Mathematical questions and solutions

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MATHEMATICAL QUESTIONS,

WITH THKIH

SOLUTIONS.

FROM THE "EDIJOATIONAL TTHES."
VOU XLVII.

\MoJ^2^'6%.%^

SCIENCE CENTER LIBRARY

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Google

MATHEMATICAL il^:^

QUESTIONS AND SOLUTIONS,

FEOM THE "EDUCATIONAL TIMES/'

WITH MAVT

PAPEES AKD SOLUTIONS

nr ABDinov to thobb
PUBLISHED IN THE "EDUCATIONAL TIMES,*'

AND FOUR

APPENDICES.

BDITBD BY

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

BBGIBTBAB

OV THB

QBNBEAL MBDICAL COUNCIL.

VOL. XLVIL

c

LONDON:

FRANCIS HODGSON, 89 PARRINGDON STREET, B.C.

1887. Digitized by GOOglC

\* Of this series there have now been published foriy-seyen Volumes,
each of which contains, in addition to the papers and solutions that
have appeared in the Educational 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.

AlTAB, Prof. SWAMIVATHA: Pros. Coll., Madras.

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

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

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

Anthoht, Edwyn, M.A.: The Elms, Hereford.

Abmevakte, Professor; PeSaro.

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

Bastov, W. J., M.A. ; HiKhgftte, London.

Basu, Satish Chandba ; Presid. Coll.,CaIcutta.

Battaglini, Prof. Giuseppe ; Univ. di Soma.

Batliss, Geoboe. M.A. ; Kenilworth.

Bebvob, B. J., B.A. ; B«gent's Park, London.

Bbltbami, Professor ; University of Pisa.

Bebg, Professor F. J. van dew ; Delft.

Besakt, W. H., D.Sc., P.E.S. : Cambridge.

Bbyevs, Professor Igvacio, M.A. ; Cadiz.

Bhattachabta, Prof. Mubibdbakath, M.A. ;
Delhi.

Bhtjt, Prof. Ath Bijah, M.A.; Dacca.

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

Bibch, B«v. J. G., M JL. ; London.

Blackwood, Elizabeth, B.Sc; Boulogne.

Bltthe, W. H., BJl. ; B^ham.

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

BOBDAOB, Pirof. Edmovd ; 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, Capitaine H. ; Montpellier.

Bbooks, Professor E.; Millersville, Pennsylvania.

Bbown. a. Cbum, D.Sc. : Edinburgh.

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

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

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

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

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

Cabb, G. S.. M.A. ; 3 Eudsleigh Gardens, N.W.

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

Catalan, Professor ; University of Lidge.

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

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

Catlby, a., F.R.S. ; Sadlerian Professor of Ma-
thematics in the University of Cambridge,
Member of the Institute of France, &c.

Chakbayabti, Prof. Byom., M.A. ; Calcutta.

Chase, Prof., LL.D. ; Haverford College.

Chbistib, E. W. D^ B.A. ; Newmarket.

Clabee, Colonel A. £., C.B., F.B.S.; Bed Hill.

CocHBZ, Professor; Paris.

Cockle, Sir James, M JL., F.B.S. ; London.

CoHEN, Abthub, M.A., Q.C., M.P. ; Holland Pk.

CoLSON, C. G., MJL.'j university of St. Andrews.

Constable, S. ; Swinford Kectory, Mayo.

COTTEBILL, J. H., M.A. ; E. N. Coll., Greenwich.

Cbbmona, Prof. LuiGi : Bome.

Cbofton, Prof. M. W., F.B.S. ; Dublin.

Cboeb, J. O'Bybnb, M.A. ; Dublin.

ClTLLBY, Prof«MJL. ; St. David's Coll.,Lampeter,

ClTBTis, B., M.A., S.J. : Univ. Coll., Dublin.

Dabboux, Professor; Paris.

Data, Prof. Peomathanath ; Calcutta.

Dayibs, D. O. S., M.A. : Univ. Coll., Bangor.

Dayis, B. F., M.A. ; Endsleigh Gardens.

Dawson, H. G., B.A. ; Christ Coll., Camb.

Day, Bev.H. G.,M.A.; BichmondTerr., Brighton.

De Longchamps, Professor ; Paris.

Db Wachtbb, Prof., M.A. ; Schaarbeck.

Dby, Prof. Nabendba Lal, M.A.; Calcutta.

DlCK,G. B., M.A. ; Victoria University.

D'OCAGNE, Maubice ; Bochefort-sur-Mer.

Dobson, T., B.A. ; Hexham Grammar School.

Dboz, Rof. Abnold, M.A. ; Porrentruy, Berne.

DUPAIN, J. C. ; Professeur au Lyc6e d'Aiigoul6me.

Eastwood, G.. M.A.; Saxonville, Massachusetts.

Easton, Belle, B.Sc. ; Lockport, New York.

Edwabd, J., M.A. ; Head Mast., Aberdeen Coll.

Edwabdes, Dayid; Erith Villas, Erith, Kent.

Elliott, E. B.. M.A. ; Fell. Queen's Coll., Oxon.

Ellis, A lexandeb J., F.EJ.^ Keinaington.

Ethta^u], W. T. a.; Pmnbroke CclL, Oifopi.

EssENNiiLL, Em HA; Coventrj.

Ev ,iNs, Hrofmssor, M.A.; Look port. New York.

EvFRKTt, Pmr. J. D., D.O.L.: Qu. Coll.. Bt^lfast

Fi L K LIN, Jose Pff; Prtaf. in Univ, of Migaouri*

FofijiiY. H.^ M.A.; BollaTV, Madrjia Prt-sidfitic:f

FcKs TFia^F. W„ B, A , ; OllRlf^ea.

FosTiLii. l*rDf.G,CAa£v\ P,R,S.; Ur»iv.Coll.,Lond,

FBA.NH;LI?f^Cllllf3TINELA.DD;M.A.;?Ptjf.ofNat.

^i^^. and Math., Union Spriuga, New York.
FUGHTSS. E. ; Uiiiverisity of ^n.^Jen.
Ga LitHAiTiT,R^vJ..M.A. . Fell,Trin.Coll.,Dub!in.
GALE.Kitir; Wrtrc*(?j5t^r Piirk, Surrey.
Gallatt.t, W., B.A.J Earl's Court, Lon<loii.
Gailiebs, flin,T.,>LA.; Ei.-Foll.C.GolUC^m,
Galton, Filancie, M.A., F.R.S.; Loudon.
Geni;se, 9mL, M,A.; Univ. Coll.. Aberyafcwith,
GKHRAN'a. II. T., B.A. s Stud, of Ch.Ch., Oxford.
Gla ISHES, J. W.L.p F,a.S.; FeU.Tnu.Coll.. Ciunh.
GoLJtEJfBERa, Pfofe^Hrsr* M.A.^ Moacow,
GoRDoy, Alice, B.Sc.; Gloucts-ster.
GouDiE, W. P., Irvine Houses Durby.
GiiAMAM, E, A.* M.A.- Triiiity Coikjje, Dublin.
Ghheptfteld, Rev. W. J., M.A. t Duiwich College.
GKEiE.-fSTREET. W. J., B.A.; HulL
Gee r^^ WOOD. Jami^s M.; Kirksvilk, MisaourL
GEir FJTEr:!. G. J.. M.A. ; Poll. Cli. ColL, Catiib.
GKiFFiTtta, J.,, M.A. i FfjUrnvrif JtisujiColUOiOn,
GRavE. W. B.. B,A.; Perry Hur. HirmmEliam.
Hadamaru, ProfcHHDr, MA., Paris.
Ha T'lU, E.. B.A.. B.Sc. i Kltii^s Sdi.. ^""arwiok.
Hall. Pm3V*sor ASAPir, UA.i \Vnatuu|SftQU,
H A 3J M OS II . J ., M.A,; Bu L;kh ui-at H i 1 1 . E iisei.
Hariteu^^ Q.; Uiiiveraity of ^t. Petersburg.
HA.BKEB, ALPKBD, M.A. ; Cheltenham.
Haekis, H. W., B.A. ; Trinity College, Dublin.
Habt, Dr. David S. ; Stoninfcton, Connecticut.
Habt, H. ; R. M. Academy, Woolwich.
Hau&hton, Rev. Dr., F.R.S. ; Trin.Coll., Dubl.
Hendricks, J. E., M.A.; Des Moines, Iowa.
Heppel, G., M.A.; The Grove, Hammersmith.
Hebmav, R. a., M.A. ; Trin. Coll., Cambridge.
Hebmite, Ch.; Membre de I'Institut, Paris.
Hebvby, F. R. J., M.A.; Worthing.
Hill, Rev. E., M.A. ; St. John's College, Camb.
HiNTON, C. H., M.A. ; Cheltenham College.
Hiest, Dr. T. A., F.R.S.; London.
Holt, J. R., M.A. ; Trinity College, Dublin.
Hopkins, Rev.G. H., M.A.; Stratton, Cornwall.
HoPKiNsoN, J., D.Sc, B.A. ; Kensington.
Hudson, C. T., LL.D. ; Manilla Hall, Clifton.
Hudson. W.H.H.,M.A.;Prof.inKing*sColl.,Lond.
Inglbby, C. M., M.Ay LL.D. ; London.
Jenkins, Mobgan, M..A. ; London.
Johnson, Profo M.A.; Annapolis, Maryland.
JOHNSTONjjW. J., M.A.; Univ.Coll.,Aberystwith
Jones, L. W., B.A. ; Merton College, Oxford.
Kennedy, D., M.A. ; Catholic Univ.. Dublin.
KiBKMAN, Rev. T. P., M JL., F.R.S. ; Croft Rect
KiTCHiN, Rev. J. L., M.A. ; Heavitree, Exeter.
KiTTUDGE, Lizzie A. ; Boston, United Statea
Knisely, Eev. U. J. ; Newcomerstown, Ohio
Knowlbs, R., B.A., L.C.P. ; Tottenham.
KoEHLEB, J. ; Rue St. Jacques, Paris.
Lachlan, R., B.A. ; Lewisham.
LAVBBTY,W.H.,MJ^..:lateExam.inUniv.Oxforu
Lawbence, B. J. ; Ex-Fell. Trin. Coll., Camb.
Leidhold, R., M.A. ; Finsbury Park.
Leudesdobp.C, M JL.; Fel.PembrokeColl.,Oxon.
Leyett. R.. M J^. : King Edw. Sch.. Birmingham.
London, Rev. H.,MJl.; Pockiington.
LowRY, W. H., M.A. ; Blackrock, Dublin.
Macdonald, W. J., M.A. ; Edinburgh.
Macpablane, Prof. A., D.Sc. ; Univ. of Texas.
Mackenzie, J. L., B.A. ; Gymnasium, Aberdeen.
MacMahon, Capt. P. A. ; R. M. Academy.
MacMubchy, a., B.A.; Univ. Ci^U.. Toronto.

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IV

McALiSTEa, Boviut, M.A., J>M.i C^rohrEdge.
McCat, W. S., M.A. 5 Fell. Trin. ColL. DiibUa.
Mc Clella )?p, W . J„ B. a . ; Prin . of Santry School.
McGOLL. H,, E.A. 1 73, RLie-SiljIiqumH BcmloNTHe.
McIntosh^ AlK3t,,B.A,; BodfurdR^JW, Londqn.
McLEOn, j.,M.A.i II,Mh Atjulpoiy. Woolwich.
Mapisoit.Ibabml, B.A.J Cardiir.
Malbt, Prof.. MA,; Queen 'a Colh, Cork.
Ha'NN, M. p. J., M,A.; Kel^Hi^^J^t-on.
MAy^irrni, M : Vr.-.f, ri, ^'|^v,-.i,.i Poly tech,, Paris.

Martin, ASTEMAB, M.A., Ph.D.; Washinffton.
Mathbwb, G. B., B.A. ; Univ. Coll., N. \^es.
Matz. Prof.. M JL. ; Kind's Mountain, Carolina.
Mebrifield, J., LL.D.,P.B^A.8. ; Plymouth.
Mbbbimait, Maitsfield, M.A.; Tale College.
Mbtbb, Mabt B. : Girton College, Cambridge.

MlLLBB, W. J. C, BX, (BdITOB);

The Paragon. Bichmond-on-Thames.
MiircHiir, G J(.,M JLj Prof, in Cooper's Hill Coll.
Mitohbbon, TMB.Aa<.C.P. ; City of London Sch.
MoircK, Prof. H. St., M.A. ; Trin. Coll., Dublin.
MoNCOUBT, Professor ; Paris.
MooK, RoBBBT Jf JL. ; Ex-Pell. Qu. Coll., Camb.
MooBB, U. K., B.A^ Trin. Coll., Dublin.
MoBBL, Professor; Paris.
MoBOAV, C, B.A.; Salisbury School.
MoBLBT, Pbahk, B.A. ; Bath CoU., Bath.
MoBBiOB, G. G., B.A. ; Weymouth.
MniB, Thomas, M.A., F.B.S.B. ; BothweU.
MnKHOPADHYAT,AsnTOSH,M.A.;Bhowanipore.
Nash, Prof. A. M., M.A. ; Calcutta.
Nbubbbo, Professor; Univ. of Li^.
Nbwcomb, Prof. Simon, M.A. ; Washington.
O'CONNELL, Major-General P. ; Bath.
Openshaw, Iti'v.T. W,. MA,' ClilicuL
O'liKDiN, John; ^'l'W Street, Limorick,
Ohcitarti* H. L., M.A.,L.C.P. ; HamfKitoad.
OyfMis, J. A., B»iSc. ; Temiyaoti Bt., Llverpt*ol.
PA5T0N. A. W., M,A. ! Fell, of Trin. ColL^Dablin.
PaMULKBPST, C, M.A. ; Ijondoii*
FBRRiar, E!!j:h,t: Gtrton Callep, Cambridj^.
PHILLIPB. F. B. W. J Ballinl Collegp, OjJurd.
PiLLAi, C, K,, M;A. ; THohy, Madras.
PiBlE. A., M.A. \ Univeraity of St. Andniwa.
Pl^AMBBBWBKI, H., M.A.; Diipihtiitim.

Pooui, GisiBtJtJK, B,A. i Clif^ltenham.
FocBiiNGTOB-j H. C.,M.A. ; York« Coll., Leecis.
Polio N AC, Pnnire CAMrLLB de i Paris.
POLLBltFEN. H., HA.} Wiiidennerc College.
PoTT^B, J B.A.; Ki chill ond-ou-Tharoeji.
Pbujjde.s, FfiANCEB E.; LookpoFt, Niiw York.
Ptt^eh. Prof. F. , M.A, ; Queen's College. Belfast.
PuTNASi, K. S.. MA.; Eome.New York.
H.A1?, Prof. B. IlAJfL^MANTA. B.A.; Madraa.
Eawbok, Robert ; Hnvwit. Hants*
Rait, Pfot M. Nath. M.A., LL.B. ; Howrah,
Hat, Prof. SabaiiaranjaxV, M,A. ; Dneoa.
Hbad, H. J., B.A* ; Brajericiae C<3lJ,, Oiford.
Bektks, G. M„ M,A. i Lee, Kent.
EETNOLue, B.J M.A.; Nottinp: Hill* London.
RiCHAkHB, LiAViDj B.A.; Aberystvrith.
EjcnASimoN, Kev. G., M.A.^ WincKester.
EoESET5>R, A„ M.A.; SchoLof 'f Hii,Coll,,Dublin.
RoBEETB, S., M.A., F.R,S. ; London.
Roberts, W. IL, M^A. ; Fell, of Trin. OoU., Dub.
Eon&ON, B.C., B.A.; Sidney JSusaex Coll., Cam.
ItOQKES. L. J., B.A, ; Oxford.
Rosenthal, L. H. ; J^lcholnrof Trin. Coll., Dublin.
Boy, Prt>I, KAUPEAHA?iNA, M.A. ; Agiu.
RirGGBIlO^^lMONKLLI; Univi^rsitA di B^ma.
EU85EU„ J. W., M.A, ; Mertoii Coll., Oxford.
RtfBBSLL, R„ B.A.i Trinity Uoilogti, Dublin,
EUTTEB. l^Di^AED ; Suiiderknd.
SAiaioaf, Rev. G.. D.D., F.R.S.; RcKJtis Professor

tif Divinity Id the University of Dublin.
Sakpbbs, J. B.; BioomLni^ton, Indiana^

Baitdbbsov, Ber. T. J., MX ; BoyBtoxi, Cambt*

Sabkab, Prof. NiLEAKTHA, MJL; Oalcutta.

ScHBVVBB, Professor ; Meroersbury Coll., Pa.

ScHOUTB, Prof. P. H. ; University, Groningen.

ScoTT, A. W., M.A. ; St. David's Coll., Lampeter.

SooTT, Charlottb a., B.Sc.: Professor in
Bryn Mawr College, Philadelphia.

Scott, B. P., M.A.; Fell. St. John'sColl., Camb.

Serrbt, Professor ; Paris.

Shabp, W.J. C. M.A.: Greenwich.

Shabpb, J. W., MJL ; The Charterhouse.

Shabpb, Bev. H. T.. M.A. : Cherry Marham.

SHBPHBBixBev. A. J.P.,B.A.; Fell, Q.Coll.,Oxf.

Simmons, Bev. T.C., M.A.; Christ's Coll.,Breoon.

SiBCOM, Prof., MJL.: Stonyhurst Colle(|;e.

SiYBBLY, Waltbb; Oil City, Pennsylvania.

SKBIM8HIBB, Bcv. B., MJL ; Llandaff.

Smith, C, M..A. j Sidney Sussex Coll., Camb.

Stabbnow, H., M.A. ; New York.

Stbgoall, Prof. J. B. A., MJL.; Dundee.

Stbin, A.; Venice.

Stephen, St. John, B.A.; Caius Coll., Cambridge.

Stbwabt, H., MJL. ; Framlingham, Suffolk.

Storb, G. G., B.A. ; Clerk of the Medical Council.

SwiBT, C. A., B.A. ; Grammar Sch., We/bridge.

Stlvebtbb, J.J., D.C.L., P.B.S.; Professor of
Mathematics in the University of Oxford,
Member of the Institute of France, &c.

Symons, B. W., M.A.; Fell. St. John's ColUOxon.

Tait, Prof. P. G., M.A.; Univ.^dinburgh.

TANNBB,Prof, H.W.L.,M.A.: S.Wales Univ. Coll.

Tablbton, P. a., M.A. ; Pell. Trin. Coll., Dub.

Taylob, Bev. C, D.D. ; Master of St. John's
College, Cambridge.

TaylobTh. ml M.A7Fell. Trin. Coll., Camb.

Taylob, W. W., M.A. ; Bipon Grammar Schooi.

Tebay, Septimus, BJL. : Famworth, Bolton.

Tbbbt, Bev. T. B.,M.A.. Fell. Ma«i. ColL.Oxon.

Theodobius, a. F., M.A. ; Bath College.

Thomas, Bev.D., M.A.: Garsington Bect.Oxford.

THOM80N,Bev.FJ).,M.A.;Ex-Fel.St.J.Coll.,Cam.

TiRBLLi, Dr. Francesco ; Univ. di Boma.

ToBELLi, Gabbibl; University of Naples.

TOBBY, Bev. A. F., MJL. ; St. John's Coll., Camb.

Tbaill, Anthony, M.A., M.D.; Fellow and
Tutor of Trinity College, Dublin. ,

TucKBB, B., MA. ; Mathematical Master in Uni-
versity College School, London.

TuBBiFF, Geobob, M.A. ; Aberdeen.

ViGABiA, Bmilb ; Hotel Sue«, Paris.

YiNCBNzo, Jacobini; UniversitIL di Boma.

VosE, Professor G. B.; WashhijKton.

Walenn, W. H. ; Mem. Phys. Society. London

Walkbb, J. J., MA., F.B.S. ; Hampstead.

Walmslby, J^.A. ; Eccles, Manchester.

Wabbubton-Whitb, B., B.A., Salisburv.

Wabben, B., ma. ; Trinity College, Dublin.

Wathbbston, Bev. A. L^M.A. ; Bowdon.

Watson, Bev. H. W.; Ex-Fell. Trin. ColL, Camb.

Wbbtsch, Fbanz ; Weimar.

White, J. B., B.A. ; Worcester Coll.. Oxford.

Whitb, Bev. J., MA ; Boyal Naval School

Whiteside, G., M.A. ; Bccleston, Lancashire.

Whitwobth, Bev. W. A, M.A. ; London.

Williams, C. £., M.A. ; Wellington College.

Williamson, B., M.A.: Fellow and Tutor of
Trinity Coll^, Dublin.

Wilson, j. M., M.A. ; Head-master, Clifton Coll.

Wilson, Bev. J., M A^ Beet. Bannockbum Acad.

Wilson, Bev. J. B., MA. ; Boyston, Cambs.

Woodcock, T.,B.A.; Twickenham.

WoLSTENHOLME, Bcv. J., M.A, Sc.D. : ProfcssoT
of Mathematics in Cooper's Hill College.

WooLHOirsB, W. S. B., F.B.A.S.,&c.; London.
Wright, Dr. S. H., M.A.; Penn Yan, New York.
Wright, W. B., B.A.: Heme Hill.
Young, John, BJL; Academy, Londondeny,

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CONTENTS.

£Mf)tmntit&l ^apersf^ ^t.

Pa«e
Note on Queetion 8376 (Vol. xlv., p. 98). Professor Nash, M.A. 97
Note on Anallagmatic Curves. (Professor Wolstenholme, Sc.D.). 61

Note on a Probability Question. (The Editor.) 72

Note on the Brocardal Ellipse. (B. F. Davis, M.A.) 134

Geometrical Construction for the Brocard Angle, &c. (R. F.
Davis, M.A.) 135

Note on the use of Common Logarithms in the Numerical Solution
of Equations of the Higher Orders. (Major-General P. O'Connell). 166

(Bm&tion& ^oVat^.

1866. (Professor Sylvester, F.R.S.) — Prove that the Jacobian of the
time of a planet's describing any arc, the chord of the arc, and the sum of
the two extreme distances from the sun, in respect to the eccentricity and
two extreme eccentric anomalies, is zero ; and hence deduce the time for
a planet or comet in terms of the said chord and sum 163

2391. (Professor Sylvester, P.R.S.) — Let fi points be given on a
cubic curve. Through them draw any curve (simple or compound) of
degree y; the remaining Zy—fi (say f/) points may be termed a first
residuum to the given ones. Threugh these f/ points draw any curve of
degree t/ ; the remaining Zv'—iif points may be termed a residuum of the
second order to the given ones ; and in this way we may form at pleasure
a series of residua of the third, fourth, and of any higher order. If /i is
of the form 3»— 1, a residuum of the first or any odd order, and if /x is
of the form 3i + 1, a residuum of the second or any even order in such
series, may be made to consist of a single point, which I call the residual
of the original /x points. Prove that any such residual is dependent
wholly and solely on the original /i points, beiu^ independent of the
number, degrees, and forms of the successive auxiliary curves employed
to axriye at it 137

a

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VI CONTENTS.

2810. (Professor Sylvester, F.R.S.) — Let Si be used as the symbol
of the sum of »-ary products ; required to prove that, if « is <n,

SS, (O^ 0,, ... , On) (^-^>| (^-^) "• (^-°")

— S,+i (a,, Oj, ... , an)Si^i (oi, oj, ... , On).
[For example, let » » 3, » » 2, then the theorem becomes
^, («-«)(«-iB)(a-7) ,,^ (^~«)(^-/3)(^-7) ,^fr (g-a)(^-3)Cg-7)

a abe—afiy, which is obviously true.] 21

2832. (Professor Sylvester, F.R.S.) — ^Prove that the curve of inter-
section of two right cones with parallel axes is a spherical curve, and that
a third right cone may be drawn through it. Prove also that a plane
circular cubic may be found such that the distances of any two fixed points
on it from everi/ point in the curve of double curvature above-mentioned
shall be in a fixed linear relation 90

2866. (Professor Sylvester, F.R.S.)— Given the simultaneous equa-
tions f>«+i = f>«+(«2— fl>)rx_i, w«+i = w«+(a?'— a?)tt«-i + (2« + l)r,*i, prove
that the general solution is of the form v« — Ac^ + fi/3«,

w. = iX {a;« + 2a; + i (-)«} o. + iX {a:3 + 2a;-i (-)«} /3, + w« + »/3„
where A, /x, y, «- are arbitrary constants, and detennine the values of
Ox, fix 37

2934. (Professor Sylvester, F.R.S.) — If «i, 8^ «8, s^ «5, a^ repre-
sent the sum of xyztuVf and of their binary, ternary, quaternary,
quinary, and sextic combinations respectively, and if E stands for the

symbol of Emanation a--+*--+c-- +rf--. +tf^+/4-, prove that
dx dy ds dt du dv

the resultant of «|, E«2, s^ E44, s^, E«e is the product of

(a-*)»(a + i-tf-rf)»(a + i + tf-rf-tf-/)W,

and of the similarly formed powers of products of the analogous linear

functions oidbedef, 101

2936. (Professor Sylvester, F.R.S.)— If /(a?) be a rational integral
function of a; of a higher order than the second, prove that it is impossible
for (fxY+ (f^Y to be a perfect square unless /(ic) contains at l^st two
groups of equal factors , 85

3427. (Professor Sylvester, F.R.S.)— If ^, i(^ are quantics in x, y,
each of degree /x ; F a quantic in ^, i|^ of degree m, and consequently in
Xf y of degree mit, ; and if J denote the Jacobian of ^, i|^, that is,

^ ^_ ^ ^.

dx dy dy dx*
D^,^ F the discriminant of F treated as a quantic in ^, i|^ ; Dx, y, F the dis-
cnniinant of F, treated as a quantic in x, y ; and if R be used as the
symbol of ** resultant in regard of or, y " ; prove that

D„, F = 2 [R {tp, rp)]-*'-^ R (F, J) (D^,^F)^.
As a particular case of the foregoing theorem, show that the discri-
minant of F, any symmetrical quantic of an even degree inx, y, is of the
form F (1, 1) F (1, — 1) Q,^, where A is a rational integral function of the
ooefEidents in F.

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CONTENTS, VU

Find also what the general formula becomes when <l>, ^ are taken linear
functiona of a?, y 146

3636. (Professor Sylvester, F.R.S.)— 1. Euleb has shown that the
number of modes of composing n with i distinct numbers is equal to the
denumerant (that is, the number of solutions in positive integers, zeros
included) of the equation

x + 2f/ + Zz+ ... +t«=sn-i(t2+f).

(1) Show more generally that the number of modes of composing n
with i numbers, all distinct except the largest, which is to be always
taken/ times, is the denumerant of the equation

>x+C; + l)y+ ... +t« = n-J(t-y+l)(t+y).

(2) Show also that, if all the partitions of n into i parts are distinct ex-
cept the least, which is to be taken/ times, then the number of such par-
titions is the denumerant of the equation

x + 2p+ ... +(i-»^ + iV» = n-J{(t-y)8 + 3f-y} 138

3661. (Prof essor Sylvester, F.R.S.) — ^If through 3« + l given points
on a cubic curve a curve of the order N + « be drawn, and through the
remaining 3K— 1 intersections of the two curves a third one be drawn of
the order K ; prove that this will intersect the cubic at a fixed point.

[This point may be called the opposite of the 3n + 1 given points ; it
becomes Dr. Salmon's opx)Osite, as defined by him in tibe Fhilosophieal
Transaetiont for 1858, when n » 1, K » 1, and is independent of the
value of N.l 137

6271. (Professor Cayley, F.R.S.) — If w be an imaginary cube root
of unity, show that, if

^" l-«2(«,-«2)ar»' (l-y2)»(l+«,y2)» (1 - ic?)* (1 + -^^ '
and explain the general theory 141

5305. (Professor Sylvester, F.R.S.) — Dtf/^m^ww. — Six right lines
along which six forces can be xnade to equilibrate are said to be in Statical
Involution.

Prove that any six right lines lying on a ruled cubic surface are in
statical involution ; and, vice versd, S. six right lines are in statical
involution, a ruled cubic surfiEtce can be made to pass through them... 166

6420. (Professor Sylvester, F.R.S.) — From the expansion of
{log (1 +a;)}* , in a series according to powers of », prove that S<,> [the
coefficient oft^ is the developed product of (1 + ^(1 + 2^) ...... (l+t<)]

is divisible by every prime number gpreater than j'+l contained in any
term in the series t + 1, i, »— 1, , i—J+l 144

5640. fW. J. Curran Sharp, M. A.) — Salmon says [Higher Flane
Curves, p. 98] that the curve parallel to a given curve may be obtained (1)
as the envelope of a circle of given radius whose centre moves on the given
curve ; (2) as the envelope of the parallel to the tangent to the given curve
drawn at a constant distance. Prove that these processes are equivalent.

146

5643. (W. S. B. Woolhouse, F.R.A.S.) — Any two triangles being
given, the first may always be orthogonally projected into a triangle
similar to the second ; determine the magnitude of the projected triangle
geometrically by an easy construction^with the ruler and compasses... 26

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YIU OOKTENTS.

5765. (W. J. 0. Sharp, MJL) — H, tfaioii«^ any point of infleotfon

O on an n-io, there he drawn three straight lines meeting the onrye in
Ai, A4...A».i ; Bi, Bs...B».i ; C,, C,...C»-i, respectively ; prove that eyery
curve of the n^ degree through the 3ji-2 points O, Ai, Aj...A«-i;

Bi...B»-i ; Ci...C«-i will have for a point of inflexion. ..".'.. 146

6809. (D. Edwardes.)— If O be any point within a triangle ABO
prove that OA«8in2A + OB«8in2B + OC»sin2C is least when O is the
centre of the oircnmscribedcircle , 14g

6828. (Professor Darbonx.) — On coupe une pyramide triangulaire
SABOparun plan parallMek la base ; ceplan rencontre les ar^t^sl^rales
SA, SB, SO en A', B', 0' ; on m^e ensuite les plans OA'B', AB'O* BCA'
Soit P leur point commun. Determiner le lieu d^crit par le point P lorsl
que le plan A'B'O' se d^place en demeurant parall^e k la base 148

5966. (Pwfessor Sylvester, P.B.S.)— By a Oartesian Oval in spaca
let^us understand a curve the distances of whose points from three fixed
points are linear functions of each other, or, which comes to the same
thing, is the intersection of two surfcices of revolution, described by two
plane Cartesians having a focus in common. Conversely, when two points
can be found whose distances from any point in a space-Oartesiiui are
linear fimctions of one another, let them be termed focL Required to
prove, that the locus ol such foci is a plane curve of the 3rd degree. It
will be observed that this curve for Cartesians of double curvature is the
exact analogue of the three foci in a straight line for plane Cartesians. . . 69

6686. (W. J. 0. Sharp, M.A.)— If a bar naturally curved be strained
the bending moment at a point, whose natural curvature is r-^ and
strained curvature p-*, isE(p-i-r-i) {l + Ar-> + Br-* + ic} 143

6827. (Prof. Cayley, F.R.S.)— Consider a triangle ABC, and a point
P; and let AP meet BO in M, and BP meet AC in N (if , to fix the ideas
P is within the triangle, then M, N are in the sides BO, AC, respectively'
and the triangles APN, BPM are r^^arded as positive) ; find (1) the locus
of the point P, such that the ratio ( AAPN+ aBPM) : a ABC may have
a given value ; (2) drawing from each point P, at right angles to the
plane of the triangle, an ordinate PQ of a length proportional to the fore-
going ratio ( A APN + aBPM) : A ABC,trace the surface which is the locus
of the point Q, a surface which has the loci in (1) for its contour lines •
(3) find the volume of the portion standing on the triangle ABC as base •
and (4) deduce the solution of the following case of the four-point prob-
lem, viz., taking the points P, F at random within the triangle ABO
what is the chance that A, B, P, P' may forma convex quadrangle P. . . 149

7060. (D. Edwardes.)— If P be any point in the plane of a triangle
ABO, and d its distance from the circumscribed centre, show that
PA*sin2A+PB2sin2B + PC»8in2C = 4(E2+<P)sinAsinBBinC... 147

7069. (D. Edwardes.)— If a?, y, « be the distances of a point P from
the angular points of a triangle, prove that the mean value of jb* sin 2A
+y«sin2B+«*sin2C, as P ranges over the circle about ABC, is three
times the area of the triangle 147

7169. (W. J. Greenstreet, B.A.)— The sum of the three sides of a
right-angled spherical triangle is a quadrant : im>ve that (1) the minimum
value of the hypotenuse is cos-^f, and (2) in this case the spherical excess
issin-i^ , , 5j

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OONTENTS« IX

7270. (W. J. 0. Sharp, M.A.)— Eliminate a, b, t, a', *', e' from

a+b+e = d
ab + be+ea » 8
abe = 8

a' + b' + i/^df

aa' + W + c*/ — ^,
a'be + aye+abe' :^ e,
ab'e' + afbc' + a'b'e^e',,,. 162

7801. (Professor Cayley, F.R.S.) — If^ the function ?5!Lt^ « ^

is periodic of the third order {<ffiu = u) : given that the cubic equation

(ft, b. e, dSx, 1)8 = has two roots w, v such that v =■ **"'*'^ , find u as
^ "^ ' 7M + 8

a rational function of a, d, 0, <^, a, /3, 7, 8 ; and examine the case in which

u is not thus expressible 163

7366. (Professor Wolstenholme, M.A., Sc.D. Suggested by Ques-
tion 7286.) — At each point P of a given curve is drawn a straight line
U, making a given angle with the tangent at P, and a straight line Y,
such that U, V are equally inclined to the ordinate at P ; prove that the
point of contact of U with its envelope is the projection upon U of the
centre of curvature at P, and that the point of contact of Y with its en-
velope is the projection upon Y of the image with respect to the tangent
at P of the centre of curvature at P. [That is, if be the centre of cur-
vature at P, and OPO' be a straight line bisected in P, then, if OL, O'M
be let fall perpendicular to U, Y respectively, L, M will be the points of
contact of U, v with their envelopes.] 64

7469. (Professor Wolstenholme, M.A., Sc.D.) — Prove that if , in a
tetrahedron, any one or two of the equations a±a: = ^±y = c±«be
true, tlien will also the corresponding one or two equations of the set
AiX«B±Y = C±Z also be true 86

7661. (Rev. T. 0. Simmons, M.A.)— If

8in(a + iB) + 8in(/3 + 7) + 8in(7 + a) -0,

show that cos ^ (o — jB) cos i (jS - 7) cos i- (7 — o) cannot be greater than J.

124

7707. (Professor Malet, F.E.S.)— Prove that
fi'log sin B (log tan 6-log A (6)}
Jo AM ^— ^^-iK{log*log*' + iir«}+i,K'log^',

)*«log cos e {log cot 6- log A (e) }
A^6) ^ ^ *'' ^
fl«log A (a) (log A (e) -log sin 6 cos e\
Jo AW ^rf6 = iKlog^log^ + i,K'log^,

where A(e)s(l-A;2 8in2e)*, F + A;^=l, and Z and K' are complete
elliptic integrals of the first kind with moduli respectively k and hf... 68

7987. (Rev. T. R. Terry, M.A.) — Four spheres, whose radii are a, ft,
tf, d respectively, are such that each touches the other three externally.
Li the space between these four, another sphere of radius r is described
touching all four externally. Show that

^-7'(l)"(ir)-(i)-« «•)■

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X • COKOTNTS.

8036. (B. Lachlan, B.A.)— If four spheres; radii a, b^ e, d, touch
^ one another externally ; and r be the radius of the sphere whLoh cuts them

orthogonaUy,theni— 2a/JL\ -2 f-^) 46

8117. (Professor "Wolstenholme, M.A., Sc.D.)— Two coincoids S, S'
have two common plane sections, and the poles of these planes with
respect to S are the points P, P' ; prove that (1) if S' pass through P, it
will also pass through R ; (2) also, in this case, the following relations
must hold (02-3#A) (ee'-6AA0 + S^aa'^ 0, e» = 27a« (ee'-6AA0.
ThediscriminJantof A?S + S' is A)fe* + e/fc5 + #A;3 + e'^ + A'.] 79

8135. (Hev. T. G. Simmons, M.A.)— If G be the centroid, I the in-
entre, of a phme triangle,^ prove that

IGP = 4R2(i + cosAco8BcosC)-jRr + fr« 86

8204. (A. Gordon.) — ^If A, B, G, D, E are five points on a sphere,
Vi2 the volume ACDE, V,3 the volume ABDE, &c. : prove that, with a
certain convention as to sign,

Vm(AB)2 + Vi3(AG)«+Vi4(AD)« + Vi5(AE)2- 165

8207. (W. J. G. Sharp, M.A.)— -If A be the angle contained in the
half of a small Gircle of a sphere, angular radius a, by arcs of lengths
b and e ; i.e., if ABG be a spherical tnangle having the angle A » B + G ;
show that (1) cos ^ + cos c = 1 + cos a ; (2) cos A = —tan ^* tan ^c ; and (3)
hence deduce Euclid I. 47 and III. 31 62

8236. (Rev. T. R. Terry, M.A.)— A uniform lamina bounded by the
arc of a parabola and its latus rectum is revolving with angular ^locity
» about the latus rectum : suddenly the latue rectum becomes free and the
vertex becomes fixed. Show that the angular velocity about the tangent
to the vertex is |« 116

8237. (W. J. G. Sharp, M.A.)— If nOr denote the number of combi-
nations r together which ean be formed out of n things ; show, from
d priori considerations, that,

nCr - nG„-r, nCr = ^n-lGr-i, nOr = n-l^r + n-lOr-l (1, 2, 3),

or, more generally, nOr — ii-j»Gr + pGj n-pGr-i +p02 »-pGr-2 + &c. ... 166

8242. (Professor Sylvester, F.R.S.) — If by a Simplicissimum of the
nth. order be understood a figure in space of n dimensions formed by
the indefinite protraction of the series of which a linear segment, a tri-
angle, and a pyramid are the three first terms, prove that (1), when each

edge is unity, the squared content is .^ ^^"^ ' -, and hence (2) de-
duce that the Gayleyan Persymmetrical Invertebrate Determinant of the
squared edges by which such squared content is imaged must be
diminished in the ratio of negative unity to ( -2)» (1 . 2 . 3 ... «)2, in order
that it may represent its absolute value. JSx, gr,^ the determinant

(abY {ae)^ 1

{bay . (bcY 1

{caf {eby . 1

11 1 .

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CONTENTS. XI

images the squared content of a triangle whose edges are^ {ab)^ (ae), {be)f
for the triangle vanishes when this determinant yanishes, hut the actual
value of the squared content is this determinant diminished in the ratio
of negative unity to 2^ (1 . 2)', i.e., multiplied hy — ^ = (2) 53

8298. (E. Knowles, B.A.)— If Pr denote the coefficient of of, in the
expansion of (1 +^)'*, where n is a positive Integer, prove that

119.

8302. (A. Gordon.) — Prove that the surface a:' +y« +«' = <?« will

represent a surface of revolution if oo^xy « cos*y« « cos' 2^, where xy
denotes the angle hetween the axes of x and y, 117

'8311. (Prof. R. Swaminatha Aiyar, B.A.) — Find how many numhers
can he formed, having n for the sum of the digits (zero not heing used
as a digit) 64

8336. (Asparagus.) — From a point are drawn two chords OPF,
OQQ' of a given circle, and two tangents OA, OB ; a conic is drawn
touching the straight lines PQ, PQ', P'Q, P'Q' and passing through A
or B ; prove that this conic will toudi the circle in A or B 117

8380. (P. C. Ward, M.A.)— Prove that (1),

a + *— c, 4a, 6fl, 4a

4i, a + *— c, 4a, 6a

6i, 4i, a + b—Cf 4a

4kby 6d, 4d, a-^-b-c

s (a + i+c)4-8 (a + i + c)2(ic+ (?a+ a3) - 128 a*<j(a +i + <?) + 16 (*<? + ca + a*)»

= result of rationalizing a^ + 6^ + e;^ = ; and hence (2) the ahove deter-
minant is symmetrical with respect to a, d, 107

8386. (J. Brill, M.A.) — Three paraholas are drawn having a common
focus ; from a point T, external to all three, tangents TP and Tp are
drawn to the first parabol^, TQ and Tq to the second, and TE and Tr to
the third ; prove that Qr . Ej? . P^ == ^E . rP ,pQ, 95

8406. (F. 0. "Wace, M.A.)— If aj + aj + «8 + . • . + «n «= «, prove that

(^')"(t-)'' (i-'r<<-)'

66

8420. (Emily Perrin, B.Sc.) — Prove that the axis of perspective of a
triangle and its pedal triangle is the common radical axis of the circum-
circle, nine-point circle, and self-conjugate circle 67

8423. p. Edwardes.)— Prove that

(1) f*'sina;(logsina;)aefo? = 2 + (log2)2-21og2— ^ir2,

(2) (*'sin2a;(logsinar)2c?a;=: Jir{2(log2)2-21og2 + ^ir2-l}. ... 119
Jo

8452. (Satis Chandra Esly, B.A.)^Two rods, of lengths a and 3, are
jointed hy a smootii hinge and rest on the convex side of a parabolic

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XU OOKTBNTS.

arc whose axis is vertical ; if, in the position of equilibrium, tbe rods in-
clude a right angle, show that Ihe angle which the chord of contact
makes with the axis ifl given by tan iO B a^/ ^ 62

8473. (Professor Booth, M.A.) — If a, 6, ^ be the sides of a plane
triangle, prove that the diameter of the circumscribing circle is a root of
the equation

••■• o9

8498. (AsfLtosh Mukhopddhyay, M.A., F.B.A.S.)— The equation

\dxl dfi dx'dt'dxdt \dtldx^ \dx I

is intepfrable when (1) Q— 0, (2) Q «» dx/dt. Hence, obtain the complete
primitive 61

8501. (R. Knowles, B.A.) — ^In any triangle, show that
icos^Acos'B— cos(A— B)(3co8Aco8B— sinAsinB) scos^c... 75

8602. (B. Hanumanta Rau, B.A.) — Through any point K are drawn
the straight lines B"KC', C"KA', and A"KB' respectively parallel to
the sides BC, CA, AB of a triangle ; prove that (1) A A'B'C'= A A"B"C" ;

(2) parallels through A, B, C to A'B', B'C, C'A' meet at a point 0';

(3) parallels through A, B, C to A"C", A"B", 0"B" also meet at a point
(y ; (4) if the coordinates of and 0' are (a, fi, y) and (a', jS', y') then
aa : ey'^ bfi : oaf— cy i bfi' \ (5) if K is the Symmeoian point, O and (Y
become the Brocard points 49

8621. (Professor Wolstenholme, M.A., Sc.D.) — The circle of curva-
ture is drawn at a point P {am^, 2am) of the parabola i/^ = iax, RE' is the
common tangent of the puabola and circle, and meets PQ their common
chord in T ; prove that

(1) TR : TR' = 1 +4m3 : 1 ; (2) RR' = 16anfi (1 +m2)t « PQ8/ 32aa;
(3) TQ : TR' = TR' : TP « «»2 : l + m2;

and (4) the locus of T is a seztic having no real rectilinear asymptotes.

114

8537. (For Enunciation see Question 7169.) 61

8643. (R. Curtis, M.A.)— Prove (1) the following formula of trans-
formation for the equation of a conic to a triangle of reference
the sides of which are the polars of the vertices of the former tri-
angle (a, b, c, /, ffy h) (ar, y, «)« - A- 1 (A, B, C, F, G, H) (X, Y, Z)2 ; and
hence (2) show that the equations of a conic referred to an inscribed tri-
angle and to one circumscribed at the points of contact will be

fyz^-gzx^hxy = and (/X)* ± (^Y)* ± (AZ)* = Ill

8648. (Asparagus.) — ^Two straight lines turn about two fixed points
0, O' with angular velocities which are as 1 : 3, both straight Hues coin-
ciding with 00' initially ; if P be their point of intersection, prove that
the envelope of a straight line drawn tiirough P at right angles to OP
will be a parabola whose directrix passes through 0' and is at right angles
to 00' and whose latus rectum is twice 00' 65

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CONTENTS. XIU

8557. (Frofeesor Mathews, M.A.)— If

^ + i^ = l and ^+i^^l

are two confocal eUipses, such that polygons of r sides can be simnl-
taneously inscribed in the second ellipse and circumscribed to the first,

prove that a - a'sn tl^^^, b - *'cn ?5,

r r

where the modulus of the elliptic functions is equal to the eccentricity of
the inner ellipse. Verify the above when r = 3, 4, 6 respectively.... 66

8659. (Professor Wolstenholme, M.A., Sc.D.) — ^From a fixed point
are drawn OP, OQ tangents to one of a system of confocal conies (foci
S, S^ centre C), and from C are let fall perpendiculars on the normals
at P, Q ; prove that the envelope of the straight line joining the feet of
these perpendiculars is the conic (parabola)

{X(a:coso-hysina)+Y (a:sina— y cosoj-fiScosa}'

— 4XY(a:cosa + ysino)(«sina— ycosa),
where is origin, OS axis of x, (X, Y) the point 0, SS'= 2c, and a is
the sum of the angles which SO, S'O make with the axis of x»

[If X«-Y3 « c«, the straight line is fixed.] 128

8671. (Asatosh Mukhopadhyay, M.A., F.R.A.S.) — Show that the
reciprocal polar of the evolute of the reciprocal polar of the evolute to

the conic -^ + -^ = 1, with respect to the circle described on the line
joining the foci as diameter, is the curve

mHmm'(i)'-(i)'mHf-ir-'>

8686. (B. F. Davis, M.A.)— If TP, TQ be tangents to a parabola, and
PQ meet the directrix in Z, prove that ZT will be a mean proportional
between ZP and ZQ 60

8600. (Professor Mahendra Nath Ray, M.A., LL.B.)— Through an
indefinite point of a given hyperbola straight lines are drawn to meet the
asymptotes ; show that the hyperbola itself is the envelope of the locus
of the middle points of the straight line 50

8607. (R. Curtis, M.A.) — Between two curves, f{x, y) — and
^ (^f y) — 0) <>f degrees m and n, show that there can be drawn mn straight
lines of a given length I, parallel to a given straight Une, from the curve
/{xy) « to the curve ^ {xp) = 0, and mn othera from tp (i^y) = to /(ay)
= 0. [The lines must be drawn in a given direction, say from left to
right.] 66

8623. (N'Importe.) — From the centre of similitude S, common radii
vectores SPP', SQQ' are drawn to similar curves PQ, P'Q'. Having
given the centre of gravity of the area SPQ, find that of PQQT' ; and
its limiting position when PQ, P'Q' ultimately coincide 60

8631 . (Professor Sylvester, F.R.S.)— Find the discriminant of

a^ + y^ + sfi + Ze^^ + 6exyz 118, 166

8637. (Professor Hudson, M.A.) — The tangent and normal to
y/e =• ^ (€*/« + €-*/«) cut the axis of « in T, O, respectively ; find tiie mini-
mum VBlue of GKT. ...► 161

b

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XIV CONTENTS.

8639. (Frofeflsor Wolstenholme, M. A., So.D.) — The distances of a
point O (within a giv^n triangle ABC) from the angular points are
«, y, z respectively; prove that the volume of a tetrahedron in which
a, bf e are the lengths of three conterminous edges, and x, y, z the lengths
of the edges respectively opposite, is -^ {{(^-sc') (*'— y^) (c*-**)}*... 30

8643. (Professor Gtenese, M.A.)— Four complanar forces Pj, Pj, &o.
are in equilibrium. Their lines of action (omitting one at a time) deter-
mine four triangles whose areas are A^, A3, &c., and circum-radii Bi, B3,

&c. Prove that Pi : Pj, &c. - ^ : ^, &c 35

Ki B,

8646. (Emile Yigarie.) — On donne deux points A et B sur une drcon-
fi^rence de centre O ; trouver sur cette circonference un troisi^me point
P, tel que les droites PA, PB coupent un diam^tre fixe CC en des points
M, N, tels que I'on ait OM - ON 36

8647. (R. W. D. Christie, M.A.)-— Prove that 2 + S » 2«», if

« = 18+28+... +n», S- 16 + 2« + ...n6, 2 - l7 + 27+...n7. ... 33

8650 (E. M. Davys, M.A.) — ^Prove that, in any plane triangle,
cos"A + 4 cosM> A sin«A+6 cos* Asia* A-5 cos* A sin8 A-4 oos» A sinw» A
-sin«A = cos2A 46

8656. (W. J. Greenstreet, B.A.) — ^In a spherical triangle, prove that

8in*R= (coso-co8»R)* + sin«asin3Rsin«(S-A) 118

8668. (W. J. C. Sharp, M.A.)— Boolb obtains the result

[[ ... X<fe»s is^-^ f X(te--(n-l) «*-« {Xxdx

^ (n--l)(n-2) ^^,f.^^^^ ^f^^.1^7

by a symbolical method ; show that it may also be obtained by ordinary
methods 71

8660. (B. Hanumanta Bau, M.A.) — Investigate the singular sdn-
tion of the equation 4t Ix + p J^\ +25 [ y^—xf/ ^ j « 0, and show that

it is the envelope of a series of circles described on the subnormal of an
ellipse as diameter 83

8661. (A. Q-ordon.) — ^Any curve of the 4th degree intersects a conic
in 8 points. These are joined, forming an octagon A^As... A«. Show
that the 8 intersections of Aj with At and A«, As with A^ and Ag, &c.,
&c., will lie on a conic 118

8671. (D. Edwardes.) — Prove that the laiua rectum of a parabola
is half the harmonic mean between any two focal chords at right angles
to one another 123

8676. (Professor Bordage.) — The three sides of a triangle forming an
arithmetical progression, (a) being the shortest, (a') the longest ; if the
distance of the centres of the inscribed and circumscribed circles is desig-
nated by t, and the diameter of the nine-point circle by D, prove that
tfa'- 3(D2-f3) 107

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00NTBNT8. XV

8678. (Profeflsor (^enese, M.A.) — AB is a obord of a. conic, BP, BQ
are parallels to the asymptotes, APQ a variable transversal meetiDg the
curve at E. Prove that the ratio PB : EQ is constant 46

8681. (Professor Hudson, M.A.) — In the epicycloid of two cusps,
if P be the describiug point when Q is the point of contact, prove that '
the tangents at P, Q to the epicycloid and the fixed circle respectively
meet on the line joining the cusps 46

8683. (Professor Ignacio Beyens, M.A.)— Si dans un triangle la pro-
jection du c6t^ BC sur BA qui est BH, et la projection du BA sur ACTqui
est AH', sont egales ; la bissectrice du A, la hauteur du et la m6diane
de B se rencon^ent en le mdme point 96

8685. (The Editor.) — Given an angle at the base of a triangle, the
sum of the two sides, and the distance between the given angular point
and the point of contact of the escribed circle touching the base ; con-
struct the triangle 75

8692. (B. F. Davis, M.A. Suggested by Question 8559.) — If, from
any point in a given fixed straight line passing through the focus of a pa-
rabola, tangents be drawn to the curve, prove that the envelope of the Hne
joining the feet of the perpendiculars on these tangents from a given
fixed point on the directrix is another parabola 129

8696. (Emile Vigari6.) — Dans un cercle donn6, par un point P pris
sur la circonf6rence on m^ne trois cordes PA, PB, PC ; sur chacune
d'elles comme diam^tre on d6crit une circonf6rence ; d^montrer que les
trois points de rencontre sont en ligne droit 32

8697. (Captain H. Brocard.)— Les droites qui joignent le centre du
cercle circonscrit aux milieux des segments interceptis par les perpen-
diculaires aux milieux des cdt6s sur les hauteurs du trian^e sont perpen-
diculaires aux m^dianes correspondantes 49

8700. (R. W. D. Christie, M.A.) — Given the sum of the expres-
sion 1*" + 2*" + S*" + 4*' +«•', to find a method of writing down at once

the sum of l»'+i + 2^+1 + 3*'+^ + 4*'+ * +!•••♦*, where r is even or vice

verad 96

8708. (Maurice D'Ocagne.) — Si, dans le quadrilat^re ABCD, les
angles opposes B et D sont droits, la droite qui joint les pieds des perpen-
diculaires abaiss^es du sommet C sur les bissectrices int^rieure et ext^rieure
de Tangle A passe par le milieu de la diagonale BD 34

8713. (Professor Steggall, M.A.) — Show that the solution in ra-
tional quantities of the equation z^+y^-^-z^ =^u^, is x ^k{a^-¥b'^—i^,
y^2kae, z ^ 2kbe, w « A; (a2 + 62 + c2) 31

8714. (Professor Genese, M.A.) — S is a focus of a conic, PN a
fixed ordinate to the diameter through 8, PQP' a circle with centre S ;
a variable radius SQ meets PN at L and the conic at B. Prove that the
cross ratio {SLQB} is constant 107

8715. (Professor Hudson, M.A.) — Prove that

tan37i°--v/6+ V3-V2-2 88

8716. (Professor Mathews, M.A.) — Prove that the real common
tangents of the circles ic^+y^— 2aa: = 0, afi-¥y^ — 2by = are represented
by 2a*(af'+y2— 2<M?) « (*y— fla? + a*)', or, which is the same thing, by

2ai(»« + y»-2iy) =-(iy-a«-a4)2 6X, 83

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XVI oonxRB.

8718. (PwfiMnr VmSbm^.}-J/mm^ BAG ^^mat ^n^ aatfere, et
(1) le emtn de gnTit^ (2) Fortkocaotie, (3) le eoatee dm ooRfe cnoon-
icnt pMComi BBe dlioite do niife . Troowl'eBvdoppedmoM BC... 116

8719. (ProfiBMor Wolrtenbofane, ILA., ScJ).)— Tlie kngtiis of the
edge* OA, OB, OC of a tetnhednm OABC an denoted bj a» *, r, thofle
of tltere^ecthreljoppoaiteedg«8BC,CA«ABb7x,jr,s; and tiie dihedral
aii£^ i;p>poM^ to then hj A, B, C ; X, T, Z ; prove that, if Y be the

Tolmne expraMed in terms ofa,^^, ^ 9$ *» ^* |«jpoot A, &c. ;

and tiienoe tiiat (1) when a, s ; ^, jr, and c+s are /»•«», Y will he a
niaTimom when C » Z ; (2) when a» s ; I, jr, and «~s are girai, Y will
heaniazimiimiHMnC-i-Z s 180°; (3) iHbena, s; ^-fjf; c-fs are given,
Y will be a TnaTJimim irilien B » X and C » Z (i.«., iri^ ^ ««, « « i) ;
(4) iHien a, x ; b+f, e—z are giren, Y will be a maTJirnim when B » X
C+Z » 180*" ; and (5) inyeiti^te if Y can have a mazimnm when a, s ;
3— y, e^s are given, or iritien a^a, ^— 9f» «— s are given; the given
qnandlxties being sappoeedahrays real and finite.

In case (4), prove also that 180** is the mazimnm valoe ol C-l-Z for
Tariations inb,p (sabject to the conditions stated), and that aero is the
mininram value of B— Y for variations in ^ s 39

8724. (B. Tucker, M.A.)— If a constant line AB moves with ends
on OX, OT, two rectangular axes, and on it a semicircle is described,
then locos of mid-point <^ arc is one of two straight lines. Find locus of
any other fixed point on the arc. [If G is Ihe mid-point of AB, and P the
mid-point of the arc, then the question may be enunciated for a bar rigidly
connected with the bar AB.] 36

8725. (8. Tebay, B.A.)— If the sines ol opposites dihedral angles of a
tetrahedron be ren)ectively proportional to the edges formed by these
angles, the areas of the four &oes are eqnaL 48

8726. (8. Boberts, M.A. Analogous to Quest. 3068.)— Show that, if
some four of the roots of a quintic form an harmonic system, then

J»-27.3«.JK+ 2B.3».L = 0,
where J, K, L are the three fundamental invariants of the orders 4,
8, 12 [see Salmon's Higher Algebra, 3rd ed., p. 211] 33

8727. (B. Hanumanta Bau, B.A.)— The sides BC, GA, AB of a tri-
angle ABU are produced to a, b, e, such that Ga, Ab, "Re are respectively
equal to BG, GA, AB. Prove that the centroid of the triangle abe coin-
cides with that of ABG 84

8728. (Gaptain H. Brocard.) — De cheque sommet d'un triangle on
m^ne des perpendiculaires aux c6t6s adjacents, jusqu*^ leurs rencontres
aveo le c6t6 oppKOs^. Les centres A', B', G' des cercles circonscrits aux
trois triangles ainsi formes sont les sommets d'un second triangle homo-
logique avec le propose. Le centre d'homologie est le centre O du cercle
circonscrit au triangle donn6. En d'autres termes, les nouveaux cercles
iont tangents au cercle circonscrit, aux points A, B, G 121

8729. (B. W. D. Ghristie, M. A.)— Of the series

1.1 + 3.6 + 6.13 + 7.26 + &c.
to n terms, show that, (1) the sum is a square ; (2) the square root is
m 1 + 3 + 6 + 7 + &C. to n terms; (3) each term is the product of two num-

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bers which may be taken to represent the shorter side and hypothenuse of
a right-angled triangle ; (4) if unity be added to the nth term, it is
divisible by 2», and H unity be subtracted it is divisible by tt— 1 ; and (6)
all terms are odd numbers 106

8737. (W. J. Greenstreet, B.A.) — ABC is a spherical triangle, the
mid-points of its sides are the angular points of a triangle DEF ; prove
that (1) cosEF / cos |a = cosFD / cos ^b » cosDE / cos ^c ;
and (2) if cos^ ia » coB^{b + e) coa^(b^e), the angle D is a right
angle. ;. 108

8746. (W. W. Taylor, M.A. Suggested by Question 7938.) —
Prove that (1) the areas of the TAYLOR-circles of the four triangles
ABC, PBC, PCA, PAB are together equal to the area of the circum-
scribed circle, P being the orthocentre of the triangle ABC ; also (2) the
lines joining their centres are the SiMSON-lines of the middle points of the
sides or of AP, BP, CP with respect to the pedal triangle 100

8749. (F. Purser, M.A.) — Find (1) the cubic locus of the centre
of a coDic passing through the three vertices A, B, C of a tri-
angle, and such that the three normals at these vertices are concurrent ;
and prove that (2) this cubic passes through the in-centre I, the three ex-
centres la, Ifr, Ic, the circum-centre O, the orthocentre H, tiie centroid G,
and the symmedian G', and cuts the three sides normally at their middle
points L, M, N ; also (3) tangents at (I, !«, Ij, Ic), (A, B, C, G), (L, M,
N, G') are respectively concurrent in G, G', ; and (4) the lines joining
L, M, N meet the curve again in their intersections with the respective
perpendiculars from the vertices 35

8764. (Professor Mahendra Nath Eay, M.A., LL.B.)— Prove that

(1) r^^^=^(i,-l), (2) ri«lB!£^ = ,(^-|) 123

^ ' Jo l + cos^a: ^* '» V / J^ i + cos2a; ^ '^

8766. (Professor Neuberg.) — On prolonge les hauteurs du triangle
ABC au deU, des sommets des quantit^s AA'=BC, BB'^CA, CC'^AB ;
d^montrer : (1) les triangles ABC, A'B'C ont m^me centre de gravite ;

(2) si a, a' sent les angles de Brocard de ces triangles, on a

A'B'C=2ABC(2 + cota), (A'B')2+(B'C')3+(C'A')« = 8ABC (3 + 2cota),
cota'= (2coto + 3)/(cota + 2) ;

(3) les points A, B, C sont les centres des carr^s construits, int^rieurement,
sur les c6t^ du triangle A'B'C ; (4) les milieux des c6t^s de A'B'C sont
les centres des carr^ construits, ext^rieurement, sur les c6t68 de ABC. . . 56

8767. (Professor Edmund Bordage.) — If two fractions aJa'f bjb' are
such that ab' ^baf = I, prove that the simplest fraction included between
the two given fractions is {a + b) j (a* + b*) 46

8763. (H. Stewart, M.C.P.)— Prove that

(l)if w(l-a;2)* + a;(i-w»)* = fl2, then ^ + tlr:!^* = 0;

^ (1-a^*

(2) if (l-«2)i + (i.^)i = a («-«), then ^==(ir:^*.... 100

^ ' ^ ' ^ (!-«»)*

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XVIU OONTBNTS.

8765. (L. J. Bogers, M.A.) — If :r|-(-02 + ^-l-ai + as + as« 0, prove
that (1) tan (^i + a^) , tan {±^ + aj) , tan {x^ + oii)

tan (a?! + oj, tan (arj + oj), tan (a:, + a,)

tan (a?! + 03), tan(a;3 + 08), tan(a>8 + aa)
and (2) the same is true if tangents are replaced by sines 88

8768. (R. F. Davis, M.A.) — Required an Analytical Proof of Feuer-
bach's Theorem.

[An analytical proof is given by the Rev. J. J. Milnb, M.A., in the
Introduction to his "Weekly Problem Papers, but this proof may be modi-
fied and improved.] 131

8769. (J. Brill, M.A.) — An ellipse is inscribed in the triangle PQR
so as to touch the sides QR, RP, PQ at P', Q', R' respectively ; prove
that, if C be the centre of the ellipse,

aQOR : aQ'CR' - aRCP : aR'CF - aPCQ : aP'CQ'.... 34

8778. fW. J. 0. Sharp, M.A.) — If wy—vz, uz^wx, and vx—uy be
cogredient to x, y, and z, the rectangular Cartesian coordinates of a point
respectively, «, v, And to are so 44

8783. (Captain H. Brocard.) — ^Les droites qui joignent I'orthocentre
aux milieux des segments intercept's par les hauteurs du triangle sur les
rayons du cercle circonscrit aboutissant aux sommets du triangle sent
perpendiculaires aux sym'dianes correspondantes 87

8784. (R. W. D. Christie, MA..)— Prove that, if

« = 1+2 + 3 + .. .+n, S = 13+28 + 3«+...+n2, S' = 18 + 28 + 3»+. ..+»»,
2 = l* + 2* + 3*+. ..+«*, <r = l« + 26+3«+... + fi«,

then '-^-1^ 99

8786. (R. Tucker, M.A.)~Pp, Qq, Rr are focal chords of a para-
bola ; if P, Q, R have a conormal point, then the centres of curvature
for j3, qt r are collinear. (This is another way of putting the first part of
Question 8693.) Find the equation to this Ime. Prove that this central
line envelopes (1) a hyperbola when tangents atjt?, q meet on the directrix,
(2) an ellipse when the tangents meet on the latus rectum 61

8789. (Professor CuUey, M.A.) — Find by a geometrical construc-
tion the equation 8S' = P* of the pair of tangents from T (a:', y') to a
circle S s a?*+,v2— «« =s 0, whose centre is ; and show that, if perpen-
diculars RL, RM, RK be drawn to the tangents and their chord of con-
tact from a point R not on the circumference, RL . RM differs from RK*
by TO*^ cos* ^ cos* 0, where 2e0 and 20 are the angles subtended by the
circle at T and R respectively 37

8793. (Professor Neuberg.) — Soient AB, CD deux droites, qui se
coupent en O. Si Ton fait toumer AB autour de O, le point double
de deux figures semblables construites sur AB et CD d^crit une circon-
f^rence de cercle 30

8797. (Professor S§irad§iranjan Rly, M.A.) — GAP and OBQ are
two fixed straight lines intersecting at 0, and C is a circle touching them
both at P and Q ; prove that the perimeter of the triangle OAB, circum-
scribed by ani/ circle passing through O and touching C externally, is
constant , , 89

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CONTENTS. XIX

8799. (Professor Hudson, M.A.) — ^Find the envelope of the straight
line flw =s/(o)cos(e— o)+/(o)8in(e-ai) 41

8801. (Rev. T. P. Kirkman, F.K.S. Suggested by Quest. 8326.)
— Upon the most sacred of the undiscoyered Egyptian pyramids,
whose vertex P is at unequal distances from the angles of its scalene
base ABC, is cut in the face PAG an ascending flight of steps from A to
b, in PC ; from b another flight mounts to in PB ; and a third leads up
from c to a, At a. given altitude h in PA. These three flights are equi-
gradient. A second path from A mounts to ^' in PC, and thence pro-
ceeding to (/ in PB, ascends from (/ to the above point a in P A . These
three flights also are equigradient. By Abca Pharoah and his Grandees
of Churdi and State were wont to mount to a platform at a for solemn
religious rites ; the admitted vulgar used only the path A&Va. Required,
a plane projection of the pyramid and its six flights of stairs, and proof
from that prelection that the first three are equigradient, and the second
three equigradient, up to the given altitude h 24

8804. (S. Tebay, B.A.) — If (a, ft, c) be conterminous edges of a
tetrahedron, {x, y, z) the respective opposites, (A, X ; B, Y ; C, Z) cor-
responding dihedral angles, and Y the volume ; show that (1)
ax by ^ cz .

sinAsinX sinBsinY sinCsinZ'
also (2), if the areas of the four faces are equal, then

a - a:, ft - y, <j - «, A = X, B - Y, = Z.
V = tV {2(ft2+(^-a?) (c2 + a2-fta) (a^^b^^(^)}\

(ft-<>)sinA+(<j-a)sinB+(a-ft)sinC — 31

8807. (For Enunciation see Question 8746.) 100

8808. (F. R. J. Hervey.) — ^Find (1) in how many ways n lines of
verse can be rhymed, so as to have r different rhymes, and no line un-
rhymed ; and (2) show that, in the case of the sonnet, the numbers of ways
with 2, 3, ... 7 rhymes are, respectively, 8177, 731731, 6914908, 12122110,
4099096, and 136136 76

8810. (Rev. T. R. Terry, M.A.) — Find the equation to the straight
lines through the origin and the intersection of the conies

«»-3a;y— 4a: + y-l « 0, 2a;3 + a;y + 4y3 + 2a;+13y+8 =* 0. ... 68

8813. (J. Griffiths, M.A.)— If we have
f{x) -Ao+2A3a;3 + 2A4r* ■•-... adinf,, ^{x) ^\-2.B^-2^^x^-^,..adinf,

where Ao - 1 + 22 (-)*cn2 ?5, Bj = 2 (-)* cn^*^ -i- sns*-^,

n n n

A, « ifesi :i(-)' cn2 !5sn2?5, B4 « 2(-)' cn«^ + sn* *-?,
n n n n

A^^k^% (-)• cn^^sn*^, Be» 2(-)*cn2?5 + Bn«?l,

and « is a nvmber from 1 to n— 1, show that/(a;) ,^(x)^ A«. 78

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XX CONTENTS.

8814. (The Editor.) — Prove that the triangle formed by joii

the mid-points of the altitudes of a triangle is one-fourth of the pe

triangle, and that the theorem holds good for any other three concurrent
lines drawn through the vertices of the triangle 59

8816. (Captain H. Brocard.) — n, Of d6signant les points de Brocard,
S le milieu de Xlfi', S' le point correspondant k S par droites sym6-
triques, It le point de Steiner ; d^montrer que ce point It se troave sur la
droite SS'. (R est une des intersections du cercle ABC avec le diam^tre per-
pendiculaire k I'axe d*homologie du triangle ABC et du premier triangle de
Brocard. Ce point est aussi I'intersection commune des paralldles auz
c6tes du premier triangle de Brocard, menses par les sommets corres-
pondants du triangle ABC.) 22

8820. (Charles F. Lodge.) — A mirror, measuring 33 inches by 22
inches, is to have a frame of uniform width, whose area is to equal that
of the glass ; show that the width of the frame is 6\ inches 78

8823. (Professor Crofton, F.R.S.)— Two discs of any form are move-
able in a plane round two fixed points A, B, respectively. Show that,
when they are in such positions that the lengfth of an endless band en-
veloping both is a maximum or a minimum, the portions of the band
which form the common tangents will meet on AB if produced, and are
equally inclined to AB 26

8830. (Professor Bordage.) — Given the equation

sin 2a . 0?^ + 2 (sin a + cos o) a; + 2 — 0,
find, and solve, the equation of the second degree that has as roots the
squares of the roots of the given equation 110

8833. (Professor Neuberg.) — Un angle de grandeur constante toume
autour de son sommet A, ses c6tes rencoutrent une droite donn^ XY aux
points B et C. Trouver (1) les enveloppes de la m^diane BB' et de la
sym^diane BH ; (2) les points oil ces droites touchent leurs enve-
loppes 38

8836. (Professor de Longchamps.) — On consid^e un triangle ABC
et le cercle circonscrit A ; soit /a une transversale rencontrant les cdt^s
de ABC aux points A', B', C. Par A' on m^ne h. A une tangente que
Ton rabat sur BC de telle sorte que le point de contact vienne occuper sur
BC une certaine position A". Demontrer que la droite AA" et les deux
droites analogues BB", CC" vont passer par un mSme point M 94

8837. (The Editor.)--OD the sides of a regular «-gon, n points are
taken at random, one on each, forming the apices of inscribed M-gon ;
again, inside the n triangles that He outside this last polygon, n points
are taken at random, one in each, forming, when joined, another n-gon ;
find the general average area of this last n-gon, and show therefrom that,
in the cases of a hexagon and a square, the averages are fj- and -|g- of the
original figure respectively 41

8840. (W. S. M'Cay, M.A.)— Prove that (1) the locus of the mean
centre of the four points, in which a line of given direction meets the
faces of a tetrahedron, is a plane (diametral plane) ; and (2) if A, B, C,
D be the areas of the faces of the tetrahedron, aU the diametral planes
envelope the quartic (Aj?)*+(By)* + (Ca)* + (Dio)* = 40

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8843. (Bey; T. P. KirlnnaTi, F.R.S.) — The angles nabc ... /m of any

formed: first, nai>aia; and next, if ^«i/i be any three consecutive
points of it, the distances in the directions /j^i and eidi of ei from the lines
fe and ed in N are equal. The line nitni meets the edge nm of N in r.
Prove that, if from s in an produced we draw snj — r»i, «»i is parallel to
the first portion ««! of L 68

8847. (W. Mills, B. A.)— Given the focus of a parabola, one tangent,
and a point on it through which another tangent passes ; prove that the
locus of the point of intersection of the variable tangent with a diameter
through the point of contact of the fixed tangent is a circle which touches
the fixed tangent at the given point on it 105

8861. (J. J. Walker, P.R.S.)— Show that either segment of a focal
chord of a conic section is a mean proportional between its excess over
the semi-latus rectum (or viee vertd) and the whole chord 41

8864. (Rev. T. R. Terry, M.A.)— Solve (1) the equation
w«+2= (2ic + 6) u>,+i— (or^ + 4a: + 3) Wg I
and show (2) that, if «« and v« both satisfy this equation, and if Mi » 2,
u^ — lO, while 1^1 — 1, i>3 = 2, then 2ux = x{x + d)Vm 44

8856. (B. Hanumanta Rau, B.A.) — AB is the diameter of a semi-
circle; AG and BF tangents. CF cuts the semicircle in D. DOt is
drawn at right angles to CF, meeting AB in G-. Prove that

AC.BF = AG.GB 112

8868. (Professor Gochez.) — ^R^soudre les Equations

(ay + 1) (a; + y) = axy, («V+ 1) (^^ + 1/^ = *'^'-
Montrer que ce syst^me est quadratique. Appliquer les formules au cas
particulier suivant: a = J^, J = -»^ 60

8862. (R. Curtis, 8. J., M.A.)— Find (1) the locus of a point in a
material lamina, such that if the lamina were moving without rotation in
its own plane, and that point were suddenly fixed in space, the ensuing
rotation of the lamina would be given ; (2) where the point should be if
the rotation produced is a maximnm ; and (3) where the point should be
in order that the loss of kinetic energy would be given. 67

8863. (Ch. Hermite, Membre de I'lnstitut.) — Determiner les int4-
gralesd^finiee T^-^n' f'*'-^— n'

en supposant que p soit une quantity imaginaire quelconque 53

8866. (Amiral de Jonquils.) — Soient trois nombres entiers, a, b, c,
premiers entre eux, deux h, deux, et v6rifiant T^uation o^ + ^^s^. D^-
montrer que Q'exposant n premier et sup^eur k 3) : (1) a et b ne
peuvent dtre, simultan^ment, premiers ; (2) si a, suppose inf^rieur k b,
est premier, ^ *- d + 1 56

8866. (Professor Catalan.) — ^Dtoontrer les contributions au th^or^e
de Fbbmat :—« suppose premier, (1) «-l-aR(»); (2) «*- 1 = 3R (n*) ;
(3) tout diviseur premier, de 0— a, divise^ a— I; (4) a + b et 0— a sont

C

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ZZU OONTBKTS*

premiers entire enx; (6) 2a— 1 et 2^ + 1 sont premiers entrt enx ; (6) le
nombre premier, a (s'll existe) est oompris entre

!7) a et b surpassent n; ^8) le nombre b, qui satisfcdt k Filiation
4+ 1)*— i* « a*, est compris entre

a (a/nf*"^ et - 1 + a (a/n)^*"^ ;
(9) soit b un nombre entier, sup&ienr au nombre entier n, Entre

ilya, tout an pins, un nombre entier; (10) aucun des nombres a-k-b,
e^a, e—b, n*est premier; (11) chacun d'euz a la forme N, on la forme
{I In) N, N ^tant un nombre entier ; (12) soient, s*il est possible,

a+b^ tf"», e-a = A"», c— * « a'* ;
alors tf = 3R (n) ; (13) (ic + y)''-^*-^** - fwry (a;+y) P,

P = Hia:'*-a + H5a^-*y+ ... +Hiy*-»,
les coeflBcients sont donn^ par la formule Hp « (1/n) [0»-i^ ± 1],
le signe + repondant au cas oii /? est pair, et le polyn6me P est ivisible
par a;2 +xp + y\ et mSme par («» + ary + y^\ si « = 5K (6) + 1 ; (14) la
difference des puissances n^*°^ de deux nombres entiers consecutifs,
a, a+1, ^tant diminu^ de 1, est divisible par na(a+l) (A^ + a+l) ;
les facteurs a, a + 1, a^ + a + 1 sont premiers entre euz, deux It deux ; en
outre, le troisi^me 6gale le produit des deux autres, augmente de 1 ;
(15) si, dans T^quation de Ferhat, le nombre a est premier, on a, par le
th^or^me 8865, de M. de Jonqui^res, a»»-l - QR [ni(A + l) (A^ + J + l)];
et (16) e est compris entre a + ^et^(a + d) 56

8869. (Professor Steggall, M.A.) — A shot of mass m is fired in vacuo
from an air gun of length /, with a charge of air that at norma] pres-
sure p would occupy a volume v ; this air initially occupies a length b of
the barrel of the gun. Show that the time of passage along the barrel
and the velocity with which the shot leaves the gun are given by the

equations T = (/+f 1**1) + (5iH;/m)*, V = (1 -b\l2ll)(fipvlm)\

where b is small compared with /, and the ratio of the specific heats of

air is taken as 1*4 , SO

8873. (Professor Edmund Bordage.) — Given two relations,

(2S = a + ft + c),
dp + ycosg-i-gcos* __ y + gcosg-i-a?cosg __ g + a? cos 3 -i- y cos a ^ «
cos(S-a) "" cos(S-A) cos(S-g) '^ '

deduce therefrom -£- = _y_ « _£_ = ^ 91

sina Bind sing sinS

8875. (Professor Nash, M.A.) — Professor Casey's Quest. 7839 may
be enunciated as follows : — DD', EE', FP' are the intersections of the
sides of a triangle ABC with the cosine circle, the order of the letters being
such that E'F, FD, IXE are diameters. The circle round AE'F cuts the
circles on AB, AC as diameters in the points a, a' ; and b, b', e, if are
similarly determined upon the circles BFD, CB^E. To show that the
circles roimd ACa, BA*, CBc pass through « the positive Brocard-point
of ABO, and the circles ABa', BO^', GAg' through the negative ^rocaxd-

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CONTENTS. XXlU

point. ProTO the following additional propexties : — (1) The circleB ACn,
ABaf touch at A the rides AB, AC, and intersect agun in a vertex of
Brocard'B second triangle ; (2) the taogents to these circles at aaf bisect
AB, AC ; (3) the points B, C, a, of are concyclic, and the circle through
them touches OB, OC ; (4) AaD and A<i1)' are collinear ; (6) Ca, Ba' in-
tersect upon the symmedian AP, and Ca', Ba upon the perpendicular AL;
(6) aE', aT meet BC at the foot of the symmedian AP ; (7) ««', E'P,
BO, and the radical axis of the circles ABO, AE'F, are concurrent;
(8) the three points of coucurrence of the three sets of lines in (7) lie upon
a line parallel to »<u\ and therefore the triangle formed by the lines «ia%
bb\ ctf is in perspective vdth ABC ; (9) the p<«le of aa' with respect to the
circle AE'F lies upou the median of ADD'; (10) the pairs of points ae',
ba\ eV are isogonal, and the infcribed conic whose foci are atf touches
CA at the foot of the perpendicular from B ; (11) one directrix of this
conic is the line joining to the intersection of FD' and DE', and the
other the line joining A to the intersection of EF', FD' 131

8876. (Professor Hudson, M.A.) — Prove that the normal chord which
subtends a right angle at the focus of a parabola is divided by the axis
in the ratio 2 ; 3 108

8886. (Rev. T. P. Kirkman, M.A., P.R.S.) — Two 28-edra have
the following triangles, the summits being marked ahe „, npq. Required,
all the symmetry of both.

abfff aic, acd^ ade^ aef^ afg, bj'e, bih, bj\ big, ^7, elk^ ekd, dkm, dfm, drf,
fmq, fgq, ghq, ghi, hpj\ hpq, jlk, jkp, mnq, mnp, mpk, pqn,

abg, afg, afe^ aed^ abc, acd, bhg, bhi, bet, qji, edj\ c{jk, dkl, dfl, def, fkl,
fknif jfgtn, ghm, ifq, ihpf ipn, inq, j'kn, jnq, hnfif hmrtf hpn 126

8887. (W. J. Greenstreet, B.A.)— In the "ambiguous case" of tri-
angles, ^ven a, b, and A ; r,r' being the radii of the in-circles, p, p' of
the escribed circles to a ; prove that

rf^{b + a) ^pp'{b^a), rp(b^a) ^ t'p'e 87

8888. fS. Tebay, B.A.) — Dj, D,, D, are the shortest distances of
opposite edges (a, a ; b, b\ c, e) of a tetrahedron, Y the volume, and A
the area of one of the equal faces ; show that

V « i (D1D3D3), a2 - ^ (Di2a2+D3«ft« + Da'cS),

and Di» + D2«+D32-i(a2 + «2 + c2)-Di« + a« = &c 161

8903. (Professor Oenese, M.A.)— If /(r, 0) be the equation to a conic
referred to any pole 0, the value of (df)l(dr) at any point P of the plane
varies as (PQ + PR) / OQ . OR, Q, R being the intersections of OP with the
conic ; thus, for a point Q of the conic, (df)l{(fr) « QR/OQ . OR; and
in particular, if O be a focus, {df) / {dr) is constant over the curve. [In
this case /=(/-<T cos e)2-rs.] 93

8905. fProfessor Neuberg.) — D6montrer qu'i tout point / de la dis-
tance des foyers F, F d'une ellipse E correspond une droite D ext^rieure
an plan de E et parallMe an petit axe, telle qu'il existe le rapport con-
stant e : a entre les distances d*un point quelconque de E It / et D.
Lorsque/occupe toutes les poritions sur FF', D se d^place sur un cylindre
qui a pour base ime ellipse semblable k E 113

8908. (Enunciation included in Question 8866.) 66

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Xnv CONTENTS.

B912. (Professor Bordage.)-— A square ABOD and a fltraigl^ line A
in its plane are given, and through the points A, B, 0, D perpendiculars
AA', BB\ CC, DB' are drawn on A, A and G heing two opposite sum-
mits; prove that (1) (BB')«+(DD')''-2AA'.BB' is a constant quantity
for every position of A ; and (2) deduce therefrom an application for the
envelope of the straight lines, such that the sum of the squares of the dis-
tances of one of them from itwo given points is constant 109

8916. (Bev. T. 0. Simmons, M.A.)— A point P being given in the
plane of a triangle ABC, it is Imown that, with P as focus, five oouics
can be drawn, four of them circumscribing the triangle, and one inscribed
in it. Show that for certain positions of P a sixth conic can be drawn,
also having P for focus, and touching the other five ; and find the locus
of P when this is possible. fThe Proposer suggests that conies with one
focus common should be called eofoeal, reserving the term confocal for
conies having both foci common.] 96

8920. (B. Enowles, B.A.) — From a point O {x, y) tangents are drawn
to meet the ellipse a V + *'^ = ^^^ ^ P ^^^ Q» * tangent parallel to the
chord PQ meets OP, OQ in^? and q respectively ; prove that

pqlTQ^ (aV+*^')* : aft + (flV + ^«^* 116

8924. (Captain H. Brocard.) — Former r^quation des paraboles tan-
gentes aux deux bissectrices de chaque angle du triangle et aux per-
pendiculaires aux cdt^s de cet angle en leurs milieux. Ces coniques
admettent, comme on sait, pour directrices les m^dianes et pour /oyers les
sommets A", B", C" du second triangle de Brocard 70

8927. (S. Tebay, B.A.) — If A^, A,, A3, A4 be the areas of the
faces of a tetrahedron, B the radius of the circumscribing sphere, and
Bj, B,, Bg, B4 the radii of spheres passing through the centre of B and
the angles of Ai, A3, A3, A4 ; the volume

^"*^'(l+i:*l^l) "»

8928. (Bev. T. B. Terry, M.A.) ~ Find the value of the definite

rdx
— sini?ic(acos2a: + dsin'rr), whenjE? > 2 157

8929. (B. Tucker, M.A.)— Find (1) the equation of the circle through
the images of the centroid of a triangle wil^ respect to the bisectors of
angles ; and show (2) that the simi of the squares of tangents to it from
the angles (taken once) is \{c^ + 6^ + c*) 88

8931. (Emile Vigari6.) — On donne deux points O et A et on con-
sid^ toutes les paraboles ayant O pour sommet et qui passent par A.
Trouver g^om^triquement le lieu (1) du point de concours des tangentes
en A et O ; (2) du point d'intersection de la normale en et de la tan-
gente en A (ces deux lieux sont tangents en O) ; (3) du point d'inter-
section de la tangente en O et de la normale en A ; (4) du point de con-
cours des normales en et A (ces deux lieux sont tangents en A). ... 81

8933. (B. Hanumanta Bau, M.A.) — If a body describe an ellipse
about a centre of force in the focus, the time of describing the arc between
the extremities of the latera recta on the same side of the major-axis is to
tlie|>erio<}|ctinieassin~V(tf) : *..,......? ,.•:•••••••••?• vv m H^

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CONTENTS.

XXV

8946. (W. J. 0. Sharp, M.A.)— Show that

-a, -a,
a?, -a,

—a, — fl
— a, — «

—a
b

=■ *(« + «)»,

122

^y d) bf ... .Ty

ft, A, b, ... ft,

where the determinant has it+ 1 rows

8948. (W. J. C. Sharp, M.A.)— Show that

fi— llsm^eV d$J ^ '

-(srn'e;^) (sinecose) 123

and

cosnO :

f»— 1! sin^j

del

8977. (Ptofessor Haughton, P.R.S.) — The form of the Terrestrial
Badiation Function has been proved to be A (0—60)** » 0, where A, 69,
n are unknown parameters, and 0, a are given by observation. The mean
monthly observations at Greenwich, ext^ding over thirty-six years, give

January, A (38-9-eo)" = 21-4 ; February, A (40-4 -60)* = 36-6 ;

March, A(42-8-eo)» - 66-9. Find e^, A, and n 69

8978. (Professor Sylvester, F.R.S.) — Show algebraically that, if
41, ft, e are the three sides of a triangle of reference, and A, B, C, the l^ree
perpendiculars on a variable line from the angles of that triangle, are
regarded as its inverse coordinates, then the equation to the two circular
points at infinity is

a«(A-B) (A-.C)+ft«(B-A)(B-0)+<?«(C-A) (C-B) - 0.... 167

8979. (Professor Brunei.) — Soient ABC un triangle, A,BiCi un autre
triangle d^uit du premier en menant par les sommets A, 6, C des droites
faisant avec les c5t4s du triangle et dans le mSme sens un angle ^. Du
triangle AiB^Ci Ton d^duit, de m§me, un triangle A3B2C2 et ainsi de
suite, toujours avec le mime angle ^. D6montrer que les points
A, A|, A3 ... sent sur trois groupes de spirales logarithmiques ayant pour
pdles les points de Brocard. Pour queUes valours de ^ le triangle d6riv^
A|B|C] est-il ^gal au triangle propose ? 109

8980. (Professor Steggall, M.A.) — In a rectangular hyperbola OY
is drawn at right angles to the tangent at P ; prove that, if YX produced
cuts SP in R, and Y be joined to X', then XR = YX', where X, X' are
the feet of the directrices 92

8981. (Professor Schoute.) — Given two conies; find (1) the locus of
the vertex of a right angle circumscribed to these curves ; and (2) con-
sider the particuls^ case of two homo-focal conies. [The problem in its
general form may be otherwise stated thus : — ^Find the locus of points
such that one of the tangents from it to a conic (2), together with one of
the tangents to a second conic (3), form with the two tengents to a third
eonic (1) a harmonic pencil.] 102

8985. (Professor Byomakesa Chakrayarti, M.A.) — ^A cylinder, weight
W, radius r, is placed on a rough horizontal plane ; a uniform plank,
weight P, len^h 2Q, is inclined at an angle 9 to |hd horizon, and res^

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XXVI CONTENTS.

with one end on the ground, the other on the cylinder (the pUnk being
at right angles to the axis of the cylinder) ; if ifr be the angle made with
the verticaf by the re-aotion of the ground on the cylinder, prove that

cot4^-oot4- + - .?-Beca 104

8994. (Charlotte A. Scott, B.Sc.) — A rectangular sheet of stiff paper,
whose length is to its breadth aa V2 is to 1, lies on a horizontal table
with its longer sides perpendicular to the edge and projecting over it.
The comers on the table are then doubled over symmetrically so that the
creases pass throuffh the middle point of the side joining the comers, and
make angles of 45 with it. The paper is then on the point of falling
over ; show that it had originally J| of its length on the table 112

8996. (Hahendra Nath Bay, M.A., LL.B.) — If x, y, z denote the
req>ective distances of any point in the plane of a given triangle ABC
from the angular poiots ; show that the following relation subsists among
them, («« + y« + 2* + a* + A«+c»)(a*a?« + iV + <j»a») ^2a*x^{ifi + a^

+ 2ftV(A34-y«) + 2ca«a(c2+«8)+ay«3 + Wt«*a+c««V+«'^<^. ••• 1^8

8998. (Rev. T. E. Terry, M.A.)— Solve the equation

^-.6a;^+(12«2-6)-^4(2a^-3)a:y = 108

ax* dx^ ax

9002. (E. M. Langley, M.A.) — If a circle and a Simson-line of one
of its points be both inverted with regard to that point, the two inverses
will have the same relation to each and to the given point that the
originals have 109

9003. (R. F. 'Davis, M.A.) — If upon each side of a triangle a pair of
points be taken so that the pairs on any two sides are concyclic, prove
that all three pairs are concyclic 93, 158

9008. (S. Roberts, M.A.) — Q-iven three circles Cj, Cj, Cs, determine
a circle cutting C| orthogonally, bisecting C^y and bisected by Cs ; and
show that in general there are two such circles which may coincide or be-
come imaginary 159

9009. (Rev. T. P. Kirkman, M.A., F.R.S.)— P and Q are a regular
20-edron and 8-edron. KLMN are 4-edray each on a base that covers a
face of P or of Q. Required the number of polyedra, of which none is
either the repetition or the reflected image of another, that can be made
by laying one or more of KLMN on as many faces of P or of Q, with an
account of the summits and symmetry of the constructed solids 125

9010. (J. J. Walker, F.R.S.) — Prove that the area contained by two
tangents to a central conic and the semi-diameters to points of contact is
equal to b'^x^+aY-a'^bK [See Quest. 3099, Vol. xiv., pp. 74, 76.]... 92

9011. (R. Lachlan, B.A.) — Show that the product of the three
normals drawn from any point on a conic is equal to the product of the
perpendiculars from the point on the asymptotes and the diameter of
curvature at the point 106

9013. (Emile Vigarie.)— Les projections orthogonales de denx points
inverses Mi, Mj {xf, y^, «') sur les trois cdt^s d'un triangle ABC, sent six
points d'une m^e circonf^rence dont I'^uation en coordonn^es nonnales

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CONTENTS. XXVll

est: (y£iinA-i-«a;smB + «y8mC) (d;'sinA+^BinB+j{'8inO)

C BinA

, yy" (a?' -fg" COB B)(g^ + a^ COB B) ^ gg'(a/-t>y^C08C) (y' + a/coflC) ')
sin B sin G 5*

[M. YioABi^ remarks that " on appelle pointi inverses en France ce
qae M. Casbt appelle uogonal eo^yugate points (Sequel to Euelid^ 1886,
p. 166)."] 1 93

9017. (A. Russell, B.A.)— If a?i, iCj, ..., a?„ «- ai, Oj, ..., fln, prove that

«! (ai-oj) ... (oi-aj «i(«j-«3)-(«j-«i)

9020. (F. Morley, B.A.>-^ABDO is a parallelogram ; is any point
on the line bisecting the angle A ; CO, BO meet BD, CD in E, F ; prove
that BE - OF 104

9024. (Professor Sylvester, F.R.S.) — For greater distinctness, the
name of Hyper-eartesianijioi to be confounded with a hyper-eartesie) being
given to that particular form of the bicircular quartic in which four con-
oyclic fod become collinear ; prove that, if four points are given in a
plane, the locus of the curve in space whose distances from any three of
them are subject to a given homogeneous linear relation is a curve of the
4th order. (This space curve may be termed a Hyper-cartesic.) 160

9026. (Professor Catalan)— Soit

fl(a+l)(fl + 2)...(a + c)±3(3 + l)(* + 2)...(3 + <j)-(a + 3 + tf)^(«, *).

(Le signe +, si est pair). (1) ^ (a, h) est un polyndme entier, k coeffi-
cients entiers ; (2) si a, 5 sent remplac^ par des nombres entiers, ^ (a, h)
devient un nombre entier ; (3) pour ces valours de a, by

(« + * + <?) ^ (fl, 3) = S» [1 . 2 . 3 ... (c + 1)] ;
(4) si, en outre, a + 3 + est premier, ^ (a, ^) i- 9)t [l . 2 . 3 ... (0 + 1)]. 101

9032. (For Enunciation see Question 8876.) 108

9060. (Maurice D*Ocagne.)— AB et MN 6tant deux diam^tres d'un
mdme cerole, si ime parallMe quelconque k AB coupe la corde NA en B',
la corde NB en A', et que les droites MA^ MB' rencontrent le cercle
respectivement auz points A'' et B", les droites AA", BB'' se coupent au
pied H de la perpendiculaire abaiss^ du point N sur la droite A'B'... 113

9063. (Mahendra Nath Bay, M.A., LL.B.)— If Oi, a^ az'"fhn~\ be
2»~1 positive numbers connected by the relation aia^,.,a2n-\^l\
show, by elementary algebra only, that the minimum value of

(l+fl,)(l+iij)(l + aJ(l + a4)...(l + ««H-i)ifl22«-i 161

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APPENDIX I.
Solationa of Qaestions 8886 and 9009, by Rev. T. P. Kirkman,
M.A., F.R.8 126

APPENDIX n.

MiflceUaneoos i^lutions and QuestionB, by R. F. Davis, M.A 128

APPENDIX ni.
Solations of some Old Questions, by W. J. Curran Shaip, M.A. ... 137

APPENDIX rV.

Note on tbe nse of Oommon Logarithms in the Numerical Solation
of Equations of the Higher Order, by Major-General P. O'ConneU 166

CORRIGENDA.

S 8'

p. 99, line 6 from bottom, for ~ read -^.

p. 161, line 7 from bottom, for ej (...) read ie (...).
p. 162, last line should be thus : —

,VD,«D,»D,»x4«(iDiD2D,)».
p. 163, line 10, readr »...»» a(l—tfcosu).

„ „ 2Z, for Tread J,

„ „ 24, readi.^anu^y,

„ „ 31, read 2e (sin Ui— sin u^) times,
p. 164, line 3, read (. . . cos m^), (. . . + sin u^) .

„ „ i, for T read J,

„ „ ll,n^...(...) )-'^("»-" | ) .

'l + C0S(tl2-lli)

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MATHEMATICS

PBOM

THE EDUCATIONAL TIMES.
WITH ADDITIONAL PAPERS AND SOLUTIONS.

2810. (By Professor Sylvester, F.R.S.)— Let Si be used as the
symbol of the sum of i-ary products ; required to prove that, if t is < >»,

(«i-«2)(«i-«a) •..(«!-«»)

« Si + i («i, 02, ... , «n) — Si+i (Oi, 02, ... , a«).

[For example, let » » 3, t == 2, then the theorem becomes
. Aa-a)(a-&)(a-y) . (b-'a)(b^0)(b-y) . . {e^a) (e-$) (e^y)
(a-^)(a_c) {b-a){b-c) ^ {c^a){e-b)

= abe—afiy, which is obviously true.]

Solution by W. J. Curran Sharp, M.A.
Assume (a?— ai)(j?— «2) ... («—«n) — (^p— aj) (a:— oj) ... (a?— On)

sPi(a;-«2) (a;-ffj) ... (a;-«n) + P2(^-«3)(^-«4) — (a?-an)(ir-«i)
+ ... + P„ {x-a{) (ar-fla) ... (ic-^n-i) ;
then, by putting x = «<, it appears that

-(fl<-ai)(a<-a2) ... (a<-o«) = Pi (»<-«<♦ i) ... («i--«i).
Hence we have P,=.-^ (^>-ai)(«»-^aV. (^>-^n)

and equating the coefficients of «— t— 1,

(-.l)<+i |Si+i (a„ ^2 ••• «»)-S»>i (ai, 02 ... On)}

= (-l)»2?Pi8.(«i,a2...«,-i,a<^i, a«) ;

therefore 2Si (a,, a, ... «„) (^i-«i^ (^i-o.) ... K^o. )
(ai-Ojj) («i-«a) ••• («2-«J

= Si+i («!, fla ... a»)-Si+i (oi, os ... a»).

VOL. XLVII.

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22

8816.

aurd, 8l6
triqoet, R
droiteSS'.

brocard.
cAUm da I
poiuUntoS

(Bj Captain H. Brocard.) — n, Of d^gnant lea points de Bro*
nuliea de CiGfy 8' le point correspondant k 8 par droitee sym^-
le point de Steiner ; d^montrer que ce point R se trouve snr la
(B est nne des intersections du cercle ABO avec le diam^tre per-
'e k Paxe d'homolo^e du triangle ABC et du premier triangle de
Ce point est aussi Tintersection commune des parallMes aux
remier triangle de Brocard, men^ par lea sommets corres-
u triangle ABC.)

BoUUiom by (1) D. Biddlb ; (2) Professor Db Lonoohamps.

1. In the triangle ABC, let n, ti' be the Brocard-points ; AECB, the
circumscribed circle, with centre H ; Hnn', the Brocard-circle ; AiBiGi,
the first triangle of Brocard ; GGj, the axis of homology' of the triangles
ABC and AiB^Ci ; EN, the diameter of the circumscribed circle perpen-
dicular to GG] ; 8, the mid-point of AA', and 8' its corresponding point.
It is required to prove that 8, 8', E are in one straight line. This will be
thecaseif (Ra-8i) : (8l-8a)=(R6-8i) : (Si-86), in which E«, Eft... are
the distances of the points from BC, AC respectively. Let A — area of ABC ;
a3+i« + «j2-m2; aJ^ + aV+i^cS - n* ; a< + A* + «* « i7<. Now 8., St, 8^,
S^ are well known, and are as follows : —

- 2A(g^ + ^«)(ag4-ga)

s* — •

3(«2+c«),

8'

Si'

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23

In order to find B^ and E^, we have H, the circum-oentre of ABO, with
H^ and H^ known, HE the radios of the circam-circle, and D, the centre
of homology of ABC and AjBiOi, with D^ and D^, known.

HB=^; H =-4^, «K-a2); H. = --?—:- M*""- 2*2);

Now B. = H,.(D,.HJHB/HD= ?M^^^^^zi^
B,-H,.(D,.HJHB,HD=-?-I^^=^^
therefore (Ra-^'J/CS^-Sa)

and (R6-Sa/(Si-S^

^ 2w< {33(j?<~n<) (gg + ^g) (^ + c3)~ [gg^^c^-^^ (m»-2y»)] (3n<+i?^) *}

On the first reduction, (R^-S;) / (S^- S^ : (B,-S;)/ (SJ- S^)
_ Pc8-g2(w8-2g8)](3w<+p^)--(jp<-n^) (g3+^ (g2+g8) .

2«4 (a2 + ^i) («2 + c5) « (3„4 +jp4) «2 (62 + ^2)
2f»4 (a2 + *2) (^2 + ^2). (3n4 + p*) ft3 (a2 + <j2)

^ 2»iM2g^~g2y~g«c24-2*V)-2p<(ggcg + a^yO .

f»4(2a4-a2^2_«2^ + 2*?c3).jp4(flf2^2 + a2^2) '

2n4 (2^~ gg^'- ^gg + 2g2g2) - 2jp< (y^g^ -f g^^^ ) ^0.2

«4(2*4«a2^S«^c2+'2g2c2)-i?4(*^g2 + fl52^) *" ' *

Wherefore, B, S', and S are in one straight line. Moreover, since
2»^ / (,1^— M*} was the common factor left out on the first reduction,
BS':S'S = 4«*:i;4«„4.

2. Othenoiae : — Le point S, milieu de la droite qui joint les points de
Brocabd, est repr^ente en coordonnees barycentriques par

a^(b^ + c^, ^(g2 + a2), <j8(g2 + *2).
Le point S', le point inverse, suivant la d^omination du Colonel Mathibu,
a done pour coordonnees (b^+(^-^, (c* + g^) - ^ (g^ + ^2) -1 .
Enfin, le point de Steiner B a pour coordonnees, comme I'on saity

(i2-<j2)-l, (g2-g2)-l, (g2-^2)-l.

On conclut de Ik que 1* Equation de BS' est ,

o fi y \

(i2_^-i^ {di^a^y\ (g«-^-> = 0,

(62 + C2)-1, (c2 + «2) -1, (g2 + J2) -1 |

ou, apr^ developpement, 2a (g^— ^c^j (^-^4) = (1),

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D'apr^ cda, on Toit que la droite A qui correspond k cette ^nation
pawe (1) par le point S, parce qae Ton a

(2) par le point J dont lee coordonn^ sont :

Cette demi^ propri^t^ a 6U signal^ d^j^ par H. Brocard an Gongr^ de
Bouen. [Voyez VAnnuairt de CAnociatUm Franfoise pour Vanancement de»
SeieneeSf 1883, p. 104.]

Un cinqui^me point, sitn^ snr A, apparait imm^diatement d'apr^
r Equation (i) ; nous voulons parler d*nn point U dont lee coordonneea
■ont (a*-W;-)-S (A<-«»a«)-i, (<?*-a«6«)-i.

Ce point IT se rencontre fr^nemment dans la gi6om4trie triangnlaire e^
il est certainement remarqu6 de tons cenx qui la connaissent un pen.

Iiorsqu^on cherche T^uation de la droite G, axe d'homologie du triangle
dereference ABC et du triangle de Bbocakd AjB|C|, on trouve

a/(a*-AV)+/3/(A*-c«aS) + 'y/(«*-a«*2)= 0.

Cette droite est harmoniquemmt tusoeUe au point XJo dont les coordonneea

■ont («*-*V), (d*-c»a2), (c«-a5*2);

TJo eet done, dans le sens que nous avons donn^ autrefois 2k ce mot, sens
qui est aujourd*hui gen^ralement adopte, croyons-nous, le rieiproqtie de TJ.

[Le terme harmoniquement assoeii a ete propose par nous pour repre-
•enter une droite fiy associee k un point donne M, dans les conditions
suivantes : — La droite AM rencontre BC en un certain point A' ; soit A'^
le point qui forme avec BA'C une ponctuelle harmonique ; les trois points
A'', B'', C sont Bitu6s sur une droite fi que nous appelons V harmonique
mtaooiie de MJ]

L' Equation (1) donne ainsi 5 points en ligne droite: le milieu de la
droite qui joint les points de Broca&d, le point inverse de celui-ci, le
point de Steinbr, le point de concours de la droite <^ui joint le centre du
cercle circonscrit k 1 orthocentre avec la droite qui joint le reciproque
du point de Lemoine k son inverse (le point J), et enfin le point Uq, ci-
desBus defini.

8801. (By Rev. T. P. Kirkman, P.R.S. Suggested by Quest.
8326.) — Upon the most sacred of the undiscovered Egyptian pyra-
mids, whose vertex F is at imequal distances from the angles of its scalene
base ABC, is cut in the face PAC an ascending flight of steps from A to
by in PC ; form h another flight mounts to in PB ; and a third leads up
from c to a, at a given altitude h in PA. These three flights are equi-
gradient. A second path from A mounts to A' in PC, and thence pro-
ceeding to c'in PB, ascends from c' to the above point a in PA. These
three flights also are equigradient. By khca Pharoah and his Grandees
of Church and State were wont to moimt to a platform at a for solemn
religious rites ; the admitted vulgar used only the path Kh'c'a. Required,
a plane projection of the pyramid and its six flights of stairs, and proof
from that projection that tiie first three are equigradient, and the second
three equigradient, up to the given altitude A.

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Solution by the Profobeb.

Let the distance *A of every point
h of PC in the projection from A he
greater than *B, and also than bG,
"We can with the ruler find through
any point b in PC the equigradient
path Ahca^ so that if Va — x and
rb=^ y^ X is given with y, one
value of X for every value of y, and
that X and y vanish together ; for, if
we take P for our A, the lines bD and
bO in the projection coincide, and P
is also our point c, t.^., x = Q follows
from y = 0. And evidently there
cannot he two equigradient descend-
ing paths from a to A, ahout the
pyramid, for no two lines ac and etc'
across the triangle APB can be equi-
gradient with the same Ab, That is, for every x there is one y, and one
only. To find x when y is given is very easy. To find y when x is given
is Mr. Biddlb' 8 problem in Quest. 8325, and to work out his solution of it
is not easy. We must content ourselves with the method of interpolation.
We can write y = ^a; = Ar + Bx^+ ... + Na;".

Next, taking any n points ^i, dj ... ^n on PC in our projection, we get with
the ruler the corresponding Oj, a^-.^an on PA, and write (P^i = yi,
Pill = iTi), yi = ipXij yg = ipX2 ..., yn =* ^Xn ; from which n equations we
obtain AB ... N in <px. Then, if Pa = X be our given distance from P,
Y = <^X is our required distance Pi, and when the point b is given, the
path Ahca is easily found by the ruler, and this solves our problem with
accuracy increasing without limit with n, the order of the parabola so
found in the face CPA, if PA and PC are our axes of x and y. This
parabola passes through (0, 0) and {x = PA, y = PC).

The second path ac'b'A. cannot be all through a descending path ; but
nothing prevents its being equigradient, if c'b' is an ascent, while a& and
b'K are downwards. Let b' in our projection be any point of PC. In CB
produced, at G make b^Qc « A'A, cutting PB in e'. In AB produced at
F make c'¥ = <j'G, and produce it to a' in AP. This a' may or not be the
a before found ; although in the figure, by a few trials, a and a* are made
to coincide. It is evident that for every b' in PC we can find an a'
in PA.

The lines on the pyramid whose projections are iV and b'K are the
sides of an isosceles tnangle drawn to the base in the planes CPB and
OPA from b' in both, and are therefore equigradient. For a like reason
e^a' and </b' in the planes BPA and CPB are equigradient, and the up and
down path Kb' (/a' is equigradient to the altitude of a'. Putting Pa'= u
and P^' = Vf we have the right, as above shown, to write

r = l|^ = A'M + B'tt2+...+N'M»».

Taking any n points rfj, d^.-.d^'in PC. we get easily the corresponding
points aiya2..,an in PA; and by n equations {Fd^ — v-^y P^i = Mi),
t?i = t|rt<i, Vg = 4^, &c., we obtain the coefficients of if/ij, so that, if our
given distance Pa = X, we get V = if/X, our required distance
Prf = Pd' :a V on PC, which gives by the ruler the requisite up and

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down path A&Va to the given altitude of a in PA. The two flights he
and bV most have a common step at some point p ; on which, before or
after the descent of the vulgar, a fitting carpet would be laid for the foot
of ascending majesty.

If the descent of the « vulgar " took place on the third fisice instead of
on the second, the two paths would not cross each other, and the gradient
of the ** conmion " steps might be lessened.

From the solution of Quest. 8843, it is evident that, with the concession
first above asked, the solution of this problem of two equigradient paths
about the pyramid from A to a in FA, is equally in our power, whatever
be the polvgonal base of it, and whether the vertex P is vertically over a
point inside or outside the polygon.

8823. (By Professor Cropton, P.R.S.) — ^Two discs of any form are
moveable in a plane round two fixed points A, B, respectively. Show that,
when they are in such positions that the length of an endless band en-
veloping both is a maximum or a minimum, the portions of the band
which form the common tangents will meet on AB if produced, and are
equally inclined to AB.

Solutions by (1) Professor Stbooall, M.A. ; (2) Professor Sibcom, M.A.

1. Let the discs be rotated through infinitesimal angles dd, d<f> about
A, B ; let Pu p^ ; qi, q^ be the perpendiculars from A, B on the straight
parts of the string. Then the amount of slack is easily seen by a diagram
to be (Pi—Pj) d$ + (qi—q^ dip. This must vanish for a minimum length,
and therefore Pi = pg, qi = jj, since de, dip axe independent, whence the
result immediately follows.

2. Otherwise : —Let the discs be smooth, and the band elastic. In the
position of equilibrium the length of the band will be a maximum or a
minimum. But then each disc is acted on by equal forces at its rim ; the
common direction of the resultants will therefore pass through A, B, and
be equally inclined to the directions of the forces. Hence the theorem.

5643. (By W. S . B. Woolhouse, F.R. A. S.)— Any two triangles being
given, the first may always be orthogonally projected into a triangle
similar to the second ; determine the magnitude of the projected triangle
geometrically by an easy construction with the ruler and compasses.

Solutions by the Profosbs.

1 . Let a triangle ABC, having the sides a, b, c, be orthogonally projected
vertically into the triangle A"B"0", having the sides a", 0", c", in a hori-

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Bontal plane. Then, considering the neutralizing result of changes of alti-
tude in passing first from B to C, then from C to A, and thence from A to
the starting point B, there arises the following relation,

^/(a2-«t"2)+ -v/(i2_y/2)+ ^(CS-^'S) =. (1).

This relation can suhsist only when the three terms of the expression are
either all real or all imaginary. In the latter case the imaginary char-
acter is removed hy writing it thus,

A/(a"2-aa^+ A/(y'2-^+ ^((j"s-ct) = (2).

But the form (1) implies that ABO is projected into A"B"C" ; and (2) im-
plies that ABC is projected /row A"B"C'

Hence, when (1) and (2) are cleared of radicals, the involved algehraical
relation amongst the sides a, b, c, and a", l/\ c", will he symmetrical and
identically the same whether the triangle ABO be projected into or pro-
jected from the triangle A"B"C".

According to the problem, A"B"0" is recnured q

to be similar to another given triangle. Upon ^^J^

AB construct a triangle ABC similar to this
given triangle ; and let the sides of the triangle
80 constructed be a', b\ e. The sought triangle >

A"B"C" being similar to this, we can assume /^

that a"-A;«', b"^kb% e'^kc, \

in which the common factor k remains to be
determined. The relation (1) now becomes

a/(«2-A;2^'2)+ ^/(^^A;2y2)+^(tfa-.A;8(j2) «0 (3).

If A, A' denote the respective areas of the two triangles ABO, ABO' ;
then 16A2 =2a2i2 + 2«»V + 2*2<j2-a<-M-<j* ") ...

16A'2 = 2a'2y'' + 2«'V + 2i'V-a'*-y*-tf*3 ^ ^*

To free (3) from surds, conceive the three terms to be the sides of a sup-
posed triangle ; then, according to the stated condition, the area of such
triangle is obviously zero. Hence, substituting, in (4),

= 2 (a»-.Aj2a'2)(*»-A;2d'2) + 2 («2- kH"^){i?--k^(^ + 2 {)f^-m'^){,^'kH^)

-(«2-;fc2^'2)2-.(^-A;2*'2)2- {^^^k^^\

which, arranged according to powers of A;, is, by (4),

-16A'2;fc4«2{(*2 + <j3-a2)a'2+(a3+c2-i2)^'2+(«S + ^.^c2}^2+16A2,

or i^. (^ + ^~a2)^/2^(^s^,2^^2)^/2^(a2^^2,^)e^ ^^^ -^!,-0-(f^).

But, joining 00' and denoting this line by 7, then

2cV ^ 2c2 {*2 + ^'2_2W'cos (A=fA')},
according as 0, 0' are on the same or on opposite sides of the base AB.
That is, 2^272 « 2tf2 (^2 ^ y^ _ 4^2^^' (cos A cos A' ± sin A sin A')

= 2c2 (i2 + i'2)_(i2 + c2-a2) (i'2+^_a'2) :p 16aA'

. (^2+c2_rt2)«/2+(a2 + c2-^)i'2+(flp2 + i2_^g2:p 16^A'.

Denoting OE, 00', the two values of 7, by 71, 72 respectively, we have
therefore 62(712 + 722) « (^ + c2-a2)a'2+ (a2+(,2_^2) y2 + («2 + i2_c2)c2,

c^M-yi) - 16AA'.

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Hence (6) may be stated thus,
the solution of which is

8A'3

266A'^

.0,

therefore

**-'^^-^'=^<-^=i^^(-=^-^'=

* = 4l. (---«) -'-^'

and the perpendicular of the sought triangle iap" = kp' = ^ (Ti^Ts)-
We have therefore the following
Construction. — Let ABC be the

first of the given triangles ; and on the

base AB describe ABC' similar to the

other given triangle. Then the pro-
jection of ABC is required to be similar

to ABC, and is constricted as follows :
From G* demit the perpendicular

CD', and in continuation of it make

D'E equal to CD'. Draw a straight

line through the vertices C and C,

and on it take CF equal to CE ; then on CE take CH equal to CF ; and

lastly bisect CH by the perpendicular line ab, thereby cutting off from

ABC the required projected triangle.

It may be observed that C and £ need not be joined, and also that the

problem admits of three essentially different projections, for BCA, CAB

may each in like manner be projected into tnangles respectively similar

to ABC.

2. OtherwisAy Oeometrieally : — Take the
plane of the paper for the plane of pro-
jection. Let ab be the projection of
AB, the triangle being turned round the
perpendicular CD as an axis so as to
bring the point A vertically above a and
B vertically below b. Then, denoting by
c the angle of inclination of the side AB
with the plane of projection, we shall
have Da = DA cos e, D* = DB cos €,
ah = AB cos c.
To completely generalize the arrangement, now conceive the triangle
ABC in its new position to turn about the side AB as an axis. The
vertex C will describe a circle having its plane inclined to the plane of
projection at an angle equal to the complement of €. This circle will
therefore project into an ellipse the semi -diameters of which are a — CD,
j3 = CD sin € ; and the excentricity is hence e — cos c.

The periphery of the ellipse is the locus of the projection of C. Let e
be the actual projection of C, and abc will then be the resulting projection
of the triangle ABC from the position it has finally acquired. As this
projection is required to be similar to the triangle ABC, the projected
vertex c must fall on the straight line DC, making Dc = DC cos € =« . DC.
Let/,/ be the foci of the ellipse ; draw C'H perpendicular to AB, making

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D'H = CD' ; ed perpendicular to CE ; and other lines joining various
points as represented in the diagram. Then D/ = « . DC, D/*= e . DE ;
therefore cf, cf are respectively parallel to C'C, CE.

By Conies, De?(- x) = ^£^ ^ \ (CE-C'C) - \ (CE'-CCO,

which is the perpendicular of the sought triangle abe, and determines its
magnitude by a simple and easy construction.
3. From the preceding article we have

Drf
'' CD''

cosE =

a be

Bd . ab Bd

ABC CD . AB CD

= «^ cos € =

Drf«

CD . CD''

which determine the respective inclinations of the side AB and of the
plane of the tiiangle ABC with the plane of projection.

4. Another Geometrical Solution
may be obtained thus: — Let ABC
be the triangle to be projected
from, and or the base A B let ABC
be constructed similar to the tii-
an^le into which ABC is to be pro>
jected. The plane of the paper
and of the triangle ABC being
taken as the plane of projection,
the position of the triangle in space
may be assumed to be suitably
located so as to cause the projection
of AB to fall on the line AB, and
such that the perpendicular of the
projected triangle shall fall on C'jy,

In effect, as regards the projected triangle, the point D is removed to D',
the segments DA, DB are proportionally shortened by projection, and
the perpendicular is the same as CD' when diminished in a like proportion.
Now, if we retain the various positions indicated and in either case con-
ceive AB to be an indefinite line, and assume in it any two points A' and
B', the distances DA', DB' will obviously be reduced by projection in
exactly the same ratio. It hence follows that, wherever the points A', B'
are taken in AB, the triangle A'B'C will be projected into a triangle
similar to A'B'C. Taking advantage of this circumstance, such points
may be conveniently chosen so as to make the angles at C and C both right
angles ; thus, join CC, bisect it with the perpendicular MO meeting AB
in O, and with O as a centre describe the circle CC'A'B'.

A consideration of what has been advanced establishes the general pro-
position that any projection of the figure which makes A'B'C similar to
A'B'C will, at the same time, make the triangle ABC similar to ABC as
required. As the angle C of the triangle A'B'C after projection must
remain a right angle, the most simple and direct mode of projection is
arrived at by drawing B'A" meeting A'C, and making the angle A"B'G
equal to A'B'C. For, if A'B'C merely turn round B'C as a horizontal
axis until A" becomes the projection of A', then the projection of A'B'O
will be A"B'C, which is obviously similar to A'B'C.

Produce the perpendiculars CD, CD' to meet the circle again in E, E',
and join B'E' ; also join CE' intersecting A"B' in D" and bisect it in
m. The angle E'CB' is equal to E'C'B' or the complement of A'B'C or
the complement A"B'C. Therefore CE' is perpendicular to A"B' and

VOL. XLVII. D

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CD" 18 fhe peipendiciilar of the projected triaagle. By the georoetiy
ofth diagnun,

C«i-iCE', mir-icxr;
therefore CD"- i (CE'-CCr), EOX'^ i{CE'+C(y).

These are wholly independent of the position of the points A, B in the
line AB. The former is the perpendicular of the projected triaogle ; the
latter is the perpendicnlar of the triangle similar to ABC from which
ABO may he projected.

Note$. — If E denote the inclination of the plane of the triangle with
the horizontal plane, then

cosE A"c ^ ciy^ ^ CE'-CCy

"a'C ED" CE' + CC

The line CA' bisects the angle CCE' ; the line CB' is perpendicular to it,
and therefore bisects the exterior angle G of the triangle CCE'. It has
been shown that the latter of these lines determines the direction of the
intersection of the projecting plane with the horizontal plane of pro-
jection.

[An attempt at a solution, the only one heretofore leceiTed, is g^ven on
p. 24 of Vol. XXX.]

8793. (By Professor Neubbbo.)— Soient AB, CD deux droites, qui
Be coupent en 0. Si Ton fait toumer AB autour de 0, le point double
de deux figures semblables construites sur AB et CD decrit une circon<-
f^nce de cerele.

Solution hy Professors Schoutb, Beyens, and others.

For any position of AB the centre of similitude of the two similar
fig^es on AB and Gi) is evidently the second point of intersection P of
the two circles A OB, COD. Therefore transformation of the problem by
reciprocal radios with the centre O brings it in the following form : —
Two lines A'B', CD' intersect at O ; to find the locus of the point P'
common to A'C' and B'D', when A'B' rotates about 0. Considering
A'C'P' as a transversal of the triangle OBO)', we find

(A'B' / A'O) . (CO / C'DO . (Fiy / P'BO = 1,
and as A'B' / A'O and CO / CD' are constant, the same may be said
of P'D'/ P'B'. Thus the locus of P' is a circle, for which D' is a centre
of similitude with respect to the circle described by B', and the locus of P
is also a cirele.

8639. (By Professor Wolstenholme, M. A., Sc.D.) — The distances of
a point O (within a given triangle ABC) from the angular points are
a?, y, t respectively; prove that the volume of a tetrahedron in which
a, bf e are the lengths of three conterminous edges, and Xy y, z the lengths
of the edges respectively opposite, is ti {(<**— ^) (*'— y') (<^— «^}*-

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Solution by Professor Nbubb&o ; Professor Gbnbsb, M. A. ; and others,
Soit y le volume du t^tra^dre en question ; on salt que

Soit y le volume d'un t^tra^dre dont x, y, z sont les ardtes d*un angle
Bolide, et a, b, e les ardtes respectivement oppos^ ; on aura

144V'3 - 2a?«8(A2 + y' + tf2+aS«a^--a:2)-a2^JVj3--aV82-W5^2_^>p3yi,
D*oii 144 {V»- V*) = (««-«») (y»-y2) (c2-,s).

Dans la question propose, on a V » 0.

8713. (By Professor Steooall, M.A.) — Show that the solution in
rational quantities of the equation j^^+y^' +2^ — u', is ^ » ^ (a^ + d^— ^),
y » 2kae, s -> 2kbc, u ^ k (a^ + b^ + d^).

Solution by W. J. Babton, M.A. ; Sa&ah Marks ; and others.
Let y/z =s afb ; substituting for y, we get (a^+i') «2 _ ^2(,,i_j,jj^
This equation is satisfied by u—x — k<P, u + x '^ k{a^ + b^ ; which g^vo
u =» iAr(a2 + ^ + cS), x = iA;{a2+^-(j«), z = kbe, y - W.

8804. (By S. Tbbay, B.A.) — If (a, b, e) be conterminous edges of a
tetrahedron, {x, y, z) the respective opposites, (A, X ; B, Y ; C, Z) cor-
responding dihechtd angles, and Y the volume ; show that (1)

_h

sin A sin X sin B sin Y sin C sin Z '
also (2), if the areas of the four faces are equal, then

a-af, d-y, <?-«, A = X, B-Y, C = Z.

V = A {2(ft2+c«-a^ (c3 + a2-.^ («2 + J8-.«jS)}*,

(J— <j)sinA+(o— a)sinB + (a— i)sinO «■ 0.

Solution by Professor Wolstbnholme, M.A., Sc.D.
(1) If ai, jSj, 7i be the plane angles at the point where a, b, c meet, X
the dihedral angle opposite «, cosX = cos a,- cos ^, cos 71

Bmi8iSm7i
, a* X » (^ ~ CQ"^ «i •" cos^ i^i — cos^i + 2 cos ai cos jSi cos 7t)*

sin jSi sin 71
but volume (V) = J abe (1 - cos^ ai — cos^ fi^ - cos^ 71 + 2 cos oi cos jSj cos 71)* ;

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also, if E), B,, B3, R4 be the radii of the circles circumscribing the tri-
angular faces, sin /3i - ^, sin 71 = ^, &c.,

whence sinX - ?iYKs?a • and -^ ^*"^^ ;

and, similarly, -^ = ^^""^^ - ;

or

hy __ cz _ (abcxyzy

sin A sin X sin B sin Y sin C sin Z 676V2R1R2RJR4
If Ai, A3, A^ A4 denote the areas of the faces,

T» ryz T» xbc t» ayo -p abz .
^=4T.' ^-4^' ^-4A,' ^^'TI,'

by et _4 Aj^AgA^^

sin A sin X sin B sin T sinCsinZ 9 V^

(2) In an equifacial tetrahedron, a ^ Xj b ^ y, c==g; A»X, B — Y,
C n Z ; and oi + i8| + 71 = ir, so that the volume

« J abe (1 — cos' ai - cos' jSi—cos' 71 + 2 cos ai cos fii cos 71)*

B ^ a^0 (cos ai cos jSi cos 71)^

The equations of (1) become -r^r = -^Ku " -^"Tri * 4 ^^
^ ' sin' A sin'B sin'C 9 V^

*

where A is the area of any face ; whence

a h ^ _c_ ^ 2_A^

sinA ^sinB "" sinC 3 V*
and (*—<') sin A + {e—a) sin B + (a— A) sin C — 0.

In such a tetrahedron, the radius of the circumscribed sphere is
4[2(a' + *' + c')]*, a result I sent to the Educational Times long ago. I
cannot discover any single length connected with the tetrahedron whose
square is equal to o^ / sin A sin X, and I do not know that we have any
right to expect it. I have much admiration of this question.

The equations

( ax ^ by _^ ez _ _4^ A|AgA3A4 \

sin A sin X sinBsinY sinCsinZ 9 V /

are strictly analogous to the equations .^ , = -r— :;r^ = -r^ ** ^,
•^ * ^ sinA sinB sinO 4S

and, writing the numerical coefficients (|)', (|)', it would suggest that the

general form in geometry of n dimensions is {(»— 1) /»}'. This I leave

to those who are experienced in such high matters.

8696. (By E. ViGARife.) — ^Dans un cercle donn6, par un point P pris
BUT la drconference on m^ne trois cordes PA, PB, PC ; sur chacune

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d'elles comme diam^tre on decrit une circonf6rence ; d^montrer que les
trois points de rencontre sont en ligne droit.

Solution by H. F. Davis, M.A. ; H. Brakspeab ; and others.

For, if a, b, c be the feet of the perpendiculars from P upon the sides of
the triangle ABC, the circle described on FA as diameter passes through
b, Q ; while a, b, e are collinear.

8647. (By R. W. D. Christie, M.A.)— Prove that 2 + S = 2«', if
«= 13+23 + ... +n3, S» P + 26 + ...W*, 2= l7 + 27+...f|7.

Solution by Rev. D. Thomas, M.A. ; Professor Beyens ; and others.
Assuming the relation to be true for n, it will be true f orn + 1 if
5 + S + («+ 1)'^+(« + 1)7 = 2 [«3 + 2« (»+ 1)3+ («+ 1)«],

or if 4« = (»+ 1)4+ (« + 1)2-2 (« + 1)8 = [n (n + l)p,

and this is the case, therefore, &c.

8726. (By S. Roberts, M.A. Analogous to Quest. 3068.)— Show that^
if some four of the roots of a quintic form an harmonic system, then

J»-27.32.JK+ 2»8.38.L = 0,

where J, K, L are the three fundamental invariants of the orders 4,
8, 12 [see Salmon's Migher Alyebra, 3rd ed., p. 211].

Solution by the Proposer.

In the Proceedings of the London Mathematical Society, Vol. xiv., Prof.
M. J. M. Hill gave the equation determining the anharmonic ratios of
the roots of a quintic. This is of the order 24 in the coefficients, and, sub-
stituting therein any value for the variable, we get the condition that
some four of the roots of the quintic may form an anharmonic ratio having
that value.

But in our particular case the condition is of the reduced order 12, and
is therefore of the form J« + A;JK + /L «= 0. Taking the form

z {ax* + lOftr^ + be)

(which we can do, since it only implies the evanescence of Hermitb's
Skew Invariant of the order 18), we have for the T of the quartic factor

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e (M~{^t And, sabstitatiiig {«* for «# in
J - 16e(M« + 3«*),

giving (9-512)> -i- ib 2432 -i- 2/ - 0. Bat abo, if a » 0, the roots of

form an harmonic system, so that 4d» +* 18 . 48+4/ = 0. The required
invariant condition is therefore J» - i (48)« JK + 2 . 48» L = 0.

^f"".

8708. (By Mauricb D'Ocaokb.)—
Si, dans le quadrilat^re ABCD, les angles
opix)8e8 B et D sent droits, la droite qui
joint les pieds des perpendicolaires
abaisseesdn sommet sur les bissectrices
int^rieore et exterieure de I'angle A
passe par le milieu de la diagonale BD.

SoUUion hy R. F. Davis, M.A. ;
G. G. Mon&iCB, M.A. ; and others.

Bisect AC, BD in O and E respectively ;
then O is the circnmcentre of the quadri-
lateral. If H, K be the extremities of
the diameter OE of this circle, AH and
AK are the bisectors of the angle A, and
the angles CHA, GKA are right angles.

8769. (By J. Brill, M.A.) — An ellipse is inscribed in the triangle
PQE so as to touch the sides QR, RP, PQ at P', Q', R' respectively ;
prove that, if G be the centre of the ellipse,

AQCR : aQ'CR' - aRCP : aR'CP' - aPCQ : aP'CQ'.

Solution hy Isabel Maddison ; Belle Easton ; and others.

Let x^/a^-i-y* I i^ = I be the equation of the ellipse; (ar^y^), {x^y^,
and (x^ yj the coordinates of P', Q', R'. Then the coordinates of P are
«^ (^3-^2) / Xi» ^ (^2-^3) / Xi> where Xi == x^^—x^^, Xg = x^y-x^y^
X3 « Xii/2—^\i <^<^ ^6 coordinates of Q and R are found by substituting
in cyclic order. After reducing, we find

AQCR : aQ'CR' = (Xi + X2 + X3) / X1X2X3, which is symmetrical ;
hence AQCR : AQ'CR' = aRCP : aR'CF = aPCQ : aFCQ'.

t

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8749. (By F. PuBSBB, M. A.)— Find (1) the cubic locus of the centre
of a conic passing through lie three vertices A, B, C of a tri-
angle, and such that the three normals at these vertices are concurrent ;
and prove that (2) this cubic passes through the in-eentre I, the three ex-
centres la, Ibi Ic, the circum-centre O, the orthocentre H, the centroid Or,
and the symmedian G\ and cuts the three sides normallv at their middle
points L, M, N ; also (3) tangents at (I, !«, 1*, Ic), (A, B, C, G), (L, M,
N", G') are respectively concurrent in G, G', ; and (4) the lines joining
L, M, N meet the curve again in their intersections with the respective
perpendiculars from the vertices.

Solution by W. J. Greenstebet, B.A. ; Professor Beyens ; and others.

(1) Let the chord joining the points a, /3 make an angle with diameter

which bisects it, then we find cotfl — ——r (sina + /3), therefore normals

*2ao

are concurrent if cot 0i + cot $2 + cot ^3 — 0, where sides of A are at angles $1,

62, 63 to diameters bisecting them. Now let a, /3, 7 be perpendiculars

from a point on the sides of the triangle, then

^j _ fl8inB-yrinC + arin(B-0) ^ . g^ ^ ^^

2o Sin B Sin C

/BsinB— 78inC4-asin(B— C) ^ a

• • r : ;;r — z — ^z — + ^ v,

2a Bin B sin

*" ri^ ('*'->') +S^('^-"^ + -<*'

Bin A sin x>

the cubic locus required.

(2) This obviously passes through the centroid, the ex-centres, the in-
and circum-centres, and symmedian point. Put a=fi — y - 0, we find it
cuts the sides at the same points as the lines a sin A = /3 sin B * 7 sin C,
viz., the mid-points. [The other parts of Quest, not solved.]

8643. (By Professor Genese, M.A.) — Four complanar forces
P], P2, &c. are in equilibrium. Their lines of action (omitting one at a
time) determine four triangles whose areas are Ai, A3, &c., and circum-

Ai . A^^

radii Bi, R2, &o. Prove that Pi : Pj, &c. —

&c.

Solution by Aspahaous ; D. Edwardbs ; and others.

Let the forces P], P2, P3 act along the sides
BC, CA, AB of the triangle ABO, and P4
along the straight line A'B'C^ The con-
ditions of equilibrium will be satisfied if the
moments of all the forces about A, B, C
vanish. Taking moments about A, we get

AARC
BO

"^1-

AAB^'
• B'O' •

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But AAnC = A4, BC = 2R4 8in A, AAB'C= A„ and B'C'= 2B, sin A,
hence P, : P, = ^ : ^^

Similarly, by taking moments about B and 0,

wegetP,:P,:P,-|2:^:^.

Jtij rvj £(4

8724. (By R. TucKBR, M.A.) — If a constant line AB moves with ends
on OX, OY, two rectangular axes, and on it a semicircle is d« scribtd,
then locus of mid-point of arc is one of two straight lines. Find locus of
any other fixed point on the arc. [If C is the mid-poiut of AB, and P the
mid-point of the arc, then the question may be enunciated ior a bar rigidly
connected with the bar AB.]

Solution by C. E. Williams, M.A. ; Professor
Mathews, M.A. ; and others,

A circle passes round AOBP, therefore angle
POA = PBA = ^iTf therefore P must lie on one
of the bisectors of OX, OY. If P be any fixed
point on the arc, we still have

POA =s PBA = constant,
and therefore P lies on oHe of two straight lines.

8646. (By E. VioARife.) — On donne deux points A et B sur une
circonf^renie de centre ; trouver sur cette circonference un troisidme
point P, tel que les droites PA, PB coupent un diam^tre fixe CC en des
points M, N, tels qiie I'on ait OM = ON.

Solution by Professor Mathews, M.A. ; Professor Schoijtb ; and others.

Produce AO to meet given
circle in A' ; through B, A'
draw a circle BNA'N' so that
th« segment BNA' may be
similar to BA'A; produce NB
to P, etc. , as in figure. Then
Z A'NB = supplement of BPA,
hence NA' is parallel to PAM ;
and, since OA'=OA, ON = OM.
Similarly, if BN' be produced
to P' and P'A meet CC in M', M'= ON'.

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2866. (By Professor Sylvestbb, F.R.S.)— Given the simultanebiui
equationa Vxf i = v,+ (x^—x) fx-i, ««+i = Ug+{x^—sc) tt,_i + (2a: + 1) r,+i,
prove that the general solution is of the form Vg =» \ax -t- ii$x,

where A, ft, u, x are arbitrary constants, and determine the values of

Solution by Professor Sebastian Sibcoic, M.A.

The values of a*, fix are given in Quest. 2845 [Vol. xiii., p. 60], and
may be written «, = n {a;-i-(-)*i}, /S, = n{a;-i+ (-)*^}.

The solution of the equation in Ux is easily verified.

Writing a? (a; +1) in the form {a;-i-(-l)*i} {a?-l + i-(-)«-i i},
and assuming «« = \yx a*, the equation for 7, will be

= (2x + l){a:+l-i-(-)«*ii},
whence yx must be of the form a* + ( — I)* c. Substituting, and equating
the terms that do, and those that do not, contain ( — )* separately,

(» + i)«r,.i-tf,-(a;~i)a,., = (2a:+l)(a; + i) (1),

««+i— «;c-l — 4c « 2x-¥ 1 (2).

EMminating a^^i, we have iix-dx-i — 4c (a? + i) and ax^,l~ax = 4c(» + f),
whence an*. 1— «» = 8c« + 8c, which agrees with (2) if c « J ; then
««-«x-i=-a; + i, and a;p = it-*^ + 2a;)+y, 7, - i {«*+2j?+ (-)*i} +y.
Similarly for «, = /ia5, /B*.

8789. (By Professor Gullet, M.A.)— Find by a geometrical construc-
tion the equation SS' = P* of the pair of tangents from T {x\ y') to a
circle S s x'^ + i^^-z^ = 0, whose centre is O ; and show that, if perpen-
diculars £L, KM, RK be drawn to the tangents and their chord of con-
tact from a point R not on the circumference, RL . RM differs from RK*
by TO- cos2 » cos* 0, where 2« and 20 are tiie angles subtended by the
circle at T and R respectively.

Solution by Professors Wolstbnholme, Betens, and others.

If TP, TQ be tangents from the point (XY) to the circle x^ + y^^ r*,
and (x, y) be any point P on either tangent (say TQ),
PQ« = 2:2 + y2_r2, TQ2 = X2+ Y^-j-s, and PQ2 ; TQ2 = PM^ : TN2,
where PM, TN are perpendiculars on QQ'. But the equation of QQ'
being xX.-\-yY^r\ PM : QN « a:X + yY-r2 : X- + Y^-gS^
or «2 + y2_^2 : x^i- Y2-r2 = (a-X + yY-r»)2 ; (X^ + Y^-r*),
giving the required equation

(a,-2 + y2_,.2)(X2 + Y«-r2) = (j-X + yY-T^'.

[Almost exactly the same proof applies to the central conies.]

vol. xLvn. B

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8833. (By Professor Nbvbbrg.) — ^Un angle de grandeur constante
toume autour de son sommet A, ses cdtes rencontrent line droite donn^
XY aux points B et 0. Trouver (1) les enveloppes de la m^iane BB' et
de la sym^diane BH ; (2) les points oil ces droites touchent leurs enve-
loppes.

Solutiont hy Professors (1) Schoute, (2) di Lonochamps.

1. As the points B,
B' (Fig. 1) determine
on the parallels XY, xy
two homographic divi-
sions, the line BB' en-
velopes a conic that
touches the parallels
and has A for a focus.
When BiACi and
B2AO2 are the positions
of the angle BAG, in
which ACi and BjA
are parallel to XY, Bj
and B2' are the points
where XY and xy are
touched hy this conic.
And by applying the
ry of Briancb

theory

x—B c y

Fig. 2.

hexagon to the triangle YBB'y circumscribed to the conic, the point E,
where BB' touches it, is easily found.

When BE (Fig. 2) is a symmedian of triangle BAC, the angles ABE and
B'BC are equal, and the triangles A HE and ADB are similar. Then the
corresponding medians AF and AB' of these triangles are isogonal with
respect to AB and AE. Thus Z FAE ^ L BAC = constant. This proves
that all the lines BE pass through the fixed point F. And as in the same
manner the symmedians with respect to C pass through another fixed point
G of xy^ and there is evidently a correspondence (1, 1) between these two
pencils F and G, the locus of the point of Lemoine K of the triangle ABC
will be a conic through F, G, touching AF, AG. Only when the rotat-
ing angle BAC is right the points F and G coincide, and all the triangles
BAC have the same point of Lemoine. (N.B. — Symediane is introduced
by M. D'Ocagne for antim^diane.)

2. Otherwise : — Gdneralisons la question en supposant quo les cot^s de
Tangle A rencontrent : Tun, une droite fixe A, au point B ; I'autre, une
seconde droite fixe A', au point C. Soit B'le milieu de AC ; le lieu d^crit
par B' est une droite i parall^le a A', et les points B, B' d^crivent sur A et
sur H deux divisions homographiques. On sait que, dans les conditions,
BB' enveloppe une conique tangente li A et i 8 en de certains points
M, «», que Ton sail determiner.

Si Ton iniagine le point R commun ^ BB' et ^ Mw, la polaire de R coupe
BB' au point cherche p. II est visible que p est le conjugue harmonique
de R sur le segment BB'.

Remarque,— &\, dans P^nonce donne, il fautsupposer que BH designe la
hauteur du triangle ABC ; alors, en observant que BH fait avec BA un
angle constant, on reconnalt que BH enveloppe une parabole. Le point
de contact de BH avec son enveloppe se trouve, comme Ton sait, sur la
circonference qui passe par A et par B tangentiellement k A.

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8719. (By Professor Wolstbnholmb, M.A., Sc.D.)— The lengths of
the edg^s OA, 0B» OC of a tetrahedron OABC are denoted by a, b, e, those
of the respectively opposite edges BO, CA, AB by x,y,s; and the dihedral
angles opposite to these by A, B, ; X, Y, Z ; prove that, if Y be the

8V

volume expressed in terms of a, d, e, x, y, z, -— -• ^ax cot A, &c. ;

oa

and thence that (1) when a^ x; b, y, and e + z are ffiven, Y will be a

maximum when C = Z ; (2) when a, x; b, y, and e—z are given, Y will

be a maximum when C + Z = 180°; (3) when a, x ; d + y ; tf + ^are given,

Y will be a maximum when B = X and Q ^2i (t.^., when i = a;, c ■■ «) ;

(4) when a, « ; b + y^ e—z are given, Y wiU be a mairimum when B ■> X,

C + Z = 180° ; and (6) investigate if Y can have a maximum when a, x ;

b—y, e—z are given, or when a— », *— y, e—x are given; the given

quantities being supposed always real and finite.

In case (4), prove also that 180° is the maximum value of C+Z for

variations in 6, y (subject to the conditions stated), and that zero is the

minimum value of B—Y for variations in e, z.

Solution by Professors Mathbws, Bbtens, and others.

Draw AD perpendicular to BO,
and AE perpendicular to OBC.

Let the vectors DA, OB, 00 be
represented by a, /3, 7.

Then, if A be the volume OABO,
6a = Sai87.

In order to find ~ , turn the tri-
oa
angle ABO through a small angle
about BO, so that A comes to A';
this changes a, and leaves d, «, x, y, s
unaltered. AA' is ultimately per-
pendicular to ABO, so that we may put

da « vector AA' « h (Y/Sy + Y70 + Yo^) (1),

where A is a small scalar quantity.

Hptic«6?^ „ d^aBy ^ Bfiyda ^ ^ Sj3y (V0y + yya + Ya$) ,5.

«a dTa Soda Safiy ^^'

Ta
by (1), observing that Sa (V$y + Y7a + Ya$) = 80/87.
Now S/37 (Yi87 + Y70 + Yo/3) = S . Yi87 (Vfiy + Y7a + Ya0)

= 40B0 . ABO . cos A = 20BO . BO .AD cos A
» 20B0 . BO . AEcot A =. 6Aa;cot A,
hence (2) becomes

8A^^ 6A^cotA ^^^^^ therefore ?^-Jaa: cot A.
8a 6A ' 9a ^

If A is to be made a maximum or a minimum by the variation of a, b,

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or a« (oot A. itf + cot X.dir) + ... + ... =o (8).

Case (1) gives da '^ dx ^ db ^ dy =: 0, de-\-dz » 0, so that (3) reduces
to et (cot 0— cot Z)»0, e = 0, z^O make A « a minimum; hence
oot B cotZ, or ■■ Z makes A a maximum;

(2), (3), and (4) may be similarly treated.

8840. (By W. 8. M'Oay, M.A.)— Prove that (1) the locus of the
mean centre of the four points, in which a line of given direction meets
the faces of a tetrahedron, is a plane (diametral plane) ; and (2) if A, B,
C, D he the areas of the faces of the tetrahedron, all the diametral planes
envelope the quartic (Aa?)* + (By/ + (Cz)* + (Du;)* = 0.

Solution by Professor Schoxjtb.

1. When a, iS, 7, 5 represent the four g^ven planes, L the point at in-
finity indicating the direction of the lines, and ir, ^, if the planes that are
separated harmonically from L by the couples of planes a and /3, y and S,
IT and % respectively, then it is evident that in the determination of the
locus in question the planes a and /3 may he replaced by ir counted twice,
the planes y and 8 by ^ counted twice, and the planes « and ^ each of
them counted twice by ^ counted four times. This proves that ^ is the
locus.

2. If F represents a vertex of the given tetrahedron, Q the point of in-
tersection of the line LP with the opposite face of the tetrahedron, and R
the point on PQ for which PR = iPQ, the locus of the mean centre
passes through R. When a, &, c^ d are the coordinates of F, which, as P
IS at infinity, satisfy the condition Aa + B^ + C(; + D<^ — 0, the locus
contains the four points

(—3a, bj Cf d)f (a, —3*, e, rf), (a, b, —Ze, d), {a, b, e, -3<Q.

Therefore its equation is — + ^+— + — = 0.
abed

By differentiation the coordinates of the point, where this plane touches

its envelope, are found to be proportional to

Aa2, Bi^ Ccj2, Dd^;

which, in due combination with the condition Aa + Bd + O + Dd ^ 0,

proves that the equation of the envelope is

(Ar)* + (By)* + (Cz)2 + (Dtr)« « 0.
Thus this envelope is the Roman surface of Steiner, the quadric surface of
which the three Hues, that join the centres of the opposite edges of the
given tetrahedron, are double edges.

[If a system of parallel lines pierce the faces of the tetradedron at
angles a, /3, y, 8, the equation of the corresponding diametral plane

is obviously -: — + -:^— + -: — + - — - = ; the projection of the faces

sma sin/3 sm7 sin 8
gives A8ina + B Bini3 + C8in7 + Dsin8s 0, whence the result.]

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8799. (By Professor Hudson, M. A.) — Find the envelope of the
straight line au^^f (o) cos (6 — a) +/ (o) sin (0 — a) .

Solution by Aspabaoub ; Professor Matz ; and others.

The equation au —/(a) cos {0 — a) ■¥/* (a) sin (0— o) represents the tan-
gent line to the curve au = f{d) at the point where = a, and of course
the envelope of the straight line is the curve au =f(B). By the usual
rule, we get = [/" (o) +/(o)] sin {B-a), Hence, unless /" (o) +/(a) = 0,
we shall have for the point of contact with the envelope 6f = o, giving for
the equation of the envelope au =f(d). In the particular case when

/" (o) +/(a) = 0, /(o) = A cos a + B sin a, f (a) « — A sin o + B cos a,
and the equation becomes aw ss A cos + B sin 0, or represents a definite
straight Ime, and no envelope exists.

8851. (By J. J. Walker, F.R.S.)— Show that either segment of a
focal chord of a conic section is a mean proportional between its excess
over the semi-latus rectum (or vice versd) and the whole chord.

Solution by W. J. Grsbnst&bbt, B.A. ; Professor Betens; and others.

The semi-latus rectum is an harmonic mean between the segments of
a focal chord ; hence

SP . 8F = RR' . PP', or SP (PF-SP) - RR' . PP',
or SP»-PF(SP-RR');

similarly for the other segment SF.

8837. (By the Editor.) — On the sides of a regular n-gon, n points
are taken at random, one on each, forming the apices of inscribed »-gon ;
again, inside the n triangles that lie outside this last polygon, n points
are taken at random, one in each, forming, when joined, another n-gon ;
find the general average area of this last n-gon, and show therefrom that,
in the cases of a hexagon aud a square, the averages are f f and -^f of the
original figure respectively.

Solution by D. Biddle.

If the mid-points of the sides of the original n-gon be first joined, and
then the centroids of the resulting triangles, a polygon is formed whose
area represents the area required. Accordingly, its comparative value
may be thus defined (} + i cos of, where a is the angle subtending a side
of the original n-gon. But the following solution gives an independent

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Tiew of the case, and the concluding equation has an interesting bearing
on the value of ir, when the n-gon is regarded as carried to its limit in the
circle.

Through the angle A draw AD, a diameter of the circumscribing circle,
as a line of reference, and reckon perpendiculars to it as positive when to
the right, negative when to the left of it ; also distances measured on it
as positive when upwards, negative when downwards, from the several
^t of the perpendiculars drawn from the angles of the original »-gon.

Taking the case of the hexagon, let Pi, Pa, ... Pj be the points taken
on the perimeter, and Q^, Qa, ... Q« be the points taken in the triangles.
Then the w-gon formed by joining QiQg, Q^Qat &c.

+ Q3?3 (H^3-H?4)-Q5y6 (H?6-H^4^ + Q4^4 (H^S-H^s)

+ (02^2 + Q«?3) (H^a -H?3) -(05^6 + QUq,){B.q,-Kq^)]

From this it is evident that the area of any n-gon so formed is

i [Q'lqi (D^n-Dy-i) + ^2^2 (I>?l-I>?3) + .- + Q*»^» (Dqn-l-'Dqd] -(A),
in which the y-terms within brackets are respectively the one before and
the one after that corresponding to the factor outside.

Now it is easy to see, Irom an examination of the shaded triangle in
the diagram, that the average position of Q is the centroid, X, and this
can be given in terms of P. For, it' we suppose Qj to be at X, then

Q3y2-i{Pli?l + P2i^2)+i{Ba-HPl/'l + P2/'2)}-4(BG + P,i;i+P2i?2),

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and D^, = i (Dp^ + Dpa) +i {DG-i (Dp^ + Dj^a)} = J (DG + Dpi + Dp,).

Let ?i, ?2, ^3, &c. = DA, DG, DH, &c., and *i, k^, k^y &c. = (0), BG,
CH, &c. Also, let a;, = APi / AB, r^ = BPg /BC, &c. ' Then

and Dpi = ^i + (^— ^ a^i, T>Pi = ^ + (?3 - ^ a^j, &c.

Consequently,

Qafi-a = H*2 + [*i + (^2 - ^i) ^i] + [^2 + (^3-^2) ^2] }» &c.,

and 'Dq,^^{h+[ln + {li-ln)Xn] + [h + {l2-h)^l]}y

Now, bearing in mind that Ar„ =s — Atj, A:„_i = — X-g, &c., and In = ^2»
^n-i = ^3» &c., by symmetry, let k^ — ki^* mj, Atj— Atj = wtj, &c., and let
h-h=fi^h-h^ U &c. Then,
Qij-i =. A:i+|{mia;i-w„(l-ir„)}, %q<i^k^'^\{in^x^-m^{y-x^'\, &c.,

and D^i = ^i+i {/« (l-^»)-/i^i}» I>y2 = ^2 + i {/i (l-«i)-/2^2}, &c.,

where /« = -/i, /«. 1 = -/g, &c.

Consequently, i [Qi^-i (Dj-n-D^g) + Q2$'2 (I^fi'i-I^^a) + — ]

« i [[[{^i + i [♦«i^i--»»»(l-a:,,)]} {/» + i[/«-i(l-a:„.i)-/„i-n]

-^2-i [/i(l-^i)-/2^2]}+&c.J
= i||[[*i^n-A;i?2 + iV«-i(l-^n.i)-iAri/,.a:n

- i^l/l (1 - ^1) + 1*1/2 ^2 + i^n »»i a?! - \l^ IHn {I - Xn)

+ i*nifn.iXi {l-Xn.ii-imifnXiXn - 4m„/n.i (1 -a:„)(l-a;„_i)

+ JWn/n Xn-imnfn x\ - \l^ m^ X^ + ^/^ »»„ (1 - a:„ )

-i^ifi ^1 + 4wi/i jri2 + ^,^^^2 a:i x^ + 4w„/i (1 -flTn ) (1 ~a;i)

-i»Wn/2a*2 (l-ir„) I + &C.

= i [[*l(^n-y + J*l(/„-l-/n-/i+/2)+J(/n-y K-mn)

+ inr K-*»") (/«-l +/2) +7B K/l-'Wl/n) +Ti^y K/n-fWi/l)"! + &C."1

= -siir r[(3^2^«-i) + (12^-1 ;„-! + 12A:2g + (ilk^ln + Vi + 3^„?n-i)

L +{13W„-13V2)-(47A;i;2 + 3*2^ + *?i) •) -,

^(Uk^h+nknh)- (3W3)j + &C-J-

From this formula we are able [by joining a series, the several terms
of which are made to correspond with the respective terms in (A), by
being all raised one degree at a time, k^ to Atj, k^ to A-j, /„ to l^ l^ to l^t
&c., until the n terms of (A) are satisfied] to calculate the numerator of
a fraction of which the area of the original n-gon is the denominator ; and
this fraction is the mean area required.

The above formula is easily reducible to the following, in which the

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44

Greek p at the foot of the several factors represents each figure horn 1 to
n in succession,

7V*'('.-8 + 8/,.2+l7/,.l-l7/,*l-8/,^j-/..3).

In the case of the hexagon, taking the radius of the circumscrihing
circle as unity, and the area of the original figure consequently as
3v/3/2, A:i = 0, *, - ix/3, A', - i^/S, A:^ - 0, /tj - -iv/3, h'^-W^'.
li =^2y Z, * 1|, ^s ' i) ^4 "" 0» ^ "== i) ^ = 1^ ; c^d the required area is
Hv/3/W3-f#.

In the. case of the square, taking the diagonal « 2, and the area also as
2, k^ = 0, Xtj - 1, Atj = 0, A:4 - -1, ^ « 2, i, « 1, /j = 0, (4 - 1 ; and
the required area » f^/2 « ^ « ^g.

8854. (By Rev. T. R. Tbret, M.A.)— Solve (1) the equation
w*+2 = (2:c + 6) Wa, ♦ 1 — (ar* + 4a: + 3) tr, ;
and show (2) that, if u^ and Vg both satisfy this equation, and if Ux =» 2,
«3 = 10, while Vi « 1, ^2 « 2, then 2w, « jc (j? f 3) c?,.

Solution by Professors Sircom and Bbtbns.

Dividing by (a; + 8j ! the equation may he written

_i_ f .Jffiil. . _!^£±J_| « _J_ ( ^-1 - Jf? ^ = constant 5
a; + 3t(a- + 2)! (a:+l)I) a? + 2t(a;+l)! x\ S

whence iff - JfiLiL^ = C(a:+ 1), and m>. - a;! Ca^i (a? + 3)+ar!Ci;
a:! (a;-l)!

then, if «x = ar ! Cx J (a; + 3). r, = ar ! Ci, the given values of « and r make
C » 1 «- Ci, then 2tf« » 4; (a; + 3) i;«.

8778. (By W. J. C. Shabp, M.A.) — If wy^vz, uz-wx, and
vx^uy be cogredient to a?, y, and z, the rectangular Cartesian coordinates
of a point, respectively, «, r, and «; are so.

Solution by D. Edwardes ; Sarah Marks ; and others.

We have WY- VZ = l^ (wy-vz) + /j (wz-w^a;) + /g (f;a:-«y)

= W (Wja; + Wo y f W3 c) - V (wj a; + «2y + fj3z),
whence iWjW — «iV = /,p - l^w, m^W - WjV = liw — /jW,

i«3W — WjV = /« « — /i» ;
multiplying these by mj, Wj, w.,, and adding, we have
W = (wi3^- W2/3) w + (wii^3 - W3/1) V + (?iW2~ ^2*«i) '''j or W - Wi« + «2t' + n^w.
Similarly for U and V.

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8650. (By E. M. Davys, M.A.)— Ptbve thit^ in atoy J)latie triangle,

fcofii^ A + 4 cos^° A sin^ A f 6 cos^ A sin* A — 6 CDS'* A sin^ A — 4 cos^ A sin^® A

-sini2A = C08 2A.

Solution by Rev. T. R. Tebey, M.A. ; Hannah MotLBK ; and others.

Quotient of left-hand side by rigM-hand side

^ (cos^ A + cos* A sin* A + sin^ A) + 4 sin^ A cos^ A (cds* A + sin* A)
+ o cos* A sin* A = (cob^ A + sin^ A)* =« 1.

8678. (By Professor (^enese, M.A.) — AB is a chord of a conic,
BJP, BQ are parallels to the asymptotes, A'^Q, a varitibl^ transversal
meeting the curve at R. Prove that the ratio PR : RQ is constant.

Solution by R. F. Davis, M*A. ;
Professor Cha&ravarti ; and others.

Let parallels through R to the asymp-
^tes meet BP,' BQ in M and N respec*
tively ; and let -or, W be the points at oo
on the curve through which BP, BQ re-
spectively pass. Then

R {ABtr-'or'} i^ constant;

dr, estimating upon BPiw,

•[PBwM} is constant ;

whence tflso so is PM : BM or PR : RQ.

d757. (By Professor Edmund Bordaoe.) — If two fra'ctions ala\
bj b' are such that ab' — ba' = 1, prove thai the' sirnplesi fraction inclnd'ed
between the two given fractions is (a + b) / {of + b') .

Solution by H. L. Orchard,' B.Sc, M.A. ; Professcw^ B'fitBlTtf ; aitd others.

If aV-ba^ = 0, then 4- -= ^ = 4±4.
a b' a' + b'

a+^a «+ — »'

Kow

a + b

a' + b'
Vol. XLVii.

/ ■ ,, < -7-, and IS >.
a' + 0' a'

a' + lf ^ If •

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8681. (By Professor Hudson, M.A.)— In the epicycloid of two cusps,
if P be the describing point when Q is the point of contact, prove that
the tangents at P, Q to the epicycloid and the fixed circle respectively
meet on the line joining the cusps.

. Solution hy R. F. Davis, M.A. } Rev. J. White, M.A. ; and others.

Let O, O' be the centres of the fixed and moving circles respectively ;
B the extremity of the diameter of the latter through the point of contact

Q, so that PR, PQ are tangential and normal to the curve at P. Then,
if A be the cusp, or point on fixed circle through which P was originally
coincident, arc AQ — arc PQ, and OQ =» 20'Q ; hence

Z AOQ = i Z PO'Q = Z PRQ.

But OQ = OR, hence QT (perpendicular to OQ) meets PR on OA pro-
duced.

7987 & 8036. (By the Rev. T. R. Terry, M.A.)— (7987.)— Four
spheres, whose radii ar« «, by e, d respectively, are such that each touches
the other three externally. In the space between these four, another
sphere of radius r is described touching all four externally. Show that

^-f'(i)"(i)-(i)-«-

R. Lachlan, B.A.) — If four sp]
emally ; and r be the radius of tl

.orthogonally, then -i- = 22 i^ - 2 f-^) .

(8036.) — (By R. Lachlan, B.A.) — If four spheres, radii a, *, c, d, touch
one another externally ; and r be the radius of the sphere which cuts them

Solution by W. W. Taylor, M.A.

Inverting with respect to the point of contact of one pair of the four
spheres, this pair of spheres becomes two parallel planes (1, 2) ; the other

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47

pair of spheres and the two spheres that touch the four original spheres
become four equal spheres (3, 4, 6, 6), whose centres form two equilatertd
triangles ; and the sphere of Quest. 8036 becomes the plane (7) passing
through the four points of contact. This inverse system can be represented
by the equations

y ± A = 0, a;2 + y2 + «2 ±2Ax« (1, 2, 3, 4),

ic2 + y2 + -5Si2>v/3Ai5 + 2A2«0, z =0 (6, 6, 7).

Now, the radius of the inverse of the sphere /(a?, y, z)

= a:2 + y2 + 2;2 + 2Aa; + 2By + 2C« + D = 0,

with respect to the point h, k, I (constant of inversion «= 4), is

Invert with reference to the point h, k, l, and call the radii corresponding
to these seven equations a, b, c, rf, r^ r2, r^ ; then

i. « i (A + ^), -i = i (A-A;) (8, 9),

JL « 1 (A2 + A:2 + ^s + 2A;0, 4- = — (*' + 't* + ^-2AA) (10. 11),

Q 4A a 4A

_L« -L(A2+A;2 + /2 + 2^/3A^+2A2) (12),

-1 = -L(A2 + ;fc2 + ^«2y3A?+2A2), Jl ^ i. (13, 14).

r2 4A ^a 2

From-(8), (9), A«i-+i-, Ar = ^ - i- (16, 16) ;

a a

from (10), (11), we have

»-(7-i) '"■'«'

from (10), (12), and (15), (18),

WZ-h = 2 f i ^ ^ -A, /^/3 = — -2-i- (19) ;

\r^ c I ri a

also, by (16), (17), (18),

(20) ;

therefore, by (19), (20), ±^ - ± 5 1 +2^ +22^ = 65 ^ -S^l,

therefore 1 - l-^ J_ ^.5 1 .^ 1 = q.

ri2 Ti a a^ ao

For fj substitute for I in (20), from (14), -t :*. 22 ^ - S ^.

[These results are most readily obtained from the general formula given
by the late Prof. CLirroiiD (on p. 335 of his Mathematical papers) for

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finding the radloi of a sphere which cuts five given flpheres at givei^
angles: — via., if the spheres whose radii are r,, r, cut at the angl^
^M,p, then

0. ~

1
77'

1

1

— 1, COSftrj 2) COSWjg, COSCtfj^^, COSWj g,
C08(|f>2,j, —l, &C„

&C., ... -I,

-1,

-1,

=

By taking m,2 = m,i^ &c. = 0, a result equivalent to (7987) is £it once
obtained ; and My taking ei^i,2= m,3 = «i,4 = w-i.s = <9i,4 = W3,4 = 9>
and «i 6 = «2,6= «3,6 = «4.6= i^, a result equivalent to (8036) is
()btainea.

The above theorem is a particular case of a more general theorem given
h^ the Boyal Society's Froceedinga for March 11th, 1886.]

8725. (By S. Tebay, B.A.)— If the sines of opposite? dihedral angles
of a tetrahedron be respectively proportional to the edges formed by these
^gles, the areas of the four faces are equal.

Solution by Pyofessprs Mathews, Beyens, and others.

Denoting by (CD) the dihedral angle of /^

which the containing planes intersect in
CD, we have

8i»(CD)-AF/AE

^ 3 vol. ABCD / 2 area ACD
area BCD f pD
Bin rCD) 3 vol. ABCD
CD " 2 . ACD . BCD'
60 also

sin (AB) _ 3 vol- ABCD «
AB " 2 CAB. DAB' *

Hence, if ^^^^^ = ^^°/^^\ &c., ACD . BCD ^ CAB . DAB ;
Cp AB

■^ith two similar equations : whence BCD = CD^ = D^-B = ABC,

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8697. (By Captain H. Brocard.)— Les droites qui joignent le centre
du cercle circonscrit aux milieux des segments interceptes par les perpen-
diculaires aux milieux des cotes sur les hauteurs du triangle sont perpen-
diculaires aux medianes correspondantes.

Solution by Professor Schoute ; R. F. Davis, M. A. ; and others.

Let O l?e the circum-centre and D, E, P
the middle points of the sides of the tri-
angle. Let OE, OF meet the perpendicular
AH on BC in the points Cj, i,. aud let dy be
the middle point of biCi. Then the tri-
angles OdiCi, ABO are obviously directly
aimUar, corresponding lines being at
right angles. Hence the median line
Oil is perpendicular to the corresponding
line AD.

8502. (By B. Hanumanta Rau, B.A.) — Through any point K are
drawn the straight lines B"KC, C"KA', and A"KB' respectively parallel
to the sides BC, CA, AB of a triangle ; prove that (1) A A'B'C'= A A"B"C" ;

(2) parallels through A, B, C to A'B', B'C, C'A' meet at a point O';

(3) parallels through 4., 5, C to A"C", 4"B", 0"B" also meet at a point
D' ; (4) if the coordinates of O and O' are (o, i3, 7) and (o', /8', y') then
aa ; cy'=b^\ aa! — cy : b$' ; (6) if K is the Symmedian point, O and Q'
become the Brocard points. '

Solution by R. F. Davis, M.A. ; Professor Sab^ar, M.A. ; and others.

Let Xy y, s be the trilinear coordi-
|iates of K. Then it will be seen that
each side of the triangle ABC is
divided similarly into three parts
^ ax '.by I cz,

(1) AA'B'C=sum of As B'KC
C'KA', A'KB'=s semi-sumof par^^llel-
ogramsA'A^^B'B'^C'C"= AA^'B' a'.

(2) Since BB' : B'C"= cz : rtr, a line
through A parallel to A' IV divides
BC in the same ratio, and its equation ia fi j y ^ ax I bz. Similarly the
equations to lines through B, C parallel to B'C, C'A' respectively ar^
7 / o = by lex and aj fi = cz j ay. These meet in a point 0,

o : i8 : 7 = czx : axy : byz,

(3) Similarly for 0', a' : i8' : 7' « bxy : cyz : azx.

(4) Whence aa : cy' = &c.

(5) 11 X \ y : z =i a \b \ c then a : fi : y = / b : a j c : b / a, &c.

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8585. (By E. F. Davis, M. A.)— If TP, TQ be tangents to a parabola,
and PQ meet the directrix in Z, prove that ZT will be a mean proportional
between ZP and ZQ.

Solution hy Bev. T. R. Tebrt, M.A. ; Prof. Bbtbns, M.A. ; andothert.

If the cooi'dinates of Z are
(—a. A), and equation to ZQP be

t^-^±i?«c (I),

Bin«^ cos a

then

ZP.ZQ-(A2 + 4a')co8ec2a...(2).

Also, from (1), coordinates of T

are a + Acot0, 2a cot 0,

therefore

ZT2 = (A2 + 4a2) cosec''a...(3).

[Since SZ bisects externally
the angle PSQ,

ZP.ZQ - SZ^ + SP . SQ « SZ« + ST3 « ZT».]

8600. (By Professor Mahendra Nath Ray, M.A., LL.B.)— Through
an indefinite point of a given hyperbola straight lines are drawn to meet
the as3rmptotes ; show that the hyperbola itself is the envelope of the
locua of the middle points of the straight line.

Solution hy Professors Scott and Ignacio Bbyens.

Let xy -» k^ be the hyperbola.
The line is xa + y$= 2a$t where
(o,i8) is the middle point ; it passes
through a?', y', therefore
x'a + y'fi = 2ai3,

also y' « — ,

hence ax^- 2oj3a/ + /8A;2 = and o^jB* '
is the envelope.

■■ a$k'^ or 0)3 = A;2 ; therefore xy = A*

8623. (By N'Importe.)— From the centre of similitude S, common
radii vectores SPP', SQQ' are drawn to similar curves PQ, P'Q'. Hav-
ing given the centre of gravity of the area SPQ, find that of PQQ'P' ;
and its limiting position when PU, V'Qf ultimately coincide.

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51

Solution by Rev. H. London, M.A. ; Professor Ignacio Bbybns • and others.

Let G be the centre of gravity of _ /-,'

SPQ, and let ^^ ^^^

SP : SP' = SQ: SO/ = 1 : A:,
/. area SPQ : areaSP'Q' = 1 : >fc2.

Let G' be centre of gravity of SP'Q'
and X that of PQQT', and m = mass of
SPQ. Therefore, taking moments about S,

k^m .[SG' = m . SG + (k^- 1) m . SX ;
but SG' : SG = * : 1, SG {Jfi- 1) = SX (&2- 1),

SX = %p^ . SG = ^!±A±1 . SG « fSG in the limit.
k^ — \ k-k-l

8716. (By Professor Mathews, M.A.) — Prove that the real common,
tangents of the circles x^-^-y^—lax — 0, x' + y^-1by « are represented
by 'lab (z^ -k- y'^—^ax) =» (by-^ax + abYy or, which is the same thing, by
2ab (a;2 + y2_2*y) = {by-ax-aby.

Solution by C. E. Williams, M.A. ; W. J. Grebnstrbet, B.A. ; and others.
Two tangents from (a:', y') to the circle x^-^y^ — 2ax = are given by
lxx''¥yy'''a{x + x')Y = (x^ + y^-2ax){x'^ + y'^-2ax').
But {x*, y') is in this case the external centre of similitude whose co-
ordinates are easily found to be a3/(^— a), abl{a—b). Substituting these
values for (a?', y'), we get (by^ax^ab)^ = 2ab {x^ + y'^^2ax)f and the other
expression is obtained similarly from the other circle.

7169 & 8537. (By W. J. Greenstreet, B.A.)— The sum of the

three sides of a right-angled spherical triangle is a quadrant : prove
minimum value of hypotenuse is cos-^ f , and that in this case the spherical

Solution by Rev. T. R. Terry, M.A. ; J. Young, M.A. ; and others.

Since cos c = cos a cos i, a.nd a + b + e =^ itr (1»2),

therefore, for minimum value of c,

tan a da + tan b db = 0, and da + db = 0,
therefore a ^ b, whence, by (1) and (2), cos a = cos d = 2/^/6, cose =» |.
This is obviously a minimum. Also

sin E = sin (2A-iir) =» J, since cos A = f.
[For another solution, see Vol. xxxviii., p. 89.]

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8453. (By Satis Chandra Ray, B.A.) — Two rods, of lengths a and 5
are jointed by a smooth hinge and rest on the convex side of a parabolic
arc whose axis is vertical ; if, in the position of equilibrium, the rods in-
clude a right angle, show that the angle 6 which the chord of contact
makes with the axis is given by tan id — a^l IP,

Solution by W. J. Barton, M.A. ; Sarah Marks, B.Sc. ; and others^

Let CA, CB be the rods,
of lengths a and b ; Gj, Gj
their middle pOints ; fia, fib
their weights ; 8 the fociis,
MOM' the directrix of the
parabola; P, Q the points
of contact of the rods ; ST
the axis ; Ri, R> the reactions
at P and Q. Then TSP= d ;
and iiherefore MCA = id,
M'CB = iir-^d.

(1) Taking moments about.
C for the equilibrium of
each rod separately, we ge£

fia . iacok^a = R^^i,

fitb . J* sin Jfl^ = R2^2»

where CP = ^i, CQ = ^j.

whence

Ri _ tj

<«fl2

Resolviirg horizontally for the equilibrium of the system,

"R

Rj sin id = Rg cos id, whence ^ =* cot id

Ri

.(2).

From (1) and (2) we get ^ / /j = a" / R

If CB meet the axis in O, since SO = SQ and OSP = 0, the angle
CQP = id J therefore ^1/^2 = tan id ; whence tan ^d == ^2/ b^.

[The equations to EF, FD, DB are

— acosA + /8cosB+7CosC = O-, acos A— )3cosB +7C0i?C = 0...(1, 2y,

acosA + iScosB— 7COSC = (3).

Hence the points of intersection of (1), (2), (3) with a = 0, jB = 0, 7 = 0,
respectively, lie on the line

ocos A + i8cosB + 7C0rfC = (4).

If S = aBy + bya + ca$, then the equations of the circuniscribed, nine*
point, and self-conjugate circles are, respectively,

S = 0, 2S-(rto + ii8 + <?7)(acosA + i8co8B + 7COsC) = 0,

S — (ao + *i8 + cy) (a cos A + i3 COS B + 7 cos C) = ;

hence (4) is the common radical axis of these circles. }

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53

(By Cb. Hermits, Membre de I'lnBtitut.)— Determiner les

int6grale8d6finies f . f ^ . f**'-^^ r,

jo sin (a? +p) J - J, sin (x +p)

en Bupposant que p soit une quantity imaginaire quelconque.

Solution hy the Proposer.

La determination de ces integrales definies s'obtient comme conse-
quence des r^sultats suivants auxquels conduisent faciloment les methodes
eiementaires.

Soit a une quantity r^elle, positive, et diff^rente de zero, on a :

Jo sin (x - ia) Jo sin {x + ia) 1 - c-« '

I^- -=—7 ^=~" -^—f ^ ='4tarctg(f-«.

Eemarquons ensuite qu'en rempla9ant a par une variable imaginaire
o + tjS, dont la partie r^elle soit positive et diff6rente de zero, et par
consequent dans toute la region au-dessus de I'ate des abscisses, les
diverses integrales definies sent, des fonctions holomorph.es de cette vari-
able. Observons maintenant que z etant k Tinterieur d*une circonference
de rayon ^gal k 1' unite et dont le centre est S Torig^e, la fonction
log(l +2/1 — 2) et arc tg 2 sont aussi'holomorphes. Or on obtient en
posant z » e-^., des valours qui remplissent cette condition, et 1' extension
des relations (I.) et (II.) k toutes les quantites a^ a.-h tjS, oil a est positif
et different de z^ro, est la consequence immediate de la proposition bien
connue de Biemann : Deux fonctions, uniformes, holomorphes ou n'ayant
dans une aire donnee que des discontinuites en nombre fini, sont egales.
en tons les points de cette aire, si elles coincident le long d'une ligne de
grandeur finie.

8242. (Professor Sylvester, F.R.S.) — ^If by a Simplicissimum <tf
the f>th order be understood a figure in space of n dimensions formed by
the indefinite protraction of the series of which a linear segment, a tri-
angle, and a pyramid are the three first terms, prove that (1), when each

edge is unity, the squared content is \!LL-1 _ and hence (2) de-

^ ^* ^ 2»»(1.2.3...«)2' ^ '

duce that the Cayleyan Persymmetrical Invertebrate Determinant of the

squared edges by which such squared content is imaged must be

diminished in the ratio of negative unity to ( — 2)" (1 . 2 . 3 ... «)2, in order

that it may represent its absolute value. £x, gr,^ the determinant

{obY {acY 1

{baY . {bo)^ 1

((ja)2 (cby . 1
11 1 .

images the squared content of a triangle whose edges are (ab), (ac), (be),
for the triangle vanishes when this determinant vanishes, but the actual
value of the squared content is this determinant diminished in the ratio
of negative unity to 22(1 . 2)'\ t.«., multiplied by — ^^ = (2).

VOL. XLVII. o

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54

Soluium by W. J. C. Sharp, M.A.
If orthogonal ooordinates be employed, the contents of the successiye
Simpliciflsima are

a?i» Pu «i» 1

i I ^1' M J_
l.'K, ir 2!

^1, yi. 1

1

X2 y^y 1

' 3!

a?8» Viy 1

and in n-dimension space

«i» yu
«2. yai

•''•ij .'a* *2» ^

a-a* y8» «8» 1

X^y Va, «4» 1

.... 1

, &c.,

V suppose.

^»+b y»+ii 1

Now, if the origin be taken at the centre of the hyper-sphere radius
R whidi passes through the n + 1 vertices, and if the edges be denoted by
(1.2), (2.3), &c.,

x^ +yi« + «R2, 2 (a:iarj + yiy2+ ) = 2R2_ (i . 2)2,

x^ +ys«+ = R^ 2(a;jaJs + y,y8 + ) - 2R2-(2. 3)«,

^i.i+yi.i -^^^

and

v«2

1, 0, 0,
0, «i, Vu

0, «!, y„

0,

^ (^lyn^i

2»*n!

0, 0, 0,

1, 2^1, 2^1,

1, 2^, 2<„

0, «n+b yn+1 tn*h 1

therefore y « (^:ili?^'

2»*(«!)*

X 0, 1, 1,

1, 2(a;i«+yi« ... ^i«), 2 (x^x^-^y^yt ... ^i^,)

If 2(ari«j + yiy,...<i^,), 2 (i-j^ + y,^... ^«)

1, 2a?n+b 2<n*i,

1, 2(a;iar„+i+yiyn+i...^i<«+i), 2 (araa?„+i + yiy»+i ... <i^„+i).

0, 1, 1,

1, 2R», 2R«- (1.2)2, ...
1, 2R2- (1.2)2, 2R2,

(«1J2«+1

' 2»(fi!)2

(~l)2H^l

2»»(n!)2

^ (-l)»»^g
" 2»(n!)2

1, 2R2-(l.n+l)2, 2R2-(2.n + l)2

0, 1, 1,

1, 0, -(1.2)2, ....
1, -(1.2)2, 0,

1, -(l.n + m -(2.n + l)2,
0, 1, 1,

(1.2)2,
0,

1, (1.2)«,

1, (l.n+l)2, (2.n + l)»,

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which prorea (2) ; also, if (1 . 2)* - (2 . 3)> - &c »> I, the determinant is
- (-1)"*> (»+ 1), and T^ = g^,, which proTes (1).

[If the n+ 1 points, which detennine the Simplicissimum in space of n
dimensions, lie on a linear locus in space of (m— 1) dimensions, the deter-
minant V vanishes, and therefore the corresponding Cayleyan Persymme-
trical Invertebrate Determinant ; and this equated to zero gives a relation
between the mutual distances ofn + l points on a linear locus in space of
n—l dimensions, as of three points in a straight line, four in a plane, five
in a space, &c.

Mr. Sharp considers this to be an epoch-making qmstion, as he believes
that it will lead to the application to space of higher dimensions of methods
analogous to those of tnUnear and quadriplanar coordinates.

8755. (Professor Nbubero.) — On prolonge les hauteurs du triangle
ABC au deli des sommets des quantitis AA'=BC, BB'=CA, CC'=«AB ;
d^montrer : (1) les triangles ABO, A'B'C ont mdme centre de gravity ;"

(2) si a, a' sent les angles de Brocard de ces triangles, on a

A'B'C = 2ABC (2 + cot o), (A'B')2 + (B'0')2 + (C'A')« =- 8AB0 (3 + 2 cot a),
coto'= (2coto + 3)/(cota + 2) ;

(3) les points A, 6, sent les centres des carr^s construits, int^rieurement,
sur les c6tes du triangle A'B'C ; (4) les milieux des c6t6s de A'B'O' sent
les centres des carr^s construits, ext^eurement, sur les c6t68 de ABO.

Solution by R. F. Davis, M.A. ; Prof. Aitab, M.A. ; and others.

Let fall the perpendiculars A'm,
B'n upon AB. Then, since the tri-
angles AA'wi, CBF are equal in all
respects, A'm =» BF and Aw =« OF.
Similarly, B'n = AF, and B/» = OF.
Hence <?, the middle point of AB,
is also the middle point of mn ; and
if ec' be drawn perpendicular to AB
meeting A'B' in c', </ will be the
middle point of A'B', and

cc'= i (A'm + B'w) = Ac or Be,
Thus (4) c' is the centre of the
square described externally on AB.

Since c</ is parallel to and one-half
of CO' ; Ccy Co' intersect in a point
G which divides each of them in the
ratio 2:1, and therefore (1) the triangles ABC, A'B'C have the same
centre of gravity G.

Since BB', Be' are equal and perpendicular to AO, Atf', respectively,
BV i& also equal and perpendicular to Cc', and (3) is the centre of the
square doacribed internally on A'B'.

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(2) Since HA'- a + 2Rco8A « a(l + cot A), HB'= &(l + cotB),
/A'HB'«ir-C, area A'HB'« A(l +cotA) (1 + cotJB).
Hence area A'B'C= A {(1 + cotA)(l + cotB) + ... + ...}

= A(4 + 2cota) = 2A(2 + coto),
■mce cot B cote + ... + ... B ly cota»cot A + ... + .. ..

Again, (A'BO' « (A'm-B'«)« + mn« » (BF-AF)2+4m«

«4{<yF»+(Ac + CF)3}«4{Cc9 + Ac« + 2A}=-2(a- + ft») + 8A,
hence (A'B')« + (B'CO' + (C'AO« - 24 A + 4 (a« + A^ + c^) = 8 A (3 + 2 cot a),
whence cot a' as stated.

It is to be observed, when ABC is equilateral, then so also is A'B'C, and
oota- cota' - V3.

8865. (Amiral db JowQUitRES.) — Soient trois nombres entiers,
0, by e, premiers entre eux, deux h deux, et verifiant T^quation a" + ^=<?*.
D6montrer que (I'exposant n premier et superieur k 3) : (\) a et b ne
peuTent Stre, simultan^ment, premiers ; (2) si a, suppose inf erieur k b,
est premier, e » b+l,

8866 & 8908. (Professor Catalan.) — Demontrer les contributions au
thlor^e de Fehmat:— a suppose premier, (1) a— I = 9JJ (,/) ; (2) «»— I
=» 9}? (nb) ; (3) tout diviseur premier, de «— a, divipe a— I; (4) a + b et
tf— a sent premiers entre eux; (5) 2a— 1 et 2b + I sent premiers entre
eux ; (6) le nombre premier, a (s'll existe) est compris entre

(nb^'^f"^ et {n (* + l^«-i}^^*;

17) a et 6 surpassent n; (S) le nombre b, qui satisfait h Tequation
^4. Ijn.^n ae a**, est compris entre

aialn)^^*'^'^ et -l + «(fl/«)^^*"^

(9) soit b un nombre entier, sup^eur au nombre entier n. Entre

{nb^-i)y» ei {n(* + l)*-^}^^*,

11 7 a, tout au plus, un nombre entier; (10) aucun des nombres a + b,
c—a, c^b, n'est premier; (11) chacun d'eux a la forme N, ou la forme
(l/«) N, N ^tant un nombre entier; (12) soient, s'il est possible,

a + i « <?'»», <?-a = b'\ e-b^ «'»' ;
alors c « SK (») ; (13) (a; + y)»»~a;»»-y»» « nxy {x+y) P,

P = Hia:»»-» + H3a;»»-*y+ — +^i!f*"\
les coefiBcients sent donnas par la formule Hp = (I/')) [Cn-i,p ± 1],
le signe -I- r^pondant au cas oii p est pair, et le polyn6me P est divisilde
par x^ + xy + y^f et mdme par {x^ + xy + y^)\ si n = 3)2 (6) + 1 ; (14) la
difference des puissances ft**°^» de deux nombres entiers consecutifn,
a, a+1, ^tant diminu^e de 1, est divisible par na{a+l) {a^ + a-\-\) ;
les facteurs a, a + 1, a' + a + 1 sont premiers entre eux, deux k deux ; en
outre, le troisi^me egale le produit des deux autres, augmente de 1 ;
(16) si, dans 1' equation de Febiiat, le nombre a est premier, on a, par le
th^r^me 8866, de M. de Jonqui^res, a»- 1 » 3R [w* (A + 1) (^ + i + 1)] ;
et (16) ff est compris entre a + b et ^{a + b).

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Solution by Samubl Roberts, F.R.8.

It seems convenient to take these three questions together. The par-
ticular results are so numerous, that brevity must be specially consulted
in each instance. They are, however, in general, simple deductions from
the considerations indicated by Baklow (Theory of Numbers, Cap. vi.)
and by Abel ( Works, Holmboe's Edition, p. 264, Vol. n.).

We cannot have a » 6 ; but H a, b are prime the conditions «— «
e—b= 1 are necessary, and c cannot be prime, since c* =» (a + ^) ^, ^ > 1.
If a (the least number) is prime, <?— 3 = 1.

(1) By F£KMAT*8 theorem applied to ^— a*— i* — 0, we have a+b—e
s-gK« = a-1.

(2) a»-l = «»-4*-(<?-3)»=:SW*(* + l)fi.

(3) {c—a) + («— 1) == * ; («— «) ^ =■ *" ; (c— a) isof theformr»orr"fi»->,
r or m being a factor of 6, therefore a— 1 contains the prime factors of
{c-a).

(4) (a-k-b) ^ k {c~a) makes a not prime to b, contrary to hypothesis.
(6) 2a-l = (a + *)-(<J-a), 2*+l = (a+*) + (tf-a).

(6) By (1) since a is > 1, 4 > a.

(7) (* + l)»-ft- >ni»-S (A + l)*»-{(*+l)-l}*'< n(* + l)»-i.

(9) If a Hea between («i*-»)^^* and {«(*+l)*-i}^^*,
we have a + 1 > (nft»-i)^* + 1 > {« (* + l)«->}^^*,
or (»*»-') + f 1 j, since 4 > n.

Similarly, (a-1) < {« (ft + l)-»}^*-l < [«{(*+ l)-l}->]^^*
<{«(* + l)-»}'^*-(l.l).

Therefore a + 1, a— 1 lie outside the limits.

(10) Each of these quantities is of the form r" or t^n^'^,

(11) As above, see Baklow, Cap. vi. (Ba&low's mistake occurs
further on).

(12) I do not understand this. If e contains n as a factor, a-^b con*

tains ft as a factor and is of the form r"w*->. For ^ "^^ is prime to n

a + b ^

if + 6 is prime to ft, and, if a + d contains n, contains ft. It has

a-^o

been assumed all along that a, 6, are prime to one another.

(13) (« + yy»-«"-jr= («+y) {(a?+y)»-*-(«*-i + ««*V + ...+y"*^)}

= *y(«+y){[(«-i) + i]^-'

^^(ft-l )(n-2) ^^j^,.^^_^^^^_^^^^jy,,3^

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-A^ (' -^-J^)* -^~S^* ^^l^ci^ * is <>^ ^^ fonis 6i:L 1, ifl made aero by piittii^
tpy for X, w betng an imaginarj cube root ol unitr. If s is of the form
4M;-I-1, tibe derired fimctioa n 'x-t-y *-'— ai*-** is ande aaro by the
aame aobstitiitioa. The fimctioa -x-i-jr)*— x*— jr* TirntufTtft therefore
(*— wy) (x— ii^y) or ** + j^f +y2, and in the Lut case twice orer.

(14), (15) lliefle are dn«ct conseqaences of the fovegoing.

(16) («-«) + {«-^) > 0, «• = «- + 4^ < (• + *)-.

8810. (Ber. T. R. Tkut, M^.) — Fmd tiie equation to the
straig^ linies throu^ the origin and the intenection of the conies
j?»-3xy-4x + y-l«0, 2x9-4-^-i-4f>-l-Sjr<i> ISy^^S = 0.

^2k<umi 3y H. FoBTBT, M Jl ; <A#PBOPoaK; mmiothen.
In the most general case, the equations to the oomos are of the form

•!» + »i + «*« =0>»
whence u^^—vm^ = — {u^v^— ri«t}» ^ ^i "* ^i J

therefore the equation to the four lines required is

In the particular case this reduces to {x—yY (2x +y}' = 0, t^., the four
lines coincide two and two with the lines x- y = and 2x -i-y = 0.

8843. (Kev. T. P. KntKMAN, F.R.S.) — The angles fia*c.../« of
any convex plane M-gon N are joined to a point P in its plane, within or
without it. In Pa, P^, ... Pit are taken points a^biCi ... m^n^^ such that
ffOi^i ... miMi is a broken line L beginning and ending on Vn. L is thus
formed: first, f}a|>aj<i; and next, if dye^^ be any three consecutive
points of it, the distances in the directions /]«i and e^d^ of «i from the lines
/e and «^ in N are equal. The line ftjti»i meets the edge nm of N in r.
Prove that, if from « in an produced we draw Mi =r rn^^ «fij is parallel to
the first portion nai of L.

Solution by the Pboposer.

The described fig^e is the orthogonal projection of a pyramid whose
vertex P is anywhere over the base N, and on which the line over
nai ... »t|»i crosses every face about P, in a path equigradient from n to nj.
For, as the lines drawn from ^i are equal, they are projected sides of an
isosceles triangle drawn from tf, in P^ to edges of N, in the faces about P
whose intersection is Ve ; and t^e sides of this triangle make equal angles
with the base, or are equigradient. Hence «w, and nax are equigradients
directed upwards from s and n in the same plane aPm, and are therefore
parallels.

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Thus we can find (Quest. 8325) with the ruler the fly's upward path
from n through aj to a point «i in P». A few trials with the ruler in
piano, using a,, 03, &c. instead of aj, will hring us to n^, n^y &c., one of
which is sure to he near enough for engineering purposes to the starting
point E.

As the solution of Mr. Biddle's interesting Quest. 8325 is not likely to
be completely worked out, this easy approximation to it by plane geometry
may be worth having.

8814. (The Editor.} — Prove that the triangle formed by joining
the mid-points of the altitudes of a triangle is one -fourth of the pedal
triangle, and that the theorem holds good for any other three concurrent
lines drawn through the vertices of the triangle.

Solution by Professor Db Lonochamps ; D, Biddle ; and others,

1. Qeometrieally : — Soient A', B',
C les milieux des c6t^s du triangle
consid^, PQR le triangle p^al
correspondant k M. Les droites AP,
BQ, CE rencontrent les cdt^ de
A'B'C respectivement en des points
P', Q', R'. Consid^rons aussi les
points P", Q", R" isotomiquea de
r\ Of, R' ; c'est k dire des points
tels que B'F'=P'C', etc.... La droite
AT" est parall^e k AP, et d'une
fa9on plus g^n^rale, on pent dire que la figure A'B'CT"Q"R" est
homoth6tique k la figure ABCPQR ; le centre de gravite de ABC 6tant
le centre de I'homothetie et le rapport d'homoth^tie ^tant 6gal k i. De
Ik, nous concluons d'abord que I'aire du triangle P"Q"R" est le i de celle
du triangle PQR. D'autre part, on sait (Journal de Mathhftatiques 614"
mentairea, 1877, p. 224), par une propriety que nous avons indiqu^e autre-
fois, que deux triangles isotomiquea P'Q'R', P"Q"R" sont equivalents,

Ainsi, en resume, le triangle FQ'R' est le \ de PQR.

2. Analytically : — Si nous consid6rons trois points quelconques Mq, M^,
H], dont les coordonn^es barycentriques sont, respectivement,

la surface S du triangle MqMiM^ est donnee, comme Ton sait, par la

formule ±825-

oo A) 7o
oi ^1 7i
02 ^ 7^

S d^signant Taire du triangle de reference

ABC; oq, /So, 70 ••• repr^ntant lea eoordonniea ahaoluea des points con-
siddr^s. Prenons un point 11 dont les coordonn6es soient (a', /8'> 70 ; les
droites A/u, B/u, C/u rencontrent les cotes du triangle ABC en des points

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Mq, Mj, Ms ayant pour coordonnees

0,

S3'

sy

/3' + 7' i3' +

., etl'ona: ±S^Z =

0,

B»' sy

/3'+y' i3'+y
-?^

;' + y' ' o' + 7'

Sa' Sis'

2 d^signant maintenant Taire du triangle p6dal Mq Mi Mj. Cette fonniile

a'3V

.(1).

donne, finalement, ±S = '-.-, , ^,. ,^, ... , ...
(o' + i3')(i8' + y)(y + a')

Consid^ons maintenant les points /io* Mi> A4 milieux des segments AMq,
AM|, AM]. Les coordonnees barycentriques absolues de ces points sont

s. s fi' 8_ y

2' 2 fi' + Y' 2 /8' + 7''
La surface :;' da triangle juq Mi Ms est donn6e par la formnle

^ s y

±S«2' =

1- «''
2 o' + 7''
S a'

2 a' + i3'' 2 o' + /8"

2 /S' + y' 2 i3' + y

2* 2 ^3^ + 7'

8 iS' S

2

done ±2' =

S^ 1

8 (a' + i3')(/8' + 7')(y + «0

/s'+y, is', y
«', «'+y, y

/3', a' + /8'

Le determinant qui entre dans cette formule ^tant d^velopp^, ou trouve
facilement (mais ce r^sultat pent dtre vu aussi par des voies plus directes et
plus elegantes) qu'il est ^gal k 4a'/3'7'-

S g'iSV

Ecrivons done

±2' =

2 (a' + i3')(i8'+y)(y + aO
La comparaison des formules (1) et (2) prouve que 2 = 42'.

(2).

8858. (Professor Cochez.) — B^soudre les Equations

{xt/ + l) {x + y) « axi/y (xY+ 1) (x^ + y^) = b^^Y-
Montrer que ce syst^me est quadratiqtte, Appliquer les formules au cas
particulier suivant : a = J^, ^ = ijf.

Solution by Professor Stbggall ; R. F. Davis, M.A. ; and others.

Let ic + 1/a; = I, y + 1/y « ?7 ; then the equations become ^ + ?; ■= «,
|2 + ^ = ^2 ^ 4^ whence |, ti are easily found. In the particular case
I = xo or 2 ; ij = 2 or ^f- ; and we have a; = 1, y = 3 or i, y -« 1,
a; = 3 or ^.

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8498. (AsOtosh MukhopAdhyAy, M.A., F.R.A.S.) — The equation

\dx) dfi dx'dt'dxdt \dt / dx'^ \dx I

is inte^ble when (1) Q—O, (2) Q » dx/dt. Hence, obtain the complete
primitive.

Solution by Professor Lloyd Tanner, M.A.

The proposed equation is a transformation of — - + Q » 0, where x, y,

dt^

are independent variables. The solutions required are, therefore,

X = <f>y + t}l/ff when Q = ; and x = <f>y + e-*^ when Q ■» dx/d(,

[See Messenger of Mathematics, New Scrips, Vol. v., p. 71.]

Note on Anallagmatic Curves. By Prof cssok Wolstbnholme, Sc.D..

In a curve which is its own inverse with respect to a point 0, if any
straight line through meet the curve in two points P, Q, so that the
rectaugle OP . OQ is constant, a circle can be drawn touching the curve
in P, Q. Hence, when we suppose Q to coincide with P, the bitangent
circle will in the limit have four-point contact with the curve at P,
and will therefore be a circle of curvature of maximum or minimum
radius. The points of greatest or least curvature (ottier than the vertices)
in any such curve will therefore be the pointa of contact of tangent?
drawn from a ceutre of inversion of the curve.

I do not remember ever to have seen this simple deduction drawn. I
ought to have myself noticed it years ago, as I have two particular cases
of it in my book of problems ; but in both I obtained the result in a mudi
' more laborious way. No doubt many interestiHg particular cases might
be considered. I have not much hope of doing anything at it myself, so
should be glad to have the above note published, pro bono publico.

8786. (R- Tuckeb, M.A.) — P;o, Qy, Rr are focal chords of a
parabola ; if P, Q, R have a conormal point, then the centres of curvature
for p, Qy r are collioear. (This is another way of putting the first part.of
Question 8693.) Find the equation to this Ime. Prove that this central
line envelopes (1) a hyperbola when tangents atjp, q meet on the directrix,
(2) an ellipse when the tangents meet on the latus rectum.

Solution by R. Enowles, B.A. ; Prof. Beyens ; and others.

In the solution of Quest. 8693 it is shown that the centres of curvature
of three points on a parabola are collinear when 2y ~^ == ; if a;|, ^i, &c.,

VOL. XLVH. H

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be the coordinates of PQR, those otp, q, rare

— , , &c., and 5y-i = -4a^ {y^ + yj + ^3) « 0,

^1 yi
«ince F, Q, E have a conormal point.

The coordinates of the centres of curvature aip, q, r are
da^ + 2ax. 16a* «

and if the tangents at p, q meet in the directrix ^1^3 *— 4a'. The
equation to the central line

16a* ^ ^aW + yit/2 + yJ^ /^_ 3a2 + 2a£3\

3yiy2(yi+y^ V x^ I

becomes yi + ya = -^8 (1)>

-2ayj«+ 3ay,«-48a6 = (ys» + 4a») (yj»a:-12a8^12ay32) (2),

if the tangents at^^ meet in the latus rectum yjyj = 4a2, and (1) becomes

3ay3y-48a6 = -(y32-4a2)(y3«a?-12a8-2ay32) (3),

and the envelopes of (2) and (3) are respectively the hyperbola

9y2-16:rS+ 112aa;~ leOa-^ » 0,
and the ellipse 9y> + 1 6^^ — 1 6aa; + 32a2 » 0.

"- „.»

8207. 0^- J- C. Sharp, M.A.) — If A be the angle contained in
the half of a small Circle of a sphere, angular radius a, by arcs of lengths
b and ; i.e., if ABC be a spherical triangle having the angle A = £ + C ;
show that (1) cos d + cos « = 1 + cos a ; (2) cos A = —tan ^b tan ic ; and (3)
hence deduce Euclid I. 47 and III. 31.

Solution by Rev. T. Gallibrs, M.A. ; Prof. Matz, M.A. ; and others.
Let AO = OB = OC == R ; then, from triangles AOC, AOB,
cos b = cos AO . cos OC + sin AO . sinOC cos AGO

= cos2R + sin2RcosAOC (1),

cos e = cos AO . cos OB + sin AO sin OB cos AOB

= cos2R + sin2RcosAOB (2).

From (1) and (2) by addition, and since AOC + AOB = 180°,

cos* + cos<?= 2cos?R = 1 + C0S2R = 1+cosa (3) ;

also, from triangle ABC,

A cosa— cosJcosc cos3 + cos<?— 1 — cosicosc

cos A = r— r-: = :— T":

2 Binm 2 A^Hc taniJtaai. (4).

4 Sin ^6 COS ^d smi(; COS |0
Now, the formulsp in Plane Trigonometry corresponding to (3) and (4)
are found by writing a — ajr, b — fi/r, c — y/r and making r infinite,
when, a, iS, y become the sides of a plane triangle.

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Thus (3) becomes cos— + cos-^= 1+ cos— (6).

But

cos-^sl^ — . ■^-+— - . j5 .... cos— «1 — - .-«--+— .-3^ — ,..,

cos^«l-±.^+JL.i5l-

Substituting in (6) , we have, after reduction, jS^ + T^^o^ + A — ,

where A is a quantity which remains finite when r is made infinite.

Thus, when r is infinite, 0^ + y^ ^ aK a result which agrees with •

Euc.1.47.

Also, when * - ^, c » ^21, in (4), we have cos A = _^}I^[ ^r).in(yf2p

2r 2r 3!V2ry ' 2r 2r SlVir/ *

also

cosA^l-J-(AV + J_(J.V-..., cosX.i-lf^V + J-fJLV
2r 2 V2r y ^4 ! V2r j ' 2r 2 U^ / 4 ! \2r I

Therefore cos A » ^ ' /rL » where \' and u remain finite when r

l+/*.(l/r2)'

is made infinite ; therefore, when r is infinite, cos A — 0, or A = 90**, a

result which agrees with first part of Euc. III. 31.

8571. (As<^T0SH MukhopIdhtIt, M.A., F.R.A.S.) — Show that
the reciprocal polar of the e volute of the reciprocal polar of the evolute of

the conic ^ + ^ ^ 1, with respect to the circle described on the line

joining the foci as diameter, is the curve

{{•.HmiiniY-dniYHf-i)'-

Solution by Profs. Sibcom , Nilkantha Sarkab ; and others.
The reciprocal polar of the evolute of ^ + ^ =h ^^ respect to
s3+y« a o'-^s is the locus of a;y, where xx^ +yyi = a?—**, -^ « "p^

*^^ "^ + ^ •= ^» whence ^ + -^ =1. The reciprocal polar of the

a* ir X* y*

evolute of the last locus is the locus of xy, where

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and, from (2), _-.*-(-) y, -_-y.^-j ^.
Substituting in (3) and reducing, we obtain the required result.

whencoi from (1),

8311. (R. SwAMiNATHA AiYAB, B. A.)— -Find how many numbers
can be formed, having n for the sum of the digits (zero not being
used as a digit).

Solution by G. G. Stobr, M.A. ; the Proposbb.
Let Pm represent the tiumber of numbers that have n for the sum of the
digits. Of these P», those that have 1 for their first digit are P«-i, those
that have 2 for their first digit are Pn.2, those that have 3 for their first
digit are P»-s, and so on. Therefore

PH-P»-l + Pn-2 + PH-8 +P»-».

ThiiB Pn is the coefficient of a;** in the expansion in powers of s of

, r, that is, of - — "^ -^ .

r(l-»){l-aj(2-«»)}-

l-2a;+a;W

= (!-«) {l+«(2-a^)+a^(2-«»)«+ «»(2-a;»)* + }.

The coefficient of a:" is 2"-(»- 3) 2*-^^ + (n-18)(n-19) ^n-ao^ ^^

_ j2»-«-(«-io) 2-"+ ("-laHn-ao) 2— >-&c.} .

It will be noticed that, if n be not greater than 9, or if the radix of the
scale of notation be not less than n + 1, Pm = 2**~ \

7356. (Professor Wolbtenholmb, M.A., Sc.D. Suggested by
Quest. 7286.) — ^At each point P of a given curve is drawn a s^aight line
U, makinga gven angle with the ttrngent at P, and a straight line Y,
such that U , v are equally inclined to the ordinate at P ; prove that the
point of contact of XT with its envelope is the projection upon U of the
centre of curvature at P, and that the point of contact of Y with its en-
velope is the projection upon Y of the image with respect to the tangent
at P of the centre of curvature at P. [That is, if O be the centre of cur-
vature at P, and OPO' be a straight line bisected in P, then, if OL, OIVI
be let fall perpendicular to XT, Y respectively, L, M will be the points of
contact of U, V with their envelopes.]

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Solution by Profs. Nash, M. A. ; Nilkanta Sarkar, M.A. ; and oth^t,

1. Let P, P' be two adjacent points on
the curve, PR, P'R the positions of the
line U, and O the centre of curvature at P,
then ultimately P, P' may be considered as
points lying on the circle of curvature.
Now, since Z OPR = Z OFR,
therefore PR'OR are concyclic, therefore

ZPRO«ir-ZPFO.

But in the limit PP' is a tangent, and PFO
a right angle, therefore OR is at right
angles to PR.

2. The line Y makes with the normal an
angle iir~ a — 2^, where <p is the inclination
of the tangent to the axis of x.

In the figure,

POP'- 9<p, QPR - iir-o-24»,
QT'R = iir-a-24> + 25^,
PEP' = RFQ'-RPQ-POF = 5^,
RP _ sinRFP
PP
and PF

therefore

sin84>
OP5^==OPsin8<^,
therefore RP - OP cos RPQ ; therefore, if PO' - OP, R is projection of
upon V.

8548. (Asparagus.) — Two straight lines turn about two fixed
points 0, O' with angular velocities which are as I : 3, both straight lines
coinciding with OO' initially ; if P be their point of intersection, prove that
the envelope of a straight line drawn through P at right angles to OP
will be a parabola whose directrix passes through 0' and is at right angles
to 00' and whose latus rectum is twice 00'.

Solution hy R. F. Davis, M.A. ; G. G. Stobb, B.A. ; and othen.

Produce 00' to S, so that OO'-O'S ;
and let zPOS-0,

Z PO'S - 36, Z OPO' - 20.
Draw PZ, O'Z perpendicular to OP,
00' respectively. Then, since O, O',
P, Z are concyclic,

ZPZO' = POO'-e,
and zSZ0'-0Z0'-0P0'=2a.

Hence PZ bisects the z SZO', and therefore envelopes a parabola with
focus S and directrix O'Z.

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8607. i^- Curtis, M.A.) — Between two curves, /(a:, y) — and
^ (Xf y) s 0, of degrees m and n, show that there can be drawn mn straight
lines of a given length /, parallel to a g^yen straight line, from the curve
f{xy) « to the curve ^ {xy) — 0, and mn others from ^ {xy) = to/(a;y)
= 0. [The lines must be drawn in a given direction, say from left to
right.]

Solution by Rev. T. B. Te&ry, M.A. ; Professor Bbtbns ; and otken.

As the degree of the curves is not altered by changing the axes, we may
take the axis of x parallel to the given straight line.

Since /is of the m^^ degree, and of the it^, there are mn points of in-
tersection of the curves f(x, y) = and ^(2:—/, y) ■» 0, therefore mn
straight lines can be drawn from f to (p parallel to axis of x and frx>m
left to right. Similarly, the other result.

8405. (F. 0. Wage, M. A.)— If aj + flj + Oj + .. . + a^ « «, prove that

(t-')"(T-'r {i-)"<<"->--

Solution by R. F. Davis, M. A. ; Rev. T. R. Terbt, M.A. ; and othert.
If a|, Oj, a„ be positive integers, take 01,03, an quantities each

equal respectively to - — 1, ~ — 1, — — 1 ,

€ti 02 an

in all « quantities whose sum = tw— » = (»— 1) ». The inequality of the
question is the expression of the fact that the Arithmetical mean of these
quantities is greater than the Geometrical mean.

8557. (Professor Mathews, M.A.)— If

^+i^ = l and ^+C -1

are two oonfocal ellipses, such that polygons of r sides can be simul-
taneously inscribed in the second ellipse and circumscribed to the first,

prove that a - g^sn ^^"^)^ , i - *' en — .

r

where the modulus of the elliptic functions is equal to the eccentricity el
the inner ellipse. Verify the above when r » 3, 4, 6 respectively.

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Solution by John Griffiths, M.A.

The solution of this question is contained in my paper "On the
Derivation of Elliptic Function FormulsB from Confocal Conies," given in
the Proceedings of the London Mathematical Society, Vol. xiv., p. 47.

8420. (Emily Perbin.) — Prove that the axis of perspective of a
triangle and its pedal triangle is the common radical axis of the ciroum-
circle, nine-point circle, and self-conjugate circle.

Solution by H. 0. S. Davibs, M.A. ; W. J. Babton, M.A. ; and othen.

Let ABC be the triangle, H its
orthocentre, DEF the pedal tri-
angle ; and let EF meet BO pro-
duced in P. Then, since BFEO
is a cyclic quadrilateral,

PB.PO-PE.PF
and PA2-PB.PC + AE.AC.
Hence

PB.P0=PA9-AH.AD=PH2-HD2 + AD2-AH.AD-PH2 + AH.HD.

Thus P is a radical centre of the three circles of the question. Similarly,
if FD, OA meet in Q, and DE, AB in R ; then Q, R are also radical
centres and the axis of perspective PQR is the common radical axis of the
three circles.

(R. Curtis, S.J., M.A.) — Find (1) the locus of a point in a
material lamina, such that if the lamina were moving without rotation in
its own plane, and that point were suddenly fixed in space, the ensuing
rotation of the lamina would be given ; (2) where the point should be if
the rotation produced is a maximum ; and (3) where the point should be
in order that the loss of kinetic energy would be given.

Solution by Professors Steggall and Betens.

Taking the axes of x, y through the centre of gravity, and that of x
parallel to the direction of motion, u being the initial velocity, tv the
final velocity about the point a:, y, arrested ; then uy^w {k^ + ar^ + y^),
whence locus (1) is a circle ; point (2) is given by a; = 0, y = k, as may
be seen geometrically ; locus (3) is given by «*— (Jk^ + aj'^ + yv^ m;2 « con-
stant, or -TJ"* la^ ~ ^» where b is arbitrary.

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7707. (Profeasor 8. Malbt, F.B.8.}— Prove that

rlog Bin e jlog tan e-log A (9)}

J^iogoos e {log cot ^-log A (fl)}
AW de^i^K,

rlOgA(e) {log A («) -log SID e cose}
AW ^*>-iKlog*log*' + i»K'log*',

wliereA(a)s(l-Jt«8in«a)*, 1«+*^=1, and K and K' are complete
elliptic integrala of the first kind with moduli lespeetiYely k and i/.

Solution by D. Edwabdbs ; Sabah Mabks, B.Sc. ; and oihert.
Let « « amu, A = ~^, B = 2ife^4-» C = it^; we have

log-£-mM--i::^coe^-il^co8?^-4i=2?cos?^-&c..

IK 1

log an f* A* « 4K log — — i»K',

logcn»-log*B.l±fcoBf-i;-^,co.f.i;-±Jcoa«^-&c..

logdnit = logC+4 / j£_coe^ +^_^coe^+&o. j.
Hence I [(logsn<()'-log8ni(logcn«dni(](ft( >-— E I log ^]
-21og^j^logsni«<i« + KlogtBClog^+K(l + -l. + l- + &c.),

rK

or, substituting for A, B, 0, and I log sn m <^ their values, this reduces to
the stated result. In the same way we find

i>K

I [(log en «)'— log en u log sn m dn m] <2m — ^ifi K.
Jo

And putting K~ m f or m in this, and subtracting from the first integral,
we get

-log;fc»riog ^"^''" tfm- r[aogdnf»)«-logdnf#logsnf#cnf#]«ff#.
Jo cn« Jo

But ^\og''^^^^^du^[^\og-^^du(K^uiotu)^(i,
Jo CUM Jo sntidntf

and therefore, &c.

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8977. (Professor Haughton, F.R.S.)— The form of the Terrestrial
Radiation Function has been proved to oe A (B—Bq)'* = a, where A, e©,
n are unknown parameters, and e, a are g^ven by observation. The mean
monthly observations at Greenwich, ext^ding over thirty-six years, g^ve

January, A (38-9-eo)" = 21-4 ; February, A (40-4-eo)*» = 36-6 ;

March, A(42-8-eo)»» - 66-9. Find e^,, A, and n.

Solution by A. B. Johnson, M.A.

log4«lr:®o /iogl?l=e. . log ^±^ /w^,

"38-9-eo/ *40-4-e„ *21-4/ ^3o-6'

1*5 / 2*4 21981

or, as a tentative approximation, — -— / — -- — -— = — - — ,

' *^*^ ' 79-3-2eo/ 83-2-2eo 19718*

or 162-6-400 - ^|f|] x 3-9 ; therefore %^ - 3716.

23177

Substituting this value in the equation

^log g'^^e^ - 1^8^ i^l' ^« ^^« « = ^'^^^ - [l"-91341].
Lastly, find A from the equation

log A+n log 6*66 = log 66*9, whence A = 13'63 = [1-13129].

On using these values for A, %y ft, there result the values 21*40, 36*63,
66*90 in the left-hand sides of the observation equations, so that, to judge
by the standard of accuracy taken, the results are sufficiently near the true
ones to dispense with further approximation ; summarily,

A = 13*63, n - 0-819, % = 37*16.

[The Pkoposbr remarks that this question involves, for the first time,
60 as an unknown quantity equal to the control temperature that guides
radiation.]

5955. (Professor Sylvester, F.R.S.) — By a Cartesian Oval in space
let us understand a curve the distances of whose points from three fixed
points are linear functions of each other, or, wluch comes to the same
thing, is the intersection of two surfaces of revolution, described by two
plane Cartesians having a focus in common. Conversely, when two points
can be found whose distances from any point in a space-Cartesian are
linear functions of one another, let them be termed foci. Required to
prove, that the locus of such foci is a plane curve of the 3rd deg^e. It
will be observed that this curve for Cartesians of double curvature is the
exact analogue of the three foci in a straight line for plane Cartesians.

Solution by W. J. C. Sharp, M.A.

Let the Cartesian in space be the intersection of the surfaces generated
by the revolution of the plane Cartesians tnr + r^ ^n and mV + /' ■» n'
about the lines joining their foci ; let the axes be taken so that the quanti-

VOL. XLVIl. I

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70

ties r, r', r" are measnred from the points (0, 0, 0), (A, k, 0), and (A', 0, 0)
respectively. Then, if (a?, y, z) be any point on the curve,

r"2«a:S + ya + «2.2yar + A'2,
and therefore 2cc ^ ^^^^^/^^ ^ r»(l-m-»)4-2mVr.^A-2-n-^

= Ar2 + 2Br + C, say,
where A, B, and are constants ; and

2 _ r^-r'^ + h^-^k^-2hx {il'-m^)r^ + 2mnr + h^ + Ji^''n^^2hx]
^ r^ " k

— AV + 2BV + C, say, where A', B', and C are constants.
H p8 = a;2 + y2 + g2_2|a:-2i7y + p + i7«,

p«-r»-|(Af3 + 2Br + C)-u(AV2+2BV + C0+p + i,a
= r»(l-A{-A'i7)-.2r(B{ + B'i,) + p + i,2-C|-C'u
■■ (Hr + K)2 (where H and K are functions |, 17 and constants),
if (l-A^-A'„)(? + ,r'-C{-C'i,) = (B{+B'i,)»,

and therefore if ({, ij) lie upon this plane circular cubic, p = Hr + K, and
if (^', ij')be any other point upon the plane curve, o* = H'r + K', where
H' and k' are functions of (I', V)* ^^d constants, ana therefore

H'p-Hp' = H'K-HK',

and therefore, &c. The three given foci lie upon the circular cubic.

8924. (Captain Bbocabd.) — Former I'^quation des paraboles
tangentes aux deux bissectrices de chaque angle du triangle et aux per-
pendiculaires aux cdt^s de cet angle en leurs milieux. Ces coniques
admettent, comme on sait, pour directrices les medianes et pour foyers les
sonmiets A", B", C" du second triangle de Brocard.

Solution hy Professor Boubals ; B. F. Davis, M.A. ; andothert.

Prenons pour axes les c8t6s A Car, ABy du triangle. L' Equation
g^n^rale des coniques tangentes aux deux bissectrices AE', AE'' est

X (y2-a;2) + (y + fxxi-y)^ = 0.

Elle repr^ntera une parabole si x = /i-— 1, ce qui donne

(jii/ + xy+2v(fAx + t/) + p^^ 0.

Les perpendiculaires aux milieiix des c6t^s AC, AB (ou mediatrices) ont

Sour Equations ar + ycosA— i* = 0, arcosA+y— ^u = 0. La parabole
evant Itre tangente h ces deux droites, on a les conditions

v8in»A + d(/i-cosA) « 0, i^ sin^ A + c (1 — /i cos A) = 0,

d'oii Ton tire ^ « ^±A£2L4, ^ « _J^*£_.

b + c cos A b + c cos a

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L'^quAtion de la parabole est done

[(tf + * cos A) y + (* + c? cos A) a>]*

— 2bc[{c + i cos A) a? + (b + tfcos A) y] + b^e^ = 0.
Si maintenant Ton prend le triangle ABC pour triangle de r^f^rence,
les f ormules de transformation

^ ^7_ M_

aa + bfi + ey aa + bfi + ey

donneront pour 1' Equation en coordonn^es trilineaires de la parabole

L' equation aa— (efi + by) cos A = repr^sente la corde de contact des
deux tangentes rectangulaires AE', AE'^ D'oti Tidentit^

^ + c2— flS « ^i,c + be) cos A « 2be cos A,
qui exprime que cette corde passe par le point A".

Autre verification : — La symediane AA" doit dtre perpendiculaire k la
corde de contact. Or, la symediane a pour equation efi—by ^ 0. On
doit done avoir identiquement (Casey, A Treat, on Conie Sect,, p. 63),

(^2- c2) cos A- cos A (be- *«) - cos B (— oi) — cos C {ae) — 0,
ou (J2— (j2)cosA + a(ico8B— <?cosC) — 0,

ou (i2-(jS)(*2 + ^_^2 + «2j + ^_J4«0,

ce qui est bien une identity.

8658. (W. J. C. Shabp, M.A.) — BooLB obtains the result

^(n-l)(|--2)^_3j3.^^

by a symbolical method ; show that it may also be obtained by ordinary
methods.

Solution by Professors Bbyens, Stone, and othen,

Cette formule est certain pour » =» 2, n — 3, etc., et nous aliens k
demontrer que si elle est verifi^e pour le cas (n) elle sera pour le suivant.
Supposons-nous que pour (n— 1) integrales on aie

III

,..Xdx^-^

nous aurons f f f f ... Xdx « [ ^* f (( **~ — X^^**~* "

[^3? / ix*'-^{xdx-{n''2)x''-^{Xxdx...±{Xx»'^dxl

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Mais, appliquant la m^thode d'int^g^tion par parties, nous aurons

-(«-2) (dx.z^'^iXxdx:^ -x'^-^lXxOx^iXaf^'^dx,

(n-2)(n-3) C^^^n-i [Xx^dx ^*^ s^-^{Xx^dx^^ [Xx^^-'dx,
1.2 J J 1.2 J I.2J '

et ainsi de suite nous aurons finalement,

:h{dx{Xx*"^dx '»±x{Xa^'^dxT{Xa^-^dx.

Bemplafant ces valeurs en (1), le coefficient de | Xi:**'^^^^ sera f apr^

avoir s^par^ le facteur commun -— ; — — J ,

1.2. 3... (rt— 1)/

-l.(n-l)- ("-;)(-^) ...=F(n-l);

mais, d'apr^s une propri^te bien connue,

_H.(„_1)_ <JLrlK^ ... =P(„_1)±1 = 0,

1 • ^

suivant (n — l) pair ou impair, et par suite la formule (1), deviendra le
th^r^me propose.

fLet T^''^' therefore f f ...Xdx** = y,

and, by integration by parts,
{XxPdx={xP^dx^xP'^^^pxP''^J^^p(p-l)xP'^^-&c.

Now, if all the values 0, 1, ... « — 1 be given to p, the result of eliminating
the successive differential coefficients of y will be the given formula. I

Note on a Probability- Qubstion. By the Editor.

The problem in Question 7624 having excited much interest, we give
space to the subjoined additional remarks and developments in regard
to the solutions and notes that have been published in our previous
volumes.

Mr. Simmons' solution seems to Mr. Putnam to be ** only applicable when
A is the first man to enter the train, and B the second. In that case there
are g—l vacancies in the car A has entered, and pg—l vacancies in the

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train. The chance is then {g-l)l ( W— 1) . It is impossihle, however, to
find any other condition in which this will prove true. Instead of throw-
ing out tn, Mr. Simmons' solution assumes m ^ pq—l. By the conditions
of the prohlem, the chance under this assumption is the least possible.
Although Mr. Biddls's solution is intricate, I do not see how it can be
avoided, if the solution is to follow mathematical principles.

" If we may eliminate any of the data, let q be thrown out. Then, if
there are vacant seats anywhere, the carriage in which A is seated will
have an equal chance with the rest, and the number of vacancies belong-
ing to it will be w/p, and the chance of B*s taking that carriage will be
l/nt X m/p = l/pf or the same chance there will be that B will enter the
same carriage A is seated in without reference to vacancies.'*

In further reply to Mr. Simmons, who, in Vol. xlvi., p. 37, has g^ven
what he considers to be test examples to prove his case, — viz., that m, in
the question, is an immaterial quantity, — Mr. Biddlb wishes to say tiiat
he is equally convinced that ^e subject cannot be so summarily dis-
missed." A very similar (not to say identical) question occurs in Mr.
Whitwobth's work on Choice and Chance (No. 229, p. 234) ; but, although
answers are g^ven to most of the questions, none is given to this, which
would scarcely have been the case if the author had considered the matter
so simple as Mr. Simmons does.

'* One thing is clear, the particular passenger. A, must not be re-
garded as being selected at the time the new-comer, B, enters the train ;
nor, on the other hand, must he be regarded as invariably the first pas-
senger who takes a seat. So far as our information goes, he is simply
one oipq—m passengers, who, before the train began to be filled, had an
equal chance of occupying any carriage and any seat therein. We must
therefore consider the number of ways in which pq^m passengers can be
distributed in p carriages, each containing q seats, and the probability of
each kind of arrangement. We must next take the total number of
occupied carriages given in the full tale of these arrangements, and find
what proportion of them have one vacant seat, what two, what three, and
so on. These wiU be the chances, as regards vacant seats, for any par-
ticular occupied carriage, such as that which A will occupy, because the
a priori chances are the same for all. And here it may be observed, that
it is only a posteriori that we can say that A has twice as much chance
of being where there are two passengers as where there is only one.

The rest is easy, and the examples given by Mr. Simmons will equally
serve the purpose now in view. " Let us take the particular instance of
3 compartments, each of 3 seats, and test the numerical results."

Here, taking to represent a vacant seat, and X a passenger, the pos-
sible contents of any one carriage will be

(a) {b) (c) {d)

XXX XXO XOO 000

(i.) *' Suppose that, besides A, two other passengers, P and Q, are
already seated."

Then the essentially different ways in which the train may be made up
can be represented as follows, the probability of each being given at the
end: —

{a + 2d) XXX 000 000

(b + c + d) XXO XOO 000

ti

(3c) XOO XOO XOO ^ff.

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*' In ihefirst arrangement^ the first passenger entering decides the carriage,

2 1 1
then — ^ s — ^ the probability that the other two join him. In the

8.7 28 *^ "^ ^

second arrangement, the two first passengers may enter as they list ; if
they go together, of which the probability is f , the chance that the third
enters another carriage is f ; if, on the contrary, they enter distinct
carriages, of which the probability is f , the chance that the third joins

one of them is 4. Therefore r-^ + ^ = -^ — the probability of the

^ 8.7 8.7 28 *^ *^

second arrangement. In the third arrangement, the chance that the
second passenger does not join the first is f , and that the third does not
join either isjf. Therefore f . f =« -^ = the probability of the third ar-
rangement. We now see that, where the first arrangement occurs once,
the second occurs 18 times, and the third 9 times. Consequently, the
state (a) of a single carriage occurs only once, whilst {b) occurs 18 times,
(e) 45 times, and {d) 20 times. Disregarding (d), because A cannot be in
a carriage where the seats are all vacant, and sdso (a}, except for addition
to the denominator, because B cannot enter where no seats are vacant,
we find that, of occupied carriages, the probability that the state shall be
(fi) a I) ; that it shall be (e) » ff . Therefore, since there are, in all,
6 vacant seats for B to choose irom, the probability that he joins A
'^ H'h+H*i=^-ht instead of -^ or i, the probability given by Mr.
Simmons."

(ii.) ** Suppose that, besides A, there are already thre^ other passengers,
P, Q, and R."

Here the specific arrangements are also three in number, and their pro-
babiHties as follows : —

{a+e + d), XXX XOO 00 0, ^

(b + 2e), XXO XOO XOO, if

(2* + (^, XXO XXO 00 0, ^•

From which we gather that the chances of any carriage being as (a), (fi),
(c)f (d) are as 4, 30, 40, 10. Consequently, as the vacant seats are 6 in
number, the probability that B joins A » ^J.| + f$.f « if, instead of
ii o' i» the probability given by Mr. Simmons.

(iii.) " Moreover, if there had been only one passenger P besides A,"
the specific arrangements would be two in number, and their probabilities
as follows : —

(* + 2rf) XXO 000 000 I
i2e + d) XOO XOO 000 f.

" Hence we see that the chances of any carriage being as («), {b), (c), (d)
are as 0, 2, 12, 8. Consequently, as the vacant seats are 7 in number,
the present probability that B joins A = A*f + Tl-f — tv> instead of
il or ^, as given by Mr. Simmons.

"In fact, the only case in which the same probability is certainly
arrived at by both methods, is where A is the sole passenger in the
train when B enters. But if it was a mistake, as Mr. Putnam intimates,
to say that the probability would be the same according to either method,
in the case of only one vacant place remaining, the mistake arose from
judging of the probabilities of A's position by an a posteriori process."

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8685. (The Editor.) — Given an angle at the base of a triangle, the
sum of the two sides, and the distance between the given angular point
and the point of contact of the escribed circle touching the base ; con-
struct the triangle.

Solution by the Proposer.

Let L and / represent the two given lines.
Draw the Z EBD = the given Z , making EB
= L and Bl) = I. Produce BD to F, so that
DF = L + ;; thenBD + DF = BF = L + 2/.

Also produce BE to H, so that BH = BP.
Bisect BH in I : draw IF ; and from I apply
IM = IH to meet FE produced ; also draw
EN II MI, and through N draw KC || BH, and
CA 11 EN. The A required is ABC. For,
by construction, Z ABO = given Z , and BD = I,

Moreover NK : IH=FN : FI=NE : IM=NE : IH, .-. NK«NE=«NC;

but AC = AE; therefore AB + AC = EB « L.

Again, CF = CK = 2CN - 2AC, .-. BO + CF - BC + 2AC = L + 2/;

but AQ + AC = L; therefore, adding, AB + 3AC + BD + DC - 2(L + 0;

but AB + AC -I- BD = L + / ; therefore, subtracting,

2AC + DC = L + /= AB + AC + BD;

therefore AB + BD » AC + DC.

Now, it is well known that the angular point opposite the base of a
triangle and tht point of contact of the escribed circle with the base bisect
its perimeter. Therefore D is the point of contact of the escribed circle
with the base BC.

8501. (H. Knowles, B.A.) — In any triangle, show that

4coi^Acos'B-cos(A-B)(3cosAcosB-sinAsinB) =cos«C.

Solution by Isabel Madison, B.Sc.

Identically, Ax^- {x + y) (3ar~y) = {x^y)\ and putting x •
y — sin A sin B, the required theorem is proved.

cos A cos B,

(F. R. J. Hervet.) — Find in how many ways n lines of
verse can be rhymed, so as to have r different rhymes, and no line un-
rhymed ; and show that, in the case of the sonnet^ the numbers of ways
with 2, 3, ... 7 rhymes are, respectively, 8177, 731731, 6914908, 12122110,
4099095, and 135135.

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76

Solution by the Proposeb.

Let ^ (ft, r) denote the required number ; we shall obtain all the case
as follows. Take aU the cases of » - 1 lines with r rhymes, and append
to each a line rhyming with some that precede; this gives r^ (n — 1, r)
ways. Again, take all the cases of n ~ 1 lines with r distinct endings one
of which is unrhymed, and append to each a line rhyming with the odd
line. The number of these is — (n — 1) 4> (n — 2, r - 1).

Hence 4> («, r) == r4> («— 1, r) + (n— 1) ^ (n - 2, r — 1).

Form a table, as below. Write the first row (of units), and first cypher
of each subsequent row (forming an oblique column) ; and continue these
rows both ways, satisfying the relation ^r (« + 1) = r4>r« +«<^r-i (»— 1) ;
^rW denoting throughout the number of r^ row which falls under the
number n : thus ^r^ = ^ (»> ^) so long as n > r. The values of ^r^ from
« = r to » =s follow an obvious law, which, it is easy to show, must
hold for each subsequent row ; we have, then, 0^0 = (-l)*""^/r!.

1

2

3

4 6

6

7

8

9

10

1

1

1

1

1 1

1

1

1

1

1

-/2

-1

-1

3 10

26

66

119

246

601

/3!

/2

1

1

16

106

490

1918

6826

-/4!

./3!

-/2

-1

-1

106

1260

9460

The complete solution of the equation ««♦!— rtt« == ^, ^f being a
rational integral function of the k^^ degree, is

ii^=:Cr«-{/(r-l) + A/(r-l)2+...+A*/(r-l)**»}if^«Cr--x«,8ay;
whence, if an integral function be required, the only solution is Un— —x***
If the given equation be Un+i—run = «***^4^, we have Un/a" as before,
giving

Un = 0*»-a*»x'»> where x» «= « (1 + «A +...+«* A*) \pn and « = al{r^a).
Again, if Kn+i— rw« = the sum of several terms of the form a'»*^if^
(which denote by 'Xa*^*^^)i we may find Mm by solving as many equations
of the preceding form and adding the results ; giving a solution of the
form «n = Or** — 'Za**xn,

To find ^r-¥\n we must solve the equation «n+i — (»*+ 1) «n = n^r (♦•—I).

Thus 4»in- 1, ^»«2'»-i-(fi + l), 2<p^n » 3'»-»-2»-»(« + 2) + »2+„+i^

64>4» « 4»»-i-3 . 3»»-8 (» + 3) + 3 . 2'»-» (ff2+ 3» + 4)-(n3 + 2n+ 1) ;

generally, it will be shown that

r ! iprn - r~-Cr,i(r-l)»-iP + 0r.2 (r-.2)»-2Q-... ± Or.r-iT,

[otherwise, (r- 1) ! 4>r « = r«-*-Or-i,i (r- 1)»»-*P + ...], where Or,m is
the coefficient of x^ in (1 +«;)♦■, and P, Q, ... T are functions of n with
unity for coefficient of highest power ; these must first be determined.

Unity, in the first term of the above expression, may be regarded as
the starting-point of a new series of these functions, each formed from
the preceding, which are to appear in the 2nd term of ^rfi» the 3rd of
^r+2> &c. ; the law of formation being

/*« = {l+rA/A:+...}{nA-i(n-l)},

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17

which, inverted, gives {l^rAl k)fkn ■» w/i.i («-l) (1).

Thus /o» = 1, /i » = w + r, nfi (»- 1) « n'' + (r- 1) n, giving

giving /j n = w' + (3r— 3) «*- + (df^-Zr + 2) w + r^, and so on.

Thus written, in powers of «, the law is not evident; hut we may
observe (and, for the moment, assume) the relation

/k« = ^*-i(/j-l)+r/A.i« (2).

From (1) and (2), A/**» *= ^/t-i«> whence A%« = k (A;— l)/*.2ft, and so
on; also, from (2), /tO = r/^-iO =» ... == r* ; hence

...+w(n-l)...(ft-A;+l).

This is the function sought ; for it will he found to satisfy (1), which,
/A_iWng supposed known, is an equation of differences for finding/*,
having only one such solution. Denoting it in future by/(r, A;), assuming
the form previously suggested for <pr, and applying the general process
to form <t>r*h ^3 have
rl iprn « r'»-Cr,i(r-l)«-i/(/-l, 1) + a,2(r-2)«-2/(r-2, 2)-...

... ±Cr,^_i/(l,r-l),

r!4»r*in = A(r + l)~-r.r-2/('', 1) + ^ C.,i(r- 1)^-8/(^.1, 2)

-^a.2(r-2)«-V(r-2, 3) + ... T -^C.,..i/(1, r),

(r+1)! ^r*i« - B(r+l)»-a.i,ir»-i/(r, 1)

+ Cr+i,2(r-l)'-2/(r-l,2)-... T Cr*i.r/(1, r),

(r + l)l0r*lO - (-l)*- «=B-Cr4l,l+aM,2-... =FCrU,r;

whence B = 1, and the above formula is completely established ; n "being
any integer, positive or negative. It may albo be written

(r-1) \<t»rn^ r«-i-0r-i,i(r-l)«-2/('--l, 1)

+ a-i,2(r-2)'»-»/(»--2,2)-.... ±/(l, r-1).

If we collect from the last the complete coefficient of a given fac-
torial, «(n— 1) ... (n-w+l), we shall find it (by help of the relation
Cr.mwC«w,«« Cr,fnCr-m,t) to bc ± Or- 1, mA*-"»-i l"""*-! (derived from
l»»-»M-i^ 2"-*"-*, ...). Hence, restricting n to be not < r, we may
easily throw <l>rn into the form

^r« = A(»-)0»-Cn,iA^'-»)0'»-l + On,2 A^*-*) 0«-«- ... ± On,r-l A^D O"-***!,

where A^*") means A^jrl. From this, and from the fact that (prn 'k if
«>r<2r, it may be inferred that A^''JO*'"^^, i? being const mt and > 0, is
an integral function Fr, of degree 2/?, containing the factors r(r + \)
... (r+j3), (which is otherwise known) ; and that (pr i^-^p) or ^(r+jo, r)
is the (r+^)*^ difference of the first zero value F(— ^). [This result
is easily connected with the meaning of A^**) 0" previously found.]

The case n = 2r is noticeable. By the construction of the table,
iPr{2r) = the product of odd numbers 1.3... (2r— 1). This may also

VOL. XL VII. K

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78

b^ inferred as follows. Suppose it specified that p are lines to rhyme
together, g* otiiers to do bo, and so on ; where j3 + ^ + . . . « «. The number
of ways will be « !/ {jp ! §' ! ...} if jp, q ... are all different ; but, if fc of them
are equal, we must divide by A; ! ; and so on. Now, in the above case,
the lines necessarily rhyme in pairs; hence 4> (2r, r) = (2r} !/(2''.r !).
It follows that the coefficient of r^ in FriB I (2P.p !).

2 8 4 6

6

7

8

9

10

11

12

IS

14

1111

1

1

1

1

1

1

1

1

1

8 10

25

66

119

216

601

1012

2035

4082

8177

16

105

490

1918

6826

22935

74316

235092

731731

105

1260

9450

56880

302995

1487200

6914908

915

17325

190575

1636635

12122110

10395

270270

4099095

135185

8820. (Charles F. Lodge.) — ^A mirror, measuring 38 inches by 22
iDches, is to have a frame of uniform width, whose area is to equal that
of the glass ; show that the width of the frame is 5| inches.

Solution by D. Biddle ; H. L. Orchard, B.Sc. ; and others.

Lines drawn perpendicular to the sides of the mirror, from its centre,
divide not only the mirror itself (w), but its frame (/), into quarters.
Moreover, \m^if\ iw» = 2x3x (6^)2 ; therefore ^m + i/ = 4 x 3 x (5i)-.
But 4 = 3 + 1, and 3 = 2 + 1. Consequently 1 x 6^ = the width of frame
required, in inches.

8813. (J« Gripfiths, M.A.) — If we have
f{x) - A^ + 2A2T2 + 2h^x* +... ad inf., ^ (a:) « 1 - 2B2J^'- ^B^s^^... adinf.

where A© - 1 + 22(-)*cn2 ?5, Bj = 2 (-)* en — -1- sn* — ,

n n n

Aj = A;2 2 (- )• en' '^bh^'^, B4 = 2 (-)' cn^ ^ ^ sn* *-5,
n n n n

A4 - A;4 2 (-)• cn2 !58n4?5, B« =. 2 (-)' cn^ *-5 ^ sn«^,
n n n n

and a is a number from 1 to «— 1, show that /(a:) . <f> (.r) = A©.

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79

Solution hy D. Edwardes ; Sarah Marks, B.A. ; and others.

Referring to Mr. Ghifpiths' Paper {Ptoceedinga of London Math. Soc,f
Nov. 12, 1886), we have

Jy « 8ii(Mm, a), X = 8n(w,^'), «o= — » ^o - -g^ K,

mod. A:, * = 1 to w-1, < »« 1 to «,

M = nsn^*-^ + nsn^?^ K, A - k^" nsn* ?^ k]
/» 2n 2»» j

Ma arn-irM( (-)'cn2?-. ^

* (l-A'^a^y l->l«Bn»*-5:.A

^ /» «^

= ^I1:=:^{A, + 2A^»+2A^+ }.

(1-AV)*

Changing y, a? into 1/Ay, I/Aj:, we have

M^ ^ (1-A'V>* 3 „ ^ ^ n

'"•"-''r -"T[-=/(-t)j:

= 11:^^!^{1-2B^«-2B,:.^- }.

Hence, M-A/A:* =/c.^ar, and, when a? and yare small, M'a/A;2 = Aq,
and therefore, &o.

8117. (Professor WoLSTENHOLMB, M.A., Sc.D.) — ^Two conicoids S, S'
have two common plane sections, and the poles of these planes with
respect to S are the points P, P' ; prove that (1) if S' pass through P, it
will also pass through P' ; (2) also, in this case, the following relations

must hold (02-.3*A) (ee'-6AA') + e2AA'= 0, 6' = 27^^ (ee'-6AA').

[The discriminant of XS + S' is aA^ t eA-^ + ♦A^ + e'A; + A'.]

Solution by Professor Nash, M.A.

Let S = ax'--^by^-¥cz^-¥du:^ - 0, and let P, P'he the points (0, 0, 0, 1),
(X, Y, Z, W), then the equation of S' is

flA2 + *y2+«2 + rfi^ + 2u;(a;rX + 6yY + czZ+<fM;W) = 0.

If this passes through (0, 0, 0, 1), 1 + 2W =« 0, and S' hecomes

ar' + V + <^«'+2u;(aarX + *yY + csZ) - 0,

which is satisfied by the coordinates ;X, Y, Z, W).

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80

The discriminant of A:S + S' ia

(k + l)a, 0, 0, aX

0, (k + l)b, 0, *Y

0, 0, {k+l)Cy eZ

flX, *Y, fZ, kd

»0,

or

putting

therefore

therefore
therefore

k(k + l)^abed-(k + iyabe{aX^ + bY^ + cZ^) « 0;

aX^ + bY^' + cZ^^dF,
abcd{k^ + dl<^+{Z-F)k^ + {l-'2F)k^'P} =0;

A ^ e ^ ♦ ^ e' A'

1 "" 3 3-P 1-2P" -P'

e2-3A* = 3Pa2, ee'-6AA'= zaK

From these we obtain the given relations

(e*- 3a*) (ee' - baa') + ©«aa' = o, e< « 27a2 (ee'- caaQ.

[It would be better if the coordinates of the second point P' were written
X . Y : Z : W ; so that X, Y, Z, W may involve an arbitrary multiplier ;
otherwise the equation 1 + 2'W «» will seem to limit the position of P'.
The general form of S', when S is given, is

where (XiY^ZiWi), (X^YjZaWa) are the two points P, F.]

X

rfS
dx'

...),

8B69. (Professor Stbgoall, M.A.) — A shot of mass m is fired in
vacuo from an air gun of length /, with a charge of air that at normal
pressure p would occupy a volume v ; this air initially occupies a length b
of the barrel of the gun. Show that the time of passage along the barrel
and the velocity with which the shot leaves the gun are gfiven by the

equations T = {l + il^bi) ^ (S^w/m)*, V = (1 -bil2fi){opvlm)\

where b is small compared with I, and the ratio of the specific heats of
air is faken as 1*4.

Solution by D. Edwardes ; Charlotte A. Scott, B.Sc. ; and others.

If P be the pressure at time t on the shot, the equation of motion is
mS = P = Ox'i (7 ratio of specific heats), on the supposition that there
is no communication of heat. By the Question C — pvy^.
Hence, integrating between I and b, we have

mV«.5j,e;(l-^), or V ^ ( ^ )*( 1 - ii ) approximately ;

l^Yat=: ^— ,

V m J (I~i?/u;l)»

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81

;* z5&

■fterofore ('£!L\* T ^ & r -^

- Sit (rf - j!)»— "f-i! (Jl -48)! + 1 (i!_iS)l

8931. (Emile ViOARifi.)— On donne deux points et A et on con-
sid^re toutea les paraboles ayant O pour sommet et qui passent par A.
Trouver geom^triquement le lieu (1) du point de concours des tangentes
en A et O ; (2) du point d* intersection de la normals en O et de la tan-
gente en A (ces deux lieux sont tangents en O) j (3) du point d' inter-
section de la tangente en O et de la norraale en A ; (4) du point de con-
cours des normales en O et A (ces deux lieux sont tangents en A).

Solutions by (I.) Professor Schoutb ; (II.) R. F. Davis, M.A.

I. In the diagram Pj, Qg, Rj, S4 are the
points that describe the lour loci, when
the axis a of the parabola through O and
A rotates about O ; M is the centre of the
segment OA, O of the segment A'A ; AL^
is perpendicular to the rotating axis.

(1). As the locus of Lq is the circle
described on OA as diameter and

OPi iALo,

the locus of Pi is the circle described on
OM as diameter.

(2) The locus of Q^ is the circle described
on A'O as diameter.

(3) and (4). The couples of points Pj and R„ Qj and 84 are so situated
on radii vectores through 0, that the segments P1R3 aod 0.^84 appear
from A under right angles. Therefore the loci of R3 and 84 are the curves
that correspond to the loci of Pi and Qg in the involutive quadratic trans-
formation of the points on radii vectores through and conjugated to one
another with respect to the point circle A. 80 we find for these loci uni-
cursal circular cubics with the isolated point 0, that touch in A the per-
pendicular to OA. The nine points common to these two cubics are O
counted four times, A counted twice, the two circular points at infinity,
and the point at infinity common to the perpendiculars to OA. The real
asymptotes of these curves are different ; the first intersects OA in a point
B, so that AB = OA, the second in a point B', so that AB'= A'A.

The particular kind of quadratic transformation made use of in this
solution is treated analytically by Prof. G. de Longchamps ("Etude
d'unc Transformation Reciproque," Journal de Mat/iematiques Specialea,
1882, p. 19), and geometrically by mo (**8ur la Construction des

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82

Cubiquos Unicursales," Annuaire de V Association Frangaise pour VAvanec'
ment des ScienceSy Congr^s de Grenoble, 1885, p. 169).

II. Otherwise: Let A« be the
ordinate of A to the axis of any
parabola having () for its vertex.
Then if, on this axis, On' be taken
= Ow, Aw' is the tangent at A,
and y, its middle point, lies on
the tangent at the vertex. If
yE be drawn parallel to the axis,
it bisects OA in the fixed point
E. Now (1) the locns of y is
the fixed circle described upon
OE as diameter. (2) if AO be
produced to A' so that A' = O A,
the locus of n' is the fixed circle
described upon O A' as diameter.
[These circles touch each other
in O.] (3) If a fixed straight
line through A perpendicular to
OA meet the axis in i, then
A»*^ = On . bn^ and bn = latus rectum. The normal at A will therefore
bisect bn in y. Now, the locus of n is the fixed circle upon OA as diameter,
while A* is the tangent A to the same circle. Therefore, since O, n, ^, b
are collinear and bff «= w^, the locus of ^ is a certain curve symmetrical
with respect to Ok. touching the circle in A. [This curve is a circular
cubic J its polar equation is r = a (sec + cos 6) and its Cartesian equation
y2 (x^a) = x^ (2a— ir) referred to O as origin and OAas axis of x. It has
for an asymptote the line through E perpendicular to OA. OA =« 2/i."|
(4) Since g bisects bfij it may be easily shown that e bisects hm. But
O, m, Cj h are collinear, while the locus of m is the fixed circle upon OA
as diameter, and that of c the tangent *Acto the same circle at A. There-
fore, as before, h describes a certain curve [r=2a (2 sec — cos 0)] symme-
trical with respect to OA. These two curves touch each other at A, and
lie upon opposite sides of their common tangent.

8716. (Professor Mathews, M.A.) — Prove that the real common
tangents of the circles x^ + y^ — 2ax — 0, x' -^y^-lby = are represented
by 2ab (x^-^y^- 2ax) = {by -ax + aby, or, which is the same thing, by

2ab{x'^ + y^-2by) = {by-ax-ab^.

Solution by W. J. Babton, M.A. ; Kate Gale, B.Sc. ; and others.

By elementary geometry we have, if (xq, y^j be coordinates of the
point of intersection of the common tangents,

ab

whence XQ= — yQ—-

a-b

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83

Substituting these values in the equation to the pair of tangents from
.(a^o» Vq) to x^-¥y^-2ax = 0, viz.,

and dividing by — -, we get, for the equation to the real common tan-

gents, 2ab (x^ + y^ - 2ax) = {by -ax + ab)\

8660. (By B. Hanumanta Rau, M.A.) — ^Investigate the singular

solution of the equation ^[^"^V j-] +25fy2_a-yJ^j_o, and show

that it is the envelope of a series of circles described on the subnormal of
an ellipse as diameter.

Solution by Robert Rawson, F.R-A.S.
From the g^ven equation, we have

^._:^i^+ 445 = 0).

4 y y* 4

Then, the (p) and {c) equations are readily found to be

8yi?-17a;=F5(9a;2-16y2)i^0 (2),

C-log{(9.»-16,^)* i 6x} T ^,^,^,,%,^,^ - (3).

From (2), ^ = -17ar=F(9^»-16yg)* ^ JO . (4).

dy 8y2 (9;c2- 162/2)*

Now, (4) becomes (00 ) when y = and 4y = ± 3a;, neither of which is
a solution of (1).

-Again, ^[L]^ 8{^45.^17(9.^~16,^)ny ^^^^

d^^P I (9jr3- 16^)* {17a; ± 6 (9a;2- 16y )*}

Now, (5) becomes (00 ) when 4y = ± Sa; and 4iX^ + 2by^ = 0, neither of
which satisfies (1).

The condition of equal roots of (2) and (3) fails to give a solution.
It appears, therefore, that neither of the above recognised methods
avails in this case in finding the singular solution. The present theory of
singular solutions of differential equations is not, I fear, free from the
reproach of not being in such a satiafactory state as it might be, consider-
ing the attention which has been given to it by every mathematician of
eminence since the days of Dr. Bkookb Taylor. Indeed it appears, from
the researches of Sir James Cockle and others, that

^ = 0, -^ = 0| derived from the C-cquation,

dC d\j

= 00, -^ — ) = 00, derived from the jt7-equation.
P I

dp d

dy dp

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84

The condition of equal roots of the C -equation and j7-equation can have
but a limited application. Another solution besides (3; can ie obtained
by the following consideration : —

Let C + R i/CS") - (6)

be the C-equation, where R, S are functions of x, y.
Thej^-equation, from (6), becomes

P-^ —TTl^ 1 ^ ^^^•

figs " ±/(S)'"^s

Now (6) is, therefore, the complete primitive of (7), which is rendered
tt-i

exact by the multiplier 1/«S ** . But (7) is also satisfied by

S = (8).

Hence (6) and (8) are both solutions of (7), and if (6) does not include
(8) by giving a constant value to iC), then (8; is a singular solution by the
usual definition. The solution (8) may, or may not, be of the envelope
species ; this depends upon the functions R, S.

Compare (6) with (3), then

when S = 0, then {9x'^'-l6if^^ ± 6x ^ ./9).

From (9), a;2 + y2 = q (10).

Now (10) satisfies (l),andis not included in (3) as a particular integral.
It must, therefore, be a singular solution by the definition.

Professor Cayley, a very great authority in this matter, would I
apprehend reject this solution as being singular. At all events, whether
it is singular or otherwise, it is a solution of (1), and is not included as a
particular case of (:i) by giving a constant value to (C).

The locus of (10) is two straight lines, viz., y = ± tr, where t = v^— 1,
but the relation of (10) to the complete primitive is not readily observed.
It is not difficult to show that (10) is the envelope of a series of circles
described upon the subnormal, as diameter, of an ellipse or hyperbola
whose excentricities satisfy er+l = 0.

8727. (B. Hanumanta Rau, B.A.)— The sides BC, CA, AB of a
triangle ABO are produced to a, b, c, such that Cfl, A.b, Be are respectively
equal to BC, CA, AB. Prove that the centroid of the triangle abc coin-
cides with that of ABC.

Solution by Rev. D. Thomas, M.A. ; C. E. Williams, M.A. ; and others.

Let o, /3, 7 be coinitial vectors of A, B, C respectively and o', /3', y' of
a, bj e respectively ; then a'= jS + 2 (7 — /8) and 5a' = 2a, therefore, &c.

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85

8135* (I^v. T. C. Simmons, M.A.)— If Ot be the centroid, I the in-
centre, of a plane triangle, prove that

IGP « fR2 (1 +C08 Acos B coBC)-iRr + lr\

Solution by Kev. T. Gallibrs, M.A. ; G. G. Storr, M.A. ; and others,

Ji r ^ distance between two points whose trilinear coordinates are
(«i, Pm 7i)> («2» ^i 72)> we know that

»« = ^ {« cos A (01-02)' + * cosB (i3i-j82y» + « cos C (7,-7a)«},

^ _ sin 2A (oi -fla)2 + sin 2B (3i - jSg)' + sin 2C (yi— 72)'
2 sin A sin B sin

For 1 we have o, = 2j3i - 71 = y** and for G, oj = ^p,, jSj-i/J^ 78 = fc»i»
where P|, ;D], j?, are the lengths of the perpendiculars from A, B, on
opposite sides.
Also jOi « c sin B = 2R sin B sin C, and so for p^ and p^.
Therefore 2 sin A sin B sin C . IG^ « (sin 2A+ sin 2B + sin 20) r*

— fRrsinAsinBsin (cos A + cos B + cos 0)
+ 4lt«(8in2Asin2Bsin2C + sin2Bsin2Csin«A + sin20 8in«A8in»B);
but sin2A + sin2B + sin2C a» 4sinAsinBsinC,

therefore IG' = 2rJ- |Rr (cos A + cos B + cos 0)

+iB? (cos A sin B sin + sin A cos B sin C + sin A sin B cos 0).
Now, cosA + cosB + cosC = 1 + 4 sin ^A sin ^B sin iO «= 1+r/R,
and the trigonometrical multiplier of

iB? asinGsin(A + B) + sinAsinBco8Cal-i-cosAco8Boo80;
therefore, finally,

IG2 = 2r»- Jra-|Rr +iR2 (i + cos A cos B cos 0)»

or « JR» (1 + cos A cos B cos C) -^Rr + |r»,

the required result.

2935. (Professor Sylvester, F.R.S.) — If f(x) be a rational integral
fimction of a; of a higher order than the second, prove that it is impossible
for (fx)^^ {fxY to be a perfect square unless /(a;) contains at Iwwt two
groups of equal factors.

Solution by W. J. C. Sharp, M.A.

If /(a;) be of order «, and F (x) = {f{x)Y + {/' {x)}^ F {x) is of order
2w. Now, if F(a;) be a perfect square, it measures its own Hessian
(Quest. 8296), which only differs from 2«F . F"-(2m- 1) F2 by a positive
factor (Quest. 5762) ; therefore F must measure F'^ and F' = 2/' (/+/"),
therefore /a+Z's measures /'2t/+/")2 and/'2 (/+/'7 = {P-^fW'^'^^U

TOL. :LLyu, L

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where uisa. function of x of an order not higher than 2n— 4 (and hcncoi
all that follows implies n > 2) ; therefore

f^ {2fr+f^-f^~u} =/% (1),

and either/ and/' have a common measure, and /has one group, at least,
of equal factors, or m is a multiple of /'*^ ; but/'* is of order 2/i— 2, and u
at most of 2»>— 4, and therefore there is at least one group of equal
lactors.

Now, let f{x)^{x^a)^<p{x) and f {i) == {x-a)'^'^4>i{x),
and ^ /'W = (•!?- «)•"-* ^2 W,

then 0, ^1, and ^^ ^^^ ^^^h of order n—m, and, if ^ {x) does not contain
x—a as a factor, neither do ^j {x) or <p.2 {x). From (1;,

or (^i* {2 (flj-a)*"*-^^^^^ (a; _a)2"«-4 <^^a _(a;«fl)2m-2^ja_t<} = (ar-a)2<^«M,

and, therefore, u must be divisible by {x^a)'^*'*-K Let « s (a?— a)**~* v,
then V is at most of order 2 (n—m), and

4>i2 {2 (a?-«)2<^4)3 + 4>32- (j;-a)- 0i2-r} b (j?-a)2<^,

ai^ either ^j and ^ have a common measure, or v s <p^u?, where tc is in-
dependent of Xf and

2{x-ay i^<p.2 + 4>22 - (x - ff)2 4>i2 _ ,^^^2 s (a; _ a)2 (^ V,

an identity which requires that the two leading coeflBcionts of <p should
vanish. Therefore ^j and ^ have a common measure, and therefore ^ and
fp' for ^1 = m^-\-{x—a) <i>\ and there must be a second group of equal
factors.

7459. (Professor Wolstenholme, M.A., Sc.D.)— Prove that if, in a
tetrahedron, any one or two of the eqnutioiis a :kx = b ^p = e ±. z be
true, then will also the corresponding one or two equations of the set
A±X«B±Y = C±Zalsobe true.

Solution by the Proposer.

In the tetrahedron OABC, OA = a, OB = b,
O^ ^ e; BO = ar, CA = y, AB = « ; suppose a
«phere can be drawn to touch the four edges
tf, b, X, y (none produced), then, if t^, ^j, /g, t^ be
the lengths of tangents to this sphere from O, A,
B, C, a=<i + ^2, X=t^-ltt^\ b^t^Art^, y=^t^-\-t^,
whence a^rx =^b+y = ^, + ^-^3 + ^4. Thus
such a sphere can be drawn ita + x = b + y. Let
I be the centre of the sphere ; ;;„ p., p^y p^ the
perpendiculars 10', lA', IB', IC on the laces ;
a'y h\ x\ y* the points of contact with the corresponding edges ; then the
dihedral angle A (opposite to the edge a) is the supplement of the
'angle O'lA' ; and if 2o, 2/8, 27, 25 be the angles subtended at I by

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87

diameters of the circles in which the sphere is met hy the faces, the angle

0'IA'= o + i8, or A = IT— o— /3; so X « ir— 7-8;
or A + X « 2ir-o-i8-7-5 = B + Y.

The proof that, when «-«:= *-y, then will also A— X*B— Y, is
exactly similar, the sphere then touching the produced edges. «

8783. (Captain H. Brocard.) — Les droites qui joig^ent I'orthocentre
aux mili< ux des segments interceptes par les hauteurs du triangle sur les
rayons du cercle circonscrit aboutissant aux sommets du triangle sont
perpendiculaires aux syniedianes correspondantes.

Solution by Professor Ionacio Bbtens, M.A.

Soit O le centre du cercle circon-
scrit, H Torthocentre. OA' p'-rpen-
diculaire sur BtJ coupe les hauteurs
BH. Cn en M et. Q; les triangles
AB(J, HMQ sont semblables, leurs
cotcci etant perpendiculaires ; et
ranj?le MH(i = A.

Soient M', Q' les points de ren-
contre de BH, CH avec le rayon OA,
nous avons HM'Q' = M'OB'— B, et par suite M'Q' est antiparall^le par
rapport au c6te QM du trianj<le HQM, c'est-a-dire, que la droite qui joint
U au milieu de M'Q,' sera la symediane qui part du sonmiet H du
trianjfle HMQ, et par suite elle sera perpendiculaire h. la symediane de
Tanj^le A du triangle propose, car ces lignes sont homologues en figures
semblables qui out leurs c6tes perpendiculaires.

8837. (W. J. Grebnstreet, B.A.) — In the "ambiguous case" of
triangles, given «, A, and A; r, r' boiug the radii of the in-circles, p, p'
of the escribed circles to a ; prove that

r/ (* + a) «« pp' (A—a), rp (A— a) = r^p'e.

Solution by H. L. Orchard, B.Sc.,M.A. ; Sarah Marks, B.Sc, ; and othert.
Let c and c' be the two values of <?. Then a/ = A'— a^, and
r _ e is'—ff) c r^ _ c^ b — a ,

a + 6'

a + A

whence r/(A + a) = pp'(6-a), and rp(b'^—a^)^r'p'L^f&Q.

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8715. (Professor Hudson, M.A.) — Prove that
tan37i° « V6+^/^- v'2-2.

Solution by G. G. Stobr, M.A. ; Rev. T. R. Tbeby, M.A. ; and others.

If a; = tan 37i°, then 2aj / (1 -a^) = tan 76° « 2 + -^3 ; whence, solving
a? = ±(-v/6--v/2)-(2-v'3), the upper sign gives tan 37i°, the lower
sign gives tan 127^.

8929. (R. Tucker, M.A.)— Find (I) the equation of the circle through
the images of the centroid of a triangle with respect to the bisectors of
angles ; and show (2) that the sum of the squares of tangents to it frum
the angles (taken once) is \ {0^ + 61^ + c').

Solution by the Paopobbr.

1. Since the median and symmedian lines of a triangle make equal
angles with the bisectors of angles, therefore the images in question lie on
the symmedians and the circle is concentric with the in-circle. Assume
its equation to be

•^(oiBy +... + ...) - (aa+. .. + ...) [a («-a)*a+ ... + ...] +\(aa + ... + ...)»

(i.);

then, since the centroid is on (i.), we have

a2 + ^ + <^ « 3 [(«-a)2+ ... + ...] +9x,
therefore 9 A = - 2 (a^ + *« + c^j + 3,2^ and 36a =* 65a* - b^\
The circlethen can be put under the form
abo(afiy¥..,-\-...)

= (ao+ ... + ...) \aa («-a)2+ ... + ... +,^ (62ai-63a2)(aa+ ... + ...)].

2. If ^«, tbi tc are the tangents from the vertices, we have (see Mather
maticafrom Educational Times, Vol. 46, Appendix n., p. 138)

^3 = («-a)2 + ^V(62a^-52«2) = ^^^ [9 {b + c-ay + 61ab'b:$a^]

= ^ [4Xa^ + 2Ue-'l2ab~12ac] = ^ [^a^ + ebe-Ub-Sac],

therefore <«* + ^6^ + te^ « \:ia^ = ^ (»« + ^ + e-).

8765. (L. J. Rogers, M.A.)— If a^i + asj + ^^3 + oj + oj + oj = 0, prove

that(l)

tan (xi + oi) , tan (arj + oj) , tan (x^ + oi)
tan {xi + 02), tan (x^ + 03), tan {x^ + 03)
tan(:ri + 03), tan(a:2 + a3), taniajg + oj

0;

and (2) the same is true if tangents are replaced by sines.

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89

Solution by F. R. J. Hervey ; Belle Easton, B.A. ; and others.

1. Each element is a product of three tangents which may be replaced
by their sum ; moreover, each tangent appears in two elements with con-
traiy signs : hence the whole vanishes.

2. Any element represented by sin a sin b sin c may bo replaced by
Bin o cos ^ cos <?+... + ... ; which being done thr«»ughout, the complete
coefficient of a given cosine =« ; f or instance, that of cos (Xi + oj is

fiin {x2 + oa) cos (xg + 03) + cos (^2 + 02) sin (ar^ + 03)

— {sin (x^ + oj) cos (X2 + a^ + cos (x^ + a.2) sin (0:2 + og) } = ;

and the same for any other, by interchange of suffixes. Hence, if each
term sin a cos b cos e be taken twice, the whole can be arranged in nine
groups of four mutually destructive terms, and vanishes identically.

8797. (Professor SiRAnlaANJAN RIy, M.A.)— OAP and OBQ are
two fixed straight lines intersecting at O, and G is a circle touching them
both at P and Q ; prove that the perimeter of the triangle OAB, circum-
scribed by any circle passing through O and touching C externally, is
constant.

Solution by Professor Wolbtenholme, M.A., Sc.D«

One of the trigonometrical questions in my book of problems is : '' The
radius of a circle touching two sides AB, AC of a triangle ABC and the
circumscribed circle is r sec? |A or rjSec^^A, according as the contact of
the two circles is internal or external." Hence, with the notation of the
questipn, the perimeter of the triangle OAB will be

20P cos» iPOQ / sin iPOQ ;

and similarly, if the contact were internal, the excess of the two sides
OA, OB over AB would be 20P cos^ ^POQ / sin ^POQ.

8473. (Professor Booth, M.A.)— If a, J, c be the sides of a plane
triangle, prove that the diameter of the circumscribing circle is a root of
the equation

Solution by Isabel Maddison ; Rosa H. W. Whafham ; and othert.

If a; be the diameter of the circum -circle, we have x^—d^ — «'cot?A,
&c. , and the equation becomes cot B cot C 4- cot C cot A + cot A cot B » 1 ,
which is correct.

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2832. (Professor Sylvester, F.R.S.)— Prove that the curve of in-
tersection of two right cones with parallel axes is a spherical curve, and
that a third right cone may he drawn through it. Prove also that a plane
circular cuhic may he found such that the distances of any two fixed points
on it from every point in the curve of douhle curvature ahove-mentioned
shall he in a fixed linear relation.

Solutions by (1) Professor Sircom, M.A. ; (2) W. J. C. Sharp, M.A.

1. The right cones x' + y^ ^ mz^, a:3 + (y — 6)* « h (z-c)' have their
axes parallel, and the sphere

(«+l)(a;2 + y=2-//iz2) - (m+1) {.c2+ (y -*)»_« (e-c)^}
passes through their intersection. Also the surface

a;2 + y2-,ns2 = X {a^ + (y-*)«-n («-r)2}

is a right cone if its discriminant vanishes, that is, if x = — . -^ — -.

ft O^—MC^

The circular cuhic in the plane a; — passing through the vertices of
the three cones, and having for foci the four points where the spherical
curve meets :r = 0, satisfies the given condition.

For, taking the origin and x, y for the two fixed points, and Xi yi «i a
point on the spherical curve, we miist have, hy the question,

{*i»+(y.-y)»+(»i-*)'}*+^('i'+yi»+V)*-M

for every point Xiyiti, Eliminating x^yi by means of the equations of
ihe curve, and rationalizing, we obtain

{(l-A^(/w+l)* + (w-m)y}«j2-2 {bz + ney + K/i (m + 1)^ b} Si

identically; hence, equating the coefficients to 0, and eliminating \,
fL, we obtain

{(n-in)y + (in + l)*}{*(y> + «')-.(*2-iic2)y} « (*» + '^y)'.
the required circular cuhic.

2. OtJherwite : — If the vertex of one of the cones be taken as origin of
rectangular coordinates, its axis as axis of «, and the plane of the axes as
plane of yz ; then, the other vertex being (0, Ar, /), the equations to the
cones are a:' + y« + 2- = m^^, d?=»+ ijy—k)^ + («-0* — «' («— 0' I
therefore, at intersection,

(n^- m2)(a?<» + y2 + ^2) + 2m^ky + 2m^h^m?k^--w?P - mW (2k - P),
and the curve of intersection lies on the sphere

a;2 + y3 + «2 + 2ay + 2fo « c,
if w-^ . ^ mn{\-n^ ^ ^ m^{k^^{\^n^l^] _ ^

and k and / are determined by the equations

(d: + bl=^c and tnrkl ** {l^m^l-bk,
80 that a third cone passes through the curve. If (0, k^, f) be the vertex
■ of this cone, and «' the corresponding value of w, at every point of the
curve of intersection, r = mz, r' = n{z-l), and r" = n' {z—f)^ where
r, r', r" are the distances of (Xy y, z) from the vertices (0, 0, 0), (0, k, /)»

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91

and (0, A;', 1% and nr-mr' ^ nl and n'r—mr" « nT, bo that the curve is
a Cartesian oval in space (see Quest. 5955).
Now, if p be the distance of (z, y, z) from the point (0, ty, f),

(if the point is on the curve)

and p«A3 + B,

hence, if any two points be taken on this curve, and p and p' be the radii
Tectores from them, p = Az-^B, p' = A'« + B', therefore

A>-Ap'«A'B-AB'.
The circular cubic passes through each of the vertices.

8873. (Professor Edmttnd Bordaoe.) — Given two relations,

(2S = a + A+c),
g + y cos g + g cos ^ _^ y + z cos g -f 3- cos g __ z + a: cos ^ + y cos a «
cos(8-a) ™ C08(S-^) C08(S-<r) * *"'

deduce therefrom -^ =. _iL « _!_ =. J?L-.
sin a sin 6 smg smS

Solution by "W. J. C. Sharp, M. A. ; Professor Matz, M.A. ; and others.
2 ^ (a: +.V cos g + zcos^) A + (y -h zcosa -h a; cos g) B ->- (z -f a; cos^ +y cos^i)

A cos (S — «) + B cos (iS — A) + C cos (8 —c)
If Aco8g + B4 Ccosa = 0, and Acos^-t-Bcosa + O — 0,

A : B : C = — sin'a ; cosg— cosacosA : cos &— cos a cose,
and the numerator = — z {l — cos^a-cos^i— cos*^g + 2cos acos* cosg}.
The denominator
e { — 2sin2aco8(S-a) sin S + 2 (cos g— cos a cos*) cos(S— *)8inS

+ 2 (cos * — cos « cos g) cos (S — g) sin S} -•• 2 sin 8
■a { — sin'a [sin (* •(- g) + sin a] + [cos g- cos a cos*] [sin (a + g) + sin*]

+ [cos *— cos a cos g] [sin (a + *) + sin g] } -»- 2 sin 5
Bs { —sin a (1 — cos' a— cos^*— cos^g + 2 cos a cos b cos e)

+ Bmb (— sin'acosg + cosg-cosa cos* + cos<icos*— cos'acoBg)
+ sin g ( — sin- a cos * -i- cos * — cos a cos g + cos a cos g — cos' a coe b) J
-^2sinS
— {— 8ina(l— C08'«— cos'*— co8'g + 2co8«cos*cosg} -h2BinS,

2x
therefore 2m »- -: — x sin S,

sina

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92

or -7^ *= -^ « -7^ « -7^, by symmetry,

sin a siiiS Bin^ siik;

[Obserying that 2 sin S cos (S— a) == sin a + sin (^ + c) and for brevity^s
sake writing A; for m/ sin S, we see that the given relation

{x + t/coae + z cos b) sec (S— a) = 2m
is equivalent to

{x^k sin a) + (y — A: sin b) cos e + («— A; sin <;) cos ^ = ;
and similarly, from the other given relations, we have

(y — A: sin ^) + (z— A; sin c) cos a + (a:— A: sin a) cos c = 0,
(«— fc sin c) + (a? — A; sin a) cos 5 + (y — A; sin ft) cos « = 0,
whence a?— A:sina = 0, y — Arsinft = 0, 2— A: sine =0.

Hence, if a, ft, <; be the sides of a spherical triangle, any quantities
Xt y, z which satisfy the given relations are proportional to the sines
of tiie angles, and conversely.]

8980. (Professor Stegoall, M.A.)— In a rectangular hyperbola OY
is drawn at right angles to the tangent at P ; prove that, if YX produced
cuts SP in R, and Y be joined to X', then XB = YX', where X, X' are
the feet of the directrices.

Solution fty R. Knowles, B.A. ; Sarah Marks, B.Sc. ; and others*

If a;'— y^ =• a* be the equation to the hyperbola, that of the tangent at
P {xy) is x'x^y'y = a*, that of OY x'y + y'2; = ; therefore the coordi-
nates of Y are aV/(a;'2 +3^2)^ -ay/(a;'2 + y'2); those of X,X', S are (fl/2*,0),
{'-al2^f 0), (2*a, 0), respectively; whence the equations to SP and XY are

y'a;+(2*a-a;0y = ^W ...(1),

2V^-(a;'=* + /^-2W')y = ay (2).

From (1), (2), we have

ax'{2h-Za) ^ aY

(2*a;-«)2 ' {2^x'-af

the coordinates of R, whence we find XR and YX' each
=:a(2*a;' + «)V2*.

9010. (J. J- Walker, F.R.S.)— Prove that the area contained by two
tangents to a central conic and the semi -diameters to points of contact is
equal to ftV + aV- aH^. [See Question 3099, Vol. xiv., pp. 74, 76.]

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93

Solution hy C. E. Williams, M. A. ; B. Knowlsb, B.A. ; and others^
Projecting the ellipse into the auxiliary circle, the (area)' would have

to be multiplied by -^ , and as X -■ a;, T « 4~^> ^^ expression becomes
0^ b

(area)2 - a«X«+a«Y»-«* « a« (X» + T«-a«),
or area « (rad. x tang.)* which is evidently true for the oirde.

8903. (Professor G-enese, M.A.)— If /(r, 0) be the equation to a conic
referred to any pole O, the value of {df)l(dr) at any point P of the plane
varies as (PQ + PR) / OQ . OR, Q, R being the intersections of OP with the
conic ; thus, for a point Q of the conic, {df)l(dr) oc QR/OQ . OR ; and
in particular, if be a focus, (df)l{dr) is constant over the curve. [In
thi« case /= (/-^ cos 6)2- f«.]

Solution hy R. F. Davis, M.A. ; Professor Chakravabti, M.A. ; and others.
The polar equation of the conic referred to O is of the form
/(r, e)=Ara + Br + C = 0,
where A, B are functions of $ alone, and is constant. For any definite
value of d, we have ri + r^^^— 2B/A, rir^ «= C/A.

Now Wl{dr) =- 2 (Ar + B) = 2A(r + B/A)

a {2r-(ri+r2)}/rir3 0c (PQ + PR)/OQ.OR.

If P coincide with Q, PQ = 0, &c. If be a focus, (<y) /(rfr) -constant,
which is a priori evident, since the semi-latus-rectum is a hannonic mean
between OQ, OR.

9003 & 9013. (9003.)— (R. F. Davis, M.A.)— If upon each side of
a triangle a pair of points be taken so that the pairs on any two sides are
concyclic, prove that all three pairs are concyclic.

9013. (Emilb Vioaki^.) — Les projections orthogponales de deux points
inverses Mj, M2 (a/, /, z^ sur les trois cAt^s d'un triangle ABO, sont six
points d'une mdme circonference dontl'^quation en coordonn^es normales

est: (y«sinA + «ar8inB + a?ysinC) (a:'sinA+y'sinB+«'sinO)
X (y'r'sinA+^VsinB + ic'y'sinC)

^sinAsinBsinC(^sinA + ysinB + mnC)[^^Cv^^^T^y-'^^"^)

i sinA

y/(a;" + g^co8B)(g' + a;'cosB) ^ gg" (a/ + y" cos 0) (</ + a;" cos 0) *)

sin B sin C )

[M. ViOARrf remarks that " on appelle points inverses en France ce
que M. Caset appelle isogonal co^ugate points {Sequel to Euclid, 1886,
p. 166)."]

vol. XLVn. >L

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94

Solution by A. B. Johnson, M.A. ; C. E. Williams, M.A. ; and others.

It seems best to take these together, as they are both «Eisy deductions
from the form of the equation in trilinears to any circle, viz. :

4R2 |y«8in A + »rsinB + arysinC} sin AsinBsinC

«= {a? sin A -f y sin B -I- 2 sin G} {p"iFsin A + fy sin B + 1^« sin 0} (I),

where p, q, r are the tangents from the vertices to the circle.

Let D, D' ; E, E' ; F, F be three pairs of points on the sides, and con-
oyclio in pairs. Then, if we take i?» = AE . AE' - AF . AF ;

(?«-BF . BF'-BD . BD'; r> = CD . CD'= CE . CE',

the circle represented by (1) must cut the sides in D, D', E, E', F, F ;
80 that all six points are coney clic.

Now let D, B, F ; D', E', F' be the projections on the sides of the
points x'f y*, af; k/x', A//, k/z', where

^ JsinA _^ sinB ^ sJnCJ « ^^nA + y'sinB + .'sinC (2).

Then AE. AE'= AF .AF -*{/ + «' cos A} {«' + y'cosA} /y'«'sin«A,

so that the projections of the points are concyclic in pairs, and therefore
all lie on the circle for which jp* = Ar (/ + z' cos A) («' + y' cos A) /yVsin- A,
and q^ and r* have analogous values. Substituting in (1) lor p-y g*, r*, k,
there results the equation sought.

8835. (Professor db Lonochamps.) — On consid^re un triangle
ABC et le cercle circonscrit A ; soit fi une transversale renconfrant les
c6tes de ABC aux points A', B'» C Par A' on m^ne k A une taiigente
que Ton rabat sur BC de telle sorte que le point de contHct vienne occuper
sur BC une certaine position A". Demontrer que la droite AA" et les
deux droites analogues BB", CC vont passer par un m§me point M.

Solution by Professor Ionacio Bbtbns, M.A.
Soit A'M' la tangente men^e de A' ; nous aorons
A'B.A'C- (A'M7= (A'A'O^ C'A . C'B = C'C"*, B'A . B'C = B'B' «,
etaussi BA- _ A^A^-BA- ^ (A^B . A^C)*- BA-

A"C A'C-A'A" A'C-(A'B.A'C)*'
ou ?^' » (A^ . A^O* . A^C- A ^B . A^C + A^B . AC- BA^ ( A^B . A^Q*
^"^ (A'C)2-A'B.AC '

BA^' ^ (A^B.A^C)^(AC-A^B) ^ (A^B.A^C)^ ^ (A^B)*
A"0 A'O A'C-A'B) A'C (A'C/*

Delamdmemani^Cona 5?^ ^ (B:C)_» ^^ AC^' ^ (ACO^.
AB" (B'A)* BC- ^BC')*

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95

Multipliant ces proportions,

BA^' CB;;^ AC^' ( ArB.BV.AC 'U ,
A"0 * AB" • BO" " \ A'C . B'A|. BC' / '
/ A'B B'O AC ,\-

done les trois droites AA", BB", 00" se renoontrent au mime point.

8385. (J. Brill, M. A.) — Three parabolas are drawn baying a common
focus ; from a point T, external to all three, tangents TP and Tp are
drawn to the first parabola, TQ and Tq to the second, and TR and Tr to
the third ; prove that Qr . Ep . Pg = yR . rP .pQ.

Solution by the Proposer ; Belle Easton, B.A. ; and others.
Let S be the common focns of the three parabolas, then we have
(SP) . (Sp) == (SQ) . (S^) = (SR) . (Br) = (ST)» ;
therefore (Qr) . (Rp) . (Pq) + (qR) . (rP) . (^Q)

= {(Sr)-(SQ)} {(Sj.)-(SR)} {(Sj)-(SP)}

+ {(SR)-(Sj)} {(SP)-(Sr)} {(SQ)-(Sp)}
- {(SP) + (Sp)} {(SR) . (Sr)-(SQ) . {8q)}
+ {(SQ) + (Sy)} {(SP) . (S/.)-(8R) . (Sr)}
+ {^SR) + ,Sr)} {(SQ) . (8|/)-(SP) . (8p)} = 0.
Hence it follows that Qr . Rp . Pg = gR . rP .j^Q.

8915. (Rev. T. 0. Simmons, M.A.)— A point P being given in the
plane of a triangle ABC, it is known that, with P as focus, five conies
can be drawn, four of them circumscribing the triangle, and one inscribed
in it. Show that for certain positions of P a sixth conic can be drawn,
also having P for focus, and touching the other five ; and find the locus
of P when this is possible. [The Proposer suggests that conies with one
fotius common should be called eo-foeal, reserving the term confocal for
conies having both foci common.]

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

^ If , the system consisting of a triangle abc^ its one circumscribed and four
inscribed circles be reciprocated with respect to any point P, we obtain a
corre«ponding system ( oasisting of a triangle ABO and five conies each
having P as focus, four of them circumscribing the triangle ABC and one
inscribed in it.

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The nine-pomt circle of the triangle abe tonchM each of the four in-
acribed circles; and in Uie particular case when one of the angles is a
fight angle it also touches the circmnsoribed circle.

Hence, if one of the sides of ABO subtend a right angle at F, the
reciprocal of this nine-point circle is a conic (focns P) touching each of
the above five conies. P must therefore lie on one of the^thiee cirdea
described upon the sides of ABO as diameters.

8688. (Professor Iomacio Bbtbns, M. A.) — Si dans un triangle la pro-
jection du cAt^ BO sur BA qui est BH, et la projection du BA sur AO qui
est AH', sent egales ; la bissectrice du A, la hauteur du et la m^diane
de B se rencontrent en le mdme point.

Solution by B. F. Datis, MJL ; B. Enowlbs, BJl, ; imd othen.

Since BE - AW, acos B » <; cos A. Let
the bisector of the angle A meet the per-
pendicular OH iu O, so that

OH : 00 - AH : AO - cos A : 1.
Then may be regarded as the centroid of
the masses a cos B, ^ cos A, ^ cos A, placed
at the angular points A, B, respectively. ^^^
For, since

<; — tfcosB + ^cosA »BH + AH,
the masses at A and B may be replaced by a
mass at H ; which with e cos A at is finally equivalent to a mass
0(1+ cos A) atO. Hence BO divides AO in the ratio ^cosA : acosB
■i 1 : 1, i.e.f it bisects it.

8700. (R. W. D. Ohbistib, M.A.) — Given the sum of the ex-
pression l»' + 2'' + 3'' + 4'' +»»•, to find a metljod of writing down at

once the sum of l*'** + 2*'+* + 3*'** + 4*'*i +»•'♦*, where ria even or

vice verad.

Solution by Prof. Gbnbsb, M. A. ; Prof. Nash, M. A. ; and others.
l' + 2r + dr + + ^r «^ +An^ + &c, = <>(») Say,

therefore (l+^i^Y + {2 ^xy + ... {n + xy ^ 4>(n + x)^4>(x),

This equality holding for all integral values of x (and therefore for more

than r values), is an identity. Equating coefficients of x,

r (l'-» + 2»-»+ +«»--i) „ ^/ (n)-ooeffioient of « in <>(«)

-^'(ff)-.^'(0).

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Hence, to find 23^*^ from 2«^, multiply by (r+ 1), and integrate with
respect to n. The constant of integration can always be determined by
the condition that the expression equals unity when n is unity, and is in
fact nBnt when Bn represents Bemouilli's numbers.

To find 2^ from 2s^*^ reverse the process and omit the term with-
out n.

[Prof. Gbnbsb adds that the above occurred to him last year as the
proper method of dealing with the series (a + d)^ + (a + 2d)^ + (a + 3d)r + &c. ,
obtained by putting x above equal to d/a].

Note on Quest. 8376 (Vol. xlv., p.

£t/ Professor Nash, M.A.

Most of the theorems enumerated in this Question are true, not only in
the particular case considered, but also when the lines wD, wT), &c.
make any constant angle with the sides. Let the angles GDw, AEm,

BF», BDV, CEV, AFV, measured in the direction of the letters, be
all » 9 + w, where w is the Brocard angle of the triangle ABC. There-
fore «EC « ir-(a + ») «ir— »DC; therefore wDCE are concyclic, and
similarly «EAF, «FBD, wD'CE', w'E'AF', »'F'BD' are concyclic ; there-
fore «DE - »CA « fitf, »EF = «AB = <», &c. ; therefore u is positive
Brocard-point of DEF, and u' is negative Brocard-point of D'E'F'.

Also «DF := »BA = B-«, therefore FDE =r B.

Similarly DEF = C, EFD = A, F'D'E'^ C, D'E'F= A, E'FD'- B,

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therefore DEF, D'E'P' are similar to one another and to ABC, and w is
centre of similitude of ABC, DEF, «' that of ABC, D'E'F'.

Because B, D are corresponding points, the ratio of similitude is
o>D / »B = sin « / sin (w + 0),
similarly for D'ET', ABC the ratio is w'D'/w'C = sin a» / sin (a + $),
therefore the two triangles are equal to one anotiiier. Again, the augle

CD'S'- ir-C-CE'D'== ir-C-e = ir-C-CDE = CED.

Therefore D' lies in the circumcircle of DEF, and similarly for E'F', so
that the two triangles have a common circumcircle a Tucker's circle .

Let Q he the centre of this circle, then Q corresponds to O, whether
considered as helonging to DEF or D'E'F', therefore

«Q/ «0 « «'Q/ «'0 » sinctf / sin (« + e),

therefore Q lies on line OP which hisects the angle «0a>'. If rK he the
radii of the circles DEF, ABC, therefore

r / R — sin « / sin (« + e) = «Q / »0,

therefore the radius of a Tucker's circle is proportional to the distance of
its centre from the Brocard-points. AIro, since OwQ is similar to BcvD,
therefore 0«Q = e. If a>i be negative Brocard-point of DEF, wi corre-
sponds to u'f therefore »'w«i = &, ««'»i = » = »«T ; therefore »i lies on
the fixed line wT, and similarly the positive Brocard point of D'E'F' lies
upon a»P. If p, ff be the symmedian points of DEF, D'E'F', then P«p
is similar to OwQ. therefore taVp = to.

Therefore Fp is parallel to a>«', and Vp' of course coincides with "Pp,
Since the Brocard-points are equidistant from the circumceutre, there-
fore Q is the centre of the circle wu'a>i»'i, therefore

wQ»i ■■ 2fi0w'»i =s 2c» =s wPfitfj,
therefore P lies on the Brocard circles of DEF and D'E'F'.

Again, from the figure, E'EF = ir-E'D'F'= ir-C, therefore EF' is
parallel to BC ; and because AE . AE'=: AF . AF', thereft>re E'F is anti-
parallel to BC. Also E'F subtends ut the circumference an angle E'F'F
= IT — e, therefore E'F = 2r8ind; therefore the three antipurallels
E'F, F'D, D'E are all equal, and this common length is proportional to

OQ » --- sin « . OQ. DD' subtends at the circumference an angle

o«

DE'D' = CE'D' - CE'D = e - A, therefore DD' = 2r sin (» - A) , there-
fore the intercepts DD', EE', FF' are proportional to sin (a— A),
8in(e-B), sin(e-C).

If 0= iir, the intercepts are proportional to cos A, cos B, cos C, and tho
circle DEF is the cosine circle.

In this case the sides of the triangle DEF, D'E'F' are perpendicular to
the homologous sides of ABC, and, since OwQ =^ ^t ^ OwP, therefore the
centre of the cosine circle is the symmedian point.

The radius of the circle is It tan a>.

In the case considered by Mr. Simmons, d « ^ir- «, .*. OwQ = Oaw',
therefore Q is the mid-point of ««', and the radius of the circle is R sin w.

The lines EF', FD', DE' form a new triangle A'B'C similar to ABC,
and DEF, D'E'F are Tucker's triangles with respect to A'B'C. There-
fore the symmedian point of A'B'C is the bisection oipp' as in the case of
ABC, i.e.f the symmedian points of ABC, A'B'C coincide, therefore P is

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the centre of similitude of ABC, A'B'O' and also the centre of perspectiYe.
If B' he the radius of the circle A'B'C, it is evident that

r_ ^ 8m«__ ^ sine therefore K ^^{^'^)

R' sin(ir— + ») 8in(# — a) R an{e + u)

« ratio of similitude of A'B'C to ABC.
But P»i / P» = sin Pw»i / sin P«i» = sin (6—ul)/ sin (0 + »),

therefore P«, / P«'= Pw// P« = R/R' ;

therefore «/ and «i are positive and negative Brocard-points of A'B'C.
The circumcentre of A'B'C is a point 0' on PO such that QajO'— ir— 0.
Therefore QO'/ sin e = Q»i / sin « = Qa> / sin » = QO / sin «,
therefore 0', O are equidistant from Q.

If the circle DEF is the T. R. circle, the lines EF, FD', DE' pass
through P, 80 that the triangle A'B'C reduces to a point, therefore
sin (0— w) = and 6 «= », and ti^e centre of the T. R. circle is the mid-
point of OP. Its radius is ^^^^L^ = -^— .
sin 2(u 2 cos u

If EP, F'P meet BC in d\ d, FP, D'P meet CA in /, «, DP, E'P
meet AB in f*,f, then, since E, t^'arecoirespondingpointsof the triangles
A'B'C, ABC, therefore PE / Prf'= R'/ R, and similarly for the others.

Therefore def, d'tff form another pair of Tucker's triangles, the sides of
which make, with the homologous sides of ABC, an angle ir— 0. If ^ he
the centre and r' the radius of the corresponding Tucker's circle,

Oo^ + 0»Q = IT or ^»P =■ Q«P,
therefore ^, Q are conjugate points with respect to the Brocard-circle of
ABC.

Also //r = ^»/Qa> = gP/QP = ^/Q0-8in (0-») /sin (0 + »).

If two Tucker-circles he taken whose centres are equidistant from the
Broctird- points, their radii will he equal, and therefore also the corespond-
ing Tucker's triangles. As a particular case of this, we see that two tri-
angles can he inscrihed in ABO which are equal to it in all respects, and
that the sides of these triangles make an angle ir— 2» with those of ABO.

8784. (By R. W. D. Christib, MA.)— Prove that, if
»= 1-I-2 + 34- ...+», S = r- + 22 + 32 + ...+«2, S'= 13 + 2«<-3»+... + »»,
5- l^+24 + 3H...+»S <r = 18 + 26+3* + ... + f|6,

then ?^+2^'^S^

52 S

Solution by H. L. Orchard, M.A., B.Sc. ; Professor Betbnb ; and others.

By summing the series hy the method of Undetermined Coefficients, or
otherwise, we find 65S'S-i-2«3 = 4»*(fi + l)''L*'^»(n + l)- I]; hence,

we have to show that n^ (n + 1)2 \2n (» + 1)- 1] = 12 (I* + 2* + ... + »«).

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In the left-hand memher put n^tn and m + 1 successiyely, and suhtract*
We thus obtain 2 (m + l)«[(irt + 2)»-m«]-(m + l)S[(w+2)«-m»], which
reduces to 12 (m -t- 1)*. Hence, if the theorem holds for « ■■ m, it holds
for n B m + 1, and so on, universally. Bat it evidently holds for n » 1,
and for fi — 2 ; therefore, &c.

8763, (By H. Stbwart, M.C.P.>--Prove that

(l)ii «(l-^2)» + ^(l-««)».a», then p + il^^O;

ax (l-j«)*

(2) if (l-«»)* + (l-fi«)*- a («-«), then ^ = ILz!f-I*.

«» (I-a;*)*

Solution by F. K. J. Hbrtbt ; W. J. G&bbnbtbbbt, B.A. ; and others.
Putting, for brevity, (1— a?')* - *, (1— w')* = t, du/dx - «', we have

* < x-u ViF— w </ a?— w «'

but < (»+ -f* («— w) « «^ + 1 -«u = « (« + + * (*— «')i ••• «' = */*•

8745 ft 8807. (W. W. Taylor, M.A. Sugjirested by Quest.
7938.) — Prove that (1) the areas of the TATLOR-circles of the four triangles
ABO, PBC, PGA, PAB are together equal to the area of the circum-
scribed circle, P being the orthocentre of the triangle ABC ; also (2) the
lines joining their centres are the SiMSON-lines of the middle points of the
sides or of AP, BP, OP with respect to the pedal triangle.

Solution by R. F. Davis, M.A. ; Professor Bbybns ; and others.

The four triangles of the question have all the same circum-radius ;
while their angles are (A, B, 0), (»— A, ^ir— B, ^ir— 0), &c. Hence,
employing the known form for the radius of the TAYLuR-cirde

« R {sin2 A sin* B sin« + co8« A COS* B cos* 0}*,

we get sum of squares of the TAYLOR-radii

« R2 (sin^A + cos^A) (sin* B + cos* B) (sin* + cos* 0) « R*.

The second part follows at once from the fact the SiMsoN-line of the
middle poiut of any side with respect to the pedal triangle passes through
the TAYLOu-centre. In this case the four triangles have all the same
peda triangle.

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2934. (Professor Stlybstbr, P.R.S.) — If «„ »2, »„ «<, Sj, «« repre-
sent the sum of xyztuv, and of their binary, ternary, quaternary,
quinary, and sextic combinations respectiyely, and if E stands for the

symbolof Emanation a — +*—-+«?-- +(i-^ +tf-^ +/4-, prove that

dx dy dz dt du dv
the resultant of «i, E«2, *i» ^41 Ht ^'6 ^ ^e product of

and of the similarly formed powers of products of the analogous linear
functions of abcdef.

Solution by Professor Sibcom, M.A.

The equations «i — 0, s, = 0, S5 = are equivalent to ^ = 0, Xr^ = 0,
:Sa^ s 0, whence we shall have 15 systems of solutiouB, such as d; + 1 — 0,
y + M—O, 2 + 9 = 0. Now, E«3 ^ 0, E«4 »> 0, E«c = are equivalent to
2aa;=0, S^kc^^O, 2aa;*»0, whence, substituting for tuv and writing /, m,
» for a— rf, J— tf, tf — /, we have

te + wy + fi« = 0, ir* + my' + fi«* = 0, ii:» + my* + fi«» = 0...(1, 2, 8),

the resultant of which is found by multiplying together the three factors
foj* + wyi* + n«i*, &c., where yi : x^ «i : Xy^ &c., are roots of the cubics,
obtained from (1) and (2),

w(m«-f|S)y3+3/»|SyJaf + 3/«mya^+/(^-fi?)a:'«0 (4),

»(ii«-m2)23+3^%SiF + 3/%Msr»+/(/2-mS)«»=»0 (6).

Substituting from (1) and (2), we shall find

ir,i + myj* + n«i»-iri»(|l!-l)(^-l);

from (4), (yiV*i'-i)(y»V%'-i)(y8V^'-i)

= (m'— /2)(/ + m + «)(m + «— Q(/— m+n)(/+m-n) / m»(m'— n*) ;
80 for «i : 01, &c. Whence, observing that the resultant is of the degree
3.5 + 1.3+1.6 » 23 in the coefficients, its expression in terms of
a, by Oy &c., is

(a-rf)'(*-*)'(<^-/)M« + *-<'-0(« + ^-*-«) . ..(« + *+«-<'-«-/)*•••.
There are 15 expressions of this kind corresponding to the 15 systems
of solutions, and as there are 15 factors 0^6, the degree of each is 9 in
the final resultant. Also the resultant is of the degree 6.15 in the foe-
tors a + b-e-d, and 6 . 15 terms of this kind can be formed, but half of
these differ only in sign from the rest, hence each is of the degree 2.
Similarly, the resultant is of the degree 8 . 15 » 120 in the fiictort
a-^b-Ye—d—e—fy of which 10 only are distinct, hence each is of the
degree 12 in the resultant. Hence the theorem.

9026. (Professor Catalan.]— Soit

«(«+ l)(a+ 2) ...(a + (?) ±*(* + l)(* + 2) ...(* + (?) = (a + * + (j)^(tf, *).
(Le signe +, si est pair). (1) ^ (a, b) est un polyn6me entier, k coeffi-
cients entiers ; (2) si a, b sent remplac^ par des nombres entiers, ^ (a, b)

VOL. XLVU. N

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devient on nombre entier ; (3) pour ces valeurs de a, b,

(« + A + c) <^ («, *) = 3R [1 . 2 . 3 ... (c+ 1)] ;
(4) si, en outre, a + * + c est premier, <^ (a, i) = 9)i [l . 2 . 3 ... (<?+!)].

Solution by W, J. C. Sharp, M.A. ; Rev. J. G. Bibch, M.A. ; and others.

Ita + b + e = 0y we have A = — a-r, and

a{a+l)...{a + e)±b(b + l)...{b + c)=^a{a + l)...{a + e){l±{^iy*^},
which vanishes for the + or — sign as e is even or odd ; therefore
a{a+l),..{a + c)^(--iyb{b+l).,.{b+c)s{a-i-b + c)<t>(ay b),
where ^ (a, b) is an integral function of a and b. Also, since {a + b + c)<i> (a, b)
has integral coefficient for all values of a and b, the coefficient of if> (a, b)
are integral, which proves (1) and (2).
Again, where a and b are whole numbers,

a(a+l),..{a + e) + (-iyb{b+l)...{b + c)

s((?+l)! {«*'a^i + (-ir***^Cc*i}

(where "Cr denotes the number of combinations which can be formed out
of n things taken r together), which proves (3) ; and (4) follows at once.

8981. (Professor ScHOUTB.)^Given two conies ; find (1) the locus of
the vertex of a right angle circumscribed to these curves ; and (2) con-
sider the particuhu* case of two homo-focal conies. [The problem in its
general form may be otherwise stated thus : — Find the locus of points
such that one of the tangents from it to a conic (2), together with one of
the tangents to a second conic (3), form with the two tangents to a third
conic (1) a harmonic pencil.]

Solution by A. R. Johnson, M.A.

The condition that the two lines (1'), together with one of the lines (2^),
and one of the lines (3^),

a^x^ + 2bixy + <riy8 « 0, a^^ + Ib^y + e^'^ = 0, a^'^ + 2b^ + ^^sS^ = 0,

(1', 2', 30,

may form a harmonic pencil, is

where ^n = ^i^^i — V» An = fliCj + OjCi— ^Ajig; ©tc.

Now the equations to the tangents from a point to (1), (2), (3) may be
thrown into the forms (1'), (2'), (30 respectively, and from the significa-
tions of Aji, A|2, etc. it is seen at once that, in terms of the coordinates of
the point, A^ = »n, hy^ = »i3, etc. ; where »ij denotes (BjOi— Fj') a^ + &c. ;
«i2 = (BjCa + BjCi — 2FjF2) x' + &c. in Salmon's notation ; the equation to

the locus is therefore = »„ {«^-4»3j«a3}-2»,i»,2«i3«23 + »J2*is W»

representing an octavic curve. It is seen, frt>m (A), that the conies Sy^ ~ 0,
*ii = 0, as is geometrically evident, touch the curve where they meet (1),

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and intersect it also where they meet the common tangents to (2) and (3),
the equation to these last being »^— 4*22*33 *= ^« The same is true of

the quartic '12*13 ~2'ii*23 ~ 0*

Also, from the form of (A), it is seen that (1) touches the curve where
it meets «i2 =° ^» *i3 ~ ^' Hence it is seen that the octavic has eight
double points all lying on (1), which also touches a tangent at each double
point. The other tangents at 'the double points form two groups which
touch, respectively, «,2 = 0, s^^ = at the corresponding double points.

Again, we may throw the equation (A) into the form

(*ll»2S-«12«13)'-4»* «22*33 = 0,

showing that the conies (2) and (3) touch the octavic at their eight points
of intersection with the quartic »ii«23— «i2*i8 = 0.

If (1), (2), and (3) have common tangents, these must form part of the
locus ; for at each point of them three out of the six tangents to the
conies coincide. Dividing out by the equation to these common tangents,
the degree of the curve locus becomes correspondingly depressed.

Suppose, now, that (1) is inscribed to the circumscribing quadrilateral
of (2) and (3), so that (2 = denoting a tangential equation)

2i « k^'^ + k^^s, therefore «u = k^s^-k-kJe-^^-^h^s^f
and the equation (A) becomes

« {aJi-2^23^'3523«22-2^2^s3a23«33-^2'V(4 + *'22»83)} {hz'^^Zi}*
i.e., « {V*22-^3^*33}^ {4-'**2?*83}»

representing the four common tangents and the twice repeated conic

k2's^2-ks^8s3 = (B)

passing through the intersections of (2) and (3), as is otherwise geo-
metrically evident, since the tangents to two conies at a point of intersec-
tion are conjugate with respect to any conic inscribed to the four
common tangents.

1 k

Since ^3 = -r- ^1 ~ "E^ ^i> therefore V*33 " *ii~*a*i3+ V*22f
A^3 ^3

and (B) may be written k^i^ — ^u — ^f ^' ^3*13— *ii~0| so that (B)
cuts (1) in the four points of contact of (1) with the four common
tangents.

Now, let 1i consist of two points, then »u = is the twice
repeated joining line, and Sjo = ^» *i3 — ^ ^^^^ ^^''^^ double contact
with one another at the points. If the points be the circular points at
infinity, then «j2 = 0, s^^ = are the director circles of (2) and (3), and
»ii is the square of the line at infinity ; thus leading to the well-known
result for the orthoptic locus of confocal conies.

* Hence 2«,i = ^2^*12 + ^3*13 5 «-^-> if (1)» (2), (3) be any conies inscribed to
the same quadrilateral, then the director conies of (1), (2) ; (1), (3) inter-
sect in four points on (1). It seems natural to call Sjj = the director
conic of (1) and (2). The four points are the points of contact of (1) with
the four common tangents.

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90S0. (F* MoBLST, B.A.)— ABDG is a parallelomm ; O is any
point on the line biseothig the angle A ; CO, BO meet BD, CD in E, F ;
proye that BE » CF.

Solution hy Rev. D. Thomas, M.A. ; Prof. Bbtbkb ; tmd •thir$.

Let a, jS be nnit-Tectors along AC, g g_

AB respectiTely, and AC — a, AB «« b.

So that AO»w(a + jS), / ^/^-^

AE — bfi-k-ma — (l^y) fl« + my (a + jS);

therefore b « my,

ji;»a(l-y)+my

- ft + a ( 1 -i-^ - a + ft- ^ - BE. A
\ m / m

Similarly,

CF-a + ft-^.

[If we draw EEL parallel to AB, and join EF, we have, from the
rhombus HL, BE = AL = AH ; but, clearly, LF is a parallelogram,
hence CF » AH = BE.]

8985. (Professor Btomakbsa Chakeavarti, M.A.) — A cylinder,
weight W, radius r, is placed on a rough horizontal plane ; a uniform
plaoJc, weight P, length 2Q, is inclined at an angle B to the horizon, and
rests with one end on the ground, the other on the cylinder (the plank
being at right angles to the axis of the cylinder) ; if ^^ be the angle made
with the yertical by the re-action of the ground on the cylinder, prove

that

cot4r-cot4-+- .—.secO.
^ 2 a P

SolutUm by D. Biddlb.

Let A be the point of
contact with the ground
of a section-area of the
cyclinder, in the same
plane as D the mid-point
of the plank, as B the
point of contact of the
plank with the ground,
and as C the point of
contact of the plcmk with
the cylinder. Also, let
AE - W, and DF = P.
O being tiie centre of the section-area of the cylinder, join CO and CA,
and draw DG, GF, DH parallel to CO, CB, CA respectively, and cut off
AI s iDH ; draw IE psurallel and equal to AE, and join EE (parallel to
AC and equal to AI), also join AE. Then L OAE = ^,

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Now, in the diagram, the plank is tangential to the cylinder. In this
case, cot 4^ = (EKcosiO + W)/(EKBinie) « ooti«+W/(EK8iniO)

-cotie + W/(iDHsinia)=coti« + W/(iPcos«8ecie8inie)
- cot ie + W sec e/(iP tan ^$),

which, by taking ^r tan i$ » a, reduces to cot i9 + — . ^ . sec 9.

But the plank can rest (on the cylinder) lower than 0, say C. Let
Z OAC - ^. Then

cot^- (EK'coe^ + W)/(EK'8in4>) - cot4> + W/(EK'8in4>)
= cot ^ + W/ { iP cose sec (0-^) sin ^}

=:cot4>+^ (cot^ + tane).

But, in this case,

a = rseca/} 2(cot^ + tane) + :^(cot^— cot^O) i

/Co • ii o iiPA Qsina \» P f / QsinO \*
/ (, LVr-Qsine/ 2W(. \r-QsinO/

l+CO80\ TJ

which fits neither \r tan OAC, nor \r tan OAC, nor ir tan i«.

8847. fW. Mills, B. A.) —Given the focus of a parabola, one tangent,
and a point on it through which another tangent passes ; proTC that the
locus of the point of intersection of the variable tangent with a diameter
through the point of contact of the fixed tangent is a circle which touches
the fixed tangent at the g^ven point on it.

Solution by C. E. Williams, M. A. ; Prof. Sarkar, M. A. ; and other$.

Let S be the focus, PT the fixed
tangent touching at P and T the
fixed point in it, through which
passes the variable tangent OTQ
touching the parabola at Q and meet-
ing the diameter through P at O.
Let TK, QK at right angles to PT,
TQ meet at E; then STP » SQT,
therefore STK - SQE ; therefore
S, T, Q, E are ooncydic and TSE
a right angle ; but, ST being constant in position and magnitude, therefore
TE is, and the locus of Q is a circle touching PT at T. And since
TO = TQ, therefore the locus of is another equal circle touching PT at
T on the other side.

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106

9011. (R- liACHLAN, B.A.) — Show that the product o the three
normals drawn from any point on a conic is equal to the product of the
perpendiculars from the point on the asymptotes and the diameter of
curvature at the point.

Solution by A. R. Johnson, M. A. ; Prof. Matz, M.A. ; and others.

This is one of a numerous class of deductions that may he drawn from
the theorem that the product of the m + n normals drawn to a curve of
order m and class n is equal to the product of the n tangents and m per-
pendiculars on the asymptotes from the point.

Let the point approach the curve ; two tangents and one normal he-
come evanescent, and the product of the tangents is equal in the limit to
the product of the normal and the diameter of curvature ; thus, then, the
product of the m-k-n—l normals to a curve from any point on it is equal
to the continued product of the «— 2 tangents, the m perpendiculars on
the asymptotes, and the diameter of curvature at the point.

Since the product of the perpendiculars on the asymptotes of any conic
from a point on it is constimt, the result might be stated, — The product of
the three normals to a central conic from any point on it varies inversely
as the curvature at the point. Analogous is the theorem, — The product of
the four normals to a central conic from any point on a similar and
similarly situated conic varies as the product of the tangents from the
same point.

In the cubic the product of the perpendiculars on the asymptotes from
any point on the curve varies as the perpendicular on the satellite of the
line at infinity ; and genef ally the product of the perpendiculars varies as
the quotient of the normals to the (w — 2)-ic through the m (w— 2) points
of intersection of the asymptotes with the curve, divided by the product
of the distances of the point from the real foci of the curve.

[By the aid of these considerations many new theorems may be obtained.]

8729. (R. W. D. Christie, M.A.)— Of the series

1. 1 + 3.5 + 5.13 + 7.25 + &c.

to n terms, show that, (1) the sum is a square; (2) the square root is
= 1 + 3 + 5 + 7 + &c. to n terms ; (3) each term is the product of two num-
bers which may be taken to represent the shorter side and hypothenuse of
a right-angled triangle ; (4) if unity be added to the nth term, it is
divisible by 2w, and if unity be subtracted it is divisible by « — 1 ; and (5)
all terms are odd numbers.

Solution by G. G. Storr, M.A. ; Rev. G. H. Hopkins, M.A. ; and otJtera,

The wth term is (2n— l)(2w2_2«+ 1), whence obviously (4) and (5).
This term = »*—(»- 1)S whence (1) and (2). Also (3) is the interpreta-
tion of the fact that (2«2_2// + 1)-- (2n- 1)2 = {in {n- 1)}" = perfect
square.

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8380. (P. C. Ward, M.A.)— Prove that (1),

a-k-b^Cy 4«, 6«, 4a

4i, a + A— tf, 4a, 6a

6*, 4*, « + *—«, 4a

4*, 6*, 4A, -a + A-(J

s (a + A + (?)< - 8 (a + ft + c)2 (*tf + (?a + a*) - 1 28 a^<? (a + * + c) + 16 (be + ca + aft)'

= result of rationalizing a* + ft* + o* = ; and hence (2) the above deter-
minant is symmetrical with respect to a, ft, c.

Solution by R. F. Davis, M.A. ; Satis Chandra Basu ; and others.

Let ft=A*.a; then <? = a (1 + a)*, or a + ft--tf + 4a\ + 6a\2 + 4aA3 = q.
Multiplying successively by A, A^, A^, and putting for A its value ft/fa, four
linear equations in the variables A, A', A* are obtained. Eliminating
these variables, determinant required = 0.

Or thus; if Sj = a + b + e, 23 « be + ea + ab, 23 = aftc,
2, ==« + ft + (j==:2ftM +... + ..., 2i2= 4 {ft<j + (ja + aft + 2aW{a* + ft* + <?*)},
(2r-423)'=642s(a + ft + (? + 2ftM+. .. + ...) =6423(2, + 2i) - 1282i2s.

8714. (Professor Genese, M.A.) — S is a focus of a conic, PN a
fixed ordinate to the diameter through S, PQP' a circle with centre S ;
a variable radius SQ meets FN at L and the conic at B. Prove that the
cross ratio {8LQR} is constant.

Solution by 0. E. Williams, M.A. ; Sarah Marks, B.Sc. ; and others.

Let SP = K = SQ, QSC - <p,

then SL = (K- /) / [e cos <^), SR = ^ / (1 - « cos <^),

and the ratio SL . QR : SQ . LB reduces by substitution to (K— /) : E.

8676. (Professor Bordage.) — The three sides of a triangle forming an
arithmetical progression, (a) being the shortest, {of) the longest ; if the
distance of the centres of the inscribed and circumscribed circles is desig-
nated by i, and the diameter of the nine-point circle by D, prove tluit
oa'- 3 (D2-t2).

Solution by K. F. Davis, M.A. ; E. Enowles, B.A. ; and others.

Let ft be the mean side, so that 2ft » a + a', 2s ^ 3ft, 2aa'« = 3afta'.
Then 2Rr = 2 (afta74S) (S/») = abaJ2s = «a'/3 ;

but i2 » R2- 2Rr = D'-aa'/S ; therefore, &c.

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8876 ft 9032. (Professor Hudson, M.A.)~ProTe that the normal
chord which subtends a right angle at the focus of a parabola is divided
by the axis in the ratio 2:3.

Solution by G. G. Storr, M.A. ; D. E.
Shorto, M.A. ; and others.

Draw PQ a normal to a parabola at P,
and QL perpendicular to the diameter
through P ; then

SP = PL == iNL = iSQ ;
therefore, since SG = SP,
we have

QG . QP = SQ«-SP2 « 3SP»,

QP2 = SQ2 + SP- = 6SP«;
therefore QG : QP = 3 : 6,
and PG : GQ = 2 : 3.

8737. (W. J. Greenstrbbt, B.A.) — ^ABC is a spherical triangle, the
mid-points of its sides are the angular points of a triangle DEF ; prove
that (1) co8EF/cosi« = cosFD/cosi* - cosDE/cosic ;
and (2) if cos^ ia » cos i (6 •«- c) cos ^ ip^^)* ^^ angle D is a right angle.

Solution by Rev. T. B. Tbrrt, M.A. ; A. Gordon ; and others.

1. cosEF « co8idcos^0 + sin|&sin|tf[coB0— co8aco8 6]coBecdco8ec0

« (cos^ \a -¥ cos^ \b -I- co^ i^— 1) / (2 cos \b cos \e)
whence the first result.

2. Also D is a right angle if cosEF — cos ED cos DF ; that is, if

cos' \a — COS* \b + cos* ^c— 1 = cos' \b cos* \e — sin' \b sin' \c
= cos i (* + c) cos i (i— tf).

8998. (Rev. T. B. Terry, M.A.)— Solve the equation
^ 6ar^+(12««-6)-f^4(2««-3)ary =

rf'rJ /MM air.

dx* dx^

Solution by D. Edwardes ; G. G. Storr, M.A. ; and others.

If ir= - — 2x, the equation becomes lA/ =z 0,
ax

And ir->0

Therefore, finally,

= (>«", ir-iC^«^(Car + C2).

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9002., (B. M. Lakolbt, M.A.)— If a oirde and a Simson-line of one
of its points be both inverted with regard to that point, the two inverses
w^ have the same relation to each and to the given point that the
originals have.

Solution hy the Pbofosbb.

Let be the point, OD, OE, OP
the perpendiculars from to the
sides BO, OA, AB of a triangle in-
scribed in the circle ; and let A', B',

C, D', E', F be inverses of A, B,
0, D, E, F. Then, since E, F, A
lie on a circle through O, therefore
E', F', A' lie in a straight line ; and
since A, A', P, F' are concyclic,
therefore Z OA'F' = OFA, there-
fore OA' is perpendicular to E'F'.
Similarly OB' is perpendicular to
FD', and OC to D'E'. But, since

D, E, F are in a straight line, there-
fore jy, E', F' lie on a circle
through 0, therefore A', B', C lie
on a Simson-line of circle D'E'F'.

8912. (Professor Bordaob.) — A square ABCD and a straight line A
in its plane are gfiven, and through the points A, B, 0, D perpendiculars
A A', BB', OO', DD' are drawn on A, A and being two opposite sum-
mits; prove that (1) (BB')2+(DD')*-2AA'.BB' is a constant quantity
for every position of A ; and (2) deduce therefrom an application for the
envelope of the straight lines, such that the sum of the squares of the dis-
tances of one of them from two given points is constant.

Solution byCE. Williams, M.A. ; Prof. Byomakbsa Ohakbavaeti, M.A. ;

and others,

1. This is proved in Casby's Sequel to Euelidf Book II., Prop. 8, the
constant being the area of the square.

2. If (BB')'+(DD')> is constant, then AA'.CC is constant, and the
line envelopes a conic whose foci are A and C.

8979. (Professor Brunbl.) — Soient ABO un triangle, AiBiCi un
autre triangle dMuit du premier en menant par les sommets A, B, des
droites f aisant avec les cot^ du triangle et dans le mdme sens un angle ^ .
Du triangle AiBiOi Ton deduit, de memo, un triangle AgBsOs et ainsi de

VOL. XLVII.

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110

smtOi ioujours avec le mdme anffle tp. Demontrer que lea points
A, Ap Aj ... Bont BUT trois groupes de spirales logarithmiques ayant pour
p61e8 les points de Krocard. Four quelles voleurs de ^ le triangle denve
AiBjCi est-il 6g&l au triangle propose ?

Solution by Profs. Nash, Schoute, and others.

The construction shows that ABC is a Tucker triangle of AiBiCi,
AiBjOi a Tucker triangle of the same species of Afi^^^j and so on. Hence
the triangles have one or other of the Brocard points as a common centre
of similitude, and mA/mAi » o^Ai / wAj ^ &c. » sin (» + ^) / sin a>, where »
is the Brocard angle. Also the angles Aa^Ai, AiwAj, &c. are all equal
to ^. Therefore A, Ai, A2) &c. all lie upon an equiangular spiral whose
pole is <a. to will be the positive or negative Brocard point according as
the line through A makes an angle ^ with AB or AG. The triangles
ABC, AiB^Ci, ^\\G2i &c. will all be equal if sin (» + ^) « sin «,
f.tf., if = X — 2ft>.

8830. (Professor Bordaoe.) — Given the equation

sin2a.a;^ + 2 (sin a + cos a) 2; •»- 2 * 0,

find) and solve, the equation of the second degree that has as roots the
squares of the roots of the given equation.

Solution by W. J. Barton, M.A. ; W. J. Grbenstreet, B.A. ; and others.

TM a AX. i. . « 2 (sin a + cos a) « 2

If a, /8 are the roots o + ^ = ^ — :— ', a$ = t— r-i

sin 2a Bin 2a

whence a^ + 02^—L^^ a^02 *

Bin2 2a' sin^ 2a'

and equation required is sin^ 2ax^—4x-¥i ^ 0;

of which the rooto are ^^^^f^^H or -1-, -1-;
Bin2 2a sm^a cos'a

as might have been expected, since the roots of original equation are
— l/sina, - 1/cosa.

8927. (S. Tebay, B.A.) — If Ai, Aj, Aj, A4 be the areas of the
faces of a tetrahedron, E the radius of the circumscribing sphere, and
Bi, Rs' ^8* ^4 ^^ ^^^^ of spheres passing through the centre of B and
the angles of Aj, A3, As* A4 ; the volume

^=*^'(l;+|*fc^f>

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ni

Solution by W. J. C. Sharp, M.A. ; Professor Beybns, M.A. ; and others.

If 0, Oj, O2, Oj, O4 be the centres of the spheres whose radii are R, Rj,
R2, R3, R4, OOi cuts the plane of the triangle BCD, area Aj,- in the
centre of the circum-circle Ki ; hence

OKi + OiKi = Ri, OKi^ + CKi3 = R', 0, K^s + CKi« « Ri«,
therefore (OKj + OiKi)(OKi - OiKO = il' - R^s,

therefore OKi-OiK, = ^-Ri, therefore OKi = ^,
R} 2Rj

.-. the tetrahedron OBCD - iOKi. A, - JR- ^, OCDA = ^Rs ^, &c.,

R^ Rj

and the volume = jRsf |; + ^' + |3 + 44] .

(.Ri R3 R3 ^a)

[A corresponding formula holds for all simplicissima. (See Quest.
8242.) Thus, if Rj, R2, Rj be the radii of the circles through two vertices
of a triangle, and the centre of the circumcircle (of radius R), we have
Area = iR2 (aR; » + *R - » + cR- 1) ;

and generally, if Y. be the content of a simplicissima in space of n dimen-
sions, Ri, Rj, &c. the radii of the hyper-spheres through all the vertices
but one and through the centre of the circumscribed hyper-sphere (of
radius R), and Vj, Vj,, &c„ the contents of the simplicissima (in space
of »— 1 (^mensions) whose vertices coincide with all but one of those of V,

we shall have y = — 2 ^.1

«'. Ri ■*

8543. (R. Curtis, M.A.) — Prove (1) the following formula of trans-
formation for the equation of a conic to a triangle of reference
the sides of which are the polars of the vertices of the former
triangle («, A, c,f, g, h) (r, y, ^)* « A-» (A, B, C, F, G, H) (X, Y, Z)*^;
and hence (2) show ttiat the equations of a conic referred to an inscribed
triangle and to one circumscribed at the points of contact will be

ftfz-^-gzx + hxy « and (/X)* ± (^Y)* db (AZ)* « 0.

Solution by the Proposer ; Professor Beyexs ; and others,
ax -i- hy -i- yz = Xf hx + by+fz^Y, gx+fy + ez = Zf
therefore Aar = AX + HY + GZ, Ay « HX + BY + FZ,

A«=-GX + FY + CZ;
therefore xX + yY + rZ = (a, b^ c^fj g, A)(;r, y, «)2,

and A {jcX + y Y -r zZ) = (A, B, C, F, G, H) (X, Y, Z)'

is («, ^ e, /, y, h) (X, y, 5/ = ^ (A, B, C, F, G, H) (X, Y, Zf.

A

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112

Again, if « - 0, * = 0, « =- 0, then A = - /», B = -g*, - - A«, F = ^A,
&o. ; tiierefoTe, by above, 2/ys + 2gzx + 2Ad;y » becomes

Of (/X)*+(5rY)» + (AZ) =0.

8994. (Ohablottb A. Scott, B.So.) — ^A rectangular aheet of stiff
pimer, whose length is to its breadth as V2 is to 1, lies on a horizontal
table with its longer sides perpendicular to the edge and projecting over
it. The comers on the table are then doubled over symmetrioallj so that
the creases pass through the middle point of the side joining the comers,
and make angles of 45^ with it. The paper is then on the point of falling
over ; show &at it had originally f | of its length on the table.

Solution ^G. E. Williams, M.A. ; W. J. Green
STREET, B.A. ; and others.

Let BD-.AD»DE-a, AG»2V2a. B

Taking moments about A, we have
4v'2a^3;-2a'.fa + 2«(2^/2-l)a.[a+(^/2-i)a]

3 ' 12 V2 48

A

8856. (B. Hantjmanta Bau, B.A.)— AB is the diameter of a semi-
circle; AC and BF tangents. GF cuts the semicircle in D. DG is
drawn at right angles to GF, meeting AB in G. Prove that
AG.BF-AG.GB.

Solution by A. W. Gave, M. A. ; W. J. Greenstreet, B A. ; and others.

Gircles can be drawn round the iig^es
AGDG, DGBF,
therefore Z ACG- Z ADG - 90°- Z GDB

= 90°-ZGFB = ZFGB,
therefore triangles GAG and BFG are
equiangular; therefore

AG : AG « GB : BF ;
therefore AG.BF = AG.GB.

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9060. (Maurice D'Ocaonb.) — ^AB et MN 6tant deux diam^tres d'un
memo cercle, si une parall^le quelconque k AB coupe la corde NA en B',
la corde NB en A', et que les droites MA', MB' rencontrent le cercle
respectivement aux points A" et B", les droites AA", BB" se coupent au
pied H de la perpendiculaire abaiss^e du point N sur la droite A'B'.

Solution by Rev. D. Thomas, M.A. ;
J. H. Stoops, M.A. ; and others.

Let the figure be drawn, and
join AH, HA",

NHA" = NAA"

= i(NA" + MB)

= i(NA" + AM'),

ANH = i (AM'),

therefore

NHA"-ANH - i (NA"),

•whichshowsthat AHA"are coUinear,
similarly, BHB" are collinear.

8905. (Professor Neubbro.)— D6montrer qu'Ji tout point /de la dis-
tance des foyers F, F' d'une ellipse E correspond une droite D exterieure
au plan de E et parall^le au petit axe, telle qu*il existe le rapport con-
stant c : a entre les distances d*un point quelconque de E k f et D.
Lorsque/ occupe toutes les positions sur FF', D se d^place sur un cylindre
qui a pour base une ellipse semblable k £.

Solution by E. F. Davis, M.A. ; C. E. Williams, M.A. ; and others.

Let PNP' be a double ordinate of the
ellipse, the normals at whose extremities
pass through f;x,y the coordinates of
P. With centre /aud radius /P, describe
a circle having double contact with E.
Then, if Q be any other point on E, the
tangent from Q to the cirde — ^ x perpen-
dicular QL on PP', or
C^-P/2«^.QL«, Q/T»=^(QL2 + P/2/^).

If, therefore, the line D be parallel to PP'

(or BB') at a height z above the plane of E » "Pfje, which is of course

independent of Q, we shall have Of ^ ex perpendicular from Q on D.

Since z^ - P/»/«!2 = (W«2^2) (a3-*2a;S), we have x^a^ + z^/b^ = 1/^ as the
equation of the trace oi the cylinder (swept out by D) upon the plane
tlurough the major axis of E perpendicular to E. This is obviously a
similar elli^iso passing through the feet of the directrices of E.

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114

The above semi-geometrical method is only applicable when / lies
between the limits x ^ -^ m^ and the circle is iW ; analysis woold,
however, not require any distinction to be made between the cases when
the circle was real and when imaginary.

When/ lies between x^ae^ and j:=tf«, although the circle is imaginary,,
the chord of contact is nevertheless real, lying outriide the ellipse.

The focus F itself may be regarded as a point- circle having imaginary
double contact with the ellipse, the chord of contact being the directrix.
In this limiting case, when /reaches F the line D lies in the plane of £,
and in fact coincides with the F directrix.

[Let (j-, y) be the point P, {k, 0) the point/, and (a; = X, « = Z) the line
D : then P/^ = (x ^ky + i/^ = a-'^c^x^^2xk + k^ ^ ^ and (distance from P
to D)-= {x — X)2 + Z^ ; hence, comparing coefficients, Ff : distance from P
to D « c : a, iiX^a^c-H; X"^ + Z-^ = a^e-'^ {k^ + b^) ; and, eliminating k,
we get, for the base of the cylinder described by D, a" =*X^ + ^-=^Z*=a2c- '-*.]

8521. (Professor Wolstenholmb, M.A., Sc.D.) — The circle of
curvature is drawn at a point P (aw^, 2mn) of the parabola y- = ^ax, RR'
is the conmiun tangent of the parabola and circle, and meets PQ their
common chord in T ; prove that

(1) TR : TR' = 1 + 4m2 : 1 ; (2) RR' » l^arn^ ( 1 + m^)* = PQS/ 32a2 ;
(3) TQ : TR'= TR' : TP = m^ : 1 +m^i
and (4) the locus of T is a sextic having no real rectilinear asymptotes.

Solution by R. Knowles, B.A. ; Rev. T. Galliers, M.A. ; and others.
The equation to the circle of curvature, centre O, is

a^i + yi _ 2« (2 + 3w2) X + ^aniMf — Zahn* s= ;
putting X = y^/iay this reduces to {y—2am)^(y + 6am) — ; hence the

coordinates of Q are 9am^, —6am; therefore PQ = Sam (I +m^* (1),

and its equation is x-^my—Zam'^ = (2).

The equation to the tangent to the parabola at R'(a/y') is

y'y^2a{x + x') = 0,
and, since this is also a tangent to the circle, we have

a (2 + 3w2) ^ ,;,3y _ (i + ^^jS) ^iax" + 4a2)i = ;
putting a/= y'^/4a, this reduces to (y'—2am)^ [y' + 2am (4m2 + 3)] = ;
therefore the coordinates of R' are a//*2 (4m2 + 3)2, -2a»» (4m2 + 3), hence

RR' == (0R''-.0R2)^ =. I6am^{l+m^)^ (3),

and its equation is x + m (im^ + Z) y + anfi {im^ + Z)- = (4).

From (2), (4) wc obtain x = am^ (2w3 + 3) (4^2 + 3) / (2«i2 + 1),
y = — 2am (hn* -t 6m' + 3) / (2w*2 + i)^

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coordinates of T. Whence we have

TP = 8aw(l+m2)$/(2m2+l), TQ « 8«w»(l +w2)i/(2m»+ 1)...(6,6).

TR'«8am3(l+m3)l/(2m2+l), (7).

TR = RR'-TR'=8am»(l+m2)t(l+4m2)/(2m2-Hl) (8),

From (1), (3), (6), (6), (7), (8) we obtain the results in parts (1), (2), (3)
of the question.

(4) The locus of T is obtained by eliminating m between equations (2)
and (4), and we get, after some labour, the sextic

8i^« + 162«V + 729aV + 72a»y* + lexY

+ 108a3d:V+ 324a3ay« + 48air*y3-2916a»a;

-3888a4a;2-432aV-576aV-64aa:» « 0,

whose rectilinear asymptotes are given by the equations y s oo ,
y = ± (- 2)* a? + B, and are therefore not real.

8718. (Professor Neubbro.)— L'angle BAC d'un triangle est fixe,
et (1) le centre de gravity, (2) Torthocentre, (3) le centre du cercle cir-
conscrit parcourt une droite donn^. Trouver I'enveloppe du c6t6 BC.

Solution by the Fboposbr.

V Lorsque le centre de gravity G parcourt la
droite EF, le sommet du parall^logramme GABD
d^rit une droite E'F' parallMe k EF. Par con-
sequent, BC enveloppe une parabole tangente h
AB, AO aux points de rencontre de oes droites
avec E'F'.

2° Lorsque Torthocentre H parcourt EF les ^^-^^ \ /

hauteurs BH, CH marquent sur AB et AC des ^^*n^ \ /

ponctuelles semblables. Done I'enveloppe deBC est ^'^ ^, /

encore une parabole. ^^

3® Les projections du centre O du cercle ABC,
lorsque se meut sur EF, marquent sur AB et AC des ponctuelleB
semblables. Done BC enveloppe une parabole.

8920. (B. Enowles, B.A.) — From a point O (x, y) tangents ste
drawn to meet the ellipse a^y^ + h^x^ = aH'^ in P and Q, a tangent parallel
to the chord PQ meets OP, OQ in p and q respectively ; prove that
j3^ : PQ = (aV + ^^a;2)* : «*+ (ay + *«a:2)4.

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8oluti<m by Gbbtrudb Poole, B.A
Let ON, On be the perpendiculars
from (4?, y) on PQ, pq ; then PQ
Xa; Yv ,

ON - <*V-»-^^*-«'^^ .
{a*x^ + b*y^^ '

^.isy + g^.^C^s^ + aVV,

and PQ : /^g- = ON : 0«, therefore

PQ : i»? = (aV + If^x^-a^l^ : {aV + *'«» + «* (**«* + «V)*}
» stated result.

8236. (Rev* T. R. Terry, M.A.)— A uniform lamina bounded by
the arc of a parabola and its latus rectum is revolviog with angfular Telocity
M about the latus rectum : suddenly the latus rectum becomes free and the
vertex becomes fixed. Show that the angular velocity about the tangent
to the vertex is )«.

Solution hff Rev. T. Gallibrs, M.A. ; 0. Morgan, M.A. ; and others.

Let the equation of the parabola which forms the 1/
curved boundary of the laxnina be ^ = iax, and let
PMP' be a double ordinate of the curve at a distance
AM s a; from the vertex, yAi/ the tangent at the vertex.

The only dynamical condition is that the moment of
momentum of the lamina about yA/ before the vertex
becomes fixed is equal to the moment of momentum of
the lamina about yAt/ after the vertex becomes fixed.
Let f»' » angular velocity about tangent at vertex after
the vertex becomes fixed.

Now, in the first case, the velocity ot every point ^
in PMP' = («— a?) » perpendicular to ^kne of lamina,
therefore moment of momentum of lamma about yAy' before A becomes

fixed = A: r a?(a— a?)».y«iir,

where A; Is a constant depending on the nature of the lamina. Also moment
of momentum of the lamina about yAy^ after A becomes fixed

= A; r a?.a?.«'. yeto, hence a> I [ax-^^^y d» — vf \ a^ydx^
or, by (1), w f* («P*-«8) <te - «' [* x^dxy .-. (f-f) « = ?«'> ^^ «*' = !••

Jo JQ

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8302. (A. Gordon.) — Prove that the surface it' + y* + e' « c* will
represent a surface of revolution if ooB^xy » cos^yz » oos^zx, where xy
d^otes the angle between the axes of x and y.

Solution by Professor Ignacio Bbyens ; the Peofosbr; and others^
Let x^+y^ + «2 become A (X' + Y^ + CZ' by change to rectangular axes,
therefore a^ + y' + z^—\ {x^ +y^ -^ z^ + 2xy cos xy + 2yz cos y« + 2m? cos a* )
becomes A (X« + Y^) + CZ^- \ (X' + Y^ + Z^) ;

but the latter equated to 0, and hence the former, represents two coind*
dent planes when A » A, viz.,

(0-A)Z2=«0 or (te + »iy + «a)' = 0,
therefore 1 - a = /^ - m* = »*, —A cos"^ = Im,

— Acosyas = mn, — Acosac = nl,
from which coa^xy = cos^yz — cos'za;.

8336. (Asparagus.) — From a point are drawn two chords OPP',
OQQ' of a given circle, and two tangents OA, OB ; a conic is drawn
touching the straight lines PQ, PQ', P'Q, P'Q' and passing through A
or B ; prove that this conic will touch the circle in A or B. '

Solution by H. London, M.A.; R. Knowles, B.A. ; and others.
Given a conic touching PQ, PQ', FQ, FQ', and passing through B*

Let K, L, M, N be points on the reciprocal conic whose polars are P'Q',
PQ , FQ, and PQ. Then T, R, S, O are the poles of DC, PP', QQ', RS.
But, if SR meet the conic at V, TV is tangent by the properties of a
quadrilateral ; therefore T is the pole of SV, therefore DO is the polar
of O. Hence OB is the tangent to the given conic at B.

VOL. XLVII.

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8656* (W* J* Gbbbnbtrbbt, B.A.)'In a spherical triangle, prove
that im^B » (co8a-co8>R}' + 8m*a8iiL3Bsm3(S-A).

Solution hy Rev. T. Gallib&s, M.A. ; Sarah Marks, B.Sc. ; and others.

We have tan R cos (8 — A) = tan ^a,

therefore sin' (S - A) = 1 - tan' \a cot« B,

therefore (cos a - cos' B)' + sin' a sin' R sin' (S - A)

= oos'a— 2cosaco8'R-hcos<R + 8in'asin'R— sin'asin'Rtan'^cot'B
= C08'a + 1-2 sin' R + sin^R+ sin' R-cos'a sin'R-1 -Hsin'R
— co8'« + cos' a sin' R = sin** R.

8661. (A. Gordon.) — Any curve of the 4th degree intersects
a conic in 8 points. These are joined, forming an octagon A1A3 ... A«.
Show that the 8 intersections of Ai with A4 and A«, A3 with A« and Ag,
&c., &c., will lie on a conic.

Solution by H. G. Dawson, M.A. ; Professor Nash, M.A. ; and others.
Dr. Salmon {Higher Flane Cttrves, 3rd ed., p. 19) proves that, if a
polygon of 2f> sides be inscribed in a conic, the n{n—2) points where each
odd side meets the non-adjacent even sides will lie on a curve of the «— 2^
degree ; therefore put » = 4, and we have the required theorem. There
does not appear, then, to be any necessity for the 8 points to lie on a
quartic.

863L (Professor Sylvester, F.R.S.)—Find the discriminant of
a^ + y» + gS + Se^h/ + 6exyz.

Solution by Professor Mathews, M.A.
Let ttSa;«+y»-H«» + 3«2«'y + 6«cy«s«(a; + y-H «)»-(«- l)(«»-Hy» + »»).
Then, if the curve m «> have a double point,

e(j; + y + «)'-(<?-l)«'= 0, tf (a; + y + «)»-(tf-l)y' » 0,
«(a; + y + «)'-(«-l)a'=0,

therefore ^ (^) ^ {^) ^ (J-f^ x,

and the discriminant is found by rationalising this equation.
Now, the rationalised form of «;* = ± x^±y^±z^ is

(w' + X''\-y' + «'- 2i0X'- . . . - 2y2- . . .)'- 64xy»io - 0,
and, putting w — e^l, x '^ y ^ z ^ e, the left-hand side becomes
\[(tf-l)i-«tf(<j-l)-3i,8}2-64«3((.-l)=-(l + 4e-84j')'-64«»(#-l)-l + 8r.

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(B. Hanumanta Rau, M.A.)— If a body describe an ellipse

about a centre of force in the focus, the time of describing the arc between
the extremities of the latera recta on the same side of the major-axis is to
the periodic time as sin~^ (e) : «-.

Solution by Professor Stonb, M.A. ;
C. E. Williams, M.A.; and othert.

The times are as the areas described
round the focus; but area LSL'= LCL'
and LCL' : area of ellipse » KCK' : area
of aux. circle

« angle K'CB : ir ■■ sin-* {e) : ».

8298. (R- Knowles, B.A.) — I£pr denote the coefficient of aT, in the
expansion of (1 + x)**, where n is a positive integer, prove that

Solution by Profesaor Sikcom, M.A. ; Professor Hatz, H.A. ; and othert.

--" '-^*»--*'^-i:ic<-'>*'- ■

or, writing 1— « for a?, we have the integral ^

n+l Jo l-x » + I V 2 »+l/

8423. (I>. Edwardbs.)— Prove that

(1) (*" sin X (log sin a;)3 £te = 2 + (log 2)«- 2 log 2 - -^ir*,
Jo

(2) l*'sin«a;(logsinar)«(to = Jir{2(log2)2-21og2 + Jir»-l}.
Jo

Solution by Professor Sebastian Siucom, M.A.

1. f^'sina; {logamx)^dx = i T {log (l^x^Ydx
Jo Jo

= if[{log(l-j)}-i.2log(l-a;)log(l+») + {log(l+a:)}']rf(»,
Jo

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f*{log(l+a:)}2(««-2(log2)«-41og2 + 2 (by Quest. 7797)... («),

£ {log(l^x)Ydx - £ (logy^^y - 2 (3).

Alflo log(l+a?)log(l-«) -«-(a; + iar«+ijr3 + ^a;4+...)log(l+ar),
and f x'«log(l + fl>)(;^a:

--i-.log2 L^f J---i+...(-l)"-i + (-l)«log2],

hence f log(l+a?)log(l-a?)(ia? --21og2 ^ J— + _L^+...\

-21og2aog2-l)-2(^+i- + -i- + ...)-l

+

1

- + -^ + ...-22Lii— («>m).

1.2 2.

therefore 2 [^ log {l-i-x) log (l--x) dx^ 2 (log 2)^ -A log 2 + i-\ii' ...(c),
Jo

aince (log 2)2 - Jir« + 22 ^— ^^ (» > m).

m ft

Adding (a), {b), (c), we obtain the required result.

2. We hare (log cos x)^ = {log (2 cos a') - log 2}'
■= {cos 2a)— i (cos 4aj) + ^ (cos 6x) — .. }*

-2 log 2 {cos 2aj-i (cos ix) + J (cos 6a;)- ...} + (log 2)».
Expanding, and patting

2 cos 2miB cos 2 (m + 1) a; s= cos 2x + cos 2 (2m + 1) jt,

vehave'(logcoSfl>)2 « ^^ir« + (log2)3-co8 2a; f21og2 + -i--H -i- +...'\

\ 1.22,3 /

+ terms in cosines of multiples of 2x

= ,.Vx«+(log2)2-co8 2x(21og2 + l)+ (a).

Integrating between ^t and 0, we obtain the known result

[*' (log cos xy dx - ^ir8 + Jit (log 2)2.
Jo
Multiplying [a) by cos 2x and integrating,

r cos 2x (log cos a;)2 rfa; « — iir log 2 - ^ir,

then r* sin2 x (log sin x)^ <f « « I ' co82 x (log cos a;)2 dx
Jo Jo

= J I * (1 + cos 2jr)(log cos xydx
.^ir{2(log2)2-21og2 + iir2-l}.

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[Referring to the solution of Question 8466 (Vol. xlvi., p. 76), we have
1 1 sin a (log sin 6 cos <^)' ded<f> = 1 1 sin m (log «)2 dto (f/i » ^ir [ (log z)^ dz = ir.

The limits are and ^ir throughout. Also, as is proved in the solution
referred to, I sin 6 log sin 6rf6 = log 2 — 1 ;

and we have the known results

f logcos<^rf<^ == -Jirlog2, f (logco8<|))2(f^ « i'r{(log2)« + ^ir'},
and hence the first result immediately follows.

Again, j | sin a cos a (log sin e cos ^)^d0d<t> = j j sin^ tosmfx (log cos«)'^« dfi.

And [ sin a cos 6 (log sin 6)^ dd— iz (log z^dz -« i,

f sin a cos e (log sin 6) rfa — — i, [ sin a cos Ode = i,

80 that

^ir- J |logcos<^e/^ + i r(logcos<^)V<^ « (logcosa»)Va»- fcos^(logcos«)'rf»,

giving f cos2 n (log cos w) 2 rfw = Jir { 2 (log 2)' — 2 log 2 - Jir^) } ] .

8728. (Captain H. Brocard.) — De chaque sommet d'un triangle
on m^ne des perpendiculaires aux cStes adjacents, jusqu'^ leurs rencontres
avec le c6te oppos^. Les centres A', B', C des cercles circonscrits aux
trois triangles ainsi formes sent les sommets d'un second triangle homo-
logique avec le propose. Le centre d*homologie est le centre O du cercle
circonscrit au triangle donn6. En d'autres termes, les nouveaux cercles
sent tangents au cercle circonscrit, aux points A, B, C.

Solutions by (1) C. E. Williams, M.A. ; (2) Professor db Lonqchamps.

1. As BAoj, CA^Tj are
right angles, we have the
angles BA^„ CAaj equal.

If TA be the tangent at
A to the circumscribed circle ,

TAB = ACB,
therefore

TA^i = ACB + CA«2 = Aoafli,
therefore TA is also the tangent to the circle circumscribed to A«i«2i
therefore the circles touch and their centres are in a line with A.

2. Otherwise : — Soit ABC le triangle propose. Les perpendiculaires
elev^es au point A aux c6tes AC, AB rencontrent BC en des points a^
a2i et il faut demon trer que les cercles Aa^a^t ABO sont tangents.

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En coordozmees barycentriques, et d'apr^s la notation que nous avons
propos6e (M6moire **Sur un nouveau Cercle remarquable du plan d'un
triangle, '^ Journal de Mathimatiquei Spieialei, 1886), le cercle AayO^ peut-
dtre repr^sente par I'^uation

(a + /8 + 7)(Ma + t>i8 + w;7)-f = (A).

Dans cette ^galit^ u, v, w repr^sentent, en grandeur et en signe, lea
puissances des sommets A, B, G par rapport au cercle Aaja,, et ( le
premier membre de F ^nation du cercle ABC.

Nous avons, dans le cas pr^nt,

w«0, v = - Ba, . Bflf- == -A; ( « ^Ii

' * ^ cosBV cosCr

On a, de mdme, w « -^ ^^ ^ ,

' cos B cos C

et, finalement, P equation de AaiO^ est

ou r«»— c?

cos A
cos B cos C*

cos A

,(a + ^ + 7)(c«/8 + ftV) + f-0.

cos B cos C

L*axe radical A des cercles ABC, Aa^a^ est done repr^sent^ par T^qua-
tion c^fi + b^^O ; c'est-^-dire que A est, pr6cis^ment, la tangente en A
au cercle ABC.

pr^c6dente,

immSdiatement, et presque sans effort, k ces nombreuses questions de la
g^m^trie modeme dans lesquelles, consid^rant diffi^ents cercles deriv^
du triangle, on demande d' observer quelques unes de leurs propriety.

8946. (W. J. 0. Sharp, M. A.)— Show that

X, -a, -a,
b, X, -a,
by b, X,

b, b.

-a, -a
—a, —a
-a, -a

b, ... X, -a
b, b, ... b, b
where the determinant has n+l rows.

= b{x + a)'*,

Solution by Isabel Maddison.
Multipljring the last row by ab-'^, and adding to each of the other rows,

the determinant becomes

.

* + a, b + a^ x + a .

b + Uf b + ttf b + a .
bs b. b,

0,

0,

0,

a; + fl,

*,

b

= b (» + »)».

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8948. (W. J. C. Shabp, M.A.)— Show that

n— l!Bm'*6\ dej

and co8«a = — — : (sin^tf— \ (sin 9 cos 0).

«— llsin^tfV de)

Solution by D. Edwabdbs.

/ rf.»-i j_ (w-U! Put <=.- cote, ^^--1, then the

\dt] t-^j ^ ' {t+jY ' ^

left side becomes — ( sin* ^-j-] ^^i ^ (cos ^ -»-/ sin 0), and the right side

becomes — (fi — 1) ! sin** (cos n0 +/ sin n0) . Equating real and imaginary
parts, we have titie formulsB in question.

8754. (Professor Mahbndra Nath Ray, M.A., LL.B.)— Prove that

f^xws^xdx ,, ,» r*2;sin'0e£r / -v

Solution by A. Gordon ; W. J. Grbenstrest, B.A. ; and others.
Galling Z the first' integral, Y the second, we have

Jol + cos2a; Jol+cos^a;

Jo \l + cos'iC l+sm^xj

therefore Z = ir (- 1 + ^x). Similarly, Y = »(-3 + J + ir)r=ir (ir-|).

8671. (P' Edwardes.) — Prove that the latua rectum of a parabola
is half the harmonic mean between any two focal chords at right angles
to one another.

Solution by the Rev. T. Gallibrs, M.A. ; W. P. GU>udie, BAl. ; and others.
From the equation

r=:— ^^ weget f ^ —^ and/=-~^,
1 + costf * "^ 1-cos'tf "^ l-sin«0*

therefore -r^ + -=- = —•

f f \a

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9017 & 2810. (A. Russell, B.A.)-(9017.) If «i, x^ ..., Xn - a,,
Os> •••> <<») prove that

ai(ai-«2)...(«i-a») Oj (oj-aj) ... (oj-Oj)

Solution by Thomas Mui&, LL.D.

9017. For convenience in writing, and not for simplification of the
proof, let ft *B 4. Then the common denominator of the fractions is

and the total numerator

+ «i%»4 {* (<h«3«4) [«3* — '^^ ^1 + V 2^i*a - «i Xci%a?8 + XiX^e^il

— +

This, when the coefficients of 2^i, Xt^i^s, &o. are collected, becomes

|«i'«s'«8'«4*l - ^ilV«3'«s'«4'| + 2afia?3|aiia,2a8»a4»|

The first determinant here =^ a^a^i^^ ^ iflx^^^i the last» — ^ (<>i<W4)9
and the others vanish ; hence Mr. Russell's expression is eqiuJ to

2 XxX^X^X^ ^

2810. A generalisation including Professor Stlvbsteb's theorem in
Quest. 2810 (see Vol. xlv., p. 129) may, in like manner, be readily
obtained (see Froc, Roy, Soe,, Edinburgh, 1886-87).

7661. (Rev. T. C. Simmons, M.A.)-If

sin(a + /3) + sin(/3 + 7)+sin(7 + a) - 0,
show that cosi (a— /3) cos^ (/3 -7) cos i (7— a) cannot be greater than |^.

Solution by the P&oposbb.

Let PQR be a triangle touching an ellipse in points A, B, C whose
eccentric angles are a, /S, 7 ; then, by Quest. 7663 (Vol. xm., p. 73.)

aABO : aPQR « 2 cos i (0-/8) cos i {fi-y) cos i (7-a) : 1.

Now, let a, i8, 7 be such that sin (a + i8) +sin (i8 + 7) + sin (7 + o) = 0,

then the normals at A, B, C will be concurrent, and in this case
aABO : aPQR cannot be >i ; hence, subject to the above condition,

2 cos i (0-/8) cos i (/3— 7) cos i (7 -a) cannot be > i.

[A solution is desired by a direct method.]

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APPENDIX L

SOLUTION OP QUESTIONS 8886 and 9009.
By Rev. T. P. KIRKMAN, M.A., F.R.J

8886. (By Rev. T. P. Kibkman, M.A., F.R.S.)— Two 28-edra have
the following triangles, the summits being marked abc ,.. npq. Required,
all the symmetry of both.

abg, abe, acd, ade, aef, afff, hje, bihy bfh, bigi q/l, elk, ckd, dhm^ dfm, def,
fmyfgQy 9Hy 9^*y ^Pj\ ^PiJlf^J^Pj f^nq, mnp, mpk, pqn,

abffj afff, afe, aed, abc, acd, bhff, bhi, bci, cj'i, co{j\ djk, dkl, dfl, def, fkl,
/km, fgrtiy ghm, yq, ihp, ipn, inq, jkn, jnq, kmn, hmn, hpn,

9009. (Rev. T. P. Kirkman, M.A., F.R.S.)— P and Q are a regular
20-edron and 8-edron. EILMN are 4-edra, each on a base that covers a
face of P or of Q. Required the number of polyedra,t)f which none is
either the repetition or the reflected image of another, that can be made
by laying one or more of KLMN on as many faces of P or of Q, with an
account of the summits and synmietry of the constructed solids.

Solution,

(9009.) The required solids are, one 7-acron, three 8-acra, three 9-acra ;
six 10-acra ; one 13-acron, six 14-acra, thirteen 15-acra, and forty- two
16-acra : in all 76 solids. Their summits and synmietry are as follows,
where the terms preceding the colons are the summit-signatures ; e.g,,
the 7-acron has the signature bH^'A^ showing three pentaces, three
tessaraces, and one triace ; the first 15-acron has 87*6^5^3', showing one
octace, two heptaces, two hexaces, seven pentaces, and three triaces.
One 7-acron :
1 ; 5^433 : 3-zoned monaxine heteroid.
Three 8-acra:
1, 2; 63534333, 65H33: both 2 -zoned monaxine heteroids.
3 ; 5*33 : 3-zoned monarchaxine homozone, which has one janal 3-zoned
axis, with repeating zones, and three like contrajanal 2-ple zoneless axes.
Three 9-acra:
1, 2; 7635*433, m^Z^i two monozones.
3 ; 6^5^33 : 3-zoned monaxine heteroid.

Six 10-acra :
1 ; 86H3^ : 4-zoned monaxine heteroid.
2 ; 7^5'3^ : 3-zoned monaxine heteroid.
3 ; 7363533* : 2-ple zoneless monaxine heteroid.
4; 76*53*: monozone.

VOL. XLVn. Q

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126

6 ; 6*3^ : zoned tnazine, haying three unlike janal 2-zoned axes.
6; 6*3^: homozone triaxine, having one janal 2-zoned axis,
with like zones, and two like janal 2 -pie zoneless axes.

One 13-acron :
1 ; 6S5'3 : 3 -zoned monaxine heteroid.

Six 14-acra :
1, 2 ; 7'6*5«3^ 6«5«32 : both 2-zoned monaxine heteroids.
8, 4 ; 76«6732, 6«5«32: both monozones.
5 ; 6*5*3' : 2-ple zoneless monaxine heteroid.
6 ; 6*5^3' : 3 -zoned monarchaxine homozone.

Thirteen Id-acra :
1, 2, 3, 4, 5, 6, 7 ; 8726-5733, 876^5«33, 7"6«6«33, 726»5«38, 7^75438,

7-67 '*33, 6'5'3' : seven monozones.
8 ; 73635*33 : 3 -zoned monaxine heteroid.
9 ; 6'5333 : 3-ple zoneless monaxine heteroid.
10, 11, 12, 13; 7'635633, 72655«3S, 7675*3», 7675*3': four

asymmetricals.

Forty-two 16-acra :
1, 2, 3, 4, 5, 6, 7, 8, 9 ; 973625»3'», 87563553^ 8726«5*3<, 87«6«543S

8767533S 736«533S 736«533S 736*5334, 736*533^: nine monozones.
10, 11 ; 83335*3^ 736*533-* : two 3-zoned monaxine heteroids.
12, 13, 14, 15; 827W3S 7^6*6^3^ 726*523^ 7^^6^Z*i four

2-ple zoneless monaxine heteroids.
16, 17 ; 826*5^3S 726^523^ : two 2-zoned monaxine heteroids.
18; 7*6^543*: zoned triaxine.
19 ; 736*533^* : 3-ple zoneless monaxine heteroid.
20; 726^5-3'*: 2-ple monaxine monozone, having a contra janal

2-ple zoneless axis.
21, 22, 23, 24, 25, 26, 27; 8276^563^, 873685*3'*, 873635*3*,

87-^6*5*3*, 8726*5*3*, 8726*5*3*, 8726*5*3* ;
28, 29, 30, 31, 32, 33, 34; 8767533*, 8767533*, 7*6*5*3*, 736*533*,

7*6*533*, 736*533*, 736*533* ;
35, 36, 37, 38, 39, 40, 41 ; 736*533*, 736*5*3*, 736*5*3*, 736*533*,

726*523*, 726«523*, 76>*53* : twenty-one asymmetricals.
42; 6*234. zoneless tetrarchaxine, which has four principal

3-ple heteroid, and three secondary 2-ple janal, axes, all

seven zoneless.
There is nothing tentative in the enumeration or construction of these
solids on the bases P and Q, to one familiar with the resulting changes of
symmetry.

(8886. ) — The answers are the 1 5«acra, 1 6-acra 42 and 1 3 above given. In
first 24-edron, i, e, I, n occur thrice only, and in the second elqp
thrice only. These summits are the vertices of the imposed 4-edra.
IBither figure is easily drawn on a triangle whose sunmiits all occur more
than three times.

If we construct the first on dfmy we have a zoneless tetrarchaxine whose

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127

four like 3 -pie principal axes hare for polar summits e^ i, ly n ; and for
opposite 3 -pie polar faces, the triangles hpjf dkm^ fgq, and abe. The four
janal secondary 2 -pie faces bisect the pairs of edges ag, kp ; bj, fm ;
edf hq. If we efface the twelve bases of the triaLgles whose vertices are
«, », /, w, we obtain another zoneless tetrarchaxine 16-edral l6-acron, with
signature 4^ 3'*, autopolar ; and if we remove next the four triaces ^, i, /, n,
we get a zoned tetrarchaxine 8-edron, which is obtained by cutting
away the summits of a 4-edron. The reciprocals of all these are readily
made from the hexarchaxine 12-edron. By cutting away four of its
summits so as to make twelve 6-gons, we construct the reciprocal of
No. 42, the first in Quest. 8886.

If we construct the second in Quest. 8886 on its face abg, we have a
2 -pie zoneless monaxioe heteroid, whose axis bisects the edges bg smdj'k,

I have only recently learned, in tumiug over the leaves of Vol. lxviii.
of Crellb'8 Journal f that in this aud Vol. lxvi. there is a treatise on the
symmetry of Polyedra. Above twenty years ago, its French author did
me the honour to converse with me in my house, in fluent and good
English, on Groups and Polyedra. I was more amused than surprised to
find that he seems to write, in Ckelle*s Journal, quite unaware that his
subject had been previously and thoroughly discussed in Fkil, Tran.
1862. My complete familiarity with all the aspects of polyedral
symmetry lacks the appetite required for a real study of M. Job.dan'8
Memoirs upon it ; but they ought to be useful to the student who wishes
to verify the above solution of Quest. 9009. I may here, in mentioning
Crblle's Journal, be permitted to request the student's help in the solution
of a puzzle that beats me — how did the Cambridge and Dublin Mathematical
Journal, Vol. ii., N. S. (vi. O. S.), p. 191, contrive to steal so much from
a later paper in Crellb*s Journal^ Vol. lyi., p. 326, on exactly the same
problem in combinations ?

The following letter to me is before me, gummed into my copy of the
learned and noble writer's life : —

** 41, Chalcot Villas, Adelaide Road, N.W. ;
'' April \%th, 1862.

" My dear Sir, — I have to thank you for several papers, which I have
looked at from time to time. Your excessively complicated subject ¥rill
take shape in your hands I see. You have ascertained that the French
Institute is mainly founded for the purpose of advancing the claims of
Frenchmen — either by this means or that. But why go and buy your
experience, when you might have had it for nothing from history?
By printing in England you may save yourself from having your
theorems given to a Frenchman. Yours truly,

A. Db Morgan.*'

The ** printing" is that of my Theory of the Polyedra, which is also
the ** subject *' spoken of. The subject I have long, long ago exhausted,
at least so far that I know how to answer any question that I know how
to ask in it. The printing has been a distinguished failure. Yet I am
satisfied that enough has seen the light in London and Liverpool to pre-
vent the giving either of my theorems or of my methods to another.

The Theory of Knots, which Professor Tait and I have pretty nearly
wound up in the Phil. Trans, and Froc. Edin., is a brief supplementary
chapter of the Polyedra. Every knot is a polyedron, or a polyedral
reticulation, which has only tessarace summits, and which has or has not
2-gonal faces.

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APPENDIX II.

MISCELLANEOUS SOLUTIONS AND NOTES.
By R. F. DAVIS, M.A.

8559. (Professor Wolstenholme, M.A., Sc.D.) — From a fixed point
O are drawn OP, OQ tangents to one of a system of confocal conies (foci
S, 8', centre C), and from are let fall perpendiculars on the normals
at P, Q ; prove that the envelope of the straight line joining the feet of
these perpendiculars is the conic (parabola)

■[X (a: cos a + y sin a) + Y (a: sin a— y cos a) — <^ cos a}'

— 4XY (x cos a + y sin a) {x sin a~y cos a),
where is origin, OS axis of x, (X, Y) the point O, SS'= 2c, and a is
the sum of the angles which SO, S'O make with the axis of x.

[If X2- Y2 - c^, the straight line is fixed.]

Solution.*

Let 0, tphe the eccentric angles of the points P, Q for any conic of the
system whose semi-axes are a, *, where a^—b^ = e^. To find the co-
ordinates of L, the foot of the perpendicular from the centre on the
normal at P, we have

da; cos 9 + 0y sin 9 a 0,

fla: sin 6— *y cos 6 — c» sin 9 cos 9,
therefore (a' sin^ 9 + *3 cos^ e)x — cH sin^ B cos 9,

( )y « —c^^ sin 6 cos' 9.

Similarly for M, the foot of the perpendicular from the centre on the
normal at Q.
The equation, therefore, to LM (whose envelope is required) is

^ y 1

c^aBin^ecose, — c^A sin e cos* , flSsin^e + J^cogSe

<?2a sin' ^ cos ^, — c'^sin^cos'^, a'sin'^ + ^cos-^

or Aa? + By + C= (2),

where, after reduction, it is found

A« — c2i(sine— sin^)) {^-co8''«ecos'</>— a'sinasin^ (1 +sinesin4))},

B « ^a (cos ip - cos ff) {a2 sin' a sin' </>-*' cos a cos ^ (1 + cos % cos ^) } ,

Co* C^adsin0sin^cos0cos^sin(^— 0).

* For this Solution I am indebted to the Editor.

= ...(1),

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129

The relations (useful generally) connecting 0, <p with X, Tare found to be

(sin^— 8in6) : (cos©— cos^) ; sin (^—9) « AX : aY : ab (3),

andcosacos</> : sinesin^ : I=a2(*2_y2) . ^(a2-X2) : (*2X2 + a2Y«)...(4),
where (3) follows from the consideration that the tangents at $, f pass
through X, Y ; and (4) from the cousideration that 6, ^ are the points of
intersection of the polar of X, Y with the ellipse.

By means of (3) and (4) we are enabled to substitute completely for
$, <f> in (2) ; thus,

{(i3.y2j2« (a»-X2) (63 + Y«)}Xc- {(a2-X2)3- (68- Y«) («« +X«)}Yy
+ c2(a2-X2)(62-Y2) = (6).

If we now put a'— X* « x— jc, 6-— Y* — X, where \ is indeterminate
and X2-Yi-c2«n, then a^ + X^^ x-ic + 2X2, ^2 + Y^ = X2+2Y2 ; and
(5) reduces to

^x3+{(ic-2Y«)Xc+(2X2 + jc) Yy-<j2jc} \ + 2kY^Xx-'K^i/ = ...(6).
Now, with the notation of the question,

a « cot-» -^^ +cot-i 5±f « cot-i -^,

therefore ic = 2XY cot a.

Then equation (6) becomes
<?2x' + 2Xr{(Xcoto-Y}ir+(X + Ycoto)y-<J«cota}A.

+ 4X2Y8coto(jr-ycota) = (7),

and the envelope required is

{(Xcoto-Y)a;+(Ycota + X)y-c«coto}2 = 4c2Ycota (a;-ycoto)...(8),

or {X (a;co8a + y sino) — Y(j;sino— ycoso) — c2cosa}3

= 4c3Ycosa(arsino— ycosa)...(9),
which can be easily shown to agree with Professor "Wolstbwholmb's
result.

In a former Note (see Vol. xlv., Appendix in., pp. 169 — 172) I
have shown that, if O be a fixed point from which tangents OP, OQ are
drawn to any one of a system of confocal conies, then ^1) the normals at
P, Q touch the same parabola, which is the envelope of PQ, whose focus
F is found by making the / PCS = Z OCS and taking CF . CO = OS', and
whose directrix is OG; (2) these same normals intersect on a fixed
straight line FH perpendicular to OF. I therefore re-proposed Professor
Wolstenholmb's question in the form given below, the geomeCrical
solution of which can be reconciled with the equation to the envelope
given above (8). (See Appendix, Section 17.)

8692. (R. P. Davis, M.A. Suggested by Question 8659.)— If, from
any point in a given fixed straight Ime passing through the focus of a ^-
rabola, tangents be drawn to the curve, prove that the envelope of the line
joining the feet of the perpendiculars on these tangents from a given
fixed point on the directrix is another parabola.

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Solution by the PfioposBR.

From any point T in the given fixed straight line ST, let there be
drawn two tangents TP, TP' to the parabola focus S. Let B be a fixed
point on the directrix ; draw BQ, BP, BQ' perpendicular to TP, TS,
TP' respectively, and let the diameter through T meet the directrix in Ii.
Then F is fixed point ; and Q, F, L, Q' all lie on the circle described
upon BT as diameter.

Let fall the perpendicular FK on QQ' ; then we shall show that the
locus of R is a fixed straight line, and consequently the envelope of Q,Qf
is another parabola having its focus at F.

Since from the point T two tangents are drawn, / PTS = / FTL or
/ QTF = Z LTQS therefore FQ = LQ' and FL is parallel to QQ' and
perpendicular to FR. Now FR =- FQ sin STF = BT sin STP sin STP'
= BT (SY / ST) (SY / SP) = BT (SA / ST). Produce FR to O so that
FO « 2FR, and draw SM perpendicular to BT. Then FO. ST = BT.SX
and FO/SX = BT/ST « BF/SM, hence FO/BF - SX/SM.

Again, Z BFO = complement of ZBFL, or of ZBTL= zTBL
= Z MSX, since S, X, M, B are concyclic. Thus the triangles BFO,

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131

MSX are similar, and / OBP « L XMS = L XBS. Hence the locus of
O is a straight line through B, making with BF (the perpendictdar from
B on the given fixed straight line ST) an angle OBF eqiml to the angle
XBS between BX and BS. Also, since F is a fixed point and FR = JFO,
the locus of B is a fixed straight line through the middle point of BF.
Hence the envelope of QQ' is a parabola focus F.

8768. (B. F. Davis, M. A.)— Required an Analytical Proof of Fbuee-
BACu's Theorem.

[An analytical proof is given by the Rev. J. J. Milne, M.A., in the
Introduction to his Weekly Problem Papers, but this proof may be modi-
fied and improved.]

Solution,

Let ABO be a triangle inscribed in a circle S, and t^i ^ ^3 be the
lengths of the tangents from its angular points to another circle S'. Then,
if at^ ± ht^ ± ct^ « 0,

these circles will touch one another.

For if Pn p^f i?3 be the perpendiculars from A, B, C on the radical axis
(L) of ttie two circles, ^1*, ^j^ ^3' are proportional to Pi, p^, p^ respec-
tively. Hence aVpi + b Vp^ + e ^p^ = 0, and this is the biown condition
that L should touch S, or that the points of intersection of S and S'
should coincide. Now the tangents from the middle points of the sides of
any triangle to the inscribed circle may easily be shown to satisfy the
above relation ; hence the circle through these middle points touches the
inscribed circle.

8875. (Professor Nash, M. A.)— Professor Casey's Quest. 7839 may
be enunciated as follows : — DD', EE', FF' are the intersections of the
sides of a triangle ABC with the cosine circle, the order of the letters being
such that E'F, FD, IKE are diameters. The circle round AE'F cuts the
circles on AB, AC as diameters in the points a, a' ; and d, h\ Cy (f are
similarly determined upon the circles BF'D, CD'S. To show that the
circles round AC0, BA^, QY^c pass through a> the positive Brocard-point
of ABC, and the circles ABa', BC^', CAc' through the negative Brocard-
point. Prove the following additional properties : — (1) The circles AOa,
ABa' touch at A the sides AB, AC, and intersect again in a vertex of
Brocard's second triangle ; (2) the taugents to tbese circles at aci bisect
AB, AC ; (3) the points B, C, a, a' are concyclic, and the circle through
them touches OB, DC ; (4) AaD and Ka'Ti' are coUinear ; (5) C«, Ba' in-
tersect upon the symmedian AP, and Ca', Ba upon the perpendicular AL;
(6) aE', a'Y meet BC at the foot of the symmedian AP ; (7) aa\ E'F,
BO, and the radical axis of the circles ABC, AE'F, are concurrent;

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(8) the three points of concQirence of the three sets of lines in (7) lie upon
a line paralld to w»\ and therefore the triangle formed by the lines aaf^
hh'y ctf is in perspective with ABO ; (9) the pole of aa' with respect to the
circle AE'F lies upon the median of ADD' ; (10) the pairs of points ac^^
ba\ cV are isogonaL, and the inscribed conic whose foci are aef touches
CA at ^'d foot of the perpendicular from B ; (11) one directrix of this
conic is the line joining to the intersection of FD' and DE', and the
other the line joining A to the intersection of EF, FD'.

Solution,

Let ABC be a triangle, K its symmedian point, DD', EE', FF the
intersections of the cosine circle (centre K) wiUi the sides of the triangle
so that E'KF, F'KD, D'EE are diameters.

Upon AD, AJy take points a, (^ such that

Aa.AD =- Aa' . AD'== AF . AB = AE'. AC.
Then, since F, a\ a, E' are the inverses (with respect to A) of the four
colUnear points B, D', D, C, the circumcircle of AE'F passes through
a, a'. Also, since B, F, a, D are concyclic, the Z BaD = / BFD = ^w,
by a known property of the cosine circle. Hence a lies on the circle
described on AB as diameter ; similarly a' lies on the circle described
on AC.

Since D, a, E', C are concyclic, and DE' is parallel to AB, Z AaC = ir
- Z CaD = IT- Z CE'D = w - A. Similarly Z Aa'B = ir- A. Hence
the circumcircles of ACa, ABaf pass one tlu-ough the positive Brocard-
point of ABC and one through the negative Brocard-point ; while

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(1) They intersect again in a vertex of Bbocard^s second triangle.
Moreover, since Z BAa = / aDE'= Z aCA, AB is a tangent to the first
circle.

(2) The circles ACa, ABa may be considered as the inverses (with re-
spect to A) of the perpendicular lines DB', DF. They therefore intersect
orthogonally, and the tangent at a to the first circle passes through the
centre of the second circle which is the middle point of AB.

(3) Since F, D', D, E' are concyclic, their inverses B, a\ a, C are also
concyclic. Now Z BaC = Z BaD + Z OaD = iw + A ; and if be the
circum-centre Z OBC = Jjt— A, hence OB touches the circle Ba'aO.

(4) The points a, of were assumed to lie on AD, AB' respectively.

(6) Since DE', FD' are parallel to AB, AC respectively, they intersect
in a point A' on the symmedian KA produced such that KA'=KA.
Then, since AE'. AC = AF . AB and A'D . A'E'- A'D'. A'F, hence AKA
is the radical axis of the circles CE'aD, BFa'D'.

In the remainder of this solution frequent use will be made of the well-
known theorem that the radical axes of any three circles (taken two at a
time) are concurrent.

From the circles CE'aD, BFa'D', Ba'aC we infer that Ca, Ba' meet on
the synmiedian AKA' ; and from the circles ACa', ABa, Ba'aC, that Ca',
Ba meet on the perpendicular AL.

(6) From the circles CE'aD, BFaiy, AE'F we infer that dE', a'F meet
on the symmedian AKA'.

Moreover, since Z aBD - iir- Z oDB = ^ir- Z aE'C « Z D'E'tf, the
four points B, D', a, E' are concyclic. Then, from the circles BD'aE',
BFa'D', AE'F, we infer that cE', a'F meet on BC. But they have already
been shown to meet on AKA' ; they therefore meet at the foot P of the
symmedian ; and Ca, CA' divide the angle at C harmonically.

(7) From the circles BFE'C, Ba'aC, AE'F we infer that aa% E'F, BC
meet in a point a ; which, since aB . aC = oE'. aF, is also a point on the
radical axis of the circles ABC, AE'F. If /3, y be similarly determined,
a, /3, 7 lie on a straight line which is the Pascal line of the hexagon
DD'EE'FF'.

(8) This line afiy is obviously the radical axis of the cosine- and circum-
circles, whose direction must be perpendicular to KO or parallel to «»'.

(9) The pole of aa* with respect to the circle AE'F is obviously the
symmedian line through A of the triangle Aaa' ; or the median line of
the triangle ADD', since aa\ DD' are antiparallel.

(10) Draw Bi/ perpendicular to CF', and BM to AC. Then, since
ZBaC+ ZBi/C » iir + A + Jir = ir + A ; and similarly ZAaB+zA^B
= ir+C; hence a, c' are isopfonal conjugates. Moreover, since ZaMB
= Z aAB = Z f/CB = Z (j'MB, M is the point of contact of the side CA.

(11) Since Z BMC is a right angle, and Z BMa = Z BAa == Z aDE'
— Z aCM ; therefore Z MaC is a right angle and C a point on the direc-
trix. Since CA, CB are tangents and a the focus, CA' (the fourth har-
monic to CA, Ca, CB) is the directrix.

[For information respecting the ** cosine- circle*' see the forthcoming
volume, entitled ** Companion to the Weekly Problem Papers," by the
Rev. J. J. Milne, M.A. (Macmillans), which has an admirable section
devoted to the new geometry of the triangle, contributed by our
correspondent the Rev. T. C. Simmons, M.A.]

VOL. XLVn. R

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134

Kon o9 THX Bbocabdai. EixmK.

1. If a point Q be taken on
tibe tangent at any point P of
an ellipse (Fig. 1% sach that the
angle SQP is constant (= fi\
the locos of Q will be a drde
whose centre O lies on the
minor axis of the ellipse, and
which has doable contact with
the ellipse.

For, drawing SY perpendi-
cnlar on the tangent, since
ZQSY«i»-3, and SQ =
SY coeec^, the locus of Q is a
circle of radios a cosec $. Let
O be a point on the minoraxis,
soch that Z SCO = $9, then the
triangles CSY, OSQ are simi-
lar and in the linear propor-
tion of 1 : cosec fi. Hence the

locos of Q is a circle whose Fii^l.

centre is O, and whose radius « SO/* =»the normal to the ellipse
passing throogh O, and conseqoentlj has dooble contact (real or
imaginary) with the ellipse.

It is easQy seen from the above proof that, when SY revolves in the
positive direction throogh an angle i*-3 into the position SQ, the locus
of Q has its centre O on the minor axis at a distance aecotfi below the
centre ; for the negative direction of rotation, O is at the same distance
o^ofv the centre. There is no restriction as to the value of fi, and by
varying it we get a series of circles having double contact with the
ellipse.

2. If the tangent to
the ellipse at P (Fig. 2)
meet a particular circle
in Qi, and the other
tangent from Qj to
the ellipse meet the cir-
cle again in Q^ then
Z SQ2Q1 = fi ; aud so on.
Starting from any point
P on the ellipse in this
way, we obtain an un-
closed polygon PQ1Q2Q3 Fig. 2.
...Qn of aoy number of sides we please, whose vertices He on the circle,
and whose sides touch the ellipse, such that

ZSQiP = ZSQ,Q, « ZSQ3Q2 = ... =: zSQ„Qn.i - fi;
and con8equently,ifS' be the other focus,

Z S'QiQ, = z S'QjQa -...==/ S'Q^.iQn = fi.
'4. ^il ^^Z^ ^® determine jS, by means of the invariant condition, that
It should be possible to inscribe figures of 3, 4, 6, 6.. .sides in the circle
which at the same time circumscribe the elHpse, QiQc—Qn (of the pre-

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135

yious article) is no longer an unclosed figure. The two foci become in
this case the two Brocard points, and /S becomes the Brocard angle c».
Taking the equation to the two curves in the forms

a^/«2 + y«/*2= 1 (2%

and »' + (y + a«j cot w)^ = «' cosec^ o»,

a;2 + y2 + 2a<?ycot«-(a2 + ^cot««) - (2).

The equation determining #c, which causes 2 + k2' to break up into
straight lines, is (after reduction)

(ic + fl2)2(K + ^cosec««) - 0.

Hence A', e', e, A are proportional to 1, 2a^ + b^ cosec' », a* + 2aH^coaec?a^
a*^co8ec2 w, or as 1, 2 + 2, 1 + 2z, s, where 2 = A* cosec^w / a^.

Now, the condition that it shotdd be possible to inscribe triangles to 2
that are circumscribed to 5' is e'-— 4A'e = 0.
Hence (2 + 2)2-4(1 + 2a) =0, or z^-iz = 0, « - 4.

Thus ; —i; sm«= — ; ^«1 — 1— 4sm'i».

o^ 2a d^

Quadrilateral : e'3-4ee'A' + Sa^^a « 0.

(2 + 2)a-4(2 + 2)(l + 2z) + 82-0, aj8-2«2- 0. « » 2.

^^ ^co8e^^2 or sin., --A..

fl2 v^2a

Hexagon :

{e'2-4A'e}3 = 4 {e'8-4ee'A' + 8A'2A} {e'8-4ee'A'+i6A'«A},

whence « = ^.

4. Summarizing results : — If B be the circum-radius and m the Brocard
angle of a

(i.) Triangle: the semi -major axis of the Brocard ellipse is Bsinw,
the semi-minor axis 2 ft sin- w, and the eccentricity = v^ (1 — 4 sin^ w).

(ii.) Quadrilateral: Rsinw, A/2Rsin2«, V(cos2«).

2
(iii.) Hexagon : R sin «, — — R sin^w, V (1 — 4 *^' »)•

The above note was suggested by, and is an amplification of, No. 1056
in Professor Wolstenholme's larger collection of Problems. Published
so far back as 1878, it contains (^as was first pointed out to me by the
Rev. T. C. Simmons) a precise and complete statement of the existence of
the Brocard ellipse.

Geometrical Construction for the Brocard- Angle, &c.

I. Let ABC be a triangle. Describe a circle touching AB in A and
passing through C. Let AP be the chord of this circle parallel to
BC. Join BP, meeting the circle in a, which will be the positive
Brocard -point.

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136

For, siDce AB touches the circle in A, l AAB » I fiCA, and also
« L APA b I ABC. An analogous construction gives the negative
Brocard-point A'.

II. Since L ACB - A^
L CAP and Z BAC »
Z APC, the triangle PCA

is inversely similar to
ABC. Hence a simple
ruler-construction can be
stated for the graphical
determination of the Bro-
card-'ingle of a triangle.

This construction (on
the authority of the Rev.
T. C. Simmons, M.A.) is
already known to conti-
nental mathematicians.

III. Similarly, if onBC
a triangle CQB be de-
scribed inversely similar
to ABC, ZAQB=ZAPB
SB to, the Brccird-angle ;
and, since Z ACP = B and Z BCQ = A, the points P, 0, Q are collinear ;
also PC : AC « AB : BC and QC : BC = AB : AC,
therefore PC . QC = AB^.

Now P, Q both lie on the circle 2, which is the locus of points at which
AB subtends a constant angle », supposing AB and w only given. If
AB = a, the centre of 5 lies on the line bisecting AB at right angles at a
distance therefrom = ^t^coto); while radius = ^acosecw. Moreover,
since PC . QC = a^ = constant, the locus of C is a circle concentric with
2, whose radius p is given by the equation

W C08ec2 « - p' = a' or 4p2 = a^ (cot^ « - 3) .
The locus of C under these circumstances is termed Neuberg*s circle ; and
its equation, referred to AB and the perpendicular bisector of AB as axes,
is «' + (y — ia cot w)' = ia^ (cot' «— 3) or x^ + xf^—ay cot « + J a' « 0.

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APPENDIX III.

SOLUTIONS OF SOME OLD QUESTIONS.
By W. J. CURRAN SHARP, M.A.

2391. (Professor Sylvester, P.R.S.) — Let /u points be given on a
cubic curve. Through them draw any curve (simple or compound) of
degree v; the remaining 3v— m (say ^Q points may be termed a first
residuum to the given ones. Through these /u' points draw any curve of
degree / ; the remaining Zif' — fi' points may be termed a residuum of the
second order to the given ones ; and in this way we may form at pleasure
a series of residua of the third, fourth, and of any higher order. If fi is
of the form 3«- 1, a residuum of the first or any odd order, and if /x is
of the form 3i+ 1, a residuum of the second or any even order in such
series, may be made to consist of a single point, which I call the residual
of the original fi points. Prove that any such residual is dependent
wholly and solely on the original fi points, being independent of the
number, degrees, and forms of the successive auxiliary curves employed
to arrive at it.

3651. (Professor Sylvester, F.R.S.)— If through 3fi + I given points
on a cubic curve a curve of the order N -»- « be drawn, and through the
remaining 3N— 1 intersections of the two curves a third one be drawn of
the order N ; prove that this will intersect the cubic at a fixed point.

[This point may be called the opposite of the 3n + 1 given points ; it
becomes Dr. Salmon's opposite, us defined by him in the Fhilosophical
Transactions for 1858, when » = 1, N « 1, and is independent of the
value of N.]

Solution,

(1) If /» =s 3t— 1. Since every «-ic curve through 3«— 1 points in a
cnbic cuts it in the same additional point (Salmon's H. P. C, p. 133),
any t-ic through the /u points will intersect the cubic again in the same
point A (the residual). Let this be V = 0.

Now if U a be a v-ic through the /u points and fti others,
Ui =0 „ vi-ic „ Ml >t "f ii

Us =0 „ va-ic „ /«2 i» A*« i>

U2r =0 „ V2r-ic „ M2r „ 1 „

this one point will bo A. For evidently Zvr ^t*r + f^-nt

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138

and if * + »'i + "s ••• + ''2r-i ~ "*»
3m « 3» + 3vi + 3v8+ + 3i'2r-i

= l + 3v + 3v2+ + 3^2^-2 +^''2r"'^

«» 3(^ + 1^3+ +vj

and /tt + /ui + fi2+ ... +;*2^ — 3m— 1 ;

therefore V.Ui . U3 ... U2r.i = and U .Ua... XJ2r -
are m-ic curves througli the same 3m — 1 points on the cubic, and there-
fore through the same additional point, which is A, the residual on the
first, and the remaining intersection of Ujjr = on the second ; which proves
the principle for this case.

(2) In the other case it is advisable, first, to show that every «-ic curve
through 3n — 2 points on a cubic cuts it again in two, such that the line
joining them passes through a fixed point.

For, if V — be one «-ic through the 3«— 2 points and A and B
W — be another ,, „ „ and and D

find a = be the line AB and /S = the line CD.

i8V «» and aW « are two (« + l)-ics through the same 3« + 2 points
(viz., the given ones and A, B, C, and D), and therefore their remaining
intersections with the curve must coincide, i.e., a » and /S ^ meet the
cubic in a fixed point.

If /* = 3»+ 1, and V be (i + l)-ic through the fi points and A and B,
and a = be the line AB, the third intersection of o = is fixed.

Now let U s be a v-ic through the /x points and fti others
Ui =0 „ vi-ic „ fAi „ fii „

then, if m = 1 +if + v2+ ... +^2^.2,

3m = 3 + 3v+3y2+-- + 3i'2r.2 = 3 + /« + /Ui + ^ + jU3... +A*2r.i

= 3l + 4+3Vi + 3>'3... + 3l'2r-l-l = 3 (l + » + lfi+V8... +V2,._,),

therefore a U . Uj ... U2r-2 = and V . Ui . U3 ... U2r-i = are m-ic
curves through A, B, and the /i + /*!••• +H-2r-\ poii^ts on the cubic, i.e.,

through the same 3»»— 1 points, and therefore the same additional point,
which is the third intersection of a = (AB).

This important theorem is thus fully proved. In Question 3651, the
single point is the one co-residual to the 3»+ 1, and is therefore fixed by
(2) above.

3535. (Professor Sylvester, F.R.S.) — 1. Eltler has shown that the
number of modes of composing n with i distinct numbers is equal to the

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denumerant (that is, the number of solutions in positive integers, zeros
included) of the equation

a?+2y + 3a5+ ... +»» = »-i (fS+i).

(1) Show more generally that the number of modes of composing n
with i numbers, all distinct except the largest, which is to be always
takeny times, is the denumerant of the equation

Jit + {j+L)p+ ... +«« = «- J (f-y+l)(»+y).

(2) Show also that, if all the partitions of n into t parts are distinct ex-
cept the least J which is to be taken y times, then the number of such par-
titions is the denumerant of the equation

ay + 2y+ ... +{i-j)<f^ + iu ^ n-i {(i-jy + Zi^j},

Solution,

1. The required number of ways of composing n with t numbers all dis-
tinct except the largest, which is to be taken j times, is the coefficient
of x**zi^ in

a?'*' + (1 + xz) J^^ + (1 + xz){\ -H xH) x^s^ + &c.

= a// { Ao + Ax2 + AggS + &c. } suppose.

That is, it is the coefficient of x'^-iz^-i in

l + (l+a;2)ar'+(l +X2)(l + a;'2)i;^ + &c. = A<, + AjZ + Aja* + &c.,

where Aq « ..

Let xz = y, therefore ""

l + (l+y)a;^+(l+y)(l + ay)a:^ + &c.«Ao+'^y+^y2 + &c.;

X x*

but l + (l + y)a^'+(l+y)(l+iry)ir^ + &c.

= l + (l+y);t^{l+(l+xy)a^+(l + a:y)(l+fljV)»^ + &c.}

« l + (l+y)a/{Ao + Aiy + Ajy2 + &c.},
and, equating coefficients,

^-;r^(Ao + Ai), ^ = a^CAx + Aj), and generally^ - ar'CAn-i+AJ ;

therefore Aj — ~ Aq, A^ =* r^^ Aj, &c.,

1—* 1— »

A„

(lW;(l-ar'*')(l-a?'^^..(l-a;^*")'
and therefore the number required, the coefficient s^"^ in A»_,-, is the

coefficient of ;p*-*('->*i)«*» in 1 ,

(l-ic')(l-a^*')(l-;r^^*)...(l-aO
which is the denumerant of

ya; + (y+I)y + (y+2)«+ ... +t«=.n-i(t-y+l)(f+».
2. If all the partitions of n into t parts are distinct, except the least,
which is always to be taken/ times, the number of these is the coefficient
of df"«< in a?'(»0(l+ir«2)(l + a^)... + ic*V(l + a:S2)(l+;r^2)... + &c.,

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140

or that of aj^'V"-' in

(l+a?»«)(l + »»«)... +a?^ (I + aj%)(l +a;*«)... + &c. = Ao+Ai« + A2«*+&c.,
suppose.

Uxz^y this becomes

(1 ■¥xy){\ ¥x^y){l + ajSy)... + of {l-¥a^y){\ +«3y)... + &c.

= Aq + Ai-^ +A2^ +&C.,

X X*

but the first member = (I + xy) (1 + x^y){y + ic^y) . . .

+ a;^' {(1 +«V)(1 + «'y)-.- +^(1 +^y)(l + «'V).- + &c.}

= 1+ r^-y+ -. ^. ^y3 + &c. + ar'{Ao + A^ + A2y» + &c.},

1— a? (1— flj)(l — a:-*;

A ar***^""^^
and therefore -f == — r- — + A„ar,

An = ^

a;

in(n-t-3}

'(l-a:)(l-a;S)...(l-a:")(l-a/*")'
and therefore the number required, the coefGlcieDt of xl^'i in A<.^, is the

coemcientoi a; in (i_^)(i_a;3)...(i_a;.-i)(i_a:0'

which is the denumerant of

a? + 2y+... + (t-^') <► + «« =n-i{(t-»2 + 3t-y}.

3427. (Professor Sylvester, F.R.S.) — ^If ^, ^ are quantics in x^ y,
each of degree /a ; F a quantic in ^, if^ of degree m, and. consequently in
X, y of degree m/i ; and if J denote the Jacobian of 4>, ^^ that is,

dtp d^ d<f> d}j/^
dx dy dy dx*

D^,^ F the discriminant of F treated as a quantic m ^, if^ ; IDx, yt F the dis-
criminant of F, treated as a quantic in a;, y ; and if E be used as the
symbol of " resultant in regard of a?, y ** ; prove that

D,,y F = 2 [R «►, ,^)]«--2m E (F, J) (J>^,^¥y,

As a particular case of the foregoing theorem, show that the discri-
minant of F, any symmetrical quantic of an even degree in ar, y, is of the
form F (1, 1) F (1, — 1) Q.^, where XI is a rational integral function of the
coefficients in F.

Find also what the general formula becomes when ^, 4^ are taken linear
fiinctions of a;, y.

Solution,
By the Question, F s (^ — w^) (^ _. jSi^) (^ — y^^) to m factors,
therefore I)x,y F » the product of the discriminants of ^— on^, ^p—fi^t &c.
X the squares of the resultants of ^— aif^, ^— i3>|#, &c. taken two and two.

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Now the resultant of <f>-a\lf and 0— jSif' is (o— j8)'*R (^, tf^),
(Salmon's Higher Algeh'ay Arts. 180, &c., whence this solution, due to
Prof. Sylvester, Comptes JRenduSf Lvni., 1078, is mainly derived.)
Therefore D,,y F = the product of the discriminants of ^— otf^, ^— i8«f^, &c.

x(a-i8) (a~7)^'* [R(<|», tf^) ]»»("- D,

= the above product x (D^,^ F)'* [R (<^, Tf^)J"»('"-i).
Again, the discriminant of ^— aif' is the resultant of

^-.^ = and ^-.ff:-[0 (I.)

ax ax dy ay

or, by Euler's theorem, of <>-oi^ = 0, # . ^ - ^ ^ s J = 0,

dx dy dy dx

unless ^ — 0, if' = 0, or the x or y introduced vanish, which last
suppositions only give numerical factors, and the discriminant can only
differ by such a iaetor from R (<|>— oif^, J) -f- R (<t>, ^), and therefore the
product of the discriminants = R (F, J) + [R (^, if^)]*" ;
and therefore D^, y F = [R {<t>y if^)]*"' -«'« R (F, J) (D^, ^ P)'*.

If F be a symmetrical quantic of even degree, let ^ = a;2+y2^ i|> __ ^y .
therefore J = 2 {x^-y^ and R (F, J) = F (1, 1) F (1, - 1),

R {<!>, rp) - I and (P^,^¥y = (o-i8)2'* (0-7)2- = Xi*,

where XI is a rational integral function of the coefficients of F.

If ^ = Xaj + /*y, if' = \'x + f/y, R (4), i|») = x'/*— ^/*'>

J = A'/*- A/*', and R (F, J) = (a'/*-a/)»» ;
therefore, by substitution, Dx,y (F) « (a'/*- A/)"* ("»-i) (D^,^ F),
which is the evidently correct result, since the discriminant is an in-
variant, and aV— A/i' the modulus of the linear transformation.

5271. (Professor Caylby, F.R.S.) — If « be an imaginary cube root
of unity, show that, if

(a>-ft>2)a; + ft,V , dy ^ {u^w^dx ,

and explain the general theory.

Solution.

II ^ ^ mdx ,j.

{(l-y2)(I-AV)}* {(l^x^){l^i^x^)y ^"

y must be a function of x. Let y = ^ (:r).

Now if, in (1), —y and —a: be written for y and x, the equation is
unaltered; therefore —y^<p(—x), and therefore xy and y/x are even

functions of :r. JjQi y j x = y^ {x^) (2).

(This form is chosen in order that y and x may vanish together.)

VOL. XLVII. 8

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142

Again, if, in (1), — and — be written fory and x, the equation is
unaltered: therefore — sii'f-— K

andtherefore — =^M^(-^) (3).

It therefore appears that y^ {x^ is fractional. Now, assuming that the
transformation is rational, let

and, from (3), Kfi{x^fi (-i-^) ^Aa(^s)« ( -^),

or

kB.(^-.») ft-,')...(»«-««) X ( ^--±,) ( *,- -1-,) ... (».- -1.)

^AA»(«.-.»)(«,-.^...K-.V («.-;^)(«,-^) •••{''--;^).

an identity which involves m = n (unless some of the quantities
aia2...rtm» bib^.-.bn he zero), and a series of conditions of the form
br-x^ = 0, equivalent to a^ — ^'^ = 0, or ar — [ 1 / (ic^a?-)] = 0. Each equiva-
lence of the first class serves to reduce the order of the transformation,
while each of the second class gives *, = 1 /(<»rK^)f

andtherefore tL = A h-x^) («,-^«) ■■■ (a,-:t^

■B (1-«2kV;(1-4Vj;2)...(1-PkV)

il a, = o', ffj = i», &c. &c. ;

and, from (3), -1 - ^ (a* ... /)!! k2» ; therefore ~ \ ,

'^ li' B (K'»-'A)»(«4...i)-'

« V A y (1-<j-A')(1-«2k%2)... (1-Pic2^s/

while, hy substitution in (1) and putting x = 0, it appears thatj

?_) («i...;r = m»,&c.,
therefore y = ,„:. . / " V ^ ^i—l LI.

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143

which agrees with the form given by Prof. Caylby (EUiptie'Functumi^
p. 169), and the quantities a, ^, ... ^ might be foond from the identity

(1 _a;') (1 - k2^) {dyY = (1 -y2) (1 - xVO m« {dxy.

This would, however, be a very long process, and an equivalent result may
be obtained as follows.

Since (l-y^(l-AV) must be equal to (l-ap2) (1 -«'«') multiplied
by a square factor, assume either

. ^_ {/W}'(i-^) „ {/Mp(i^») ,

"^^^'^ ■ -^[h) ""'^' -Mi^O

"'(Hi)

of which either set fulfils the condition, while the equation assumed

gives just enough conditions to determine a, 6, ... /, and g^ve a relation,
the modular equation, between k and X.

In the particular case where n = 1 — the cubic transformation (assuming
the second form that the quantities may be real),

and let ic = «*, X = t;*.

Therefore 2A-1 « f — V »* - ^ «», 2A-A2 - aV = w8a?,

therefore ?^^* - (2A-1) ««, or 2A (l-«V) - A^-fA;* ;

therefore ± 2 ~ (l-iA>«) - ^ -wV, or ± 2uv (1-«V) = «*-!;<,
the known form of the modular equation (Caylby's FunetUms, p. 188).

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144

Returning to the form y = ( — ya^x -^^»

and comparing it with the given equation y = \"~w ;^ + «-g
It appears that m = f -^V a^ = «-««, -i = ?^ ,

and aV « «» («- w^j j therefore «-=-«* = -«, and

(f)'-- -^

therefore — = w* = — k^, and k = -x, and x' = ic^ = — « ;

and therefore ^ ^ ^"-T"'^^^ .

(1 -y2)* (1 + «y2)* (1 -*2)* (1 + 0,2:2)*

a result which may be obtained from the equations which lead to the
modular equation, viz.,

In these let k = - X ;

therefore 2A- 1 = ic (-l)*a2^ 2h^h'^^K^a^, A««ic(-1)*,

therefore {2h - 1) A^ + 2A-A2 = 0,

and, since h is not zero, A^- A + 1 = 0, therefore A = - w,

therefore ir^ « — A* = — w* — — « = X^,

ind ic2«3= l-(l_^)2== -2«-<»«- l-« = «2(„.j^^

and • a2 - -ft, (0,-0,2) = - «Il5^,

«2

which give the same transformation.

5420. (Professor Sylvester, F.R.S.) — Prom the expansion of
{log (1 +a:)}*» , in a series according to powers of », prove that S,-, j [the

coefficient oiO is the developed product of (l + ^)(l + 2^) (! + •<)]

is divisible by every prime number greater than /+ 1 contained in any
term in the series »+ 1, i, »— 1, , «— y+1.

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145

Solution.
Let z « log, (1 + x), therefore 1 + a; » <?% and
*v « (1 + x)yl^

1 a 1.2 \z J 1.2.3 \ z I

therefore {log(l+.)r = «- = ^{l-f^^. + ^-;^^5^j.' + &c.],

a.

and hence ^^ r (—1)^ is the coefficient of a:*** in

(•->+2)...(i + l)^ '

{log(l+^)}W.i = :^-.>iJl-| + ^-|. + &c.]*-'*\

which proves the proposition, as the coefficient of a!**^ in this last expan-
sion cannot involve any prime factor (in the denominator) greater
thany+ 1, and will involve i—j'-i- 1 (in the numerator), if that number be

a prime, unless n (t-y+ 1) =• + 1 when t + 1 — -^ and • -^ + 1 =* -^
and i—j'+l is not greater thany+ 1.

iProf. Sylvester observes that his theorem ought to be stated as
ows:— Sf,y contains every divisor of the product (*+ 1)»(»— 1) ...
(f —j'+l) which has no prime factor less than J +2. He also observes that
the well-known theorem ordinarily associated with Wilson's, viz., that
Sj»-i,j when^ is a prime number and/<p— 1 contains^, is an immediate
consequence of his theorem, which teaches this, but much more besides ;
that is to say, his theorem is at once a generalization and an extension of
the theorem associated with "Wilson's theorem.

In general the coefficient ofa^'in (l + -^ + -«-+---+ ,• — r + &c j

consists of two groups of fractions. In one group each fraction (»— /+ 1)
appears as a factor in its numerator, in the other group the factor (i -/ + 1)
will not so appear.

But this latter group has no existence unless y contains i—j'+l as a
factor, which cannot be the case when i —J + 1 is greater than j nor d fortiori
when i— y+ 1 is greater thany+ 1. In every fraction of either group the
denominator contains only powers of 2, 3, ... {j+ 1).

Hence, striking out from the product of

(» + l)f.(i-l)...(i-y+2) (.-->+ 1)
all powers which it contains of the numbers 2, 3, ... (y+ I), or, which is
the same thing, of all the prime numbers not inferior toy + 2, the quantity
which remains must be a divisor of S,- ,.

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146

5640. (W. J. CuRRAN Shabp, M.A.)— Salmon says THiffher Plane
Curvet, p. 98] that the curve parallel to a given curve may be obtained (1)
as the envelope of a circle of given radius whose centre moves on the given
curve ; (2) as the envelope of the parallel to the tangent to the given curve
drawn at a constant distance. Prove that these processes are equivalent.

Solution.

On the first supposition, we have to find the envelope of

U-xy+{n-y)^^k^' where <p{xy)^0 or L + M^ = (1,2).

ax

Now, from (1), ({-a:) + (ij-y) ^ = 0, therefore ^ = ^Zllf,
ax Jj M

and ntp {xy) =La; + My + N=»0 and {i—xY + {-n—yf ■» ** ;

. t:f _ n^zV ^ ^ . L(^-a:)+M(tT~y) ^ k

*' L M ""a/L2 + M^' *' L^+M2 \/La + M*'

therefore L| + Mij + N = k>/'i?-¥U\

the equation to the line enveloped upon the second supposition.

5755. (W. J. C. Shakp, M.A.) — If, through any point of inflection
O on an n-ic, there be drawn three straight lines meeting the curve in
Ai, A,...An-i ; Bi, B2...B„-i ; Ci, Cg-.C^-i, respectively ; prove that every
curve of the n^ degree through the 3« — 2 points O, Aj, A2...An-i;
Bi...B»-i ; Ci...C„_i will have O for a point of inflexion.

Solution.

If we call the second line, which, with the inflexional tangent at 0,
makes up the polar conic* a sapra -harmonic polar, and if the three lines
meet this in A, B, and C, these points are also common loci of supra-
harmonic means of the point O with regard to all curves through the
3» — 2 points. This locus, then, which in general is a conic, must, since
these three points of it are in a right line, be for all those curves the
same right line ; and, therefore, the point O must be a point of inflexion
on each of them.

In the above I have called OE the supra-harmonic mean between

ORi, OE,...OR»>if ll^l == J-- + -1- + ... + 1

OK OKi OEj, 0R„_

5809. (D. Edwardes.) — If O be any point within a triangle ABC,
prove that OA* sin 2 A + OB^ sin 2B + 00^ sin 20 is least when O is the
centre of the circumscribed circle.

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147

7050. (D. Edwaedes.) — If P be any point in the plane of a triangle
ABC, and d its distance from the circumscribed centre, show that
PA2 sin 2A + PB^ sin 2B + PC- sin 2C = 4 (RS + ^) sin A sin B sin C.

7069. (^' Edwardes.) — If X, y, z be the distances of a point P from
the angular points of a triangle, prove that the mean value of x^ sin 2A
+ y2 sin 2B + z^ sin 2C, as P ranges over the circle about ABC, is three
times the area of the triangle.

Solutions.

(7050.) If the centre of the,circum-circle be taken as origin of rect-
angular Cartesian coordinates, and (x, y), (arj, yj, (a-j, yj), {x^, y^ denote P,
A, B, C.

PB2 = (aj-a?2)^+ (y-yz)^ = R2 + ^^-2 (^arg + yya)*
PC3 - {x-x^f+iy-y^-^ - R2 + rf2-2(:rd;3 + yy8),

therefore

PA2, X,, y,

^1, Vly 1

PB^ x^, y.

=:(Ra + rf2)

^2» y2» 1

PC2, 0^3, ^3

^3» ya» 1

or 2PA2. A BOC + 2PB2. A CO A + 2PC*. A AOB = 2 (Rs + rf2) . a ABC. . . ( A),
and 2AB0C « R2sin2A, 2aC0A = R=^8in 2B, 2aA0B - R2sin2C;
and therefore

2AABC =s R2(sin2A + 8in2B + 8in2C) = 4R« sin A sin B sin C,
therefore PA2 sin 2A + PB^sin 2B + PC2 sin 2C= 4(R2 + d^) sin A sin B sinC.

(6809.) And PA^ sin 2 A + PB^ sin 2B + PC^ sin 2C is a minimum when
d =0, i.e.y when P coincides with O the circum- centre.

In space of n dimensions, if V be the content of a simplicissimum ABC,
&c. (Quest. 8242), and Yj be the content of the simplicissimum OBC...,
Va of OAC..., &c., the formula

PA2.V, + PB2.V2 + PC«. V5+...= (R2 + rfS)V

may be established in the same way as the formula (A), and it follows
that the sinister is a minimum when P coincides with O, the centre of the
circumscribed spheric (hyper- sphere).

(7069.) From the above, we have

a;2sin2A + y2 8in2B + 22sin2C= 4 (R^ + rf^) gin jj^ gi^ b gjn C,
and the mean value

= 4sinAsinB8inCx4f f^'"'^V + y2 + R)2rf^rfy^ 4 pP"*''^*^^
Jo Jo Jo Jo

= 4R2sin AsinBsinC ^ (x'^-vy'^^-l)dxdy -ir \ dxdy

jqJo JoJo

= 4R2sinAsinBsinCx ?f + — « 3 x 2R2 sin A sin B sin C
8 4

= 3 A ABC. [This integration may be effected thus : —

0;'-""(-%y'+i)<^<^y = j;(f% i) (!-.=)» .i^

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148

2 f 1

« 1 1^ ar2 (1 -a;2)* ^^+jf (1 -^)*^*

24 3 "* 8 *

We may, however, more readily obtain the mean value as

= 4 sin AsinB sin C f^ C'{t^ + 'R')rdrde + T C'rdrde = &o.
Jo Jo Jo Jo

From Question 7050 it appears that, if •

PA2. ABOC + PB2.ACOA + PC2. AAOB
is constant, P lies on a circle whose centre is O and &c. in higher space.]

6686. (W. J. C. Sharp, M.A.) — If a bar naturally curved be strained,
the bending moment at a point, whose natural curvature is t"^ and
strained curvature f>-^, isE(p-^— r-^) {l + Ar-*+Br-*+d;}.

Solution.
Using the notation of Minchin's Statics, p. 510, if PP' be the portion
of any fibre intercepted between two near normal sections, PP" the same
fibre in its unstrained state, nn' the portion of the mean fibre intercepted
between these sections, and r and p the initial and strained radii of cur-
vature, we have

«« p nn' r PF' \p r J r-¥y

and consequently the longitudinal stress on the prism standing on a small

area8«r ^^( L -l.\r JL. z^

\p r/ r±y

and therefore the bending moment = E ( — j I -^ — dff

r
r=E (— - J-)I, ifthebarbeof small section I

^E^i- i^ 5"!+ l^d(r+ {^dir + &c,\ in any case

5828. (Professor Darboux.) — On coupe une pyramide triangulaire
SABC par un plan parall^lek la base ; ce plan rencontre les aretes lat^rales
SA, SB, SC en A', B', C ; on m^ne ensuite les plans CA'B', AB'C, BC'A'.
Soit P leur point commun. Determiner le lieu d6crit par le point P lors-
que le plan A'B'C se deplace en demeurant parallele k la base.

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Solution.

Taking the given tetrahedron as tetrahedron of reference, and the
faces opposite to S, A, B, C, respectively, as X = 0, ft * 0, y — 0, ir =* 0,
the coordinates being tetrahedral, so that x+/A + y + iraOis the equation
to infinity ; let x— a (a + /* + y + tt) = be the equation to A'B'C,
then at A', if = 0, ir = 0, \-a (\ + fi) = 0,
B', IT = 0, /x « 0, \-a{\ + v) = 0,
C, /i = 0, </ = 0, x-a(x + ir) « 0;
then the equations to CA'B', AB'C, BC'A' are

a{\ + ft + y) =X, a{y + ir + \) «» A, and a(ir + A + fi) = A,
and therefore at P the common point of these planes, /x = y = », and the
locus of P is the line drawn from S to the centroid of the face ABC.

A similar resuU may be obtained in exactly the same way in space of
n dimensions. It may be enunciated as follows : — If SABO... be a sim-
plicissimum (Quest. 8242) in space of n dimensions, and a linear locus
A'B'C... parallel to the face ABC... meet the edges SA, SB, SC... in A',
B', C... &c., respectively; the locus of the point P in which the linear
loci AB'C'..., BA'C'..., CA'B'..., &c., meet, wiU be the line joining S to
the centroid of the opposite face ABC...

6827. (Prof. Catlet, p. R.S.)— Consider a triangle ABC, and a point
P ; and let AP meet BC in M, and BP meet AC in N (if, to fix the ideas,
P is within the triangle, then M, N are in the sides BC, AC, respectively,
and the triaugles APN, BPM are regarded as positive) ; find (1) the locus
of the point P, such that the ratio (AAPN+ A BPM) : A ABC may have
a given value ; (2) drawing from each point P, at right angles to the
plane of the triangle, an ordinate PQ of a length proportional to the fore-
going ratio ( A APN + A BPM) : A ABC, trace the surface which is the locus
of the point Q, a surface which has the loci in (1) for its contour lines ;
(3) find the volume of the portion standing on the triangle ABC as base ;
and (4) deduce the solution of the following case of the four-point prob-
lem, viz., taking the points P, P' at random within the triangle ABC,
what is the chance that A, B, P, P' may form a convex quadrangle P

Solution,
Take A as origin of rectangular Cartesian

coordinates, and AB as axis of x, and let

4, ij be the coordinates of P.
Then N is the intersection of ^ = tana .^

(AC)andy=-^ (a:-c), (BP),
^— (J

therefore A APN - A ABN- aAPB

^ g C grytana ")

** 2 lij + (g-Otano '')

_eri( -ty + ^tang 7

"* 2 (.ii + (c-.|)tanay

VOL. XLVII. T

3 CD*

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150

Similarly, M is the intersection of

y = -(a;-c)tani8... (BC) and y - (ij/fla?... (AP),
therefore aBPM = aABM- aAPB

« £ f ^tan/8 7 s_ ^ ( (/?— tan g-t? 7.

** 2 iij + |tan/3 " > 2t ij+^taniS )*

hence, if AAPN+ aBPM : A ABC :: r : 1 and;? = isina {\ and 17 being
the current coordinates), the equation to the locus of P is

tan a ^ (g-|) fan g- ty > _ ^^

(.17 — 4tano +

f c tan a 17 + ( tan g

or (2ij + fy)(ij-4tano + ictano)(ij + |tan j8-ictana)

+ \c {ij [t^p (tan a + tan jS) + c tan o tan j8] + i rpc tan a tan jS} — s M,
a cubic through A and B, the asymptotes of which are the parallels to
AC and BC through the middle point of AB and the line 2t7 + r/7 » 0, a
parallel to AB at a distance depending on r ; and therefore, as the satellite
of infinity is also a parallel to AB at a distance depending upon r, the
asymptote 2iy + rp « touches at an inflexion at infinity.

Any node there may be will lie upon dM/d^ = 0,
or (2rf + rp) {(tanjS— tana) ly-(l-ic) tan o tan iS} = 0,

and there is no node on 2ri-\-rp; therefore, if there is one, it lies upon

(tanjS— tan 0)17- (^-^c) tan o tan /3 = 0;
but, where this cuts the curve,

{2ii + rp)rr^ + icn[rp (tan a + tan jS) + <? tan a tan /9] + i^rj^c^ tana tan /9 =■ 0,
so that the cubic is non -singular. With the same origin and axes of :r and
y, and the line at right angles to the plane of the tri^gle at A as axis of
z, the equation to the locus of Q is

(•2y f nz){y—x tan a + ^c tan o)(y + x tan jS— Jtf tan j8)
+ i<^ {y [m2 (tana + tani8) + tftanatan3] + ^ r/tB tan o tan jS} = 0;
or, denoting 2y-v^z by L, y— ^tana by M, and y+a;tani8 by N, this
may be written

LN (M + «tan .)_, (Mtang + Nt.!..^' _
tana + tan)3
90 that the planes Ls2y + /t2 ^ 0, M + ctano = y— (a?-(j) tano = 0,
N 5 y + :r tan j3 = U all meet the surface where it meets the plane
Mtan)3 + N tano=j».y = 0, and the lines thus determined lie wholly
on the surface, and any planes through these lines meet the surface on
the three systems of qnadrics,

LN = A(j (M tan j8 + N tan o), L (M + c tan o) = Ac (M tan 3 + N tan o),
N (M + c tan o) =» Ac (M tan jS + N tan a),
which are hyperbolic paraboloids. The curves (1) are the sections by
planes parallel to 2 = 0, the plane of the triangle and the plane 2y + /u2 «
will meet the surface at inflexions in these sections, and the planes
y~x tan a + ^6 tan a = and y -1- tan j8 — ^(; tan j3 = will always con-
tain two of the asymptotes to these sections ; as all these sections are
non-singular, the surface will not have any multiple point or multiple
line. If the equation be written in the form

( ~ y 4- a; tan a g tang— a? tang — y \

y — a;taua + ctana y + a:tang >*

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151

it appears that there is only one value of z for any values of x and y, and
that z cannot become infinite for any finite values of x and y unless they
make y + x tan 6 ^ 0, ot y-x tan a + e tan o = ; and the traces of these
planes are the parallels to BC through A and to AO through B, and thesd
are always external to the triangle, and therefore z does not become
infinite within the limits of the integration required for the solution of (3).

The volume required in. (3) is the integral of

__ y r — y-^-rtan a gtanj8~a;tan j8—y •)

"" /* Xy—xta.na + ct8LD.a y + xtamfi )

_J.f-2y + c(tano + tani8) + <?tan2o — r^^^fr etan^fi X

fii y-{x-c)tajia y + xtaufi)

with respect to x and y over the whole area of the triangle ABC,

rbcos* cxtauA re r(c— x) tan /?

C rbcos* fxtauA ce ("{c— x)tan/J ")

and therefore — ] \ + I I f <* *

(.Jo Jo J6C06«JO J

and \udy = — { — y^ + cy {ta.n a + ta.nfi) + c{x— c)tan*o log [y — (a? — «) tan a]
—ex tan* $ log {y + x tan )3)},

1 rb COS m r

therefore the volume = — i —x^ tan' a + cx (tan a + tan 0) tan a

M Jo C

+ c (a?- c) tan^a log -^ - (ja; tan2 iS log ^^5^±^?£i ] ete
^ ' e—x taniS J

+ — f } -{c-xytaji*fi'¥e{e''X){ta.jia + taii0)t&a$

M JiCOB* C

+ (? (or-c) tan^alog ^''^^^^^^ «^-,; tan'iSlogi- ] do?
tan a « )

(or writing a? for c— a? in the second integral)

■s J_ r *3 —ajstan^o + <?« (tan o + tan i8)tano + (?(«— (?) tan- o log— ^
M Jo C , <^-*

- „ « , tan a + tan /9 7 ,

— <?a?tan2^ log — ^ 1: l ^x

taniS )

+ - [''^""'y — a:2tan*j8 + c2;(tano + tani8)tani8 + tf(a;-c)tan2i8log —
M Jo C ^-^

. ., ^ tan a + tan 37 ^^ .

— (jartan-^alog — y ax;

tana )

and the first integral

__ b^ cos a sin^ a ^ sin 7 sin o a^tf cos* /3 sin* a , sin 7

^ 3 ~i ' cos iS 2 ' cos* a sin a cos /9

g*gcos*i8 8in*a __ ^ cos* a siV* Q |^ sin 7 _ ^ si n*a .
4cos*o 2 * C08-ia sin/3coso 4 'cos^o'

and the second is derived from it by interchanging a and b and a and /9.

Therefore the volume = — 5 -z — - * cos a

be An a • b . a*c C082j88in2o 1^ sin 7

+ — : — s»n 7 « + -IT ' 5 log ■: —

2 cos 3 2 COS* a Bin o cos /3

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152

_£wn2« ,^^ , ^c corf

4co^ft^ ^ 2

xBin*3

C08?/3

log

a'am^/8

aco6)3 +

«<? sin j8

siny
sin /3 cos a

sin' 3

a **<? cos-

sin 7 + -— . r—

coso 2 cos^^

log

amy
sin /3 cos a

4co8'/3
/*(. 3

((J«-d3coB«o)-^''

cos' j3 sin*
C08*^ a

-log

sm'y (
sin a cos /3 J

2 cos a cos 3 4

pe

pg' sin 7
4 cos a cos /3

4 cos a cos 3

(a sin o cos' 3 + A sin /9 cos' o) I

4 cos o cos i8 sin 7

[sin'(a + i3) — 2sinasini3co8aco8i3] r
= ^ { —4 + il = ~ (where /? = a sin jS or i sin a).

If, in the figure, APE'S is to be a convex quadrilateral, P' must lie in
one of the triangles APN or BPM ; and, therefore, for any position of P,
the chance for a convex quadrilateral is (AAPN+ aBPM)/ AABC = r
and fjiz =^ rp, and the chance is fiz/p ; and consequently the chance for

any position of P in the triangle = | — diccfy + I Xdzdy^ both integrals
being taken over the triangle, and by what precedes this

pxbfi

aaid the chance for a re-entrant quadrilateral may be at once deduced as
1 — ^ = f, or it might be obtained, in a similar way, by finding the locus
of P where the ratio A APB + fig. CNPM : a ABC is fixed, and describing
a surface of which the ordinates should be proportional to the ratio for
each point, and then integrating as above. This surface, if described on
the opposite face of ABC, would coincide with the surface in the question
if the latter were removed a distance unity towards ABC.

[The above result may be confirmed as follows : — The mean values of
AAPB, aANB, AAMBare^AABC, iAABC, iAABC; andthemean
valueof (AANP+ aBMP) = mean value of (A ANB+ AAMB-2AAPB)
s= -^AABC, which gives the same result as above (see §66 of Prof.
Cbopton's article on FrohabUity^ in the Encyclopaedia Britanniea.)']

P£ ^
2

7270. (W. J. C. Sharp, M. A.)— Eliminate a, *, e, a\ b\ e' from

a-^b + c

^d

ab

+ be + ca

=s S

abc

^ 8

a'b' + h'c' + cfa' =■ *'
a'b'c' = 8'

a'bc + ab'c + oM = 6,
ab'e^a'btf^-a'h'c^^.

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Solution,

(a, by e) are the respective roots of a:*— ffjr^+wr— 8 = (1),

K*i, O »» ' » »» a^-d,x^ + 8,x-9,^0 (2),

(a*„*ri, Ml) „ „ „ a^-sx^ + dSx-n' --0 (3),

(fli*i, Vi»^i«i) »» M a:8_«^;c2 + ^^8jar-8,2^0 (4).

Let G, A, Gri, Ai have their usual meanings for (1) and (2), and ff, S'A,
ffi, Ji^^i for (3; and (4) ; then (Burnsidb's Theori/ of Equations y p. 113)

a«l + Wl + CCl-tf=i<f<fl+3[i(-G+^/A).i(-C^l-^/Al)]*

+ 3 [i (-G- a/A) . i (-Gi+ a/Ai)]*,

a,J(? + «*!<? + «*<?! = d - i efi« + 3 [i (-^ + 8a/A) . i (-Gx- a/Ax)]*

+ 3 [i (-ir-8A/A) . i (-Gi+ a/Ai)]*,

with a similar equation for Oj. These three, being rationalised, give the
results of the elimination.

[If ax^ + by^ + cz^ + 2fyz + 2^3® + 2hxy = m,

a'a;2 + b'y^ + ^j'z^ + 2/^5? + 2g'zx + 2A'icy = m*
be two concentric quadrics referred to rectangular axes,
« + * + <? = <?, a' + V tc'=.d\

ab + bc + ea^P^p^-^h^^8, a'*' + *V + cV-2/2-2^'S-2A'2 = ,',
Qbc + 2fyh^aP'-bff^-ch^=ny a'bV + 2fyh' -a'f^--by^--c'h'^sZ',

aa' + **' -\-cc'-2ff'-2gg'- 2hh' = ^ ,
a'^c + aA'c + ab</ + 2/5^^ + 2fg'h + 2fgh' --a'P-b'g'^- (fh^

^2aff^2bgg'^2chh'^e,
ah'(/ + a'b</ -S-a'b'c-^^fg'h' + 2fgh' ■¥2fg'h^aP'-bg'^-cU^

- 2«'/'/- 2b'g'g-'2c'h'h = a',

are their invariants and co-variants for rectangular transformations ; and
the given values of dy 8, 8, d\ »', 8', e, $, and 6' are those which they
assume when the quadrics are coaxal {Proceedings of the London Mathe^
matical Society ^ Vol. xiii., Nos. 193 — 4), and are referred to their principal
axes, and hence the resultants so elegantly obtained by Prof. Sircom are
the invariant conditions that the quadrics should be coaxal.]

7301. (Professor Cayley, F.R.S.) — If the function ^^^^5 = ^pu

7W+8
is periodic of the third order {<ph4 = u) : given that the cubic equation

(a, by Ot d'Qxy 1)3 = has two roots u, v such that v « °" , find u as
^ "^ ' 7« + 8

a rational function of a, b, e, d, a, jS, 7, 8 ; and examine the case in which
u is not thus expressible.

Solution.
7«t5 ^^^' ^^' (o + 5)7W + (8-T)37)

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154

and

and therefore, in order that ^^ (u) = m,

a? + 20/87 + /B7« = 83 + 2^878 + 0/87, ai^d o' + «*+tt8 + /87= 0,
which are hoth satisfied if (^ + 8'+oJ+ ^87 «

(for j3=0, 7 = 0, = 8 the other conditions make ^ (i#) = t#) , and this is
the necessary and sufficient condition that ^ should he a periodic function
of the third order.

Again, if aa^-^-Zba^-k-Zexy^-k-d^ — have two roots m and v such

that « - 5?i±|,

7« + 8
the equation

a(ar + /B)3+3A(oa; + i3)S(7iF + a) + 3c(o4? + i3)(7» + 8)« + rf(7* + 8)'=«0,
or {aa^ + 3*0^7 + 3(?a7* + drf) a^

+ 3 lao^jS + A (asa + 2a)87) + <J (2a78 + /Sy*) + (i7««} a;"

+ 3 {aai82 + A (i827 + 2oi38) + <J (2)875 + a82) + £^5*1 a:

+ («/83 + 3Ai8«8 + 3tfi352 + <?«8) = 0.

Say a'a:' + 3A'a;'+3<?'a? + rf' = 6 has the root v in common with the
orig^inal equation (a, b^ e, d^x, 1)^ = 0, and therefore the resultant
+ (ad'-afd)^-9 {ad'^a'df (io'-A'c) + 27 {ad -afey{cdf-dd) •
27 {pd'^b'd)'{ab'''a'b)^%\ {aV-a'b) {bd-Vc) {ed'-ifS)
-27 {ad'^€^d) {ab'-afb) (ed'-e'd) =

must he satisfied hy the ahove values of a', b', tf^ and d*. The root may
he found hy finding the Gr. C. M. of

(«, b, e, rfjaj, 1)3 and (a', A', <?', rf'Ja:, 1)3,
which will he found to he

{9c (a*'- a'A)2- 9* {fib' - a'A) («y - o'c) + 3a (o^- <?'<?)«

-a (aA'-a'i) (flkf -a'<f)} ar,
+ {3d {ob'-afbY-Zb {ab'-a'b) {ad'-a'd)'-a{a(f -ale) {ad'-a'd)\,
and therefore v —

^{Zd{ab'^a'b)^^%b[a}/-a'b){ad' ---a'd)-a{ae -'a'e){ad'--ald)\
9c {ab' ^a'b)^-^9b{ab' ^a'b) {ad ^a'c) + 3a (a<?'-a'c)2-a(a*'-.a'^) {ad'-a'd)^

and « « ^, so that hoth u and v are expressed rationally in terms

of a, A, c, d and a, /3, 7, 8.

If, however, the two quantics (a, A, c, ifja;, 1)3 and (a', *', c', rf' Ja?, 1)3
have a common quadratic factor, v as found ahove = 0/0.

The G. C. M. is 3 {aV -a'b)x^-\-Z{ac ^a'c)x+{ad'-a'd),
and the common roots ohtained from a quadratic equation are not
expressed rationally in terms of the quantities a, b, c,d; a', b\ </, d\

In this case, if u, v, to are the roots of (a, b, e, d^x, 1)3,

r> r» are the roots of (a, 3', (t, a'Oa?, 1)',

7M + 5 7t; + 8 7m; f 8 ^ ' * ' .x. » / >

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156

andweliavenotonlyv= ^i±^, but also either w = ?l±^orelaef«-?^^;

. au + B yi*+ 9 yv + B 7W + 8

and, if ^3^(11), the roots are either w, <t>{u), and ^(w), or

w, 4>(tr), and ^^'Ctt?).

And hence, if the function he periodic of the third order, the trans-
formed equation will be identical with the original one ; so that, if

a» + 82+a5 + 37 = 0,

and ao3 + 3Aa?7 + 3ro>« + ci>3: aa^fi + b (a^Z + 2afiy) -^ e (2ai^+ By^+d'^H

: Ba0^ + b($^ + 2a$9) + e[2fiyZ + a^)-\-dyP : afi^ + Zbfi^i + Zefii^ + dh^

:: a : b : e : df
the equation («, *, r, d'^x^ 1)' = has its roots u, ^(w), and ^(«),
where ^ is a periodic function of the third order, and is unchanged by

the transformation

y =

ax-i-B

7« + 8

(where a: $ : y :Z may be found from the above rdatioos, and also the
condition which must be fulfilled.)

8204. (Alice (Gordon.) — ^If A, B, C, D, E are five points on a sphere,
Y]s ^e volume ACBE, Y,) the volume ABDE, &c. : prove that, with a
certain convention as to sign,

Vu (AB)2 + V,5 (AC)2 + Vi4 (AD)2 + Vjs (AE)^ - 0.

Solution, *

K (xj, vi, «i), (a^a, y^ z^ , (jTg, ys, «g) be the rectangular Cartesian

coordinates of A, £, G, D, and E,

^i' + yi'+«i', a^, yi. «i» 1
^s'+ya' + ^a'i «8» y* «2> 1
^3^ + y5^ + «8^ arj, ys, 23, 1 =0;

^4' + y4' + «4'. *4, y4, «4» 1
a?6' + y«^ + «6^ ^5, y6» «6, 1

therefore, multiplying the second, third, and fourth columns by

^11 2yi> ^) and subtracting the sum from the first,

-a-i^-yi'-V, ari, yj, «i, 1

x^-¥yf^'Z^-2XyX^-2yiy^-2z^z^ arj, yi, «j, 1

ar,2 + y3« + 23«_22rja:,-2yiy3-2«,«3, x^, y^, 23, 1

a^4^ + y4*+«4^*2af,a?4-2yiy4-2zia4, x^, y^, £4, 1

a^5* + y5* + «6^-2a?ia?5-2yiy5-2ziZ5, x^, y^, 25, 1

then, subtracting the top line from each of the others,

(«i-ar2)«-|.(yi-y,)2+(si-2-:)«, x^-x^ y,-yi, «,-«i

(ici-a:3^;2+(yi-y3)'+(^i-^)^ a^-^i» yj-yi* «s-«i

(a;i-aJ4)2+(yi-y4)2+ (21-24)2, af4-jri, y4-yi, S4-«i
(Xi-ar5)2 + (yi-y5)2 + («i-r5), a-j-ari, y^-yi, r^-Si
which proves the property.

-0;

^Qf,

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156

This method of proof applies equally in space of any dimensions, and
the general property may also be proved by means of simplicissimum
content coordinates, as I have done in a paper on the ** ftoperties of
SimpHcissima," &c., read before the London Mathematical Society in
April, 1887.

8237. fW"- J- C. Sharp, M.A.) — If nCr denote the number of combi-
nationa r together which can be formed out of n things ; show, from
a priori considerations, that

nOr " nCn-r> nPr =■ — n-lCr-l, n^r = n-lCr + n-lCr-1 (1, 2, 3),

or, more generally, nCr — n-pOr +pCi n-pOr-i +1.C2 »-pCr-2 + &c.

Solution.

The given equations, which easily follow from the known formula
for "Cr, may also be proved from d priori considerations.

Since every set of r formed out of n things implicitly forms a set of «— r
(those not taken), and that every change in the one set involves a corres-
ponding change in the other, "Cr = "Cn-r- If one of the things be re-
moved, **"^Cr-i sets of r— 1 things, and "-^Cr sets of rthings,may be formed
out of the remaining « — 1 ; and, since the missing thing may be prefixed to
each of the "-^Cr-i sets of r — 1 things, there are "-^Cr-i sets of r things
in which a particiilar one appears, and **"^0r in which it does not ; hence
*»Cr - **~^Cr + "~^Cr-i; also. Considering each of the n things in turn,
« *»~^Cr-i includes all the sets of r things, so that each occurs r times, once
in virtue of each of its components ; therefore r »*Cr = «"~*Cr-i. If i? of
the things be removed, the number of sets of r will be made up of those
which contain none of the p, one of the p, two of the^, &c., or of ""^Cr,
"-^Cr-i, and ^Cj, since each of p may be taken with each set of r- 1 of
the »— jP, **"'Cr_s, and^Cj, &c. ; therefore

8631. (Professor Sylvester, F.E.S.) — Find the discriminant of
«3 + y8 + «8 + 3tf ScV + 6^a;y«.

Solution.
The discriminant is the resultant of the equations,

a:2 + 2^ar(y + a) + «(y + 2)2 = 0, y^ + 2ey (z + x)-¥e {g + x)^ ^ 0,
22 + 2tj5 (a; + y) +c (a: + y)' =s 0,
which are equivalent to the system

i?(» + y + 2)2-(«-l)a;2 = (c-l)y2«(^-l)22 (A),

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157

and from the first equation

{(«« + tfir' + <f«^'-4tf«(a:V^ + y8«8 + «V)}« - 64tf^%* (ar + y + »)«
from the equations (A),

therefore {<f2 (l + 2tf)2-4<f2x 3tf2}s - 64^7 {e- 1),

therefore {(1 -*)* (1 + 4tf-8d2)}«+ 64<f^ (1 -c)» « 0,

and this expression is the discriminant ; and the invariants T and S are

(1-^)4(1+4^-8^ and e{l-e)\
results easily confirmed by substitution in the general values g^ven in
Salhon's Higher Flans Curvet, pp. 184 — 6.

8928. (Rev. T. R. Tbk»y, M. A.)— Find the value of the definite
— smpx (fl cos' x + b sin' x), when p > 2.

X

Solution,
Denoting the integral by I,

dl f*
4 -:- ~ 4 I cos 17^ (a cos' x + b eia^x) dx
dp Jo

= (a-b) I {com {p + 2) X + C06 {p~2) x] +2 {a+b) C cos pxdx

= Of unless p *= ± 2 or 0, in which case it is infinite.
Therefore, if j? is not ± 2 or 0, I is independent of p. Let i' « 1, there-
fore 21= [ ^sinar {<» + * + (a-i) cos 2a;}
Jo *

= (a + *) fsin^i:^ + i!?:i^r(sin3a?-sinar)^
' ' Jo X 2 Jo ' 'a?

«(«+*) iir + i (a-i)(itr-iir)(aaBG0RY'8 JSxampleSf p. 480),
therefore I « (« + *) i*.

8978, (Professor Sylvester, F.R.S.)— Show algebraically that, if
a, b, e are the three sides of a triangle of reference, and A, B, C, the three
perpendiculars on a variable line from the angles of that triangle, are
re^irded as its inverse coordinates, then the equation to the two circular
points at infinity is

««(A-B) (A-.C) + 3'(B-A)(B-C)+c'(C-A) (C-B) - 0.

VOL. XLVn. V

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158

Solution.
In areal coordinates the circular points at infinity are represented by

(fifUf + UhX + t^XffO, and X+ju + f-0 (1> 2),

and 'if l\ + mfi + ny ^ represent any straight line, the square of the per-
pendicular upon it from [\V/= Q (/\' + ntf/ + wi/)]*; as I have shown in a
paper read before the London Mathematical Society^ m April last ; therefore*
A:B:G::/:m:f>, and the equation to the line may be written

Ax+BAt + Cr = (3),

therefore from (2) and (3) x : m .' v :: B— C : C— A : A— B, and therefore
from (1), a2(A-B) (A-C) +^(B-C) (B-A) + e2 (C-B) (C-A) - 0,

8996. (Mahbnora Nath Rat, M.A., LL.B.) — If x^ y, z denote the
respective distances of any point in the plane of a given triangle ABC
from the angular points ; show that the following relation subsists among
them, (a:3 + y«+«5 + a2 + j2 + c2) {a^x^-\-h-y^-^<?z^) = 2aH'^{€fi + a?)

Solution.

If the volume of the tetrahedron ABCD, whose edges are DA = x,
DB = y, DO = », BC = a, OA = A, AB = c, is (Quest. 8242) V,

V2 - -JL- o» 1» 1» 1> 1
8 X 36 1, 0, x\ y^, «»

1, x\ 0, e\ h^
1, y^ c2, 0, a2
1, z^ h^, a\
and therefore, if D lie in the plane of ABC,

0, 1, 1, 1, 1

1, 0, x\ y\ «2
1, a?2. 0, ^, ^2
1, y\ c\ 0, a2

= 0,

1, z\ ^, cr,

which, when expanded, gives the result required. The same result is ob-
tained in a different way in Salmon's Gemnetry of Three Dimensions, p. 32.

9003. (R. F. Davis, M.A.)--If upon each side of a triangle a pair of
points be taken so that the pairs on any two sides are coney clic, prove
that all three pairs.are concyclic.

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•(1),

Solution,

If Aj, Aa ; B^, Bj ; Ci, Cj be the points in
which BC, CA, and AB are divided,
ABi . ABo = ACj . ACj,
CA, . CaJ = CBi . CBa,
and BAi . BAj = BC, . BCj.

Now, let the circle through Ci, Cj, B,, B^ cut
BC in Aj', A3' ; then

BA/. BAj' = BCi . BCj « BAj . B Aj .
CA/. CA2' = CBi . CB2 = CAj . CAj,
or (BC-BA/) (BC-BA2') = (BC-BAi) (BC-BA^),

and therefore BAi' + BAj'= BA1 + BA3 (2),

therefore, from (1) and (2), BA/ and BAj' are the roots of the same quad-
ratic as BAj and BA2, and A,' and Aj' coincide with Aj and Aj.

If «, b, c be the middle points of the sides of the triangle, and D, E, P
the feet of the perpendiculars from the vertices upon the opposite sides,

A3 . AE = ^bc cos A = A(j . AF
CA . CE « ia* cos C = Ca . CD
Ba.BD ^iaecosB » Be . BF

and therefore the points a, A, e, D, E, F are concyclic, or the circle about
a, by c passes through D, E, and F. And if O be the intersection of AD,
BE, and CF, the circle through D, E, and F, the feet of the perpendiculars
from the vertices upon the opposite sides of the triangle AOB (viz., D, E,
and F) will bisect the sides, and so the circle about DEF passes through
a, bj c and bisects AG, BO, CO. (I think this is, perhaps, the simplest
proof of the existence of the nine-point circle.)

The equations (A), may be proved by elementary Geometry as follows :
The circle on BC as diameter will pass through F and K (Euclid iii. 31),
therefore AE . AC = AF . AB (iii. 36), or 2AE . Ab = 2AF . Ac, and
similarly for the others.

.(A);

9008. (S. Roberts, M.A.) — Given three circles Cj, Cj, C3, determine
a circle cutting C^ orthogonally, bisecting Cj, and bisected by C^ ; and
show that in general there are two such circles which may coincide or be-
come imaginary.

Solution.
Let the axes of rectangular Cartesian coordinates be taken so that
ar2 + y2_^3=0sCi, (ar-A)2 + y8-A2=0=C2, (a:-.A')2+ (y-A-')2-c««0sC5,
and (a:— {)2+ (y— iy)2_R2_-o = S, the required circle. Then, since Ci and

S cut orthogonally, ^2 + ^2 =, ^2 + ^2 (1) ;

since S bisects Cj, (|-/0"2 + >?' = R^-*^ (2);

and, since C3 bisects S, U-'h')^ + {'n-k')' « c^-R* (3) ;

from (1) and (2), { = — ,

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160

and therefore, from (3), ilfji - A''+*^-«s+a»+2R'- 4^(<.»+is + A»)

A

sA + 2R«- ^B, say,
h

and 2A'| = ~ B ;

.-. 4R< + 4R2^A-^B-;fc^| + (A^~By+|^B2-4a2A;'s = 0,

and the yalues of R^ are real, equal, or imaginary,

according as I A- ^B-k'^Y> - < /a- ^b]V |!B«-4a2Ar^,

„ „ _2(a~^b)+A^> -< ^-4«»,

A A*

When the values of R^ are real, those of n are so, and those of { are
always so, and therefore, &c.

9024. (Professor Stlybsteb, F.R.S.) — For g^reater distinctness, the
name of Hyper -cartesian (not to be confounded with a ht/per-cartesic) being
giyen to that particular form of the bicircular quartic in which four con-
cyclic foci become collinear; prove that, if four points are given in a
plane, the locus of the curve in space whose distances from any three of
them are subject to a given homogeneous linear relation is a curve of the
4th order. ('JChis space curve may be termed a Hyper-cartesic.)

Solution,

If the plane of the foci be taken for that oixy, and the axes be so chosen
that they are (0, 0, 0), (A, 0, 0), (h\ k\ 0), and (A", A;", 0), and if ^, q, r, $
be the four focal distwices of the point {x, y, z) on the curve,

«2 « ^a_ 2h"x - 2k" y + h"^ + k"^,
and Ip^mq + nr = and rp-^m'q + n^t « 0,

or 2^y}2 + 2;n2»Vr2 + 2«2/2,.2p2_^^_^y_^V* = (1),

2l'^m'^P'q^ + 27n'^n'^q^s^ + 2n'^ns^p^-l'*p*~m'*q*-n'*f^ « 0...(2),
or ap*-j-2ljpl^ + S = and ay + 2Ly + S'= 0,

where L and L' are linear and S and S' quadratic functions of x and y ; and

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161

their resultant, which is the trace of the curve on the plane of xi/, is

(aS'-a'S)2+4 (a'L-aL') (LS'-L'S) - 0,
a quartic, to every point of which two points, one above and another below
the plane, correspond, so that these lie on two quartics, each of which is
the reflexion of the other.

9063. (Mahbxdba Nath Ray, M.A., LL.B.)— If aj, a^ a, ... <i2f.-i be
2n-l positive numbers connected by the relation aya^^„,a2n i °» 1 ;
show, by elementary algebra only, that the minimum value of
(I +ai) (1 +03) (1 + aa) (1 + aj ... (1 + aa,..i) is 22»»-i.

Solution.

If a? be any positive quantity x+l/x > ^ < 2,
according as {x—l)^> — < 0,

but (^—1)' is <t 0, therefore x+ljx <}: 2,

i.e.f 2 is the minimum value oi x + I/0.

Now, if a^a^... «n *= 1>

(1 +«,) (1 +<0 ... (! + «„) = (a.-T,...a.)» (a{ + «;«) (,» + «;•) ... {«» + «;»)

= («! + «:') ("J + «;*)-(«i +«;'),

and the minimum value of each factor is 2 ; therefore the minimum value
of product is 2»». Of course the quantities aj, a*, &c. are the positive or
signless square roots.

8637. (Professor Hudson, M.A.) — The tangent and normal to
y/c = i (€'/<'+ *-'!<=) cut the axis of a? in T, G, respectively ; find the mini-

mum value of GT

Solution,
Let the tangent be inclined to the axis of x at an angle <t> ; then
GT »« tf sec 4> (tan (p + cot4>) = <j/ (sin^^— 8in'4>) ;
this is least when sin4» « 1 / >/3, and therefore GT » ie^/Z,

8888. (S. Tbbat, B.A.)--Di, I>2» I>8 are the shortest distances of
opposite edges {a, a; b, b; c, e) of a tetrahedron, V the volume, and A
the area of one of the equal faces ; show that

V = J (DiD.Da), A' - i (Di^aH D./ J*- + D3V),
and Di^ + D-Z + Da*- i(a2 + ^ + cj2) - D^HaS^ &c.

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162

Solution,

Let ABCD be the tetrahedron, and let BO = DA = a, CA = DB = b,
AB = DC = c bisect BC in E' and AD in E ; then, since DB = AC and
BE'=CE',andDBE'=DBO =BCA=ACE',
DE'« E'A, and therefore EE' is at right
angles to AD ; similarly, it is at right angles
to BC, and it is therefore the shortest dis-
tance between them. Therefore

DE2 + EE'i = DE'S
and DE's = i (^ + c--^a^,

therefore

Di2 = EE'2 = DE'2-DE2 « i {b^ + c*^a^,
and, similarly,

D22=i(c2 + a8-*2), D32=i(a2 + A»-c2)»
therefore

Di^ + DjS + Da^ = i (a2+i2 + ^2) ^ Di« + a3 = &c.

= Tff {«^ (*^ + c2. fl2) + ^ (c3 + ejj_i2) + c3 (oS + ^S- C^)}

Sx'i

also (Quest. 8242), V^ - ^-^ 0, 1, 1, 1, 1

1, 0, c^, *2, a2

1, ^S 0, a2, ^2

1, *2, «S 0, c3

1, a2, ^2, (^,

or (replacing the 2nd column by the sum of the 2nd and 5th less that of
the 3rd and 4th, replacing the 3rd column by the sum of the 3rd and 2nd
less that of the 4th and 6th, replacing the 4th column by the sum of the
4th and 2nd less that of the 3rd and 5th, which gives 4 times the deter-
minant).

1

4 X 8 X 36

^Di^D^Da^
144

0, 0, 0, 0, 1

1, «2-i2-(j2, ^^ifi^a\ l^-<^^a\ a2

1, J2 + ^_^2^ c2-a2_^^ rt2^.c2_i2^ ^
1, d2+c2_fl,2^ fl2+^_^2^ J2_a2_^2^ ^2

1, a^^jfi^c"-^ a2^.32_c2, a2 + c2_j2^ o

1, -1, -1, -1

1, 1, -1, 1

1, 1, 1, -1

I, -1, 1, 1

_ D^2Dom,2 p^p^T)^

9 9 •

^ D,2D22D,2

36

1,

-1,

1

1,

1,

-1

-1.

1,

1

I

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163

1856. (Professor Sylvbstbb, F.R.S.) — Prove that the Jacobian of the
time of a planet's describing any arc, the chord of the arc, and the sum of
the two extreme distances from the sun, in respect to the eccentricity and
two extreme eccentric anomalies, is zero ; and hence deduce the time for
a planet or comet in terms of the said chord and sum.

Solution.
"With the usual notation (see Tait and Steele's Bynamies), we have
CQSU—e ^, J 1 . „ /, 1 — ^

COS0 :

1— tfCOSW

, therefore 1 + «*co8 tf =

1— ^COSM*

"^^"Vl \l-.cos«j j- l-.cosu '

= iSyzfL^ail^ecoBu),
1 + c cos d

Now, if <i, <2 be the times from perihelion to the ends of the arc, T the
time of describing it, «i, u^ the eccentric anomalies of the extremities, r,,
rj, Biy 02 the corresponding radii vectores and true anomalies, S the sum
of ri and r^, and C the chord, P being the periodic time.

T= t^-ti =i-{«8-«sint<2--(wi-tfsin«i)}

.(1),

S =ri + r2 = a (l^e coaui + l-e COB u^) (2),

03 = ri2 + r32-2rir2COS {di-O^ = ri^ + r^^-^Tir^ {cos a, cos 63 + sin tfj sin a.^}
« a»r(l-*^s«i)*+(l-(?cos«2)'-2(l-^cos«i)(l-^cos«2)

L f cos«i-g co8«2— g ^ (l-g2)sin«i8inM2 ^"1

^ 1 1 - g cos Ml 1 - g cos Wa (1 - tf cos Mi)(1 -e cos Wg) ) J

= a2 [2— 2cos«iCOS«3-2sin«iSinw2-^'(8ui«i-8iii«^)*] (3)-

Again, the Jacobian of 02, T, S is merely that of 0, T, S multi-
plied by; 20, and therefore they will vanish together unless « 0,
a case which may be disregarded. Now, T (0*, T, S)

'' 2ir

-2c(sinwi-sinw2), 2sinwiCOS«2-2cosMi8in«#2 . ^

— 2^2 (jQg f^^ (sin ^ — sin «J,

sin Wi - sin u^, - 1 + ^ cos Wj,

-(costti + coswg), esinwi,

-2sinMiCOS«2 + 2cos«*isin«/2 + 2c2cos«,(8in>i-sintt2)

1 — CC0S«2,

gsinM2
(adding 2e times the second row to the first)

0, sin Ui cos «3 - cos u^ sin 1*2— « (sin Wj - sin eij),

sin«i-sin «*2» - 1 + c cos «i,

- (cos Ui + cos M3) , * sm «i ,

- sin Ml cos wj + cos Mj sinM2 + * (sin ^i - sin Mg)

I — gC08M2

p^

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164

K (adding the third column to the second)
— {— 8intfiCOB«2 + co8WisinMs I sini#i— siniia, ^(cosi#i— cobkJIsO;

+ 1 (sin Wj — sin fij) } | — (cosf#i + cosmj, « (sinnj — sin Wj) |
and therefore T (C, T, S) = 0,

and T is a function of e^ M|, and u^ only as it is a function of S and C
(Boole's Biff. Equa.y Sup. Vol., p. 56), and therefore it is possihle to
eliminate e^ Mj, and t<s from the equations (1), (2), (3).

From (2), e t , ^^"^ -,

a (cos Ml + cos U3)

therefore, from (3),

C-a»f2-2coB(«.-.>.)-(gg:^V(''""l-"""'V]
I ^ ' \ a I \co8«, + cosa,y j

(. ^^ ^ \ a I l + cos(Wi-f*Jj

a quadratic in cos (tij— t«i), whence

"•<-^ - ('-^)o-^°) wn'-'^°)'{'-'-^r

C-2g«-t-(2a-S)« '>
2a« j

-('-=-^)(-5£2)^V![-('-S^)"]

« C0S2 cosz^'f sine sins' » cos(z— c^,

if 1-8^:2^^^ ^^ l-S + 0^cos«'.

2a 2a

So that z and s^ are functions of S and C. Therefore

tan r^i=^\ - tan '-=^ - Bin,-gin»'
\ 2 / 2 cos» + cosr

Now, from (1), (2),

T- ^ (tf,-ti,- ^~^ sinwj-sinjwj^
2ir (. a cos Uj + cos M3 J

2irC « 2 >'

and — Zl5 = coB«+co9«' and tan "87**^ = tan^^,

a 2 2

therefore ^ *" ^ {«—«'— sin r + sin e'}

a formula given by Lambebt, as an extension of one of Eulbb's for para-
bolic motion, viz., T - -i- ((S + 0)*-(S-C)*},
6v^ju *■

which is proved in Tait and Steele. See also Pontecoulant, Sf/sthne du
Monde, Vol. i., pp. 277 and 287—291.

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165

5305. (Professor Syltbsteb, F.R.S.J — Definition, — Six right lines
along which six forces can be made to equilibrate are said to be in Statical
Involution.

Prove that any six right lines lying on a ruled cubic surface are in
statical involution ; and, vice verady if six right lines are in statical
involution, a ruled cubic surface can be made to pass through them.

-0.

Solution,

If aj, ^1, Ci, ai, jSi, 7i ; a^t ^3> ^ 03, $^9 72* ^c* ^ ^® direction oosines,
and the coordinates of fi^ed points upon the six lines, and P|, P„ &c. the
forces acting along them ; the conditions of equilibrium are
2Pa = 0, 2Pi = 0, 2Ptf = 0,
2 P (tfjB- J7) « 0, 2 P {ay-ea) = 0, and 2 P (ba-afi) = 0,
and the necessary and sufficient condition for statical involution is

«1> <»2> ^ ^4f ^i ^

bi, ^2* hi *4» hf *«

Ci, Cj, C^t Ut ^61 ^6

eiPi-biji, e^^b^y^, <?8^8-%a> <^434-*474» <?636-*676» ^^ei^e-^cTc

«m — <^l«l> ''272— <V«2> <'878-<?808i ''474-^4«4» ^^^n/b — HHj «m- <?6««
Vi-Oi^i, b^t^-a^ ^503-03^8, *4«4-M4» V8-«6^5> *6««-««i86

Now, if a\ + b n-k-ev -¥{€$— by) p-k- {ay^ea) <r + {ba^a0) t =
be satisfied by any point in the line

a?-a y—B g-y
a " 6 " tf '
it is satisfied by every point in the line, t.^.,

aA + A/i + ttr + (cy — fo) p + (az^ex) v + (Aa?— «y) t = ;
and, if there be six sach Hues, the determinant above will be transformed
into the equation to a ruled cubic surface, upon which each of the lines
lies, by writing x for each of the a's, y for each of the jS's, and z for each
of the 7*s. Also any ruled cubic surface may be reduced to this form,
the components of each column being the coordinates of any six lines upon
it, and so the proposition is proved.

TOL. XLTtt.

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APPENDIX IV.

NOTE ON THE USE OF COMMON LOGARITHMS IN THE
NUMERICAL SOLUTION OF EQUATIONS OF THE HIGHER
ORDERS. By Major-Gbnbral P. O'CONNELL.

The problem to be solved may be stated as follows :

Given f{x) «■ aa?**» + ^a?'» + ca^+ ... + Z = 0,

where a, J, c, ... Z; m, w, p, &c. are constants ; w, », jp, &c. being larg^
numbers consisting eiliier of whole numbers, whole numbers and fractions,
or whole numbers and a few places of decimals arranged in descending
order of magnitude, i.e., m is the largest and every other is smaller than
any of those preceding it.

To find a value of x which, when substituted in /(a:), shall reduce that
function to as small a value as the extent of the tables of logarithms used
and the true value of x permit it.

When m, n, p, &c. are small whole numbers, I know no method of
solution better than Hobneb's, which has been very fully explained by the
late Professor Db Morgan, by Todhuntbb, and others.

When #», «, Pt &c., are large whole numbers, the solution may be ob-
tained in the manner shown by that able mathematician and computer,
the late Mr. Thomas Weddle, in a pamphlet published in 1842, by
Hamilton, Adams, & Co., Paternoster Row. Mr. Weddle does not use any
table of logarithms, and his computations are therefore very long.

The method here used is nearly that of Mr. Oliver Byrne, as explained
in his ** Dual Arithmetic," published by Bell & Dalby, 186, Fleet Street,
1864. It differs therefrom, however, in the use of common logarithms, and
in presenting the calculations in a more condensed form, a single page be-
ing sufficient for the solution of an equation of a very high order, as may
be seen by reference to my Solution of Question 9185 [which will appear
hereafter in Vol. 48], where an equation of the 622nd degree is solved,
and a value of x is found containing six places of decimals or seven
figures in all.

In Mr. Weddle' s, Mr. Byrne's, and in the present method, x is con-
ceived to be resolved into factors. In Mr. Weddle' s method,

«=R(l+ri)(l + r2)(l + r3)..., •

where R is the first significant figure of x multiplied by some power of 10,
and rj, rg, rg, &c. form a series of decimals each consisting of one signi-
ficant figure preceded by a decimal point, and 0, 1, or more zeros, the
number of zeros increasing as we pass along successive values of r from

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167

left to right. In Mr. Btbne's method

where M contams one or more significant fig^es of x multiplied by some
power of 10, and m^f m^ imj, &c. may have as values any suitable whole
numbers or zero. In the present method;

ar=ai(l+a2)(l + a3)...,

where ai is the first approximation to the value of x^ and may contain only
a whole number of one, two, or three figures, or a whole number and a
decimal, a^ (1 + 03) is the second approximate value of x, a^ (1 +03) (1 + a^)
is the third approximate value of x, and so on, 03 being less than unity,
as less than 03 ... , and a^ less than ^n-i ... , &c. ; one advantage of usini?
a table of logarithms being, that each approximate value will generally
be found to contain nearly twice as many correct figures as the one
preceding it.

In all these methods, the values of the factors into which x is conceived
to be resolved are discovered successively, commencing with the left-hand
one, then the one next to it on the right, and so on.

ITie first factor is usually obtained, in using Wbddlb's or Bybne's method,
by a process of guessing and trial, or by some simple transformation or
other algebraic artifice.

In using the method about to be described, it will often be found that
a value of the first factor containing three figures may be obtained by the
use of logarithmic slide rules. This facilitates its application and shortens
the whole process, and is therefore recommended to all who are accustomed
to use these instruments. It may, however, be conveniently applied by
those who do not use such aids.

The method of passing from the first to the second approximation in
the value of x, due to Sir Isaac Newton, and, as far as I know, generally
followed until it was slightly altered by Mr. Oliver Byrne, is thus de-
scribed in Toi)HUNTER*s Theory of Equations (4th Edition, 1880), page
142:—

** Let f{x) — be an equation which has a root between certain limits
a and jS, the difference of which is a small fraction ; let c be a quantity
between a and jS which is assumed as a first approximation to the required
root ; and let c-k-h denote the exact value of the root, so that A is a small
fraction which is to be determined.

" Thus, f(c + h) = 0, that is, by article 10,

f{c) + hf (c) + Y^r(c) + ... |^yw(,) ^ 0.

Now, since h is supposed to be a small fraction, A*, h^y &c. will be small
compared to A ; if we neglect the squares and higher powers of h in the
above equation, we obtain f{c) + h ./' (c) = 0, thus

/'W

Supposing, then, that we thus obtain an approximation to the value of A, we
have c^'ryfl as a new approximation to the root of the proposed equa-
tion."

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168

The simple alteration introduced into the above process by Mr. Btbnb
is, to express the second approximation to the root, not ase— "^^t but as
ell — ^^- I . This slight variation suffices to change the method to one

suited to logarithmic calculation.

The first step in the process is to find the logarithmic values of the
several terms of the given equation when a.^ (the first approximation to
its value) is substituted for 0. When this has been done, we have the
set of tenns,

log {« (ai)-}, log {h {a^^}, log {c (aj)"} ... log /,

to which, adding

^{^-i^'^- '-{-^r -{'-.^r'

we obtain

log [« {ai (1 +0,)}-], log \h {qi (1 +«4)"], log \c {«, (I +«2)}p] ... log/.

Before we can add the second line of logarithms, shown above, we must

first obtain the value of o^ »> — )^, v ' ^^^^ can be conveniently

effected by arranging the calculation in the form shown in the following
example, where t^e three operations of finding (I) the logarithmic values
of the several terms of the given equation, (2) the natural numbers cor-
responding thereto, and (3) the value of the next factor in the value of
the unknown quantity, are recorded in different parts of the same central
columns, a separate column on the left hand being reserved for the
logarithms of the required powers of the successive factors of the value of
the root and for the final value of the root, the several portions being
numbered so as to facilitate reference, as well as correction if necessary.

In this way the computation circulates from the left-hand column down
the central one, and back again to the left-hand column, and so on again
until it is complete.

The example worked out is (Weddle, No.7, page 27) an equation having
four real roots, of which two are positive. One of the positive roots is
given by Mr. Weddle, and the other is here found.

This equation in x has first been transformed to one in y J «■ f ^ j i,

in which a first approximation to the value of y is easily guessed.

It will be seen that only three factors are obtained, and that

y - 10-» X 1-5 (1-0372)(1-00064574),

an ordinary seven-figure logarithm table having been used. If more
factors be required, Vegas' ten-figure table, or Peter Gray's twenty-four-
fig^e table, may be used. When the latter table is used, a good deal of
room and calculation are needed for passing from a large natural number
to its logarithm and back from a large logarithm to its natural number,
so that it is better to keep this portion of the computation separate and
transfer the results alone to the form here used.

London : Printed by 0. F. Hodgson ft Son, Gough Square, Fleet Street, E.G.

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K'ow ready, Boyal 8yo, d67 pp., with 285 Diagrams, price 340.

A SYNOPSIS

OF ELEMENTARY RESULTS IN

PUEE MATHEMATICS,

CotUiUning Six Thousand Propositions, Formula, and methods of Analysis, with abridged
Demonstrations; supplemented by an Index to the articles on Pure Mathematics which
have appeared in thirty-two Bbitish and Foreign Jousnals and Transactions
during the present century, containing about 14,000 references.

By G. S. OARR, M.A.,

Late Prizeman and Scholar of Gonville and Caius College, Cambridge.

The work may also be had in Sections separately, as follows : —

Section I. — Mathematical Tables ; including 0. G. S. units, s. d.
Least Factors from 1 to 99,000. The

Gamma-function, &c Price 2

„ n. — ^Algebra 2 6

., III. — Theobt op Equations and Detebminants 2

IV. — Plane Tbigonometby C together ; with ") o a

„ v.— Sphebical do. c 22 Diagrams. >

„ VI. — Elementaby Geometbt ; with 47 Diagrams 2 6

„ VII. — Geometbical OoNics ; with 35 Diagrams 2

„ VIII. — DiPPEBENTIAL OaLCULUS 2

„ rX. — Integral Calculus 3 6

„ X. — Calculus op Variations -j

„ XI.— DiPPEBENTIAL Equations C 3 6

„ XII. — Calculus op Finite Dippebences )

„ XIII. — Plane Coobdinate Geometby, with 182 Diagrams 6

„ XIV. — Solid Coobdinate Geometby, with 26 Diagrams 3 6

N.B.— The Index will not be sold separately, but possessors of previous sections may complete the
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The work is in part designed for the use of University and other Candidates who may be reading for
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Mathematics are presented withm the compass of a few pages. This has been accomplished, firstly, by
the omission of all extraneous matter and redundant explanations, and secondly, oy carefully com-
pressing the demonstrations in such a manner as to place only the leading steps of each prommently
before the reader. The whole is intended to form a permanent work of reference for mathematicians.

" The book before us is a first section of a work which, when complete, is to be a Synopsis of the whole
range of Mathematics. It comprises a short bnt well-chosen collection of Physical Constants, a table of
fkctors up to 99,000, fh>m Borckhardt, &c. &c .... We may signalise the chapter on Geometrical Conies as
a model of compressed brevity .... The book will be valuable to a student in revision for examination
papers, and the completeness of the collection of theorems will make it a useful book of reference to the
mathematician. The publishers merit commendation for the appearance of the book. The paper is good,
the type large and excellent."— Journal of EdiuxAion.

•• For the most part a carefully edited print .... a well selected synopsis, well arranged too for its
chief purpose as a book of repetition, and containing a good number of useful theorems, especially in the
Algebra, which are not usually to be found in the books of elementary reading .... It is clearly printed,
and there is a good table of contents. On the whole, if the work, when complete, is supplemented by a
good index, it will occupy a place not filled up at present." — Popular Science Review.

"Having carefully read the whole of the text, we can say that Mr. Carr has embodied in his book
he most useful propositions in the subjects treated of, and besides, has given many others, which do not
o frequently turn up in the course of study. The work is printed in a good bold tyi)e on good paper, and
he figures are admirably drawn." — Nature.

** Mr. Carr has made a very judicious selection, so that it would be hard to find anything in the ordin-
ary text-books which he has not labelled and put in its own place in his collection. The Qeometrical por-

on, on account of the dear figures and compressed proofs, calls for a special word of praise. Tlte type
exceedingly clear, and the printing well done."— Educationai Times.

"The compilation will prove very useftil to advanced students."— 27k« Jowmal c/Seknee,

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AN INTEODUCTION TO GEOMETEY.
FOR TEJE USE OF BEGINNERS,

CONSISTING OP

EUCLID'S ELEMENTS, BOOK L

ACCOMTANIBD BY NuMBROUS EXPLANATIONS, QUBSTIONS, AND ExBRCIflBS.

Bt JOHN WALMSLEY, B.A.

This work is characterised hy its abundant materials suitable for the tnun-
ing of pupils in the performance of original work. These materials are so
graduated and arranged as to be specially suited for class- work. They fur-
nish a copious store of useful examples from which the teacher may readily
draw more or less, according to the special needs of his class, and so as to
help his own method of instruction.

OPINIONS OF THE PRESS.

"We cordially recommend this book. The plan adopted is founded upon a proper
appreciation of the soundest principles of teacning. We have not space to give it in
detail, but Mr. Walmsley is fully justified in saving that itjarovidea for 'a nafural and
continuous training to pupils taken in classes.' ^'—At7ien<Buin.

*• The book has been carefully written, and will be cordially welcomed by all those
who are interested in the best methods of teaching Geometry.*'— iSc^oi Guardian.

" Mr. Walmsley has made an addition of a novel kind to the many recent works intended
to simplify the teaching of the elements of Geometry. . . . The system will undoubtedly
help the pupil to a thorough comprehension of his subject." — School Board Chronicle.

* When we consider how many teachers of Euclid teach it without mtelligence, and
then lay the blame on the stupidity of the pupils, we could wish that every young
.... . <• .. . 'Itake"*""

teacher of Euclid, however high he may have been among the Wranglers, would take the
trouble to read Mr. Walmsley*s book through before he begins to teach the First Book to

g boys."— Journal qf education.

Ve have used the book to the manifest pleasure and interest, as well as progress, of

young boys."— Journal qf education.

" We have used the book to the manifest pleasure and interest, as well as pre.
our own students in mathematics ever since it was published, and we have the greatest
pleasure in recommending its use to other teachers. The Questions and Exercises are
of incalculable value to the te&cher.**— Educational Chronicle.

KEY to the above, price 3s.

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Thirteenth Edition, 12mo, cloth, price Id.

XAMPLES IN ALGEBRA FOR JUNIOR CLASSES.

Adapted to all Text-Books ; and arranged to assist both the Tutor
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Third Edition, cloth, lettered, 12mo, price 3s.

EXAMPLES IN ALGEBRA FOR SENIOR CLASSES.

Containing Examples in Fractions, Surds, Equations, Progressions, &c.,
and Problems of a higher range.

THE KEY ; containing complete Solutions to the Questions in
the " Examples in Algebra for Senior Classes," to Quadratics inclusiye.
12mo, cloth, price 3s. 6d.

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IRAOTS relating to the MODERN HIGHER MATHE-

MATICS. By the Rev. W. J. Wright, M.A.
Teact No. 1.— DETERMINANTS.
„ No. 2.— TRILINEAR COORDINATES.
„ No. 3.— INVARIANTS.

The object of this series is to afford to the young student an easy introduc-
tion to the study of the higher branches of modem Mathematics. It is pro-
posed to follow the above with Tracts on Theory of Surfeuses, Elliptic In-
tegrals and Quaternions.

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MATHEMATICAL QUESTIONS, with their SOLU-
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W. J. C. MiLLBB, B.A., Registrar of the General Medical Council.

Of this series forty-seven volumes have now been published, each volume
containing, in addition to the papers and solutions that have appeared in the
Educational Times, about the same quantity of new articles, and comprising
contributions, in all branches of Mathematics, frt>m most of the leading
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New Subscribers may have any of these Volumes at Subscription price.

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LECTUEES on the ELEMENTS of APPLIED ME-
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and Mechanics at the Royal Military Academy.

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TRACTS ON MECHANICS.. In Three Parts— Parts 1
and 2, On the Theoij of Work, and Graphical Solution of Statical
Problems ; by Morgan W. Crofton, F.R.S., late Professor of Mathematics
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Third Edition. Extra fcap. 8vo, price 48. 6d.
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ELEMENTARY MANUAL of COORDINATE GEO^
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Master of the Royal Naval School, New Cross.

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AN INTRODUCTORY COURSE OP

PLANE TEIGONOMETRY AND LOGARITHMS.
By JOHN WALMSLEY, B.A.

" This book is caareftilly done ; has full extent of matter, and good store of examples."—
Aihsnaum.

** This is a carefully worked out treatise* with a very large collection of well-chosen
and well-arranged examples."— Papw*/or the Schoolmaater,

"This is an excellent work. The proofs of the several propositions are distinct, the
explanations clear and concise, and the (ceneral plan of arrangement accurate and
methodicaL"— 2%tf Mu9«um and English Journal (f Education,

** The e^lanations of logarithms are remarkably mil and clear. . . . The several parts
of the subject are, throughout the work, treated according to the most recent and
approved methods. ... It is, in fact, a book for beainnerSt and by far the simplest and
most satisfiActory work of the kind we have met witlL"-~Education4U Times,

Price Five Shillings,

Af$d will he supplied to Teachers and Frivate Students only, on application to the

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A KEY

to the above, containing Solutions of all the Examples therein. These num-
ber seven hundred and thirty , or, taking into account that many of them are
double, triple, &c., about nine hundred; a large proportion of which are taken
from recent public examination papers.

By the same Author,
New Edition, fcap. 8vo, doth, price 5s.

»LANE TRIGONOMETRY AND LOGARITHMS.

JL FOR SCHOOLS AND COLLEGES, Comprising the higher branches
of the subject not treated in the elementary work.

'* This is an expansion of Mr. Walmsley's ' Introductory Course of Plane Trigo-
nometry,' which has been already noticed with commendation in our columns, but so
greatly extended as to justify its being regEirded as a new work .... It was natural that
teachers, who had found the elementary parts well done, should have desired a com-
pleted treatise on the same lines, and Mr. Walmsley hiEbs now put the finishing touches
to his conception of how Trigonometry should be taught. There is no perfunctory work
manifest in this later growth, and some of the chapters— notably those on the imaginary
expressAon ^^^, and general proofs of the fundamental formulae— are especially good.
These last deal with a portion of the recent literature connected with the prpora for
sin {A+Bhjkc., and are supplemented by one or two generalized proofs by Mr. Walmsley
himself. We need only further say that the new chapters are quite up to the level ra
the previous work, and not only evidence great love for the subject, but considerable
power in assimilating what has been done, and in representing the results to his readers."
—Educational Times,

By the same Author.

Preparing for Publication.

Suitable for Students preparing for the University Local and similar

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AN INTEODUOTION TO MECHANICS. With
numerous Examples, many of whidi are fully worked out in illustra-
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ALGEBRA IDENTIFIED WITH GEOMETRY, in
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1. Euclid's Conception of Batio and Proportion.

2. " Camot's Principle'* for Limits.

3. Laws of Tensors, or the Algebra of Proportion.

4. Laws of Clinants, or the Algebra of Similar Triangles lying on the
Same Plane.

0. Stigmatic Geometry, or the Correspondence of Points in a Plane.
With one photo-lithographed Table of Figures.

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"iVrOTES ON THE FIRST PRINCIPLES OF DY- /^

-L^ NAMICS. By W. H. H. Hudson, M.A., Professor of MathematicjA
in King's College, London. q/^

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