Show that for any proper cubic the Cayleyan of the Hessian is
the Hessian of the Cayleyan ; and that the discriminant of the polar
conic of any line vanishes doubly when the line touches the Cayleyan.

9804. Show that the nodes on the locus of a point rigidly connected
with the middle bar of a three bar system lie upon the fixed centrode.

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9805. If PBC be a small circle of a sphere, B and C fixed points, and
P any other point upon it, then, if the arcs BC, PC, and PB be bisected
in D, E, and F respectively, and if the arc DE meet BP in B' and B",
and DF meet CP in C and C" ; show that (1), for all positions of P, B',
C, B", C" lie on the same great circle, that of which O the pole of the
small circle is the pole, and that B'C = B"C" = w-B'O" = x-B"C' ;
(2) that these arcs are of constant length for every position of P on one
side of BC ; (3) that the values of those arcs corresponding to positions of
P on opposite sides of BO are supplementary ; (4) that the points 3', B'',
and C, C" are the poles of the arcs OF aud OE which bisect the sides
BP and CP of the triangle BPC at right angles ; (5) that the six points
in which the sides of the triangle D£F meet the corresponding sides of
the triangle PBC lie on the gi-eat circle of which O is the pole.

9806. If ABC be an acute-angled triangle, o, /S, y the circular
measures of its angles; show that P being a random point in the
triangle, the chance that the angle BPC is obtuse, is

^ .^"". {8in2i8 + sin27 + »-2a}.
2 8mi8 8m7 »■ •*

U the angle at 3 (i3) be obtuse, the chance is

— ^iBA_/27 + 6in27}.
2sin)88in7 ^ '^

9807. The perpendiculars from the vertices of a triq,ngle upon the
central axis (the line which passes through the circumcentre, the ortho-
centre, the nine-point centre, and the centroid) are proportional to

cosAsin^B — C), cos B sin (0— A), and cpsOsin(A— 3),

those on one side of the line being reckoned positive, and those on the
other negative.

9808. The mean value of the pedal triangle of a random point in a
triangle ABC is ^W (1 + cos A cos P cos 0), where A, B, and C are the
augles of the triangle, and B the circumradius.

9809. Given forces act along the sides of a tjiangle, in the same
sense. The value of the mean sum of their moments about a random
point in the triangle is the mean of their moments about the vertices t)f
the triangle. If the forces be such that, if applied at a point, there would
be equilibrium, the sum of the moments is the same for all points.

9810. A, B, are fixed points, P another point, construct the resul-
tant of forces acting along PA, PB, PC, when they are proportional to

(I) PA»,PB»,andPC2; (2) ±, -L, and ^;

(8) ^. pgi. and ±^.

9811. If masses P, Q, and It be placed at the vertices A, B, and
respectively of the triangle of reference, show that the trilinear equations
to the principal axes at the C. of G. of the masses will be

lx-tmi/ + nz =s 0, and tx -f m*y + n'z == 0,

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where I : m : n and f : m' : n' are determined by the equations

a e a b e

pi?l + Q^ + R«4'„0,and

fl- ^ c*

//' + mm' + «ff'— (iwn' + #»'«) cos A— (w/' + n'l) cos B — {hnf -f Z'm) cos C = 0,
and that at any point the principal axes are conjugate {i.e.^ each passes
through the pole of the other) with respect to the conies

— .- + -2^: — -h — ■■

fl^ b'^ c'^

and ^+m2 + n2— 2m« cos A— 2n/ cos B — 2/m cos C =

(the circular points at infinity).

9812. If p. denote j;^^.. and Q„.£-

show that P„ = -— — P„_i,

2» (1— a^^)" 2n

•a;)'*

and that

_ 1 k'^'lr2»'l 2n-l r,
dx

Jo4(l-;

){(l-a;2)(l-A:2a:2)}*

-'•-i(?)'-H(?)'-■-^^M"^--•

9813. Show that

J(l-a;-)* 2» ^ ^ 2n J(l-a;2)* '

-JL — .^«f!i_if _? dx.

0(1 -a;2)* 2» Jo(l-a;2)*

And hence show that

^■' 2«! i2» + l 1 ■ 2 ■2« + 3 1.2 2»'2« + 6

Jo {(!-*') (1-AV)}» 2i IS Vi/ 1».22 V2;
is.22.32 V 2 ^ ^"■'•j-

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148

9814. Show that, for all integer yalaes of n,

Ji J_ J_ 1 1.3 J^ _l 1.3.5 J^ 1 ,

2/1 + 1 1 ' 2 * 2w + 3 "'' 1.2* 22 *2n + 5 1 . 2 . 3 * 2^ " 2« + 7 ^'

« i ( 1- J!lri + (2n-l)(2n-3) ^^^ |
2'»i (2»-l) 2(»-l).2(»-2) )

■^(-')"2^iW^^«^('^^2).

9816. If dashes ahove the line denote differentiation with respect to ar,
and dashes helow with respect to y, and ^ and tp stand for any functions
of X and s/ ; show that the equation

+ 4>y — <^'Y + «<^''4'' + b\i^'^ =

may always be transformed into a linear differential equation with con-
stant coefdcients.

9816. If the perpendiculars PA", PB", PC" from any point P on the
conic \i/z + /xzx + yxp^K {z Bin A + y BinB + z Bin. C)^ =

be produced to A', B', and C respectively, and if

PA'=« ?i^.PA", PB'=?^.PB", PC=!^.PC",

show that the area of the triangle A'B'C is constant.

9817. If A', B', C be the reflections of any point P, on the circum-
circle of the triangle ABC, with respect to the sides ; show, by Euclid,
that A', B', C lie in a straight line which passes through the orthocentre of
ABC. Hence deduce the theorem (Quest. 2145) that ** the feet of the
perpendiculars let fall on the sides of a triangle from any point on the
circumscribing circle lie in a straight line. Show that this straight line
is equidistant from the point and from the centre of perpendiculars of the
triangle."

9818. If ABC be a triangle. A', B', C the feet of the perpendiculars
from any point upon the sides, and A", B", C" points in PA', PB', PC
(produced if necessary) such that PA" =/. PA', PB" =^ . PB', PC'^= h . PC,
respectively ; show that, when A", B", C" lie in a straight line, the
locus of P is a conic circumscribed to ABC ; and that, when the area of
A"B"C' is constant, the locus of P is a conic having double contact with
first at infinity. Also, when the conic is given, determine flgih,

9819. If the sides AB and AC of a spherical triangle ABC be divided
in F and E respectively, so that sin AF : sin BF : : sin AE ; sin CE, the
great circle FE will cut the great circle BC in a point Q such that
BQ + CQ = », and the great circles through all such divisions meet in
the same points, and conversely.

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9820. Prove the following— (i.j If abc and A'B'C be the pedal tri-
angles of the circumcentre of the triangle ABC and of any other
point P, A'B'C = iAABC x (R2^ OP'^/R^ where R is the circumradius.
(ii.) If and K be the centres of two circles whose radii are R and r, P
any point on the second circle, and PL the perpendicular from P to the
radical axis of the circles, 20K . PL = R=« *- 0P=^. (iii.) The area of the
pedal triangle of any point P on a circle, the centre of which is K, with
respect to a triangle ABC, of which is the circumcentre and R the
circumradius, is ^aABC (OK . PL)/R2, where PL is the perpendicular
from P upon the radical axis of the two circles.

9821 . Show that, if a point be taken at random in the circumscribed
circle of a triangle, the mean area of the pedal triangle is ^ of the tri-
angle.

9822. If y =« a:*»-i log oj, prove that, when r is not > w,
g=(»-l)(«-2)...(»-)^— [log.+ (^^^,4... + j^J|;
and henee show that

(«-li(«-2H«-3Jf_l_ _1_ _1_^
1.2.3 i»-l n-i »-3j

in-1 «-2 «-3 1>

+ £lfl - li± + Jl:!^ 1-2.3^ ^^^7

n \ «+l (n+l)(n + 2) (»+I)(» + 2)(» + 3) j

9823. Prove (1)

f ^ ^JL±J,tan-ifflV.tane]-^Il-^ ^"^ ;

J(acos20 + ^8in-^6)2 2a»A* \\ a I ) 2aba + bt&ii^d

(2) that Xs« + *ar»,

f a;"»-i loga:.X^^.^"^^^^^^""^)x'' - *!??fa;"»*«->(mloga:- l)XP'^dx;
J m- m^ J

(3) that (*' cos- <psinn<p.d<p = i m^^^I^Mmil^ ,

9824. If the diagonals of a quadrilateral are at right angles, the sums
of the squares of the opposite sides are equal. Hence, if «, A, c, d be the
sides in order, a, c, i, d will be the sides in order of a quadrilateral in a
circle the area of which is ^ [ac + bd).

9825. If particles be projected along lines meeting in a point, with
velocities proportional to the projections of the same v< rtiral line upon
each line ; at any time the particles will lie upon the sphere descrihtMl
upon the space traversed by the vertical particle as diameter. Hrnco find
the line of quickest transit when an inelastic particle is dropped from ()
to A (a vertical distance /*), and after impact upon an inelas<tic line AB

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(the line to be detennined) proceeds along it to B, a point on a given
circle in a vertical plane through OA.

9826. If particles be projected from a point A, with the same velocity
V, along lines meeting at A ; at any time (t) they will all lie upon the
surface generated by the revolution, about the vertical line through A,
of the bicircular quartic, which is the inverse about A of a conic whose

focus is at A, its directrix horizontal at a distance — — upwards from A

I where k is the radius of inversion, and its eccentricity -7^ ] . The
conies corresponding to different values of t envelop a circle of which the
highest point is at A and the radixis ^.

9827. If XJ — be the homogeneous equation to a curve of order n,
and ^3X,|.+y,A+^|.;

show that the discriminant of

X»Ui + X«-»/iAUi + \\** - V'AlJi + &c.,
only differs by a factor from the result of substituting
for a, fiy and y in the tangential equation to the curve.

9828. If U » be the homogeneous equation to a surface of order n,

A J. ^ ^ d , d'

and As.,- +y,- +«,-+«-,-,

- . d d d d

dxi dj/i dzi dwi

show that the- discriminant of

X»»Ui + X~ - » (/tA + vA') Ui + iX"- « 0*A + kA^s Ui + &c.
can only differ by a factor from the result of substituting

^i> Pv «i> ^i

^2» t/2* hf ^3 ^^^ <h fif 7> <uid 8 in the tangential equation to

^31 P3f «8t ^3

the surface.
Show that this may be extended to higher space.

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APPENDIX III.

UNSOLVED QUESTIONS.

1087. (The Editor.) — ABCD is a conic whose centre is 0. If the
radii vectores OA, OB, 00, OD represent in magnitude and direction four
forces, show that the direction of uie resultant passes through the centre
of a second conic which is parallel to the first, and passed through the
points A, B, 0, D.

1196. (The Editor.)— Given the vertical angle and one oi the con-
taining sides, construct the triangle, when the ratio of the base to the sum
of the other side and a given line is given, or a minimum.

1207. (Alpha.) — To determine the position of a rock (R), the angles
subtended at it by the distances between three headlands (A, B, C) were
observed, viz.,

BRC = 161° 2' 42", CRA = 133° 25' 67", ARB = 75° 31' 21" ;
and it was known, from a previous survey, that AB = 8883, BO = 9870,
and OA — 10857 yards. Find the distance of the rock from each of the
headlands.

1228. (Alpha.) — A messenger M starts from A towards B (distance
a) at a rate of v miles per hour ; but before he arrives at B, a shower
of rain commences at A and at all places occupying a certain distance z
towards, but not reaching beyond, B, and moves at the rate of u miles an
hour towards A ; if M be caught in this shower, he will be obliged to stop
until it is over ; he is also to receive for his errand a number of shillings
inversely proportional to the time occupied in it, at the rate of n shillings
for one hour. Supposing the distance z to be unknown, as also the time
at which the shower commenced, but all events to be equally probable,
show that the value of M's expectation is, in shillings,

a X z V V* u )'

1274. (The Editor.) — If an indefinite number of parallel equidistant
lines is drawn on a plane, and a reguleir polygon, the diameter of whose
circumscribed circle is le8sthan.the distance between consecutive parallels,
is thrown at random on the plane ; prove that the probability that the
polygon will fall on one of the lines is 1/ /', where I is the perimeter of the
polygon, and /' the circumference of the greatest circle that can be placed
between the parallels.

1913. (The late Rev. R. H. Wright, M. A.)— Find the condition in
order that a straight line passing through an angular point of the triangle

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of reference ohall bo a normal to the conic whose equation is
(/a)*+(mi8}* + (ti7)* = 0.

1016. (Sir R. Ball, LL.D., F.R.S.) — Show that the equation of
squares of difFerences of the biquadratic (a, A, c, rf, e) (a?, 1)* = has for
its discriminant (H being ■» b^—aCf &c., as in Quest. 1876)

(27 J2- 13)2 (4H»- aSiH - a^J)^ (65296H3J + 2Z0iaWP- 1 6632a2HIJ

-625a3P-9261«'J2)«.

1919. (The late Professor Townsend, F.R.S.) — If a system of quadrics
touch a common system of eight, seven, or six planes, their director
spheres (that is, the spheres which are the loci of the intersections of their
rectangular triads of tangent planes) have a common radical plane, axis,
or centre.

Prove the three general properties involved in this statement; and show
from them, respectively, that—

(1) The director spheres of all quadrics passing through the four sides
of any skew quadrilateral have a common radical plane with the two
spheres of which the two diagonals are diameters.

(2) The director spheres of all quadrics passing through a common line
and touching four common planes, have a common radical axis with the
four spheres of which the four connectors of the intersection of three planes
with that of the line and fourth are diameters.

(3) The diameter spheres of all quadrics touching six common planes,
have a common radical centre with those of the fifteen quadrics determined
by the fifteen different triads of intersections of the planes taken in paiis.

1928. (N'Importe.) — Given the four cones -ci/^+bz'^-'fu^ ^ 0,
ex^ - az^-gvt^ = 0, - hx^ + ay2- hw^ = 0, fx^ + gy^ + A«2 = o, and the four
conies which are the sections of these by the planes a; = 0, y = 0, « = 0,
«> = 0, respectively; prove that (1) any line touching three of the four
cones touches the fourth cone, and (2) any line meeting three of the four
conies meets the fourth conic.

1936. (N'Importr.) — (1) Three points are marked at random on a
given straight line ; find the chance that, of the four parts into which
it is thus divided, any three will be together greater than the fourth.

(2) Again, a rod of given length has a piece cut off at random ; from
the remainder a piece is again cut off at random, and the piece then left is
divided into four parts at random : find the chance that any three of the
parts will be together greater than the fourth.

1989. (Professor Cremona.) — On donne un tetra^dre abed, Con-
siderons tous les cones quadriques S qui ont leur sommet en e et sent tan-
gents aux plans cad^ cbd le long des droites ca^ cb ; et tous les cones
quadriques S' qui ont leur sommet en a et touchent les plans ode, abc
suivant les droites ad^ ab, Un cone S et un cone S' etant choisis arbitraire-
ment, se coupent suivant ime courbe gauche G du 4^ ordre ; et toutes ces
courbes C ont un rebroussement en a et sent osculees en b par un meme
plan stationnaire ebdy etc. (voir Comptes Rendm 17 mars 1862, p. 604).
Demontrer geom^triquement les proprietes qui suivent : — (1) Si d'un point
quelconquejo de I'espace on m^ne les plans osculateurs h, toutes les courbes
C, les points de contact formeront une surface P de 3« ordre et 4« classe,
qui passe par le point donne p et par les six aretes du tetra^dre abed ;
(2) Les plans osculateurs des courbes C, aux points oil celles-ci sont

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couples par tm plan quelconque donne x, enveloppent une surfece n de
3^ classe et 4^ oridre qui touche le plan -k et passe par les six aretes du
tetrafedre ; (3) II y a une correlation de figures, dans laquelle k un point
Pf donne arbitrairement dans I'espace, correspond le plan «■ qui oscule en
p la courbe C qui passe par ce point ; et» vice versdy ^ un plan donn^ x
correspond lepoint j? de contact entre ce plan et la courbe C qui est osculee
par ce meme plan. Si le point p decrit un plan x', le plan x enveloppe la -
smface n' ; si le plan x toume autour d'un point fixe p% le lieu du point
p est la surface P', etc., etc.

1999. (R. Tucker, M.A.) — (1) P is a given point on the side of a
triangle, and Q another given point in the same plane : it is required to
inscribe in the triangle a maximum triangle having P for a vertex and its
base passing through Q. Again (2), P is a given point on a circle, and
Q a point in the same plane with it : it is required to inscribe in the circle
a maximum triangle having P for its vertex and its base passing through
Q. Discuss the several cases fully.

2233. (The late T. Cotterill, M.A.) — 1. If two points are given,
there are two others in the same plane such that the distances of points
taken from each pair vanish. In orthogonal diameters of orthogonal
circles, the diameter of each circle cuts the other in such pairs of points.

The products of the distances of a point in the same plane from the
pairs are equal, and the cosine of the angle subtended at the point is
real.

2. If a triangle is given by a point on a circle and the intersections of
a line and the circle, give . a construction for the centres of the circles
touching the sides of the triangle.

2287. (W. B. Davis, B.A.) — Find what algebraical curves can be
expressed by arcs of the circle.

2293. (Professor Whitworth.)— If B, B' be two real points and 0,
O' the circular points at infinity, prove that (IJ the fourteen-points conic
of the quadrilateral BBOO' is the rectangular hyperbola whose conjugate
axis is BB' ; (2) the critical circumscribing conic of the same quadrilateral
is the circle on BB' as diameter, and the critical inscribed conic is the
ellipse whose foci are B, B', and whose excentricity is ^ a/2 ; (3) these
three conies have double contact at the vertices of the hyperbola — the
minor vertices of the ellipse ; (4) if EE' be the third diagonal of the
quadrilateral BB'OO', then the critical circumscribing conic of the quadri-
lateral EE'OO' is an imaginary circle, concentric with the circle on BB',
their radii being in the ratio a/(— 1) ; I, and the critical inscribed conic
is the imaginary ellipse whose real foci are B, B', and whose excentricity
is i a/2 ; and (5) the critical circumscribing conic of the quadrilateral
BB'EE' is the rectangular hyperbola conjugate to the former one, and
the critical inscribed conic is another rectangular hyperbola, similarly
situated to the last, and having B, B' as foci.

2314. (The late T. Cotterill, M.A.) — Taking points in a plane;
prove that (1) the sum of the squares of the distances of a fixed point
from the four centres of the circles touching the sides of any triangle in-
scribed in a fixed circle is invariable, and this holds, if two of the points,
or even if all three, coincide on the circle ; (2) more generally, if A, B, C, D
be four such centres, and we denote the square of the length of the tan-
gent drawn from a point — to the fixed circle by (M) ; to the circle on the

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diatncfter AB, by (AB), &c. ; to the circles circumscribing BCD and copolar
to it, by (A) and (a), &c. : then we shall have the following system of
identities : —

2(M) « (BC) + (AD) ^ (OA) + (BD) = (AB) + (CD)
«(A) + (a) = (B)+(3) - (C + 7) = (D) + (5)
«i{(A^) + (B) + (C) + P)}=i{(«) + (i8) + (7) + («)}.
Hence AP^ + BP^ + CP^ + D^* « a constant is the equation to a circle
concentric with (M) ; and (3) if A, B, C, D are any points in the plane, state
the nature of a pair of points, corresponding to the circular points at
infinity, such that a similar system of equations must exist between the
corresponding conies through the two points.

2367. (The late T. CoTTBRiLL, M.A.)— *
afe

1. If the determinant fbd yanish, the equations

e de
o(3«« + <jy«-2rfy2) = fi(ex^ + az^-2ezx) » y{ay^ + bx'*-'2fxy)
are satisfied by two conjugate pairs of values 'of the variables, one pair
being independent of the constants a, i9, 7. Find the equation to the
quadric satisfying the values y = « = 0;« = a?=sO; a;=»y = 0, and the
last pair of values ; and the linear equations satisfying each conjugate
pair.

2. Show that such a system of equations is the analytical representation
of the projection of the three circles described on the lines connecting the
opposite points of intersection of four lines in a plane as diameters, the
circle circumscribing their diagonal triangle, and the line at infinity.

2366. (Professor Burnsidb, M.A., F.R.S.)— Determine (1) the locus
of points such that the polar conies with reference to the curve IT shall
be equilateral hyperbolas, where

U - 2 ^^^^^ ^ ;

Ayz + j::* (_ <p cog ^ ^ y cos B + « cos C)

and show (2) that this locus passes through the vertices of the triangle
ABC and through the feet of perpendiculars of the same triangle.
2390. (The late G. C. Db Moboan, M.A.)— Prove that

f** 1/ {xtp (x-alx)} dx ^ 0,

« being anything positive ; and

J+co - r + oo

1 / {o? {x + alx) jdx ^2\ ^x/x dx,

a being infinitely small and positive, and <p being such a function that
the subject of integration is finite for all finite values of a;, however small
a may be. In the second case, if ^/a? be infinite when a? = 0, the
integral on the r%ht must be replaced by

J -00

2322. (The late Professor Townsend, F.R.S.)— Express the radius
(B) of a circle orthogonal to three others in a plane, in terms of their

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three radii {p, q^ r) and the three sides (a, b, e) of the triangle determined
by their three centres. Investigate the corresponding formula for the
radius of the circle orthogonal to three others on the surface of a sphere.

2402. (R. TucKBB, M.A.) — Prove that the locus of a point whose
distance from its polar with reference to a given conic is equal to its dis-
tance from a given point is a quartic curve, which, when the conic be-
comes a circle, degenerates into a cubic curve.

2419. (The late T. Oottbrill, M.A.)— 1. If A A', BB', CC are the
opposite intersections of a complete quadrilateral, an infinite number of
cubics can be drawn through these points and another point D, touching
DA, DA' at A and A'. Amongst these cubics, there are two triads cS
straight lines and four cubics having respectively a point of inflexion at
B, B', C, C

2. The locus of the intersection of tangents at B, B' is the conic
DAA'BB' ; and of tangents at C, C is the conic DAA'CC.

Give the reciprocal results when the class cubic degenerates.

2436. (Professor Crofton, F.R.S.)— If pi, f>2, pg, P4 are the distances
of a point from four concyciic points 1, 2, 3, 4 ; and if a, ^, 7, 5 are the
triangles formed by joining 1, 2, 3, 4 ; then the equation

Pi V W - f^ -/O) + Pz Viy)' P4 '\/(«) » 0, or p,p3 ^(ay) « p^ ^/(3»),

represents two circles cutting each other and the circle 1234 orthogonally.

2439. (The late T. ColrrBRiLL, M.A. ) — If a, *, c, d are coUinear points as
well as a;, y, t.t and«ln the same plane ; then of the 16 intersections of ax^
aj/y bxf by with ez^ ety dz, dt, 8 lie on one conic and 8 on another. The
4 points of intersection of these conies lie on a third conic through abxy,
and a fourth through cdzt^ and these conies are respectively harmonics to
the lengfths (a*, xy) and {cd, zt). Also, the tangents to the four conies at
any point of intersection are harmonic.

2440. (S. Roberts, M.A.) — 8how that the centre of a curve of thefith
degree (t\^., the centre of mean distances of the points of contact, if
parallel tangents) is the 0. M. D. of the poles of the line at infinity.

2442. (The late Professor Townsbnd, F.R.S.) — Through a given
point in a plane draw a line the simi of the squares of whose distances
from any number of given points in the plane shall be a maximum, a
minimum, or given.

2443. (J. Griffiths, M.A.) — Prove (1) that the Jacobian of the three
conies represented by the trilinear equations

8 = sin2 A . a2 + &c. - 2 sin B sin C .*)37- &c. = 0,
S'=co82A.o2 + &c. -2cosBco8C.)37-&c. = 0,
F r= sin 2A .a* + &c. - 2 sin A . fiy- &o. = 0,
breaks up into the three right lines

^ , y ,0 y . « -

8in(C-A) 8in(A-B) ' sin (A - B) sin (B - C) ^ *

+ —- ^ «0.

8in(B-C) sin(C-A)

Hence show (2) how to construct geometrically the common self-con-
jugate triangle of the three conies in question.

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2467. (Professor Cropton, F.R.S.) — A Cartesian oval, having a
given focus F, is made to pass through three fixed points on a straight line.
Show that the fourth point in which it meets the line is also fixed ; and
that the locus of the points of contact of its douhle tangent is a circle with
F as centre.

2468. (W. B. Davis, B.A.)— Prove that a curve of the fifth order
and fifth class has three double points, three points of rebroussement,
three double tangents, and three points of inflexion.

2487. (The late Rev. R. H. Wright, M.A.) — If a conic be circum-
scribed about a triangle ABC, and tangents be drawn at A,*B, C, and
produced to meet so as to form respectively three triangles having the
sides of the triangle ABC for their bases ; find forms for the bisectors of
the angles of the external triaugles, and the equations to their circum-
scribing circles in trilinear coordinates.

2488. (The late Professor ToWnsend, F.R.S.)— Apply the method of
homographic division to draw the two right lines which intersect four
given right lines in spoce.

2496. (The late Dr. Booth, F.R.S.)— Let U s <p (|, v) be the tan-
gential equation of any plane curve, t the portion of the tangent between
the point of contact and the foot of the perpendicular p from the origin.
Then generally

i being the angle between the perpendicular and the radius vector.
Now a weU-known formula for the rectification of plane curves being

1 « ^± (pd\,

V • V COS A. > J Sin \
in which =» I and ■■

P P

we shall find the rectification of a plane curve of which the tangential
equation is given generally as easy as to find the quadrature of that whose
projective equation is given. [To apply these principles, take the Caustic
of the circle for parallel rays, discussed in the Editor's Note to Question
1509 (Vol. II., p. 21).]

2601. (N*Importb.) — Find the equation whose roots are the differ-
ences of the roots of the equation («, A, c, rf, e) (a;, 1)* = 0.

2508. (Professor Ceofton, F.R.S.) — 1. A point being denoted by
(p, tr, t), its tri-polar coordinates or distances from three poles R, S, T,'
show that all circles represented by Ap- + Ba-^ + Cr^ = are orthogonal to
the circle RST. 2. If O be the centre of the circle Ap2 + Btr^ + Cr^ = D,
show that A, B, C are proportional to the triangles SOT, TOR, ROS.

2538. (The late T. Cottbrill, M.A.) — Prove that, if the equation to
a curve is of the form x^y^isT = k (where p-\-q-¥r = 0), the order and
class are the same, and the singularities are reciprocal ; (2) if the variables
denote point coordinates, the locus of a point on a tangent to the curve,
which with the point of contact is harmonic to the lines ar^ + y^ + 2axy=\)^
is of the form Pp . Q^ . R»* = jwg*", P and Q being linear and R of two
dimensions in a? and y ; and find (3) what is the reciprocal theorem, if the
circular points at infinity are the pair of points.

2555. (The late Professor De Morgan.) — The following is a theorem
of which an elementary proof is desired. It was known before I gave it

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in a totally different fonn in a communication (April, 1867) to the
Mathematical Society on the eonic octogram ; and the present form is as
distinct from the other two as they are from one another. If I, II, III,
lY be the consecutive chord-lines of one tetragon inscribed in a conic,
and 1, 2, 3, 4 of another; the eight points of intersection of I with 2
and 4, II with 1 and 3, III with 2 and 4, lY with 1 and 3, lie in one
conic section. A proof is especially asked for when the first conic is a
pair of straight lines. There is, of course, another set of eight points in
another conic, when the pairs 13, 24 are interchanged in the enimciation.

2660. (J. J. Walker, F.R.S.)— Given that either of one pair of
impossible roots of the equation 3a:* — 1 6a:^ + Z^x^ + 8a: + 09 = gives a real
result when substituted for x in bs^^l^^—lx^ it is required to find the
four (impossible) roots of the biquadratic.

2564. (The late M. Collins, B.A.) — A being a curve whose equation
is given in the usual Cartesian rectangular coordinates, B the evolute of
A, and C the evolute of B ; required a general differential expression for
the radius of curvature of C, on the usual supposition of dan being taken
constant, and likewise on the supposition of dx^ + dy^ {— dz^) being taken
constant.

2666. (Professor Crofton, F.R.S.) — 1. A convex boundary of any
form of length L, encloses an area A. If two straight lines are drawn at
random to intersect it, the probability of their intersection lying within
itUp^ 2irflL-2.

2. The probability of their intersection lying within any given area »,
which is enclosed within A, is p « 2ir(itfL~^. [An interesting but much
more difficult problem is to find the chance of their intersection lying on
a given area », extei-nal to A.]

3. If an infinity of random lines are drawn across the given area A,
their intersections form an assemblage of points covering the plane, the
density of which is clearly uniform within fl. Show that at any external
point P the density varies as 0— sinO, where is the angle A subtends
at P.

4. If A be any plane area, enclosed by a convex boundary of length L,
and e be the angle it subtends at any external point P {x, y), prove that

|J(0-sinO)</j;dy « JL«-fl,

the integral extending over the whole external surface of the plane.

2678. (Professor Crofton, F.R.S.) — If A be the difference between
the whole length of a complete hyperbola and that of its asymptotes, and
if be the angle between the tangents to the curve from any external
point (ar, y), then jj {$— sine) dxdy ^ J a', the integral extending over
the whole surface of the plane outside the hyperbola. [When the two
tangents touch the same branch, 6 is the exterior angle which they make.]

2680. (A. W. Panton, B.A.)— 1. If F and F' are the foci of an oval
of Cassini, and C its centre ; prove the following construction for the
second pair of foci. The circle through F or F' and the two points where
any line through C meets either oval (the curve consisting of two distinct
ovals) cuts the axis in one of the required points.

2. If S and S' be the foci thus found, and P any point on the curve,
prove that PS . PS' is proportional to PC.

VOL. XLIX. ' V

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2602. (Professor Crofton, F.B.S.)^1. If be the angle between the
tangents to an ellipse from an external point {x, y), then jjedxdy ^ tA,
the integ^ extending over the annular space between the curve and any
outer similar coaxal ellipse ; A being the difference of the parts into
which that space is divided by any tangent to the inner ellipse.

2. Show toat this theorem holds for any two convex boundaries, so
related that any tangent to the inner cuts off a constant area from the
outer.

3. Show that, if the same integral, with regard to any convex boundary,
be extended over the annulus between it and any outer convex boundary,
jj ddxdi/ ^ IT (e— 2A), e being the area of the annulus, and A the average
area cut from it by a tangent to the inner boundary (the tangent being
supposed to alter by constant angular variations).

2629. (The late Professor Townsbnd, F.R.S.) — For a system of
quadrics inscribed in the same pair of cones, real or imaginary, and having
. consequently double contact with each other at tiie extremities of the chord
common to the two planes of intersection of the cones, show that a
variable line touching in every position the two cones determines (a) two
systems of pointe inversely homographic to each other on every quadric of
the system, {b) four systems of points, all homographic with each other on
every two quadrics of the system, (c) pairs of variable chords cutting each
other in constant anharmonic ratios in every pair of quadrics of the system,
{d) triads of variable chords in involution with each other in every triad of
quadrics of the system.

2632. (The Editor.)— Prove (1) that 1 . 2 .3 ...n < 2*'»(»-»); and

(2) that < 4a or 4^, when a and b are both positive.

2664. (The late T. Cotterill, M.A.J — 1. Five points (no three in
the same line, and no four in the same plane) determine, by the lines and
planes through them on a plane, a system of ten points and ten lines, the
points lying m threes on the lines, and the lines passing in threes through
the points (Caylby). Show that the figure is its own polar reciprocal to a
conic ; and that, if a conic and triangle in its plane are given, the rest of
the figure can be constructed.

2. A quadric through the five points cute the plane in a conic contein-
ing triangles conjugate to the fixed conic. There is a quadric passing
through the fixed conic, to which one of the five points and the plane
are pole and polar, the remaining four pointe forming a self -conjugate
tetrahedron.

2672. (R. Tucker, M.A.) — From a point on an ellipse chords are
drawn parallel to fixed straight lines ; find the maximum triangle formed
by the three chords of section.

2683. (R. Tucker, M.A.) — To each point on the circumscribing
circle of a triangle corresponds a foot -perpendicular line ; this cute the
circle in two pointe ; required the locus of the intersection of the feet-
perpendicular lines corresponding to these pointe of section.

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APPENDIX IV.

NOl'ES, SOLUTIONS, AND QUESTIONS.
By R. W. D. Chkistie.

(A.) DIOPHANTINE ANALYSIS.

This branch of Algebra derives its name from its inventor, Diophantus
of Alexandria, in Egypt, who flourished in or about the third century.

'* No person has ever surpassed Diophantus in the solution of these
problems, and few have equalled him."

1. To find three square integers whose sutn is a square integer.
Let a?2, y', «' be the three required squares, and let x'^ + y^ ^ o^.
Assume a^+z^ = (na^z)^, then a « {2nz)l{n^~ 1).

Let z = n*— 1, then a =• 2«.

Similarly, y « m^— 1, a: = 2m, a = m2+ 1 « 2fi.

Therefore the squares are (2w)2, (w«-l)2, {(m'-l) (m2 + 3)/4}2, where
m may be anything, but, if integers are desired, then m must be an odd
number. Otherwise, let (2«y)2, {y(«'— 1)}^ and x^ be the required
squares. Assume (2«y)3+ {y {n^-l)}^ + 3F^ = {(»2+ l)y}2 + a;2 « (ny + x)^,

say. Then a: - ^(!^±illl^| y. Let y = 2«; thena; = n* + «8+l,

and the three squares are (2«)*, {2f»(«3-i)}2^ {«* + w2+l}«,
where « is anything.

It is clear the process may be extended to 4, 5, 6, &c. squares, but
the following method is simpler.

2. Find four integers in Arithmetical Progression, the sum of whose squares
is a square integer.

Let a?— 1, X, a?+l, a; + 2be the required numbers.

Assume (a;- 1)2 + {sp) + (a; + 1)2 + (a; + 2)^ = 4ar2 + 4a> + 6 = (2a;-y)2, say.

Then x = -sL , where y is anything integral or fractional.

4(y + l)

Let y = 6, then a: = ^, and the four squares are (i^)', (H )2, (f|)2,

(11)2, or,.rejecting denominator, 12 + 16- + 292 + 432 == 542,

It is plain the process may be extended to any (m2) squares, e.g,,

22 + 62 + 82 + 112+142+172 + 202+232 + 262 = 482.

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3. To find three square numbers in Arithmetical Progressum,
Let a> be the first square, and 2az -¥ ^ the progression.
Then the three squares are a^; a'+ 2<wf + «* ; a2 + 4<»j? + 2r^
The first two are already squares. It remains to make

«' + ^ax + 2x^ » a square ^ {nx- a)\ say.

Then ,«?1^.

Let a «n2-2; thenar- 2(« + 2);

2ax + x*^ com. diff. = {in) (« +!)(« + 2),
and the required squares are {««-2)2; («2 + 2ii + 2)2; («5+4« + 2)*,
when n is anything.

4. If N — a3 + ^, prove that it also equals

where a, 5, m, and n may be anything integral or fractional.
Let nx^a, and mx-^b » sides of squares sought.
Then (#M:-a)2 +(»«?-*)« - N. Therefore {n^ + m^x ^2{mb + na),

Let a be any of the following expressions, viz.,

2n, 4(« + l), 4(2»-l), 3(2» + 3), 12 («+!), 8(2« + 3), 6(2» + 5),
7(2« + 9), 10(« + 6), {m(2w + m)}, &c. &c.

Similarly, b the corresponding expressions

(««-l), {(2#f + l)(2» + 3)}, (4«2-4«-3), 2f»(#f + 3), (4n2+8f»-6),

(4n8+12#f-7), 2f»(« + 6), {2(«+ l){ii + 8)}, «(» + 10), {2«(« + m)},
&c. &c.

Then N becomes a square, and we shall have divided the sum of two
squares into two other squares, N also being a square.

6. We know, by Eulbr*8 theorem, that a sjrstem of any number of
binomial factors being multiplied together, their product is the sum of
two squares, e.g,, {a^ -i- 6^ {e^ -^ d^ i^+P) (y^+A^ «i?^ + ^.
If, now, we make ad ^ be andfff '^ eh, q vanishes, and thus we can divide
a square into four different pairs of squares, as e.g,,

1230« = 12002+ 270« = 7382+9842 - 11222+6042 = 7982 + 9362.

6. To divide a given square number which equals the difference of two
square numbers, into the difference of two other squares.
Generalising the expressions given above, we easily obtain
{2r2+2r(2« + l)+4#f}2+{(2fi-l)(2r + 2n + l)}2

= {2r2+(2»+l)2r + 4i»2+il8,
where n and r are anything.
In this equation make « — 2, « « 1 respectively, and we get

(2f«+10r + 8)2+ (6r+ 16)2. {2r2+ 10r+17}2 (1),

and (2r2 + 6r + 4)2 + (2;- + 3)2= {2r2 + 6r + 6}2 (2).

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161

Now, equate the shortest sides 2« -i- 3 ^ 6< + 15, say.
And let 2« + 3 » side of given square => 21, say, then a ^ 9, r — I.

Then in (1) make r «- 1, and in (2) r = 9.

Thus 20« + 21« = 29», or 21« - 293-202 (1).

And 220» + 21« = 2212, or 21« » 2212-220' (2).

Similarly for any other square whose side is given.

7. Draw a straight line cutting two eonemtric circles so that the part
intercepted by thetn is divided into three equal portions.

Let R, r be the radii of the outer and inner circles respectively. As
the outer intercepts are always equal, let x be the middle intercept. Then

we must have E'— r* « 2a^,

2mr

The analysis gives us

m3-2*

Let r =« m^—2; then x « 2m, and R = fn« + 2 ; e.g., if m =» 4 inches,
then r^ XyB,^ 7, 4, 9 inches. Therefore, measure ok 4 inches from the
outer circumference cutting the inner one, and produce the straight
line joining the points. The three intercepts are equal.

8. (Question 2814.) — To Jlnd three rational square integers in Arith'
metical Progression having a common difference o/ 13.

"We have

{x^+{x + 1)2} {{x+ 1)2 + {x + 2)2} db {2 (a; + 1)}2 = {2x^ + 4a? + 3)2

or (2a;« + 4a?+l)2 (1).

Also (w2 + «2)2±(,rt + «)(m-n)(4mf») = {m2±2in«-«2}2 (2).

Now 13 = 22 + 32 ; therefore let a; = 2. Thus (1) gives us
13x26 + 36 = 192 and 13x26-36 = 172.

Let i? « 13, j2 « 25, r2 ^ 36, and puti?j2 for #» and r2 for « in (2) ;
then {pY + *'*)^±(pg^-^r'^{P9^—^^{^P9^r^) = two squares.
But (^j^— r*)(4^2r2) = a square,

therefore fi^—r^ = a square, and consequently

Jo2(74 + r* 0^164568*241 . ,

— i^H -r^ ^ 3Q *^:»»^vu^:«x _ gq^are required,

(l^q^^r*)^^ 376584400 ^ « -reHiurwu,

and the other two = -_|V + »1^^13^

(P'q^-r^){^q^)
Hence it appears that the question will always admit of a solution
when the given number plus or minus a square are both squares.

9. To find a number which, being added to or subtracted from a sqtuire, the
sum or remainder shall be a square.

Use theorem 2 in (8), supra.

Let m = 2, w = 1 ; then 62db24 = 72 or I2.

N.B. — The three squares are inArithmetical Progression, thus 12 : 52 : 7'.

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10. Comtruet a parallelogram toho$e sides and diagonals may he represented
by integers.

Let a, d be the two contiguous sides, and 0, d the diagonals ; then

(« + 3)«+(o-3)2-c« + <P.

Assume p, q ^ any integer, and a, b, e, d are now easily found.

11. Tioo chords within or tvithout a circle intersect at right angles. Con*
struct it so that the four segments as well as the diameter mag be represented
by integers.

Resolve any square into four other squares ; e,g.,
D«-a« + A2 + c2 + ^ (V, infra).

Then D — the diameter ; a^b^c^d^ the four segments of the intersecting
chords.

12. Construct a regular decagon and a regular pentagon in a circle so that
the sides of each, as well as the radius, mag be represented by integers.

Let P «= side of pentagon, and D ^ side of decagon, B = radius.
Then P^ « R^ + D'. Therefore assume any of the expressions given in
A . 4 for R and D, making n » any integer.

13. Construct a series the terms of which may be taken to represent the
three sides of a right-angled triangle, and find the sum,

Wehave (2fi-l)2+(2it.ft-l)« = (2«2--2it+l)«.

Therefore assume the n^ term = (2n-l)(2f».fi-l)(2M2— 2fi + 1) ;
and the series becomes 1.0.1 + 3 . 4 . 6 + 5 . 12 . 13 + 7 . 24 . 26, &c.,
and the sum = Jw' {(4f»2_i)(nS~l)}.

14. To find a series of biquadrates equal to a series of squares.
We have 2i.„..«M)/sJ(„.«,„ = (3«' + 3«-l)/6.

If M :b 6, we have

l* + 2^ + 3H4* + 5* + 6* « 62 (12 + 22 ■|.33 + 4« + 6' + 6«).
From this value of n others may be obtained thus : —
Suppose that re — /, y = g is a solution of the equation x^—'^y'^ « a,
and let a: = A, y = A; be any solution of the equation x^ -Ny2 » 1 ; then

a?2-Ny2 - t/^-N^2)(A2-NA:2) = (/4 + Ny*)2-N {fk^ghf.
By putting x=^fh± ^gk, y =fk ±gh,

and ascribing to A, A; their values found by convergents, &c.

15. For X^ = 1* + 2' + 3* ... w*, see an interesting solution in Vol. xlix.

(B.) RESOLUTION OF SQUARES.

1 . If any number N can be resolved into the sum of n squares, then
2 (n — 1) N can be resolved into the sum of « (»— 1) squares.

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Let N = fl2 + A2, then 4N » (2o)« + (2*)2.
Again, if M = o2 + ^2^.yi^ |;hen

4M = (a + i3)2 + (a-^)2+(a + 7)2+(o-7)2 + (i8 + 7)'+(i8-y)2.
H a2 = o* + ^2, by any of the formulaa in (4), we easily get

-[i(a + )3)P + [J(a-«P+[i(a + 7)?+[i(a-7)P + [i(^ + 7)?

+ [i(i3-7)P.
Ex.gr.: 102 + 22-62 + 82 + 22- l2 + 2« + 32 + 42 + 52 + 72.

Again, let N » a2 + ^ + c2

«[i(a + *)]2+[i(a-*)]2+[i(« + .)p + [i(a-.)]2 + [J(* + ^)P + [i(*-^)?.

Ex.gr.: N=* 122 + 62 + 22-102 + 82 + 42 + 22= 92 + 32 + 72 + 52 + 42 + 22.

2. If any number N can be resolved into the sum of n squares,
\ {n— 1 . ft— 2) . N can be resolved into the sum of 4n squares if n > 3.

LetN = a2 + ^2 + ^ + ^^

Then3N«[i(a + 3 + c)]2 + [J(o + *-tf)]2 + [J(a-* + c)]2+[J(* + tf-a)]2
+ [i(a + ^ + rf)P + [i(a + 3-rf)]2 + [i(a-* + rf)p + [i(5 + rf-a)]2
+ [i(a + <? + rf)]2 + [i(a + c-rf)]2 + [i(a-<j + rf)p + [J((, + d_a)]2

+ [i(* + (J + rf)]2 + [i(* + c-rf)? + [4(*-^ + ^? + [i(^ + ^-W.

Similarly, 6N = sum of 20 squares, ION => 24 (or 30 squares), &c. &c.

And, since the sum of the first n natural members is a perfect square,
if w is = A;2 or A:'*— 1, where k is the numerator of an odd, and k' the
numerator of an even convergent of ^^2, if either (1) J(#i— l.fi— 2)— N;
or (2) N and i (»— 1 . « — 2) (a sequence from unity} are both squares, w©
can thus resolve a square into 16, 20, 24 ... 4 (» + 3) squares.

3. Weknowthat(a + *)2 + (* + c)2 + (r + a)2 = (a + i + c)2 + a2 + i2 + c2«M.
If a - i (Sb-e), we have also [i(7*-<?)]2 + K(* + c)]2 « M.

If also we make 5{b + e) : 7b— e ::»»:«, where »»2 + «2 „ ^2^ ^q g^^
M2 - [i (7b-e)y + [$(* + c)]2 - (a + 3)2 + (* + e)^ + {e + af
= (a + * + <j)2 + a2 + 62 + c2.

Or again, M2 - [| (7a + 17*)]2 + [J^ (a + *)]2=3 squares=4 squares, &c,
Ex. gr. : 1302 „ 6O2+1202 - 602 + 962 + 722-1092 + 132 + 372 + 692
— five squares, &c.

= 782 + 1042 = 402 + 782 + 962 =. 1072 + 1 12 + 292 + 672, &c.
Generally, to get N2 - a;2 + y2 ^^ (a + 3)2 + (a + c)2+(i + c)2

« (a + 3 + c)2 + a2 + ^ + c2,
let (« + *):(* + <?)::#»:«, where »»2+«2 « pi.

Then ^^m^^^^j)-nb^ ^^ (a + 3)2 + (a + c)2+(3 + c)«

n
(m2 + w2)f3 + g)2 ( (w-n)3 + (m + >»)g y
" «2 +[ ,n j

^ p(3 + g) |2 ^ ( (^~n)^ + (m + »a)g |2 ^ ^2^^3 .

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and to make x^-^y* ^ N^y assume

p{b-k-e) : (m-fi)ft + (»t + fi}c ::?:>•» where y' + r^ « «2^
and we get e ^ pr^{m - «) ^, * = (w + «) ?—!>*•, « « (m— «) y +/?r.

Thus finaUy, N^ = (2i»«)2 = (2^?^' ^ (2pr)3 = (2i«(7)2 + (2/>r)2 + (2n^)2
- {jt?r + (m + f»)y}^ + {(m-fi)^ ••'■i'''}' + {(♦» r«)y-i?r}2 ^. {jt?r- (w-n) j}2.

And since >n, » ; gt fl Pt * &re interchangeahle, we can always secui-e
four different resolutions of N^

Ex. gr. : 1302 = 1202 vSO' . 96«'r 502+722 -.- IO92 + 372+I32 + 692
-502+1 202 « 302 .;. 1 203 .> 402 « 952 ^ 552 + 252 + 652
-1042+782 = 402 + 782 + 962=. IO72 + II2 + 292 + 672
- 782+1042 « 302+ 1042 + 72 - 1032+312+12 + 732.

The following transformations of squares may here he noticed : —

where N and r may he iuf,egers > unity. Or, again

N2.N2f'^-^|%N2f-f'\V,
Cm2 + r25 (jn^ + f^y

where m and r ntay he assumed at pleasure m > r.

(2) (a2 + ^ ((j2 + rf3) - (a<? ± &rf)2 + (arf :p ^C)2 =r (flk?)2 + (Atf)2 .?- (a4)2 + {irf)2,
(a2.;.62)(^ + rf2)(^+/») „(A3 + B2)(u2+/2) « (A^±B/)2+(B^=FA/)2,

(a2 + i2)(c2 + rf2)(u2 +/2)(^2 + ^2) « (^2 + B2)(C2 + D^)

= (AC±BD)2+ (BC=rAD)2 ^p^-i-q^
where /) =• AC + BD = (a<j + W) (^^ +J%) + (*c-a(/) {fg-eh),
q - BC-AD = (itf-flr^ («^ +/A) - (a<? + i<0 (/y— *A)>
is one of eight solutions.

(3) a2 + *2j.^= [j(a + i + <j)]3 + [|(a + J-<j)]-i+[|(a-*+<j)]2

+ [i(3 + .-a)]2 (1),

and a2 + ^ + ^ + ^«[j(a + j + tf + rf)]2 + [^(a-i_c + rf)]2

+ [i(«-* + <?-^? + [i(« + *-«-<^P... (2);
also (a + * + £;)2 + a2 + ^2 + ^= (a + 3j2+(^ + ^)2+(^ + rtj2.

Thus (1), (2), ife2 + /2 + ^2 + „2«a;2 + y2 + 22 + ^ + ^.
(1), (3), i?2 + ^ + ^«a;2^.y2 + 5;2 + ^+(^ + 5 + ^)2j

(4) (fl* + 3«2+l)2« {2«(f»2+l)}8+{„4 + „2+l|2

«(2«)*+{2n(«2-l)}2+(„4 + „2+l)2;

(«) (y-«)2 + (2-a;)2+(a;-y)2+(a; + y + 2)2« 3(a:2 + y2 + j,2).

Thus

a? + J2 + c2={t(y-«)}2+{J(i5-a:)}2+{Har-y)}2+{j(^ + y + «)}*,
where a;, y, z are the sides, and a, 6, « the medians of any triangle. In
this way other identities may he made practically useful.

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(6) (ey-bz)2+(az^cx)^-¥(bx-ayy+(ax + bi/ + czy

= (ax)^ + (ay)2 + {azy + (bx)^ + (by)^ + (^2)2 + {cz)^ + ((?y)2 + (ez)^
= (fl2 + i2 + c2)(a;2 + y« + a2).

4 . By Pollock's Theorems we can resolve any number N into one, two,
three, or four (not more) squares as foUows : —

Take any trigonal number (or sequence from unity), sayi? =« 10.
We have 2;?+ 1 = 21 = m^ + m+l = (2m ± 1)^, and the squares are
(n ± 1)2, «2, «2, »2; and 21 = 32 + 22 + 22 + 22.

Again, if ^ = sum of two trigonal numbers, say 15 ; we have
2i?+l = 15=:2a2 + 2a + 2&2 + l = (^5 + i)2^ ^2^ ^2^ ^2^

whose roots are (*+!)> (~^)» (+^)> (~^) = !•

Thus 15 - (3)2 + (-2)2 + ( + 1)2+ (-1)2 (roots - 1).

Lastly, if ^ = sum of three trigonal numbers, say 47 ;
we have 2i?+l = 47 = 2a2+2a + 2*2 + 4it2±2« + 1,

and the roots are ^ a+n ± I, a—rif b + n, b—n, (—1).
Thus 47 = (-3)2 + ( + 2)2+( + 5)2+(-3)2 (= 1).

Therefore, since every number must either be a trigonal number or
composed of two or three trigonal numbers (Lbgendre, Theorie des Nombres), ,
every odd number may be resolved into integral square nimibers (not

exceeding four) whose algebraic sum will be 1, 3, 5 2»— 1, up to the

maximum. Even numbers may also be resolved into square numbers
(not exceeding four) the algebraic sum of whose roots may always equal 2.

It may be interesting to reproduce here Pollock's method of resolving
odd numbers into squares.

Eesolve 57 into four squares the sum of whose roots may equal 9.

Then2«j + 1 = 9; thus w =» 4, andw2 + «j+l = 42 + 4 + 1,= 21.

Now
57-21 = 36, and 36 + 1 = 37=(- 1)2 + ( + 2)2 + ( + 4)2 + (-4)2 (roots = 1).

2 +2 +2 +2
Therefore 21 + 36 = 57 = l2 + 42+62+(-2)2 (roots = 9).

By this theorem we can instantly transform one square into two, three,
or four squares, ad libitum ;

e.g., 181 = 12 + 42+82+102.

Then 181-42= 155 « 12 + 82+102 = 42+62 + 72+82;

thus 12 + 102 = 101 = 42 + 62 + 72= 12 + 62 + 82.

Then 42+72 = 12+32 = 65=22+32 + 42 + 62;

thus 72 = 22 + 32 + 62, &c. , &c.

5. We know, if N = a2 + ^2 + ^ + ^2^ there are two resolutions of 4N into

VOL. XLIX. X

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uneven squares, viz.,

= A2 + B-'+C« + D2, say (1),

«(A-2rf)2+(B + 2rf)2 + (C + 2tf)2+p + 2tf)2 (2).

£.ff., let N = 92 - 23 + 42 + 62 + 62 - &c., by Pollock ; then
4^-^182=42 + 82+ 102 + 122= 172 + 62+ 12+ 32 = 52 + 72 + 92+ 132, by Gauss.

6. In order to resolve M2 into N integral squares, we may make use of
the following principle : —

(« + 1)2_ (^^2 _ 2» + 1 = any odd number > 1,

therefore take any N squares whose sum = 2« + 1,
Bay 22+32= 13 = 2«+l;

thus #f = 6, then 72 = 62 + 22 + 32, or generally

(«■+«- 1)2 = (w.« + l)2+(«+l)2 + w2.
Or again, we have 12 + 22 + 32 + 42 + 52 + 62 = 91 = 2« + l and » = 45 ;
thus 452 = 442 + 12 + 22 + 32 + 42 + 52 + 62, or generally
(3«2+l5« + 28)2= (3n2+15w + 27)2 +

n2+ (» + 1)2+ (« + 2)2+ (» + 3)2+ (« + 4)»+ (« + 6)2,
where n is arbitrary, and so on.

7. The same object may be effected thus : — We know by easy analysis
that if a; = a^-b^, y = 2ab,

then «2 = a;2 + y2.

also, if a; = a2+ ^2-^2, y = ^acy z = 2*r,

then «2 = a;2 + y2 + j;2j

again if x = a2 + ^3 + ^2_^^ y = 2<w?, z = 2bd, w = 2crf,

then «2 ^ u;2 + fl;2 + y2 + ;52 .

nnd generally, if A, ^ a^-k-a^^a^ ... — «J,

and Aa = 2«, a,„ A3 = 2,a^ a^ An = 2a„-i<i„,

then U2= A,2 + A22 + A32 +Ai.

E.g., let «i = 1, ^2 = 2 flj = ^ »

thea 112= 12 + 22 + 42+62 + 32,

and flo on for any number of squares.

H . Numerous resolutions may be obtained thus : — Take the first n^
iiaturttl nimibers and arrange so. Let n = 4: ; then

1, 15, 3, 13, 6, 11, 7, 9,
16, 2, 14, 4, 12, 6, 10, 8.

Ntitc! the odd numbers in the top row and the even numbers in the bottom
TH^f also the sum of the numbers at the top and bottom = 17 = »2+ 1.

I

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Take any four odd numbers whose sum =32 from the top row, and the
remaining four even numbers from the bottom row ; e.g.^

6,7,9,11, 2,4,14,16,
8, 7, 9, 13, 2, 6, 12, 16.

The sum of the odd numbers =32, and the sum of the even numbers = 36,
and the total sum = 68 = m . »'*+ 1.

Then we have at once : —

(a; + 5)2+(a? + 7)H(a; + 9)2+(a?+ll)3+(a? + 2)2

+ (a; + 4)2+(a;+14)2+(ar+16)«
= (a: + 3)«+(a; + 7)2 + (x + 9)2 + (iF+ 13)2+ (a? + 2)2

+ (ar+6)2+(a:+12)2 + (;r+16)2,
= (a: + 1)2+ (a?+ 16)2+ (a:+ 14)2+ (ar + 4)2+(a:+ 12)2

+ (a; + 6)2+(ar + 7)2 + (a? + 9)2,
= (a?+8)2 + (a? + 10)2+(a:+ 11)2 + (a: + 5)2+(a?+ 13)2

+ (a: + 3)2+ (a? + 2)2+ (a: +16)2,

- (a:+ 1)2+ (a;+ 12)2 + (ar + 8)2+ (:c+ 13)*+ (a?+ 15)2

+ (a: + 6)2+(a;+10)2+(a? + 3)2,
= (« + 1)2+ (a; + 6)2+(a; + 11)2+ (a?+ 16)2 + (a: + 4)2

+ (a: + 7)2+ (a: + 10)2+ (a: +13)2,
= (ar+16)2+(a;+14)2+(a;+12)2+(ar + 9)2+(a; + 3)2

+ (a; + 2)2+(:F + 5)2+(a: + 8)2,

- (a:+ 11)2+ (a: + 16)2 + (a?+ 10)2 + (a?+ 13)2+ (ar + 1)2

+ (a; + 4)2 + (a: + 6)2 + (a? + 7)2,

- (a?+ 11)2 + (a; + 16)2+ (ar + 4)2 + (a: + 7)2+(a:+ 14)2

+ (a: + 2)2+ (a: + 9)2+ (a? + 6)2,
&c., &c.

But as these squares are not all different, we get, after elimination of
the identical squares, —

(1) (a: +14)2+ (a; + 4)2+ (a: + 7)2+ (a; + 9)2

-. (a: + 8)2+(a; + 13)2 + (a;+10)2+(a? + 3)2r=4a:2 + 68a: + 342,

(2) (a; + 1)2 + (a: + 15)2 + (a; + 12)2 + (a- + 6)2

= (a: + ll)2+(a? + 2)2+(x + 6)2+(ar+16)2 = 4a:2+68a; + 406,

(3) (a: + 15)2 + (a?+14)2+(aj + 12)2+(jr + 9)2

= (a;+ 11)2+ (a;+ 16)2+ (a;+ 10)2+ (a?+ 13)2 « 4^2+ I00a?+ 646,

(4) (a;+3)2 + (a: + 2)2+(ar + 5)2+(a; + 8)2

= (j;+l)2 + (a: + 4)2+(a: + 6)2+(ay + 7)2-4a;2+36a:+102,

(5) {X + 12)2 + (X + 8)2 + (a; + 15)2 + (a; + 3)2

= (a:+ll)2 + (a; + 16)2+(a. + 4)2+(a?+72) = 4a;2 + 76:c + 442,

(6) (a: + 14)2+ (a; + 2)2 + (a: + 9)2+ (a; + 5)2

= (oj + 1)2 + (x + 6)2 + (a; + 10)2 + (a? + 13)2 = 4a:2 + 60a? + 306,
&c., &c.

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It is now easy to make any of these » a square.
We have (1) 4a;2 + 68a: + 342 = (2a:+ 17)2 + 63 = y« + 63 = z\ say;
and, since 53 » 27 + 26, we have at once 2 — 27, ^ = 26, and ^ » f .
From this value of x we easily obtain

642= 172 + 232 + 272 + 372= 162 + 262+292 + 362.
Similarly (2), 4a?2 + 68a; + 406 ; if a: = V» ^® establish

1182 = 432+712 + 662 + 632=632 + 452 + 612 + 732.

(3) givesus 222 = 52 + 72+112+172=32 + 92+132 + 152,

and (4) the same. In order, therefore, to find another value of x we have
by Eulbb's metiiod a: = y + i. Thus

4a;2 + 36a?+ 102 = 4y2 + 40y + 121 = {py-Uf, say.
Then y = V, and a; = V^.
Neglecting the denominator, we have

1902= 792 + 862 + 972+1152 = 732 + 912+1032+1092.
In this way endless resolutions may be obtained.

n = 6, by a similar process, will give us a still larger variety of
equations, of which one is

(a: + 1)2 + (a: + 33)2 + (a? + 7)2 + (ar + 27)2 + (a: + 22)2 + (a; + 21)2
= (ar + 36)2 + (i; + 4)2 + (a; + 30)2 + (a;+10)2 + (ir+15)2+(ar+16)2, &c., &c.
If ft = 10, we may obtain any number of resolutions such as
(a; + 100)2 + (a; + 68)2 + (a; + 92)2 + (ar + 86)2 + (a; + 4)2 + (a; + 6)2 + (a? + 6)2

+ (a: + 7)2 + (a: + 8)2+(a? + 99)2
-.(aJ + l)2 + (a? + 3)2 + (a? + 9)2+(a;+ 15)2 + (a; + 97)2 + (a: + 96)2+(a; + 96)2

+ (a: + 94)2 +(-,; + 93)2 + (a; + 2)2,
or, if we prefer it,
(2)2 + (2«2 _ «)2 + (2» + 2)s + (2»3- 3f»)2 + (4« + 2)- + (2«2- 5«)2 + {fin + 2)2

+ (2«2- 7„)2 + (13« + 2)2 + (2«2_ 9«)2
= (2n2)2 + („ + 2)2 + (2w2 - 2»)2 + (3« + 2)2 + (2w2^ 4w)2 + (5» + 2)2
+ (2»2-6«)2 + {In + 2)2 + (2«2_ 13«)2 + (9« + 2)2,

the principle of which will readily suggest itself on a comparison of the
upper and lower squares ; and generally, for any unevenly even number of
squares, say 18, we have

(2)2 + (2«2_ w)2 + (2« + 2)2 + (2n2_ 3«)2 + (4f» + 2)2 + (2«2_ 5n)» + (fin + 2)2
+ (2it2_7„)2 + (8» + 2)2+(2f»2-9„)2+(i0n + 2)2+(2»2-ll«)2+(i2n + 2)«
+ (2«2- 13n)2 + (14» + 2)2 + (2«2_ 15«)2 + (25» + 2)2 + (2«2_ I7n)2

»(2«2)2+ („ + 2)2+ (2«2_2«)2+ (3« + 2)2+ (2»2_4«)2+ (5» + 2)2+ (2«2_6«)2
+ (7w + 2)2+(2«2-8«)2+(9» + 2)2+(2;»2_io«)2+(ll« + 2)2+(2w2_l2»)2
+ (13« + 2)2+ (2w2- 14»)2+ (15/* + 2)2+ (2W--25/02+ (17» + 2)2.

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9. To make s'/^i = a square, where

2; = !'• + (« + !)♦• + (« + 2)** («+n-l)^

and a the common difference, we have

2i/2i = l+a(«-l)+«2[i(ft.w-l)]=M2.

(a = 1.) Then »« + « = 2M2 ; thus SM^ + 1 =« a square = N2.

Thus 8M^— N^ = — 1, an equation easily solved by the even convergents
of ^/2.

(a = 2.) Then 2«*- 1 = M^, solved by the odd convergents of ^/2 ;

^ Tw 'T ^x. 13 + 33 + 53+73 + 98 -o
..^.,n = 5,M = 7;thus ^^3^,^,^, =7-

Consequently, 13 + 33 + 53 . . . (2w - 1)3 = (nM)^, where n is the denominator
and M the numerator of any odd convergent of //2.

(a=3.) Wehave 9w2_3»-4 = 2M2, or (3«-J)2 = J (8M2+ 17),
M = 1 obvious.

Therefore, let M=N+1 ; thus 8M*+17=8N2+ 16N + 25-(i;N-5)2, say.

Thus N = ^-^1^;^; i? = 4, &c. ; N = 7, &c. ; M = 8, &c. ; « = 4, &c. ;

13 + 43+73+103 Qj

'•^" 1 + 4 + 7 + 10 =^'

and so on for higher values of a.

When any one value of n is obtained, others are readily got by substi-
tution.

In a similar way we can convert 2^ / si into a square, where

2? = a'' + (a+l)'' + (a + 2)»* (a + n-l)'*,

a being the first term ; e.ff., a = 3 gives us, iin^—n-^ 2a (?e — 1) + 2a2= 2M2,

w2 + 6» + 12 = 2M2 or (« + f )2 = i (8M2 - 23), M = 2 obvious.

Therefore, let M=N + 2 ; then 8M2-23=8N2 + 32N + 9 «(i?N-3)2, say.

Then N - '^{^P-^^^\ ^ == 4, N = 7, M = 9, « = 10.

p — 8

Thus 33 + 43 + 58 + 63 +73 •»• 83 + 93+ 103+ 113+ 128 ^

3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 '

and so on for any value of a.

10. To resolve a square into two, three , four , or more squares.
Applying the results given above, we easily establish
(2w4 + 4;>3 + 4„s + 2» + l)2= (2«2 + 2« + l)2+ {(2» . «+ l)(«2 + «+ i)p
= (2;j + l)2+(2».«+l)2+{(2».«+l)(n2 + »+l)}2
= four squares, by Pollock, = &c.
Ex,gr., (w= 1) 13'-^= 5^+ 122 = 32 + 42 + 122 = 42 + 62 + 62 + 92 « &c.

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11. Again, to resolve three squares into three others, the sums of wh6se
roots are equal, we may make use of the following identity,

(b + ecCf + {hcf + rf2 = {be + rf)2 + ja + (^^2^ where *, Cy d are arhitrary,

the sum of the roots heing (* + i^(c+l); and, if * « 2», c = « + l,
<? = 2« + 1, we easily get

(2«2 + 5» 4- ly + (2» . « + 1)3 + (2« + If =x (2w2 + 5» + 1)2 + (2n2 + 2n + 1)2
-(2»2 + 4n+l)2+(2«)2+{(n + l)(2» + l)}2.

Ex. gr., 82 + 42+32 = 82 + 52 = 72 + 22 + 62,

192 + 122 + 52= 192+132= 172 + 42+152, &c.
Similarly {m (« + fH)}2+ {n («+#«) }2+ (mw)2 = {m^^mn-^-n^Y.

12. To resolve an even square into eight or sixteen squares.
Primes are of two classes, viz., 4«+ 1 and 4» — 1 form.

Now it has been demonstrated by Db la Grange (Afemoirs of Berlin,
1768), and others, that primes of the former class may be resolved into
the sum of two squares ; and, since it may be shown that every even square
> 5 equals the sum of four 4« + 1 primes, it follows that every even
square > 6 equals the sum of sixteen squares, since Pollock has shown
that every odd number may be resolved into four squares ; thus

62= 1 + 5 + 13 + 17 = sixteen squares (by Pollock)
« eight squares (by Lagbanoe),

82 = 6+13 + 17 + 29- &c.,

102= 13 + 17 + 29 + 41 = &c.,

&c. &c.

182= 37 + 41 + 97 + 149 =41+61 + 73 + 149 = 41 + 63 + 73 + 167

= 41 + 73 + 97 + 113 = 41 + 73 + 101 + 109 = &c., &c.,
= sixteen squares in six ways at least.

(C.) RESOLUTION OF CUBES.

EuLER has demonstrated that it is impossible to find any two cubes
whose sum or difEerence is a cube ; also that the formula a^:ht/^ = 2s^ ia
impossible.

There are two methods of resolving a cube into three cubes.

(a) ff two of the cubes are given, and a third be required to make the
three cubes equal a cube, we may use the following formula : —

{a(b^-a^)Y+{b{b3-(^)y+{a{2b^ + <^)Y= {*(^ + 2a3)}3.
Thus any cube may be resolved into three cubes ; thus,

lb (b' + 2a' ji lb (// + 2a«)) {b (L' + 2a»)) *

and a similar principle applies throughout.

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{b) Again, if it be required in general to find three cubes whose sum may
equal a cube, Euleb gives us

X =p + q=^ {fi + ^ffu) + (fft-'fu),
y=.p-q = {ft + Sffu)-{fft-^fu),
t> =. r + « = {kt—hu) + {ht + dku),
z = r— « = {kt~'hu)^(ht + dku),

where u =f{P + Ss^^^h{h^ + 3k^), t = Zk{hf + Zk^)^Zff{P + 3ff^),
and ffSl^hjk are arbitrary.

.-. x^ + f/^^f^-2^ = {{ft + Zffu) + {^t-fu)Y+{{ft + 3ffu)-(s/t-fu)Y
= {{kt-hu) + {ht + Uu)Y-{{kt-hu) - (fit + Uu)Y.

(c) To obtain four cubes equal to a cube, we may use the following equa-
tion:— («-a)3+(»-*)3+(*-.c)3 + 3a*<? = «3.

Assume Zahc = (3mn)', thus abc = dm^^. Now 9m^n^ = abc may be
assimied in any combination ; e.g., let a = 9m, b = m^n, c = n^. Thus,
e.ff.y if M = 2, n = 3, we easily get 13 + 63 + 73 + 123 = 133.
Other combinations are a = 9m\ b = n, e ^ n^ ;
a — 9m', b = wn, <? = f»2 ;
a = 9»j, b = m3, c = n^, &c., &c.
To secure the desired result make m<n,

(e?) If m>n, we may secure the result that the sum of two cubes equals
the sum of three cubes thus : w = 3, » = 2 gives us
133 + 413 + 363= 493 + 53.

{e) "We may secure four cubes equal to two cubes by the use of the
following identity : —

(a; + y + 2)3+(a; + y-2)3 + (a;-y + «)3+(a;-y-«)3«4a;(a?3 + 3y2 + 3552)

= 4a:3+12a:y2+i2a;22 = a:3 4.3^(a;2 + 4y2 + 4z2) = ipS + 3;c (M2),
if a; = a2 + ^-c2, y == «<?, and z = Jc, making M = a^+i^ + ga.
If, also, we assume 3a; = M, i.e. if a^ + i^ _ 2^2, by making

a^^-\-2qp-q\ b =^ p'^-lqp-q^, c=^^^-q\
we establish (a: + y + «)3 + (a; + y — 2)' + (a;— y + 2)' + (a;— y — «)3 = a;3 ^ (3^j3 .

(/) Also, if a:, y, and 2 are so chosen that any two are > the third, we
have the sum of three cubes resolved into three other cubes, e.g.,

i? = 2or3, ^=1, gives us 113+133=13 + 33 + 53+153,

p = i, q^l, gives us 7993 + 5613+173 = 2893 + 8673 + 2213.

{ff) To find n cubes whose sum may equal a square.
Let 2;|= l'' + 2'' + 3'-... +«^

It is easy to prove that 2* + (22')' =* (322)2,

(13 + 23 + 33... +^.3)^(^2^^)8^^ ^•^^^^l■^2;^+l j2^

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e,g,f if n » 5, we have

l» + 2» + 3» + 4» + 6»+30»= 1662- {3 (12 + 2* + 3' + 48 + 52) p.
[si - (si)' is well known.]

(A) Tojind two cubes whose sum may equal the difference of two squares.
Since 2* = (S^', we get from (^)

(22')'=(32,)»-(2')»,

and, since we know that «' = ( ^lli! \'_ ( ?!i:5 V,

we have at once ««+ (»'+«)• - (" " ^ Z"''^)'- (^)*-

C/) To find a number of odd cubes whose sum may equal a number of even
cubes.

We have i»-2» + 3*-4»... =-(4«» + 3«2) ;

therefore l» + 3» + 5»... +(2n-l)» + 4N» + 3N2 = 2» + 4» + 6»... + (2N)»,
or U + 3» + 5»...+N» + 3N2(N + l) -2» + 4» + 6»... + (2N)».

To make 3N2 (N+ 1) = a cube, let N - 8 ; thus

l» + 3»+6» + 7» + 9» + ll»+13»+15»»2»+4»+6»+10»+14»+16».

To obtain other values of N, we may proceed by Eulbr's mode to sub-
stitute N = M + 8 in ZW (N -i- 1), and equate it to (pm + 12)>, and so on.

(k) To resolve a cube into a number of squares.

We know by the Analysis that x^ + y* can be made « (/?2 + y2)*».
Let X = y _ zpq\ y = Zp^q-q* ;

then a?2 + y'=CP« + ^» = {i»(i^ + ^')}2+{^(i>' + «^}*;

similarly

(^2 + j3 + r2)>={p(i;2+^ + r2)}2+{y(j,2^.^ + ^p+|^(^ + y3 + r2)}2,

and generally

{il« + r + r2...««}»= {p (1^ + 52+. ..;5J)p+{^(pS + ^8+...;52)}+&C.,

C>n.fw + 1.2m + l')» r,«/m.fn-^l .2m + l\72

1 — 6 — j-ri — 1 — )i

(/) Every cube may be resolved into an Arithmetical Progression.

We have S = a+(a + *) + (a + 2*) ...a + (w-l)d = «|a+ **-! . j| .

Let f> = 2m + 1, then 2 == (2m + l)(a + ^).

If a + ^» = (2m + 1)2, then sum of n terms '= «'. Thus let

a = 2w-l, * = 2(fi-l), «=»6;
tf.^., 9 + 17 + 26t33 + 41 = 5».

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(m) To resolve a number into three cubes.

Take 6 for example.

Assume 6 = (2 - *ar)» + (ex- 1)8 + (ete- !)»,

and make ^ » (0 + d) / 4 ;

then x^ {66(c8 + eP)-16crf}/{2l(c8 + d8)_tfrf((j + rf)}.

If tf « 7, rf = 6, 3 = 3, then 6 = ( J)8 + (j)s + (j^)8.

Again, take 8 =« 2^ = {i-bx^+iex -f 1)«+ (dx^ l)i,
and make 3 — {c + d)/4;

then « » {3 (dS-u«)-6*3} / {c» + rf»-.A»}.

If d^ 7, e — 5f then ^ » 3 and a: » ;^, and we have
36» + 98'« 92» + 69»,

or the sum of two cubes resolved into two other cubes.

If rf = 6, tf = 7, * = 3, we get 3; = - f and 20* - 7* + 14» ^- 17», or a
cube equal to three cubes.

Similarly, for any number if we take care to make the given number in
the original equation vanish and to equate the coefficient of a; to zero.

(n) To find a cube equal to the difference of two squares.

Assume a;3— ^'= a^ » (ar-fty)', say;

then 2»a? = y (y+n').

If y s 2n, then x = ft'+2» and a — 2n—n^; thus

(2»)' - (»2+2ft)2-(»2«2n)«,
where n is arbitrary.

(0) To resolve the sum of four cubes or biquadrates into four other cubes or
biquadrates.

We have (a + 3+<?)'» + a* + ft* + <J» — (a + 3)* + (* + <?)'• +(c + »)•• + X",

(«=3). (a + 3 + (;)» + a» + *» + (j"=: (a + b)9 + (b + c)t + {e + a)t + 6abc.

To make 60^0 = X', proceed as in (C. e), and assume 6abe = (6mn)^
in any order, as, tf.y., a = 36m, ^ = m%, c = nK

Thus m« 2, « = 3 give us 3' + 4' + 24»+ 31» « 7' + 12»+ 27» -1-288.

(fi — 4). Again, for

(a+b + c)* + a!^+b* + c*==^ {a + by + {b + e)*+(e + a)*+l2abe{a + b + e),

Tomake I2abc{a + b+c) - X* == (nb)* b&j ;

let c = 1, a = 3^*^ thus 36 (3»2* + 3 + 1) = (w*)2.
Now assume Zn^ + J + 1 = (2«*— 1)' say ;

thus b — -^-^ , where n is arbitrary ; e,ff., « • 1 gives us

1^+2^ + 94 = 344.74 + 34.

VOL. XLIX. Y

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(D.) SOLUTIONS OF OLD QUESTIONS.

1010* (The Editor. ) — ^Pind integral yalues of oj, y , z which will make
«'-2y', y*—Zz\ aj^— 6«3 all squares.

Solution.
An obvious solution \ax ^Zz, y=2z\ but this makes two of the squares
identical. Therefore, let y - 2«, then «8-8«5 = a\ a^-6z^^^\
thus a«-»-3«2 = J8.

Let a = jf^—Zq\ z « 2py ; then b —p^ + Zq*.
To find X, we have a^ = a« + 8«2 »« ^ + 6«3 = j?* + 26/?3^2 + 9^
— (jp2— 6^)^, say therefore, 3p = 2q,
Thus a? «j93-5^2, y = ipq, z = 2p^, where 3i? = 2y ;
<f.^., 1» » 2, J = 3, a? = 41, y = 24, « « 12 ;

a2 = 23«, ^ == 31', c2 « 12.

1014. (The Editor.) — Find the least integral value of x which will
make the expression 927a;^— 12362; + 413 = a square.

Solution.

We have 103 (33?- 2)2+ 1 = p^yji.e., y2-103r' = 1.

The twelfth convergent of V^103 gives us y = 227628, z = 22419, and

the 24th y = 1035379»1667, z = 10201900464 ; but neither of these values

of z will give us an integral value of a? ; so that either we must go fiirther

or else the method of convergents does not ensure us integiul values.

If a; = I, we have 927a;2_ 1236a; + 413 = 1.
Suppose the quantity a-¥bx + ex^ ^ ff\ where a; » /, so that we have
a + bf+e/^^ff^y

., . . /m^— 3fn--927\

then we get ,-,.^ ____),

where m is arbitrary, an equation giving all other possible values of x ;
whence it is clear x can never be an integer.

1042. (The Editor.) — Find values of x which will make each of the
expressions 3a;3 + 1, a;^ + 1^ 2ar*— 3a;2+ 2 a square number.

Solution,

EuLBR has shown that a;^ + 1 can never become a square except in three
cases, viz., a; — 0, —1, and 2 ; so that, although we may readily obtain
values for x in the other two equations, it would appear that no value of
X will simultaneously satisfy the three equations.

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2814. (The late Matthew Collins, B.A.) — Can the common differ-
ence of three rational square integers in Arithmetical Progression be ever
equal to 17?

Solution.

Square numbers are of two forms, 4N and 4N + 1.

Let the three required squares be x'^y y'^y t?.

Suppose (1) a;2 = 4N ;
then y'^ = a:^ + 17 »= 4N + 17 = 4M + 1, which is possible,

«2 =a?2-17 = 4N-17 = 4M-1, „ impossible.

Again(2),leta:2=4N + l;
then y' = 4N + 18 = 4M + 2, which is impossible.

Thus, by either supposition it is impossible for a;-, y', «2 to be squares
simultaneously.

8930. (R- W. D. Christie.)— Prove that, whether (n) be odd or even,
sin»fl= sine { (2cos0)»-i-(»-2),2co8a)'»-3+ &L=:^^;^IlD (2 cos »)»•-»

>("-^K^~^H^-6)(2cose)n-7^...j'.

Solution,*
Cette formule est certaine pour les cas n = 2, n = 3, parce que I'on a
d'abord

sen 26 = 2 sen 6 cos 6 = sen a {(2 cos 6)2-1 1^

sen 30 = sen 2a cos <; + sen cos 20 = 2 sen cos^ + sen (cos^ - sen' 0)
= sen0(4co82 0-l) -= sen {{2 cos 0)2 -(3 -2) (2 cos 0)o}
= sen0{(2co8 0)«-i-(3-2)(2co8 0)3-8}.

Cela pose, nous aliens prouver que la formule qui est verifi6e pour le cas
(w— 1), lo sera pour le cas \n).
Par supposition nous avons

sen (»- 1) « sen J (2 cos 0)»»-2- (n - 3)(2 cos 0)»-*

■, («-;K«-s) (2coB6r-'...| (1).

Mais sen w0 = sen . cos («— 1) + cos sen (n— 1) ;

mais (voyez Carr, Synopsis of Fure Mathematics, page 177) Ton sait que

2cos(»-l)0 = (2cos0)^-'-(n-l)(2cos0)"-8 + ^^'~|^^^~'*^ (2cos0)»-^..,

1 . iS
d'od cos (» - 1)

= 2»-2(cos0)— i-(«-l)2«-*(cos0)«-s+ (^IllK|:=i)(cos0)«-«.2~-«...,

et substituant la valeur de sen («— 1) en (1) nous avons tr^ facilement
la formule demandee.

* This solution is due to Professors Betens and Catalan.

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9444. (R- W. D. Chwstie.)— Solve (1) in integers «* + irV* + y* =«* ;
and (2) note the result when a b 3.

Solution,

We have {a^+xi/ + ip){x^-'xy + y*) = oA.

Assume a?±xy + y^ — a or * » ar^i ary/w ;

then ay«±fiy/»-l. Lety-ii-1; then 3; = ±n; also a«n'— n + 1.
n may now he assumed any integer at leisure.

II ** + «V +5^ - «' = («*- «V)', sayi

then ««(2ii? + l)-(n*-l)y2.

then jc'=tn*— 1.

9621. (R. W. D. Christib.)— Prove that (j;**Mir*)/6 is an integer
where p is any perfect number and ir any prime number except 5.

Perfect numbers end in the digits 6 or 8 ; therefore p^ ends in 6.
Prime numbers end in 1, 3, 7, or 9 (except 2 and 5) ; therefore it^ ends
in unity, also 2^ ends in 6.
Therefore CP^-ir^)/5 is an integer, except ir — 6.

9608. (Sbftimus Tbbat, B.A.) — Find the least heptagonal number
which when increased by a given square shall be a square number.

Solution,
The general form of heptagonal numbers is \ (6x'— Zx), Let a> be the
given square, and h the number sought.

Assume a^ + A = (a- w)' ;

then h s=s n3— 2af>.

But 40A + 9 is always a square ; therefore assume
40 (n3~2an) + 9 « (6fi + 3)3.
Thus n a 20a + 9, and a may be assumed at pleasure. Let a = 1,
then ft = 29, therefore h =. 293-58 =. 783. Thus 783 + 1 = 283.

9629. (Professor Gbrondal.) — Partager 90° en deux parties ar, y
telles que la tangente de I'une soit le quadruple de la tangente de Tautre,
et prouver que tan |a? = 2 sin 18®.

Solution,

We have a? + y = 90" (1),

and tana; = 4tany (2).

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Then
therefore

therefore

tanrtany ^s 1

4tan2y - 1

tan y = ±h

t&nx =±2 = (2tan Ja;)/(l-tan2ix) ;

tania: = J(<v/6-.l) « 2siIll8^

(1);

(1), (2),

9643. (R. W. D. Christie.) — If 2n^V +2'' + Z'' ... »!»•, prove that
2n is exactly divisible by 2n when r is odd.

Solution.

Obtain 2n by the formula (n+1) 2»i« {duldx)'XH*i and separate o^^
values of r from the evens, and we shall find that 2n is a constant factor
of the odd and (2n + 1) / 2 (r + 1) of the even. Thus, let « ~ 2n ; then
r = 1 gives us »,

«^,

**'(4*-l),

i««/l6«3-20«« + 3(4*-l)},
4«2|i6^_32«» + 34«3-6(4a-l)},

--L- «» {960*»-2800»< + 4692«»-4720«- + 691 (4* - 1)}.

iA;.[4(6*-l)],

|A:a[4(12»2-6«+l)].

>A:« [^ (40«3-40»2 + 18«-3)].

r= 3
r- 6
r- 7
r- 9
r = ll

r = 13

r=. 2
r== 4
r« 6
r= 8
r = 10

r = 12

;j{48«<-80»« + 68«2_5(6«-l)}],
-L- . {3360a« - 8400«4 + 1 1480»»- 9440«2

+ 691 (6«

r= 14

-1)}],

^ks [J {l»2««-672«5+ 1344»<-1760«»

+ 1436«3-106(6»-1)}],
&c., &c.

Thus the theorem appears to be true for odd values of r only ; that.it
is not true for even values of r may easily be tested by making r ^ 4,
n = 2, 3, or 4, for example.

9668. (Professor Vuibert.) — Si Ton designe d'une mani^re generalo
ir Sm la somme dos puissances do degrc m dcs n premiers nombres entiers,
jmontrer qu'on a (3JS6 + 2Si^) /0S4 = Sj/Sj.

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9643. (R W. D. Chmstib.)— If 2: - !»• + 2'' + 3** .. . n»- ; prove that
21 is divisible by 2j».

9226. (J. White.)— Prove that

13 + 23 + 33 ... M3 is a factor of (l»+2« + 3« ... M«) x 3.
8784. (R. W. D. Christib, MjV.)— Prove that, if
«« 1 + 2 + 3+.. . + », 82 = 12 + 22 + 32 + ... +w2, S3=P + 2S+3»+...+n»,

2 - l* + 2* + 3<+...+», <r- l« + 2» + 3« + . .. + »«,
then {3<r + 2«») / 62 - Ss/S^.

9883. (R. W. D. Christie, M.A.)— If 2r = l»* + 2»*...+«»-, prove
that 726 + 624 = 1222 2s.

9042. (H. L. Orchard, B.Sc., M.A.)— Prove that 1' + 2^ + 3' + ... + a:^
is a factor of the expression Zsfi + 12x7 + 1^0^—7 x^* + 2jt^.

9102. (H. L. Orchard, B.Sc., M.A.) — Show that the series
V + 27 + 3' + 47+ ... + 97 is divisible by 27.

8647. (R. W. D. Christie, M.A.) —If «« 18 + 23 + 33+ ... +«>,
8 = l* + 2* + 3» + . ..+«*, 2 = 17 + 27 + 37+. ..+n7; prove that 2 + 8 = 2»2.

9142. (R. W. D. Christie, M.A. See Quest. 8700.)— If

2r = l'' + 2»' + 3'' «^

prove that (92ii + 302, + 92;) / 23 - (112,0+3028+ 726) /2j.

Solution.*

In 1834 Jacobi proved that Sgn^i contained 8,2 as a factor, and that
S'in contained Si as a factor. Another proof was given by Prouhet in
I80I ; and agaiu, an a priori proof by Caylby in 1867 ; and there are pro-
bably others. The simplest of those mentioned is Prouhet's; viz.,

writing (l+A)*" in the form 1 +riA + r2A + ...,

it is easily shown that (Sr = l*" + 2*" + 3'' + . . . + n**) ;

(r « odd), riSr-i + r3Sr-3 + ...+»*r-4S4+rr-2S, = J[(» + l)'- + fi'--2n-l]

(1);

(r «even), riSr^i +r,Sr-3 + ... +rr.8S3 + r,..i8i := i[(n + l)'' + n*"-l]

(2).

In (2 ), (« + 1)*" + «•' — 1 vanishes when n = and when « + 1 = ; there-
fore, by putting r = 2, 4, 6, ..., we prove that Sj, and therefore S3, and
therefore 85, and so on, are successively divisible by i»(« + l). But
Prouhet proved the full theorem. Let

K « {n + \Y+n^-'l-2rSi = (w + l)'-+W-l-m(« + 1) ;
then both K and dK/dn vanish when « =« and when « + 1 = 0, there-
fore K contains n^ {n + 1)2 as a factor. Hence, by putting r = 4, 6, 8 ...,
we prove successively that S3, 85, S; . . . contain Si'*^ (or 83) as a factor.
Similarly, the secoud part of the theorem is proved. (No. 9643.)

* This solution is due to Mr. J. D. H. Dickson.

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The following theorem is capable of proof —

(2n— l)S2n-2 = 38, (aoS2»-6 — «2S2n-7 + ^4S2»i-9— .••±<»2n-6SiT«2n-fi}

-(3).

2llS2i»-l = 483(^0^211-5— *2S.n-7 + *4S2H-9—...±^n-6Si=Fd2n-5}

.(4).

where ^q == 2«— 4, ip = 2n— 4,

oj = i (2ii-4)(2n-6), i, - i (2«-4)(2«-6),

the remaining a's and d's being somewhat complicated functions of
Bbrnoulli's numbers. The first few cases are appended —

654 = 3S,{28,-i} (5),

655 = 483 {28i-i}... (6),

786 =3S3{483-28i + i} (7),

8S7 =483 {483-48, + !} (8),

988 =883 {685-683 + ^1^1-1} ....(9),

108, =483 {685- 883 + 681-1} (10),

ll8iof= 382 {887- 1285 +1683-1081+4} (11),

12811- 483 {887-I685 + 2683-2OS1 + 6} (12),

138i2- 3S2{l0S9-20S7 + 44S5-^FS3 + i||^Si-f§i} (13),

148i3 = 483{l0S9-¥87 + i^pS5-HFS3 + Hi^i-W}... (14),
&c.
No. 9226 follows from equation (6).
No. 8784 (the same as 9668) may be written in the form
(3S5 + 28,83)/S3 = 684/82;
and by equations (5) and (6) each side equals 6S1— 1.

No. 9683 comes from (5) and (7) by simple addition.

No. 9042 is "prove that 248; = multiple of 83." And No. 9102 is
nearly the same question, with 9 written for x.

No. 8647 is 85 + 87 = 283', and, like No. 9683, follows from equations
(6) and (8).

No. 9142, by equations (8), (10), (12), and (7), (9), (11), shows that
each side of the given relation is equal to 24 (87 + 85).

The number of relations like the above maybe indefinitely extended by
the theorems (3) and (4).

9767. (R- W. D. Christie.) — Prove that «"» is the sum of n con-
secutive odd numbers.

Solution.
«»»»-i— « is always even = 2p, suppose.
Then w"» = 2iw + »2

= the sum of 2p + 1, 2p + 3, 2j» + 6, ... to « terms.
Thus also

5n= (w.m+1 .2m+l)/6 = n + 3 (w-1) + 5 (n-2) ... 2m-l.

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9876. (R- W. D. Christib.)— Prove that

2tan-»— ±tan-* « iir,

where a is the coefficient of a?*» and d of «»♦ * in the expansion of - — -.

Solution.
If 2a* ±1 -*3_a2^ then

2tan-i 4- ±tan-» -— i- = iir

becomes tan**— -^, itan-^-— ; — - = 1.

2ab±l 4ab±l

Now tho coefficient of a?* in the expansion of is the sum of

(u+1) terms of l + 2a:-a?*

f 2>' + (n~l)2''-gH- ^~^;^^"'^ 2'«-* »+l or l](-l)%

and those coefficients bear the assumed relation ; the sign depending on b.
Examples, — If n = 2, then a » 5 and 3 => 12 ; thus we have

2tan-»-^-tan-i-^ « Jir,
12 239 *

whioh lA Machin's formula.

If » = 5, then a ~ 70 and b » 169 ; thus we obtain

2tan-i — + tan-i-^— = ix.
13-^ 47321 ^

(E.) NEW QUESTIONS.

9877. DioPHANTUs' Epitaph.

Hie Diophantus habet tumulum, qui tempera Titae

mius mira denotat arte tibi.
Egit sextantem juvenis ; lanugine malas

Vestire hinc coepit parte duodecimo.

Septante uxori post haec sociatur, et anno
Formosus qninto nascitur inde puer.

Semissem aetatis postquam attigit ille patemae

Infelix subit^ morte peremptus obit.
Quatuor aestates genitor lugere superstes

Gogitur : hinc annos illius assequere.

DST8. Every number contains an even number of factors, and tftre-
l^rc tho numbers of odd and of even factors are either both odd or both

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181

even, except when the original number is a square, and then the reverse
is the case {i.e.y it contains an odd number of odd factors, and an even
number of even factors, and consequently an odd number of factors).

9879. Prove that a'^h'^tf.., (where r = 2, 3, or 4) is of the same form
as the squares, cubes, and biquadrates themselves, viz., 4N and 4N + 1,
9N and 9N±1, 16N and 16N + 1.

9880. 1. If N = a2 + ^2^ prove that it also equals

{2m»*+ (»2-m2) a}2/(w2 + «2)2+ {2i»w«+ (w2-«2) i}2/(ws + n2)2,
where a, *, w, «, are any integers whatever.

9881 . Show that the sum of » terms of the following n series

lr+2»- + 3»- n\

lr + 3r + 5r (2n- 1)%

lr + 4r + 7r (3»-2)%

l'' + (n+l)'' + (2« + l)»- («2-«+l)»*,

where 2|[ = sum of l** + 2*' + S** to n terms.

9882. Let »= 12 + 22 + 32 n^, 8 = 13 + 23 + 33 n\

5= 1^ + 26 + 35 w5, then S + 22= 3*2.

9883. Prove the following property of prime numbers. Distribute
the primes from unity together with their multiples as in Quest. 9225,
into groups oifour (having however ^^ in i\\Q first group) then

^„ = ^ + w = 6(^7-1-2),
where g means the group, t the tens, and u the units in any prime.
Ex.gr., • ^1 = (1+2 + 3 + 5 + 7) =6x3,

, (11, 13, 17, 19) = 2+ 4 +8+10 = 6x4,
(23, 29, 31, 37) = 5 + 11+4 + 10 = 6x5,
[41, 43, 47, (49)] = 5 + 7 + 11 + 13 = 6x6, and so on.

9884. Prove that the sum of the factors of any number is

S/= (2''*i-l) f ^^-^ j, a = any prime > 2,

where/ (a prime) the number of odd factors is Xjn^ of the number of
even factors. Show also that, if n is any even number, N is a square
(except when /= 2). [jEa;. — Let N have 7 odd and 14 even factors.

Then 87

vhen /= 2). [Ex.—LQi N have 7 od
= (23-1) I ^^y) » and N = a square.]

98 .5. Let <r= 12 + 22 + 32 n\ «= 13 + 28 + 33 n^,

S = 14 + 24 + 3* «S 2= l8+2» + 36 n\

Then 7S + 5S = 4« x 3<r.

VOL. XLIX.

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182

9886. Every square number is divisible into two sequences from m
any integer).

9887. Prove the following equation

«;♦' = «»(«!:)+«:-',

where a means any prime number, and a!^ — sum of factors of a^ ; hence

show, if we could solve 2 = ^ ( >)+<»> j^ integers, an odd perfect
would be found. *''**'

9388. Divide the sum of two cubes into two other cubes.

9889. Take any number of my digits (1, 2, or 3 together) , and I am equal
tf] a sequence from unity. Cast out the nines from my dozen divisors
arid you'll find the factors of each of my digits. I am a famous number,
but not a perfect number, and both myself and the sum of my digits are
ilhni^ible by a perfect number.

9890. Find a number from the remainders after dividing it by a number
of primes, say 3, 5, 7, 11, and 13.

9891. Draw a straight line cutting two concentric circles, so that the
jiurt intercepted by them is divided into three equal portions.

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