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4UTH HM ATIC AL QUESTIONS, 



Wn-H TTtEIR 



SOLUTIOKS. 



tTtOM THE -EDUCiTiUXAt. TOiH!^/ 



VOL. XLIX. 



r 



Mocih 33*g.S4», 





SCIENCE CENTER LIBRARY 



Digitized by VjOOQ IC 



-.-, .^ 



Digitized by VjOOQ IC 



Digitized by VjOOQ IC 



MATHEMATICAL 

aUESTIONS AND SOLUTIONS, 

FEOM THE "EDUCATIONAL TIMES," 



WITH MANY 



PAPEES AI^D SOLUTIOlsrS 



IN ADDITION TO THOSE 



PUBLISHED IN THE "EDUCATIONAL TIMES," 



AND POUR 



APPENDICES. 



EDITED BY 

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

BBOISTBAB OP THE GENEBAL MEDICAL COUNCIL 



VOL. XLIX. 



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

1888. 

Digitized byCjOOQlC 




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




^ I 






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

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



'Digitized by VjOOQ IC 



LIST OP CONTEIBUTOES. 



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

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

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

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

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

ABMENAifTE, Professor; Pesaro. 

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

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

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

Babu, Professor Ealipada, Dacca College. 

Battaglini, Prof. Giuseppe ; Univ. di Boma. 

Batliss, Geoegb. M.A. ; Kenilworth. 

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

Beltrami, Professor ; University of Pisa. 

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

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

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

Bhattachabya, Prof. Munindbanath, M.A. 

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

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

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

Blackwood, Elizabeth, B.Sc. ; Boulogne. 

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

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

BoBDAGE, Prof. Edmond ; Coll. de Nantua. 

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

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

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

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

Bbooks, Professor E.; Millersville, Pennsylvania. 

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

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

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

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

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

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

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

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

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

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

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

Catalan, Professor ; University of Lifege. 

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

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

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



EssENNELL, Emma ; Coventry. 

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

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

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

PiNKEL, B. F. : Ada, Ohio. 

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

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

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

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

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

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

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

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



IV 



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

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

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

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

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

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



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

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

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

Mathematics in the University of Oxford, 

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

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

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

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

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

o 



CONTENTS. 



iMatftematiial ^apersf, ^r. 

Page 

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

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

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

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

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



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

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

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

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

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

-14-63442007818570452204 « 0. ... 70 

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

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

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

x^ y^ v' 'lit 

a 



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VI CONTENTS. 

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

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

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

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

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

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

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

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

-J_ + _±_ + -J^ + ... = -^taii^ 40 

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

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

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

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




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vu 

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

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

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

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

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

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

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



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

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

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

dx (^dx 

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

dx ) 

80 

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

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

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

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



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Vlll CONTENTS. 

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

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

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

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

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

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

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

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

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

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

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

- BC . AF . DE cos (CFE « ADB + DCA) 

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

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



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CONTENTS. IX 

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

+ a^^b + c)^2ax•~a^) - 0.] 82 

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

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

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

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

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

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

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



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X CONTENTS. 

8315. (Professor Booth, M.A.)— If 

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

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

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

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

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

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

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

yU f^H rpH gtl *" -jjH yU \ J » / 

depends on the solution of 

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

51 

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

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

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

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

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

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

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

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

30 



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CONTENTS. XI 

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

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

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

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

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

17.8 

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

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

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

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

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

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



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Xll CONTENTS. 

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

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

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

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

AMN + BNL + CLM = ABC 35 

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

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

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

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

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

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

63 

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

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

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

'>)(^"-)/(lrH-(^-»)/(,^-') 

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

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

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

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

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

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

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



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CONTENTS* XIU 

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

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

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

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

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

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

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

8853. (A. Russell, B.A.)— Prove that 

Jo Jo Jo \ 4a2«3' 4^2^' ^cYf 

is a solution of the differential equation 

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

dt dx^ rfy2 az^ 

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

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

X l+y2 a^ l+y6 

X OK- 1 «A-1 

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

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

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

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

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

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

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

8940. (W. J. C. Sharp, M.A.)— If 

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

h 



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XIV 



CONTENTS. 



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

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



= A 



yii .vj, .vs 



«ii 



+ &C., 



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

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



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



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



0. 



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

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

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



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



86 



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

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

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

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



(w + l)-ic 



aa:»*>+ 'i±i(% + *23)a^'*+&c. . 



135 



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

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

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



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CONTENTS. XV 



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

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

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

3i.2» Vj, S2.31 S2.4 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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



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XVI CONTESTS. 

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

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

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

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

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

B?«4Ri»-B,« 34 

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

2r = l'' + 2^ + 3*' «^ 

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

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

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

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

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

— 1 c^^^' Bm(eQiiix)Buxnzdx = — 75 

IT Jo nl 

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

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

— = ^ , when m = 1 or when w = 2 53 

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

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

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



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CONTENTS. XVU 

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

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

96 

9226. (J. White.)— Prove that 

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

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

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

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

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

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

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

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

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

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

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



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V 



XVIU CONTENTS. 

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

by P*, P*.i- Pt = (n2-«) ^L^, 

a — p 

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

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

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

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

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

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

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

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

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

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

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

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

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

a^ Ir \ e l — ej 

(4) =(«<*), =4*{(l-^)* + 2/esin-»^}; 

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

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



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CONTENTS. XIX 

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

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

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

(m + »-l)!r! 

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

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

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

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

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

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

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

79 

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

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

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

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

dx* dJt.^ dx^ dx dy^ dy 



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XX CONTENTS* 

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

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

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

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

52 

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

that C0^,»+2 ^d ^-?i±i 65 

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

124 



^:\ 


'i\ ^3-'. 


'I' 


JT^ 


yi\ y^S 


yi' 


(*I.Vi)- 


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


, (*4y4)-' 


1, 


1. 1, 


1 



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

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

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

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

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

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

5'+ 1 2 ! 4 ! 

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



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CONTENTS. XXI 

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

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

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

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

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

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

__ n(/i-4)(n-5) «3i3^n.6cos3 (A-B) 
3! 

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

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

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

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

88 

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

/ 2^2- *2 \ «2 ^ b^t/^ , .- 

(,"''>' — ^^^V' 'i^^W^^'^ ^ 

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

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

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

77 

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

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

IK 

C 



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XXU CONTENTS. 

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

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

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

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

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

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

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

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

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

(«-3)!3! V / » 

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

100 

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

The nephew is and uncle too. 

In how many ways can this he true P 114 

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

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

109 

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

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

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

112 

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

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

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

93 



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CONTENTS. XXUl 

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

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

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

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

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

,V«2 («+ 1)2 (»- l)(w-2) 109 

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

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

prove that = = , 

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

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

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

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

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

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

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

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

90 

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



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XXIV CONTENTS. 

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

_!5_S— (i^J-'""^!- 94 

(W— 1)'^ OP*^ — (j9+l)o+i? 

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

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

94 

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

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

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

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

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

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

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

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

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

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

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

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

117 



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CONTENTS. XXV 

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

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

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

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

22«+2 + 4i-«==, 17 are a:= ±1 Ill 

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

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

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

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

.i A-x A A-x 

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

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

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

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



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XXVI CONTENTS. 

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

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

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

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

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

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

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

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

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

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

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

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

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

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



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

2 tan-14 ± tan-i ,, ^\ ;; - Jir, 

b d^ + 2ab~a" * ' 

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

l + 2j;-a;- 



180 



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CONTENTS. XXVU 

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

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

APPENDIX III. 
Unsolved Questions 151 

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

(A.) Diophantine Analysis 159 

(B.) Besolution of Squares ; 162 

(C.) Resolution of Cubes 170 

(D.) Solutions of Old Questions 180 

(E.) New Questions 186 



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



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MATHEMATICS 

VBOM 

THE EDUCATIONAL TIMES. 
WITH ADDITIONAL PApSeS AND SOLUTIONS. 



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

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

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



Solution by Professor Nash, M.A. 

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

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

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



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



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22 

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

Solution by the Proposer. 

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



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



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

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



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

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



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

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

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

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

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



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23 



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



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



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

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

Since Z QOR = 2QMR 

= OPM = OPR, 

QO touches the circumcircle of OPR and 

Q02 = PQ . QR. 




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

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

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



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24 

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

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



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

Solution hy Artemas Martin, LL.D. 

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

The required probability will be 

p^ [[[dxdydz I uAdxdydz, 

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

Jo LJo Jja-«-y Jia-«Jo J * JoJo Jo 

-mr\:..-.'''-<..r '''■]'■ 

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



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25 

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



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

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

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

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

a a a a a 
s = is known. 

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

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

Substituting after putting ^ » 0, we have 

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

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

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

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

the coordinates of the centre are given by 

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

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

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

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

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

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

(a^b) «''+ (e^b) a'^-2b'ae = (2), 

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



i 



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

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

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



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



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

Un -T (XJ„-i) ' T-i(^«-2)i "^d 80 on. 






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

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



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



(-1)" / 1 ±Y'' ( 1 \. 

6...2(»-2) \a rfa/ \a + */ ' 



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

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

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

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

" 16 a* (a + *)3 16 a» (a»- ^)» 

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

2 «! \dp) \dql \p+Vpq)' 

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

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

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

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

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



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

— . . . to » terms J ; 

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



(a2-^2)» 



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

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

Jo (i^ + jtan^a;)" 

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

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

get a finite series for r ( | . 

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



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



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



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

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

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

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

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

which gives the result as a series of m terms. 

Thus p''_tan2£ ^ ^ jr^ 1 

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

fi' tan^a- ^^ ^ ir f 1 2 •> 

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

but I think the former method preferable. 



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

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

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

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

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

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

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

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

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

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

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

therefore da^ ±{dx—2yd$), 

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

y (a»-ir2)* 

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

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

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

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

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

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

4a Jo (.«»—(«»—*') cos* ^ > 

$ ° 1-tf 



^Ua Jo la> + (6a-a«)co88^^ ^ ^j 



^ tan->— ^+1; 



^(1 -<?«)* (l-<^)* 

tr 2 

therefore li '^ "" "^"^ * ^ (!-«')*• 



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

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

a Jo 

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

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

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



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



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

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



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



Solution by the Proposer. 

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



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



VOL. XLIX. 



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Solution by ProfesBor Ignacio Betkks. 

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

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

a a a '" 

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

n n \n I n-1 

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



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



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

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

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

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

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

whence (2). 



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



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

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

Similarly, from the second equation, 

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



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

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

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

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



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

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

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



Hence, if 



(A1B3C3D4) = 



Ai B, C, Dj 

Aa Bj Ca Dg 

A3 B3 Cs Dg 

A4 B4 C4 D4 



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



.(1). 



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

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

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

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

|^(^iyj*'3> = (i y,»3); 

^(^iyj*'») = (^i 1 «,)y (4). 

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

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

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

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

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

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



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

Solution by the Proposer. 

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

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



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



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

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

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

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

where « = |, therefore, &c. 




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

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



h y^ ar* 1 + y" 



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



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

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

1+A^;c2 A aA-1 

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

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

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

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



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84 

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

Aj Aj 

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

^1 yi 



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

E« = 4Ri»-R52. 



Solution by Professors Ionacio Bbyens ; Bordaob ; and others. 

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

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

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

•*• ^^'^^ = {a^^b^^c^r • 

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

4R 

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

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



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



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

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

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



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

3a 2a 

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



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

Bar* d^ dx^ dx dy2 dy 

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

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

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

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

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



^^ff ^-«'/(ar + 2w*/i^W«. 



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



de 



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



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



Solution by the Proposer. 

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



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36 

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



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

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



Solution by Professors Ionacio Bbtens, Matz ; and others. 

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

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

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

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

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

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

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

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

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

KP^KS PS KPg ^ Kffl ^ PS» 

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

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



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

""'- ^-ii 

whence the result follows. P 

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




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

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

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



Solution by Professor Nash, M.A. 

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

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

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

.and A = bc^f*, B =» ea—y^ &c. 



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

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

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

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

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



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



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

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

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

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

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

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

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

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

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

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



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

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



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



Solution by Professor Q. J. Leobbeke. 

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

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

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



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



12»aj2 32_a;i 62-ic2 - 4^ 2 



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

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

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

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



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

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

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



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

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

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

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

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

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

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

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



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

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

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

ah b . ^ ah a 

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

whence the equation of the envelope is 

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

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

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

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

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

^V a&» m' aH 

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

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

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

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

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



acos0 

y_ 

dsin^ ' 

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



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



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Solution by the Proposer. 

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

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

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

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

c * 4. • X IOC 20A2A;2( ,., 

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



of two reentrant points < 






M 

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

MC - i Jx2y3fl?x + 3ja:yrf^Jy«rf2:, 

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

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

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

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

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

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



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

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

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

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

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

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

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

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

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

In ascertaining the integration with respect to the second component 

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

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

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

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

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



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

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

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

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

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

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

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

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

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

*iy2-^2yi» ^^vz-HVi^ 

hence found to be 




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



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

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

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



.(6). 



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

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

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

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

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

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

Hence, by substitution in (4), we find 

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

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

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

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

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

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

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

>i-i('-f) (•)■ 

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

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

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

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

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

VOL XLIX. F 



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

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

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

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

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

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

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

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

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

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

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

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

M« M2 27 V 4/108'^ ^*^^ 36 

And the probability of 

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

two reentrant points » = — (I— pp'). 

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



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

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



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

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

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

-, therefore pk^i = — , 



l-(«-l)a:-(«-l)ar2 



and 



Pa.i-P*- 



jj* + 2_g**2 



i-a*-i 



o-iS 



This seems to be the correct result. 



-i3 



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



(-^if - 



b'^ 



yi). 



*V 



a2 (2a«-*2)2 



1. 



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

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

also RM:PN = GM:GN 

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




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therefore QQ' is, by comparing coefficients, 

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



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

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

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

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

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

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

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

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

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

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

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

and with this substitution we have formula (3). 

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

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

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



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

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

may be proved in the same way. 

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



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

Solution by D. Biddlb. 

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

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

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




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



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

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



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



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

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

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

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

or 3r = 4a(2cosfl— 1) ; 

therefore A = jfrVe 

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

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

= 4a2(2»-3<v/3). 

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




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



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

therefore /- « ^^" ^^ « tan i0 ; 

2vr 1-cosifl 



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hence, by Grboory*s series, we get 



4r * 2vr 



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



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

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

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



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




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

depends on the solution of 

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

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

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

^-1 yn-\ 5;»-l > * 

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

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

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



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

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

ax—btf bu — cz ez—ax 

« ^ — <r say, 

a:— y y—z z—x 

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

therefore g — — ^, y — --^, « — -^ ; 

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



8315. (Professor Booth, M.A.)— If 

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



show that 






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

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

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



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



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

tan37i** = (^/3-^/2)('v/2 + l); tan67i°= ^/2 + l. 
Also tan22r - 'v/2-1 ; tan52i** = (^/3+ ^/2) (a/2-1) ; 

tan 82^ = (-v/3 + -v/2) ( a/2 + 1) ; 
therefore product -(2^/6+ a/2- 3)(2v/6 + -v/2 + 3) = 17 + 8 v/3. 
[By easy reductions, the product can he brought to 

(5 + 6co876*')(6-6c08 76°)/(2v/2cos76°) « the given result.] 



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9195. (Sir James Cockle, F.RS.)— Integrate 



d^u 



dt^ (aH*3^'») 



-, when m = 1 or when m «= 2. 



Solution by the Pboposer. 
1. Letm= 1; put 4^=A2^2_c2; then ^+i ' 



X* 



: is flatiflfied by 



«= Co(l + Vx)e"^'' + C, ( -^ + ^^) (»~''*^^ + C, ( J+ ^*) «"^*^*' 

where a and fi are the unreal cube roots of xinity. 
2. Let m = 2 ; put ^ — n sin and 4%' ._ _ ^2 . tj^e^ 

Icostfe^fl/ " a3(cose)3' e^flUosd^/eJ " "* a^ (cos «)* * 
whence ^.3tan«^.3 {(tan*).^}^ = ^ ; and.if »' = f" 



' ^0 aS 



rf»' 



rf3f/ 



.<^«' 



rfw' 



^+3tanfll^ + 3{2(tanfl)« + |}^ + 3{2tana[l + (tan6)2]}./-^,. 

Now put u « (cos e) y ; then -^ + -r 5-^ = 0, whereof the coeffici- 

d^ dd a* 

ents are constant and wherein a » ~6 V^— 1. The solution of case (1) is 
obtained by an analogous process. 



8743. (C. BiCKBBDiKE.) — Prove that (1) the length of a focal 
chord of the parabola is I cosec' (p ; (2) when the chord is one of quickest 
descent, cos (p ~ (|)' ; and (3) the time of quickest descent down the 
chord then is a/(3^/)/^, where I is the latus-rectum, and ^ the angle 
made by the chord with the axis. 



Solution by Geobob Goldthorpb Stork, M.A. 



1. SP + SF: 



2a 



2a 



1 — C08<^ 1+COS^ 

_ ia ^ I 
sin* <t> sin* <l>' 

2. Here t = {2Z/(8in*<^.ycos<^)}* ... (a) ; 
and, as this is a minimum, sin* <^ cos <^ =x— a:' 
(where x ^ cos (p) must be a maximum, which 
is the case when Goa<f>sx = (^)^, 

Substituting this value in (a), we find the time of quickest descent 
to be as stated in the Question. 




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9381. (Professor Sylvester, F.R.B.) — If {g and r being prime 
numbers) 1 +j0+p'+ ...p'-* is divisible by g, show that, unless r divides 
q— 1, it must be equal to q and divide j0— 1. 



Solution hy Professor Gbnesb, M.A. 

Let N( denote a number which is expressed in the scale of ^7 by 111 ... 
to t digits. Now, q being a prime, 

^-/,-M(^), !.<?., j0(/?-l)N,.i -M(^); 
and, by the question, N^ = M (q) ; thus N, . i, N^ have a common measure 
q, unless q divide (js— 1), for it clearly does not divide i?. 

1. The arithmetical process shows at once that N,_i, N,. caimot have a 
common factor unless of the form N^ where t divides q^\ and r ; but r is 
prime, therefore r divides q^\ (which is not prime). 

2. If i> — i»^ + 1, we find Nr « M (j) + (1 + 1 -i- ... to r terms), whence 
q^r. 



2353. (The late Professor De Moboan.)— The 
late Dr. Milnek, President of Queens' College, Cam- 
bridge, constructedalamp which General Pbreonet 
Thompson remembered to have seen. It is a thin 
cylindrical bowl, revolving about an axis at P, and 
the curve ABCD is such that, whatever quantity 
of oil ABC may be in the bowl, the position of 
equilibrium is such that the oil just weto the wick 
at A. Find the curve ABCD. 




Solution by D. Biddlb. 

Let ACi, ACj, AC3 represent 
the surface of the oU at three 
different times ; then in PBi, 
PB2, PB3, perpendicular to these 
respectively, will lie the centre 
of gravity of the oil at those 
particular times. Consequently, 
the level of the oil, always 
flush witii A, and the line join- 
ing the centre of gravity with 
P, describe equal angles in a 
given time. 

Taking a wedge-shap^ed por- 
tion of oil, of infinitesimal 
depth, with its apex at A, and 
its base at C|, its particular 
centre of gravity will be at 
fAOi from A, say at N, ; and, 
if Ki be the gentre of gravity 
of the mass of oil when at the level AC^, then N|E| wiU be^^tangential to 
the locus of the centre of gravity of the mass. 




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Let :ri, y^ be the coordinates of the centroid K|, x, y, the coor- 
dinates of the centroid E (that is, of the oil when the vessel is full). 

Let M a the mass having the centroid E), and a/, y* be the coordinates 
of the centroid of AC^D (= 1 — M) now empty ; also let Z DAC| » a. 

Then PKisinaM = (x'-ar) (1-M), 

and (PKiCoso-PK) M = (y-y') (1-M), 

whence TK^ ^ ^^"^'^^^ ^^^^, and PK - cota(«-a;l)-(yl-y)• 
co8a sino V v wi 

If ABCB were a semi-circle, and P its centre, these conditions would 
be fulfilled, but as the oil sank in the vessel there would be no bias to 
cause rotation. For the purposes specified in the question, it is essential 
that, as the oil is consumed and its level sinks, the layer taken oS. shall 
have been unequally divided by the perpendicular from P, the lesser 
portion being always on the side next A, so that what remains is heavier 
on that side, until by rotation equilibrium is restored. 

A very near approach to the curve required is given in the accom." 
pAnying fig^e, where r — cos* fl, dr/d$ = sin0/2r, and the area of the 
side of the vessel = JAD*. With this curve, if, as may be supposed, 
equal quantities of oil are consimied in equal times, an indicator parallel 
to AD and projecting from the vessel as if drawn from P, will nse at a 
uniform rate, because the sine of the angle DAO lengthens at a uniform 
rate. Such being the case, wo have at once not only a convenient light, 
but also a time-keeper scarcely inferior to Alfred the Great's graduated 
candle. 

The positions of E and P, though easy to find by practical methods, do 
not yield readily to the integral calculus. 

But AL==}[ cos*fl(?0 + J cosflifl, 

LE == I 008^0 sin d<^d-i- Q0%9dB. 



9386. (Professor Neuberg.) — Si suivant les perpendiculaires abaiss^es 
du centre O du cercle circonscnt h, un triangle ABC, sur les cStes de ce 
triangle, on applique, dans un sens ou dans 1' autre, trois forces ^gales, 
la i^sultante passera par le centre de Tun des cercles tangents aux 
trois cotes. 



Solution by Professors Genese, M. A, ; Betenb ; and others. 
Let L, M, N be the mid-points of the sides, P, Q, R the points of con- 
tact of any one of the four circles touching the sides. Then we have 

PL- J(PB + PC), &c.; 
and if, following Laguebre's principles, we define, for this case, the 
positive direction of the sides as that leaving the circle in the positive 
sense of rotation, we have 

PC + QC = 0,&c., .-. PL + QM + RN = 0. 



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Whence, for Boitable direcHonB of the foroee in qneeiion, the sum of their 
moments about the contra of the circle is seen to be zero. 

[Professor Gsnesb adds that this problem also occurred to, and was set 
by, him at Aberystwyth, in 1886 ; and that, if the equal forces be repre- 
sented by radii of the drcum-cirde, the lines representing the resultants 
terminal at the centres in question.] 



9316. (Professor MukhopIdhtAt, M.A., F.R.S.E.)— Prove that (1) 
' the locus of the mid-points of the chords of curvature of the conic 
4V + fl^2 « a'*2 ig the sextic SiSa-^^c^ + i-ayS = (a-«a;'-*-2y«)l pass- 
ing through the origin ; (2) the area of 2i is half the area (A) of the 
ellipse ; (3) the envelope of the chords of curvature of the same conic is 
the sextic ^2^(a-^x^-^b-^t/^-4)^+27 (a- x2-*-2y2)2 - ; (4) the area 
of 2] = f A ; (6) trace the locus 5i and the envelope Sa, and show that 
they touch each other and the conic at the ends of the major and the 
minor axes. 

Solution by Professor R. Swaminatha Aiyab, B.A. 

1. The line 
a-^a;co8 0— i"V8in0=co820...(i.) 
passes through the point (<^) on the 
ellipse, and makes the same angle 
with the axis that the tangent at 
the point does : it is therefore the 
chord of curvature at <p. The 
diameter conjugate to it is the line 
a-»ir8in^ + *-^yco8 0= 0...(ii.); 
and, eliminating <p between (i.) 
and (ii.), we have 2 as the 
equation of the required locus. 

2. From (i.) and (ii.), x = Ja (cos + cos 30), y = J* (sin <^— sin 30), 

I ydx « \ah J (sin 30 - sin 0) (3 sin 30 + sin 0) dip « \ab f F ^0. 
The required area ^ ab I ' Frf0 = ^irab = JA. 

3. Differentiating (i.), we have a-ia:sin0 + *-iy cos0=2 sin20...(iii.). 
From (i.) and (iii,), a? = Ja (3 cos - cos 30), y « i* (3 sin + sin 30) ; 
and eliminating 0, we have 2^. 

4. f ydx =« ^ab J (sin 30 + 3 sin 0) (sin 30 — sin 0) (^ » | a& f F (£0 ; 
hence the required area of Sj is \ab I Fe^0 === firad = f A. 

6. As increases from to 2ir, the locus 2i is traced in the order 
PQKSTUVW ; and the envelope Xj in the order ACBDA'EB'FA. The 
points (± (1^2, ± bV2) are cusps in the latter curve, the equi-conjugate 
diameters of the ellipse being the cusp -tangents. In the first curve the 
origin is what might be called a double tacnode. 




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8989. (ProfeoBor Wolstbnholmb, M.A., Bo.D.)— In a tetrahedron 
OABO, OA - fl, OB = 4, OC - fl ; BC - x, CA - y, AB - 2, and the 
dihedral angles opposite to these edges are respectively A, B, C ; X, T, Z. 
Having given tiie equations i — y = J (a + a?), «— ««ia— a?, B » Y, 
C + Z - 180°, prove that B - Y = 60°, 0-A » Z-X - 30° ; and find 
the relations between a, h, e. 

Solution by Sbpthcus Tbbay, B.A. 
We have x » 2d— a, y ^ by « » « + 26— 2a ; and therefore 
BO + CO = AB + AO, and BC-BO - AC-AO ; or (Quest. 8606) 

A + Z = X + C, andA-Y = B-X; andsinceB^Y, andC+Z= 180°, 

therefore A = B + C-90°, X-B-C + 90°, 

and sinAsinX — 1— cos'B— cos^C. 

Let the areas of the faces BOO, COA, AOB, ABC be denoted by Ai, 
A3, As, A4 ; then 

sinB ^ sin BOO ^ ^ Aj sin B ^ sinAOC ^h^ Aj . 
sinA ^ sinACO x ' A,' sinX " sinBOO " a * Ai * 

.-. sinAsinX = ^.8in2B- 1 -cos^B-cos'C, .•. cosC= ~^sinB. 

Also sinB ^ sin ABC _ b A4 ^ sinBCO ^ Aj 

sinC sinOBO" « * Ai "" sinACB A4* 

Therefore sin' C = ^ . sin* B. These equations give 

BinB-— A-=-^, sinC=-*:=^5 ±, (1,2). 

b + c—a c + a o + c—a c + a 

putting 6 - fl = a. 

Now the general relation among the dihedral angles of a tetrahedron is 

2 (sin* A Bin2 X) - 22 (cos X cos Y cos Z) - 22 (cos B cos C cos Y cos Z) « 2 ; 

which in the present case reduces to 

8co82Bco82C-4cos2B-4cos2C + l = (3). 

From (1, 2, 3), wo have 

48in2B = 2-sec 2C = 3+ ^^-^^ = r^«- 

(c + o)2— 2o^ (c-j-o)' 

This equation reduces to 

a* + ^(c-b)a^-2{7(^+ebc-'b^a^+i{Zc^-¥7bc^ + b^c + bi^)a') ... 

= 3A4-4^^(j+106V+12*c3 + 3<H > ^ ^' 

which is the general relation among a, *, e. 
Thus the least value of B is 60°, which makes 

C = Z = 90°, a = b =x = yy and 2a = c V3. 
In the other cases we must have 

gin B = — > -8660254 < 1. 

b-k-c—a 



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68 

We have assumed e > b > a. Now b/{b + e^a) is less than el{2e—a) ; 
and since e is the greatest value of b, B will be a maximum when 
sinB == e/{2c-a). liet b^e^l ; then, from (4), a*-24a« + 48a-24 = 0; 
from which we nnd a » '8618704. These values of a, b, e make 

B = Y - 60°34'22"-76, and C - 82°36'13"-11. 

If a be small in comparison with c, so that c? and higher powers may 
be neglected, we find 

2b^e^/Z (l+ -+^). and sinB « i^/3 f 1 + j^V 

which is not sensibly affected if a = or < 'OOl. Hence any solution 
depending upon a small value of a makes A+X =» 120°, nearly; and 
since C + Z = 180°, therefore C-A + Z-X = 60°, or C-A=Z-X=30% 
sinceC-A = Z-X. 
For the maximum value of B we have 



a - -8618704 


a? = 1-1481296 


B = 60°34'22"-76 


b= 1 


y-1 


C«82°36'13"-ll 


<; - 1 


z = 1-2962692 





A - 63°9'36"-87, X = 67°69'9"-66. 

[There being 6 equations gfiven, apparently independent, it would seem 
that the shape of the tetrahedron most be fixed, but there is certainly 
more than one solution. One obvious solution is when a = x — b ^ y^ 
when it will be found that C'v/3 = 2a=Z, C=Z=-90°, A = X=B=Y = 60°. 
The tetrahedron in which 

Ca^ 4*8023, x = 4-8044 I satisfies the conditions; and 
lb = 4-80336, y = 4-80336 A = 69°69'20"-96 = C-60°, 

(c = 6-64538, z = 6-64738 X - 60° 0'39"-04 = Z-60°, 

B = Y = 60°]. 



8701. (A. Russell, B.A.) — Resolve into quadratic factors 
(a2_ bcY (b + e)^ {b-e) {fl2 + 2a(b + c) + be} 
+ (b^-eay {c + ay {c-a) {*« + 2b{c + a) + ea} 
+ (c^-ab)'i{a + bY{a-b){(^-{-2c{a + b) + ab}. 



Solution by R. F. Davis, M.A. j Professor Betens ; and others^ 

Let A = {a^-bc)(b^e), B= ..., C = ... ; so that A + B + C -0. 

Then, since B -0 =- (* - c) {a2 + 2a (* + c) + *<?}, 

the given expression may be written A* (B —C) + ... + ... , which can easily 
be reduced to the form -ABC (B-C) (C-A) (A-B). The given 
expression (which is homogeneous and of the 18th degree) is therefore 
equal to the product (with its sign changed) of nine quadratic factors ; 
three of the form b^—c^, three of the form a^—bCf and three of the form 
a" + 2a (i + c) + be. 



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7759. (Ptofeasor Hanumanta Rau, M.A.) — From one end A of 
the diameter AB (a 2a) of a semicircle, a straight line APMN is drawn 
meeting the circumference at N, and a given straight line through B at 
M, at an angle a ; show that the locus of a point F, such that AP, AM, 
AN are proportionals, is the cubic through A, 

r as 2a sin^a sec 6 cosec^ (a - 6), or 2a sin^ a (x^ + y^ « (a; sin a— y cos o)', 
which, when a = ^ir, iir, becomes 

2a2(a;3 + y2) ^ ^^ 2a^{x^ + y^ ^ x^x-^yf. 



Solution by G. G. Storr, M.A. ; Rev. T. Gallibrs, M.A. ; and others. 
The polar equations of the line and 
the circle are respectively 

r == 2a sin a cosec (a — 6), r = 2a cos d. 

But AP.A:N'=AM2; hence the 
locus of P is given by the equations 
stated in the Question. 




8852. (J. Gbifpiths, M.A.)— If a, jS, 7, 8 be the roots of the quartic 

ax^-^^ba^-^Qcx^-^^dx-^e - 0, and if ^ - ^^^^ + ^^l3i • show that 

a— 5 i8-5 

(2-^)2(1-2^)2(1+^)2 108J2' 

where I « ae- Ud +Zc^, J = ad^ + eb^ + (^- ace- 2bed. 



Solution by D. Edwardbs ; G. G. Storb, M.A, ; and others. 

Let the quartic be linearly transformed into a' (l—nui^) {l—nx'), and, 
Bfi in the Fundamenta Nova, let 

U-7V = A(l+w^, U-5Va:B(l-f»^, U-aV-C(l+«*;c), 
U— /8V *» D (1— n*ic). Putting x « — w-i, +n-i successively in these, 

Butwehave {m^^n'^^Umnf ^V_ 

(m + nf (Z^mn^m^-n^)^ 27J3 
(Caylby's Elliptie Functions, Arts. 413 — 14) ; hence substituting for m/n 
in terms of q, we have the stated result. 



8850. (W. J. Grbbnstrbbt, B.A.) — Prove that the sum of all the 
harmonic means which can be inserted between all the pairs of numbers 
whose sum is «, is ^ (w*— i). 



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Solution by A. W. Cavb, M.A. ; W. J. Barton, M.A. ; and others, 
8am -2r!t^ + 2J«-22^3ii^^ -I 
L *• *• ** J 

-2 [1 + 2+ ... +(n-l)]-2>[l« + 22+ ... +(»i-l)*] 
«n(«-l)-i(n-l)(2n-.l)-i(n«-l). 



9340. (R. Knowles, B.A.)— In Question 9149, if BD and AC intersect 
in O, and CA meet KH in M ; prove that the lineB GM, GA, GO, GB and 
LC, LO, LA, LH form harmonic pencils. 



Solution by G. G. 
MoR&iCB, M.A., M.B. 

G{M.A.O.B} 

-G{M.A.O.C} 

-H{M.A.O.C}, 

which is an harmonic 
pencil, by the known 
property of a complete 5 
quadrilateral; similarly 

L{C.O.A.H} 

-H{C.O.A.M}. 




8300. (Professor Hanumanta Rau, M.A.) — Prom any point P on the 
circle described about an equilateral triangle ABC, straight lines PM, PN, 
PR are drawn respectively parallel to BC, CA, 
and AB, and meeting the sides CA, AB, BO H 

at M, N, and R. Prove that the points 
M, N, R are collinear. _d/ 

Solution by D. O. S. Davibs, B.A. ; 
R. Knowlbs, B.A. ; and others. 
Since ABC is an equilateral triangle, evi- 
dently ANP, PMO, PRO are angles of equi- 
lateral triangle. Hence N, A, M, P and 
P, M, R, are conoyclio. Join PA, PC, 
NM, and MR ; then 

ZPMN-PAN = P0B; 

hence N, M, R are collinear. 




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4251. (Colonel Clahkt?, C.B., F.R.S.)— If A, B, Cbe three circles, B 
being within A, and C within B ; prove that the chance that the centre 

of A is within C is -. 

7 

Solution by the Pkoposer. 

In the accompanying diagram, 
let a, b, c be the centres of A and 
of B, C, two circles lulfllling the 
condition of which the probability- 
is required. It will be convenient 
to denote such cirrles by accented 
letters B'C. Let (C')a?y represent 
the number of circles C within 
a B' of radius ijS = y and for 
which ab = x\ while (C) the 
entire number when B' has taken 
all magnitudes and positions. 
Then, lastly, if (C) denote the 
entire number of C*s, the chance 
required is (C')/(C). 

It is easy to see that the centre e 
of any C cannot fall outside the 
ellipse whose foci are a, b and major axis y = radius of B', for this ellipse 
is the locus of the centre of those circles which, passing through a, touch 
B'. Let c be on an interior confocal ellipse whose minor and major axes 
are z and v — {x^-\- «^)* ; then the number of the circles C with centre c is 
c^—ca^ y-v. And so, if c be any point in the elementary area \ird {vz) 
contained between two consecutive confocal ellipses, the number of circles 
C' whose centres are in that area is iir (.v — v) d {vz). This integrated from 
« = to « = (y2— a;2)* gives us (C%y. Now we have 

[(y-'V)d {vz) = (y — v) vz + J vzdv^ 
which between the limits, and since vdv =^ zdz, becomes simply \z^ ; that is, 

"We may include all the circles B', corresponding to x and y, by multiplying 
this by the area contained between the circle whose centre is a with radius 
X and the consecutive concentric circle whose area =rf(ira;'). Hence, 

between the proper limits, (C) = Jir^ ^^ (y^ -x^)^xdx dy. 

Taking unity as the radius of A, x < \, and y > do and < 1— a?, that is, 

{^')^is'^^^^^' (y^-x^^xdxdy (o), 

the y integration is to be first effected. After some reduction and a 
substitution 1 — 2x=u^, we get, after the second integration, (C) =ir2/7-l80. 
It is, of course, inmiaterial which integration is performed first. If we 
commence with the x integration, the integral (a) is transformed into 

VOL. XLIX. H 



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C2 



The first doable integral embraces all circlos B' for wbicb the radins is 
less than i ; the second, those having a radius greater than 4 ; the separate 
values are easily found to be 



180 2« 180 \ 7 2« y 



180' 



180 

But (a) is most easily integrated by taking two new variables such that 
1-^2 «a? + y, ij-v/2 « y— x. Thus we have, instead of (o), 

(CO =*«»£'""' J* ({!,»- {«,4)^{rf, = t-r=£'''^*(f-«i''Q. 

giving immediately the result as before, viz., n^ 17 'ISO, 
Finally the number of C*s within a B of radius y is 

2ir J r rfr (y — r) = Jiry* ; thus we have 

Jo Jo Je Jo 

This gives the required chance = — . 

[Assuming that in each case 
the position of the centre is first 
taken, and then the radins, within 
the proper limits, Mr. Biddlb 
proceeds to find the required proba- 
bility as follows : — 

** Let O, Q, V be the centres of 
A, B, C respectively ; let OD = 1, 
OQ«ar, QT(-QE)-y, 
QS(=QF)««. 
Then we readily discover the follow- 
ing limits for success : — 

Bero<a;<J, a;<y<(l— a?), 
zero<«< J (x + y). 
Make OS = ST = EF - y-«, 
and join QV (« QS) ; also join OV, and let Z OQV = ^. Then we have 

the further limits, zero< *< SQO ( - cos-' ^'^^-^^''^^ ^ 

\ 2xz J 

and OV < « (s rad. of C) < (y — «), which may be rendered 

(x« + «'- 2a« cos 4»)* < w < (y - «).** 

By this process he arrives at an integral very troublesome to evaluate.] 




7986. (J- Brill, B.A.)— ABCD is a quadrilateral, AB and DO when 
produced meet in E, and AD and BO when produced meet in F ; prove 
that AB . CE . DF cos (ABD + CEF + OAF) 

+ AD . OF . BE cos ( ADB + CFE + OAE) 
- BO . AF . DE cos (OFE + ADB + DCA) 
-OD.AE.BFcos(OEF + ABD + BCA) « AO.BD.EF. 



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63 

Solution by A. Gobdon ; Professor Nash, M.A. ; and othert. 

Let t, y, k be unit vectors along 0-c, Oy, Oz, and a, /3, y, J, &c. unit 
yeotoTS in the plane of ary, such that 

a a t cos a +j sin a (inclined at a to Oj;, &c.), j3 » t cos fi +J sin fi, &o. ; 
then 0/3 = cos o - i3) + A: sin (o— jS), &c., 

alfi.jl^ « cos(o + 7-i8 — 5) r A; sin (0 + 7-^-8) — a/5. 7/^8; 
hence, if 

oi = /i/8 + ^7 + y + ..., a,r=i»ia + »«i7 + «»88 + ..., OJ«f•li3 + «^7 + '»^'^ •» 
and 0, ^, 4^, ai, a,, 09, jS, 7, 8 ... are all vectors in the plane of ^y, we 
have aje . 02/4» . ag/if^ = «i/<^ . «4/e . oj/if^ - &c. - ^Jli;" iS/fl . y/<p . 8/if^, 

where 2 constitutes a summation of terms each formed of the product 
of 3 quaternions, such as li$l$ . fn^yl<t> . nj^l^ (one from each vector), 
(the three numerators being the same in none of the products). 
Let large letters denote vectorSf small letters lengths, so that 
AB » — BA, but ab ^ ba; then 
, _ AF -AE AF AC_AE AO 
EF " EF * AO EF * AO 
AF.AC-hAF.AE-AE.AF-AE.AC 
EF.AC 
_ AE.CF~AF.CE AE /BF-BC\ AF /CD + DEX 
** EF.AO ""acV EF j AOV EF / 

AE.BF.CD-hCE.DF.BA-i-FO.BE.AD + FA.DE.BO . , 
" AC.EF.BD * 

, ^ BA CE DF AD BE FC DE FA BO AE CD BF^ 
* BD * EF * AO "^ BD * AC * EF ^ AO ' BD * EF BD ' EF ' AO ' 
Also, by Ptolemy's theorem, CD . BE . AF = CE . BA . DF, 
and BO . AE . DF = FO . BE . AD ; 

therefore 1 - ^ . ff . ^' {cos (ABD + OEF + FAC) 

^ ""-^ ""^ +Asin(ABD + CEF + FAO)}+.... 
Hence bd . ef. ae^^ba.ec. df. cos (ABD + OEF + FAC), 

the result required, 
and =« J^a.^^J.^C/'sinCABD + OEF + FAC). 



8771. (W. J. G&BENSTBEET, B.A.) — Provc that the Beriae 
U g sina l^i -I- J^sin^a + J^sin^a + ^]'^^'^^^ Bin«a + ...| » tan^a. 

Solution by Prof. Ionacio Bey ens ; B. Enowles, B.A ; and others, 

ff «cosa[i+ L3 8in»«+ 144 sin^«+...] 
da C 2.^ 2.4.6 > 

= -r-~ { (1 - sin^ a)-* - 1} = J sec' ia ; therefore m = tan |a. 



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9327. (F. R. J- Hbkvby.) — The point O is fixed, P describes a 
straight line A ; OP and a line T passing through P rotate uniformly (in 
the same or in contrary senses) with angular velocities as 1 : 3, and be- 
come simultaneously perpendicular (or, in the limiting position, parallel) 
to A. Show that the envelope of T is a cardioid. 



Solution bf/B^F. Davis, M.A. ; G. G. Sto&b, M.A.^ and others. 

Let the circle (centre R) 
roll upon an equal circle 
(centre C), U being the 
point of contact and CURV 
the common diameter 
through U. Then, if Q be 
an invariable point on the 
moving circle, QV and QU 
are the tangent and normal 
respectively to the cardioid 
described by Q, whose 
cusp is at E, the point 
passed over by Q when in 
contact with the fixed circle. 
Since arc UE » arc UQ, 
angle 

UCE = UEQ = 2UVQ ; 
so that, if CP bisect the angle UCE, CP = PV. Draw PN, PM perpen- 
dicular to CE, CR respectively, and take CO = CV = constant. Then the 
angles OPN, CPN, CPM, VPM are all equal, and VP may be conceived 
as having revolved from PN through an angle three times as great as that 
revolved through by OP from PN in the opposite direction. 




8737. (Professor Mukopadhyat, M.A., F.R.8.E. — Extension of 
Qut^stion 8107.) — If 0, 4>, ^ be the angles of inclination of any two tan- 
gents to a conic, and of their chord of contact, to a directrix, show that, 
if « be the eccentricity of the conic, 

^^ A-^ sing + M-^ sin » .,_1-A».1-m2 

A-*COS6 + fi-*CO8 0' 



e« = 



sin' sin' 



Solution by R. Enowles, B.A. ; Professor Matz, M.A. ; and others. 
Let (x^y ^i), (A, k) be the coordinates of the ends of the chord and its 



pole, then tan ^ = ^, 
b*h 



therefore 



sin'tf* 



coi9 *» + 

a'y; 



*«x, 



COt0 B 



fl2#/2 



b'^ifi^-e^x^)^' 



and, since ^ = 



1-A« 
sin'fl' 



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therefore \ = , and u = -. 

and A-Uine + A*-'8in» _ a«(y, + y,) _aaAr_^^. 



9353. (Professor Asutosh MukhopAdhyAy, M.A., F.R.S.E.)— Points 
D, E are taken in the sides AB, BC of any triangle ABC, such that 
BD = m . DA, BE = #1 . EC. If O be the intersection of AE, DC, prove 

that C0^,n+1 ^^ AO^^l 

OD « OE m 



Solution by R. F. Davis, M.A. ; Professor W. P. Casby, M.A. ; and others. 

The point O is (by hypothesis) the centroid of masses m, 1, n placed at 
A, B, C respectively. Whence, &c. 



9089. (Emile Vioari^.) — Par leg sommets A, B, C d'un triangle on 
m^ne des parall^les aux c6te8 opposes qui rencontrent le cercle circonscrit 
en A', B', C Les droites A'B', A'C, C'B' rencontrent respectivement 
AB, AC, BC en a, jB, y. Demontrer que Torthocentre du triangle atiy 
est le centre du cercle ABC. 

Solution by R. Enowles, B.A. ; Professors Matz, M.A. ; and others. 

The equations to AA', BB', CC and the coordinates of A'B'C are 
respectively by + ez^O, ax + ez='0, ax + by»0; 1/a, [c^ — b'^jaHy 
(b^-c^)la^e; {e^-n^/ab^, l/b, {a^''C^)l{b^c); {f^-a^)lae\ {a^-^b^/b^c, l/c 
(omitting 2a in all the coordinates) ; hence we find the equations to A'B', 
A'C, B'C, and also the coordinates of a/87 • — 

{a^'-<^)la{a^^b^, {(^--b^) I b {a^^b^), 0; 

(^-«-)/a(c3-fl2), 0, {<P-b-)lc{e^-a^); 0, {l^-^a^) j b {l^-<?), 

(a^-c^)lc{b^-'<P); 

therefore the equations to afi and the perpendiculars from 7, fi on ai9, ay are 

a {c^-b^) X + b (c^-a^) y + c (a^^b^) z = 0, 

a [(^2 + ^2) cos A-*<?] a; +* (aS-^j cos Ay + <? (a2-^ cos A? = ....(1), 

a(b^-c^)coaBx + b [(a^-^c^ cos B- ac]y + c{b'^'' a^) cob Bz ^ (2), 

and (1), (2) are each satisfied by RcosA, RcosB, RcosC; hence the 
orthocentre of the triangle a/87 is the centre of the circle ABC. 



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66 

8667. (N'Importb.) — Two equal perfectly elastic balls, moving in 
directions at right angles to each other, impinge, their common normal 
at the instant of impact being inclined at any angle to the directions of 
motion : show that, after impact, the dii^ctions of motion will still be at 
right angles. 

Sohttim by F. R. J. Hbbvbt ; Rev, T, Gallibrs, M.A. ; and others. 

Let OA, OB represent the velocities, either before or after impact, and 
C be mid-point of AB. Then OC, velocity of mass centre, is invariable ; 
and so (elasticity being perfect) is magnitude of relative velocity, or length 
AB. But, if AOB be a right angle, length AB « 2 x length OG ; and 
the converse. 



9360. (B. Curtis, M.A.) — A tetrahedron ABCD is circumscribed to 
an ellipsoid, and straight lines are drawn through the centre from the 
cornel's to the opposite sides meeting them in X, Y, Z, W ; show that 

ox oy oz ow^ 
xa"*"yb zo"*^wd 



Solution by J. O'Btrnb Cb.okb, M.A. 

Let Pj, jp„ Pa, P2i P3, Pit P4, Pa* «i, «2» *i> *4 ^ t^© parallel perpen- 
diculars from the angles of the tetrahedron, and the point O, and the areas 
of the sides on which they respectively fall ; then 

Pi'i +P^3 +i'3*3 +-^4*4 = 3 times vol. of tetrahedron:* Pi»i = P2«2= P8«3= P4«4. 

Therefore '&- + $ + ^+'&==l> whence the result. 

Fi Fa Fa Fi 



8270. (^' Edwardes.) — Let ABC be an acute-angled triangle, and 
L, M, N the points where the angle bisectors meet BC, CA, and AB re- 
spectively. Prove that (1) the circles ALB, ALC cut one another at an 
angle A, the circles ALC, ANC at an angle ±| (0— A), and the circles 
ALC, BN<^) at an angle 90°- JB ; (2) the centres of the pair of circles which 
pass through L are equidistant from the centre of the circle ABC, and 
similarly for the other two pairs ; (3) if p^, p'^ be the radii and Sj^ the 
distance between the centres of the circles wMch pass through L, and 
similarly for p^, p'^, &c., p^ />m Ps = p'l Pu p'n ** 'l^m^n *» (^) ^ ^1 ^® ^® 
distance of the circle ALB (or ALC) from the centre of the circle ABC 
(radius R), and similarly for d^, rf„ R^ - R {d^d^ + d^^ + d^i) — 2d^d^2=^ J 
(5) if the base BC and the circum-circle BAC be given, the envelope of 
the line joining the centres of the circles ALB, ALC is a parabola whose 
focus is at the centre of the given circle and latus rectum 4Rsin3 ^A. 



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Solution by Professor Swaminatha Aiyar, B.A. 

(1.) Let 0, P, Q be the circum-centres of ABC, 
ALC, ALB ; then OQ, OP are at right angles to AB, 
AC respectively, therefore QOP and BAG are 
supplementary, also 

l AQO = ALC ; and I APO = ALB. 
Therefore QOP and QAP are supplementary and 
a circle passes through Q, A, P, 0. Also 
Z QAP = BAG. Again, AP : AQ = AC : AB, 
and aQAP is similar to ABAC, and the straight 
line AO bisects the angle QAP. 




B 



(2) Therefore QO, OP, the chords of the circle QAPO, are equal ; that 
is, P and Q are equidistant from 0. 



(3) 



Ag.AP^^.^j^^j^ P^^Pj,^ 



AB 



AC BC 



Similarly, 



and ?N^?N^?N. 



Therefore 



a e a c 



(4) AO.QP«AQ.PO + AP.QO = (AQ + AP)//i. 



R 



a 



Therefore 



Similarly A = ^^f and -3^ = 
R3 b-k-e e + a a + b 



did^d^ 



b + e 
a 



b 
e + a 
b 
R 
^1 



Therefore 
a + b 
e 



a b e rf, ' ' 



^ ' dz 

(5) If R be the middle point of PQ, OR is at right angles to PQ. 
The distance of R from BC = } the distance of P and Q 

- i (LCcotJA + LBcotiAJ _ j^^^^j^ 

The locus of R is thus a straight line parallel to BC. Therefore PQ 
touches a parabola whose focus is O and whose tangent at the vertex is 
the locus of R. • 

Its latus rectum = 4, distance of O from the locus of R 

= 4(JacotJA^JacotA) = ataniA=: 4R8in«iA. 



8540. (Rev. T. R. Terry, M.A.)— Show that the series 
1 ,ffl g , fn(w+l) q{q + r) ^ w (fn + l) (w-t-2) y (g + r) (y ->• 2r ) ^ 
p 1.2 p{p + r) 1.2.3 j»(P + r)(i? + 2r) 

is convergent if jo > q + mr. 



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68 

Solution by the Proposer ; Professor Nash, M.A, ; and othert. 
Denoting the hyper-geometric series 

«jB^^ ai[«4j]3(^+J)^ 
^1.7 1.2.7(7-»-l) 

by F {a, i8, 7, a;}, it is well known that F {o, jB, 7, 1} is convergent if 

7>o + i8. Now the given series —FJtn, -^, =^, 1 y, and is there- 
fore convergent if i? > g + mr. 



7244. (D. Edwardes.) — The circles of curvature at three points 
of an ellipse meet in a point P on the curve. Prove that (1) the normals 
at these three points meet on the normal drawn at the other extremity of 
the diameter through P ; and (2) the locus of their point of intersection 
for different positions of P is 4 {a^x^-¥bhf^ = {a^^b^f. 



Solution by the Rev. T. C. Simmons, M.A. 
Let a be the eccentric angle of P', the other extremity of the diameter, 
that of P being of course ir + a, and let i3, 7, 8 be the eccentric angles of 
the points of contact of the circles of curvature, then 

a;/<icos^(ir + a + i8)+y/*8ini(ir + a4iB) = cos^ (ir + o+ jB), 

or ir/a8inJ(a + /S)-y/*cosJ(o + /3) =* 8inJ(a-i8) (1), 

the chord joining P with i8, and xja cos ^ + y/i sin jB = 1 (2) ; 

the tangent at i9, must make equal angles with the axes, therefore 

sin J (o + /3) sin /3— cos i (a + jB) cos jB « 0, 
or o + 3iB - (2m+ l)ir. 

Similarly, o + 87 =» (2« + 1) ir, a + 35 = (2^? + 1) » ; 

therefore a + i3 + 748i8an odd multiple of ir, or the normals at a, jB, 7, 8 
are concurrent. 

Again, if a, jB, 7, 8 occur in this order, since their differences must be 
even multiples of ^ir, it is evident that, when they are unequal, iB — 7, 
7—5 each = fir ; in other words, jB, 7, 8 are at the vertices of a maximum 
triangle in the ellipse. Consider now the normals at jB, 7 : they are 

2<w;Bini8-2*ycosi9= {a^-h^%m2fi (3), 

and Ux sin (/3 + fir) - 2by cos (/8 + Jir) - [a^ -lP)%m (2/3 + |ir), 

which reduces to 2ax ( \/3 cos /3 — sin 0) + 2by ( >v/3 sin /8 + cos 0) 

« _(fl2_i2)(8in2/8+ >v/3 cos2iB) (4), 

whence, by (3) + (4), 2aircosiB + 2iysini9 «-(a2-i2)cos2/8 (6). 

Squaring and adding (3) and (5), we obtain for the locus of the inter- 
section of the normals 4 (aV + V^y^) = (a^^h^^ 



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9418. (Professor Sylvestek, P.R S.)— If /?, i^j are each prime num- 
bers, and 1 +j»+j»'+ ... +y-* = ^, prove that j is a divisor of q — i. 
Example : 1 + 3 + 32+ 3=» + 3* = 11*, and 2 is a divisor of 11-6. 



Solution bt/W.8. Foster. 

therefore $>-' +^'-» + ... +y+ 1 « -^— (1 +i?+i^+... +i?*-2), 

a.ndp is a prime number ; hence, as in Question 9381, y divides p—l, or 
J = p and divides q~l. In the first case, p " AJ-^-l; therefore 

1 +(A;+ 1) + {Aj + 1)«+ ... + (A; + l)-i - q^ ; 
therefore i + By = qj ; therefore B;' = q'—q + q—iy 

and, sincey is a prime number, q^—q is divisible, by j\ therefore q — iis 
divisible byy. In the second case, we should have 

l+i^+Z^+.-.-CQ^ + l)"- 1+C,i?« + C,i^+..., 
which is impossible if jn is a prime number. 



9423. (Professor Neubeho). — On casse, au hasard, deux barres de 
longueurs a et ^, chacune en deux morceaux. Quelle eat la probabilite 
qu'un morceau de la premiere barre et un morceau de la seconde, 4tant 
juxtaposes, donnent une longueur moindre que c ? 



Solutions by (1) Professor De Wachteh, (2) Professor Schoutb. 

1. Assume two rectangular axes, OX, 
OY ; take OA = a, and OB = b, and 
describe the rectangle OAMB. From 
any interior point if perpendiculars be 
drawn to OA and OB, they may repre- 
sent, in one of the possible combinations, 
the parts of a and b to be added 
together. 

The amount of possible chances will 
be represented by the area OAMB and 
measured by ab. On OX and OY, re- 
spectively, make OC = OC = given 
length c, and join CC. The sirni of the 
distances of any point in CC from the axes is = c. First of all, we must 
have a + b > c^ if CC is to cut o£E a part from OAMB. The required 
probability P is the ratio of the area limited by the axes to the rectangle 
OAMB. Draw AA.' and BB' parallel to CC Assuming a + b > c and 
b > a, three cases are possible. 

(1) b > a > c. CC falls between AA' and 0, and P — c"l2ab. 

{2) b > a. CC lies between AA' and BB', and P = {2c -a' 12b. 

{^) b > a, CC falls between M and BB', and P = 1 - (a + A - cY[2ab. 

TOL. XLIX. I 



c 


\ 




















8 


'\ M 














\. N 




C 


S X 


\ 


A 


\^^ ^\ 


^N 






, \ 






a \ 


C 


\^ '^^ 


^\ 




\ \. ' 


V 






\ \ 







X V 


C A 


C\ C 



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2. Otherwise : — Si-' Von repr^ente les longueurs des deux morceaox 
qu*on reunit par x et y, on a lea conditions generales < 2^; < a et 
a < 2y < b, tandis que les cas favorables sont soumis a la troisieme con- 
dition x-^y < c. Ea considerant x Qi y comme les coordonnees rect- 
angulaires d*un point dans le plan, on trouve pour la probabilite en 
question d^aprds les figures suivantes : 



k. 




pour c <\b (Fig. 1) P = 4r/2fl*, 

pour \b<e< Ja(Fig. 2) P = [4c2-(2c-*)2]/2a*, 

pour \a< c< J(a + *)(Fig.3) P = [4c2-(2(?-ft)=i-(2c-a)2]/2a*, 

pour i(«+*) < ^^CFig- 4) P = 1. 

On a suppose a > b\ quand a = i, le cas de Fig. 2 disparaJt. 



9406. (W. J. Barton, M.A.)-Show that, if R = 2r, the triangle is 
equilateral, without employing the expression for the distance between 
the centres. 

Solution by Professor Emmerich, Ph.D. 
If R « 2r, the in circle is equal to the nine-point circle. After 
Feuerbach*s theorem, the incircle of each triangle is touched by the nine- 
point circle, and it may easily be seen that the incircle lies entirely inside 
the nine-point circle ; for, if the bisector AD of the angle A meets the 
circumcircle at E, the incentre T lies between A and E ; therefore the 
projection of the point T upon BC lies between the projections of the 
j)oint8 A, E ; that is to say, the first projection, which is a point of the 
incircle, lies inside the nine-pcint circle. Therefore, if R = 2r, the two 
circles coincide. Hence the sides are touched in their mid-points by the 
incircle, etc. 



1898 & 4043. (Hugh MacColl, B. A.)— Find the number 
situation of the real roots, giving a near approximation to each, of 
a^ + 4-37162a:3-24-S64235876U2+ 34-129226840859882^ 

-14-63442007818570452204 = 0. 



and 



Solution by D. Biddle. 

The ordinary 7 -figure Tables of Logarithms do not permit of a near 

approach to accuracy in this investigation. But we can proceed thus : — 

Rendering the equation x^-^-ax^—bx'^-^cx — d'^ (I), 



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form another equation thus : {x^ + yx— s){a^—uz + r) =- (2), 

in which y— « = a, uy-i-z—v « i, uz-\-vy= <?, and vz = rf, whence v = iilz^ 
y = {az^-¥cz)l{^-\-d), and m = (cz—ad)/{z^ + d). Then find any one real 
root of X from the original equation. A very good method of effecting 
this, in equations like the present, is to separately record the portions of 
X as they are found, and amend the coefficients «, A, c, rf, so as to form an 
equation similar in kind, hut of which the remainder of x is the unknown. 
Thus, let ArtS the last portion of x found, consisting hy preference 
of only one figiire, and rf„+i s the error resulting from it; also let A = 
the addition necessary to raise (Ai + Aj + ... A„) to x. Then 

A* + a.Ai-*„Ai + rnA,-rf„ = -e/„,i (3), 

(A„ + A)4 + a„(An + A)3-*„(A^ + >i)2 + ^(AH + A)-rf„«0 (4); 

expanding (4) and suhtracting from it (3), we have 
A*+ (a„ + 4A,.) A3-(*,.-6Ai-3A«fl„) A« 

+ (c + 4A?, + 3A?.«„-2A„*„) A-^„*, = (5), 

which may he rendered A^ + fl'n+iA'— *»^i A2 + e?,»+i A — (f„4i = (6). 

This is a condensed form of the old method of extracting roots of the 

fourth power, and is scarcely to he surpassed for equations such as the 
present untU full logarithmic tahles to 24 places of decimals are provided. 

A a b 

1-000,000,0 4-371,620 24-964,235,876,1 

-100,000,0 8-371,620 6-849,375,876,1 

-100,000,0 8-771,620 3-277,889,876,1 

-010,000,0 9-171,620 0-586,403,876,1 

•006,000,0 9-211,620 0-310,655,276,1 

•001,000,0 9-235,620 0-144,630,116,1 

•000,900,0 9^239,620 0^116,929,256,1 

•000,080,0 9-243,220 0-091,981,472,1 

-000,001,0 9-243,540 0089,763,060,9 

-000,000,9 9-243,544 0-089,735,338,4 

1-217,981,9 = oTj 

d 

34-129,226,840,859,882 14-634,420,078,185,704,522,04 

1-315,615,088,659,882 0-097,809,113,425,822,522,04 

0-400,888,513,439,882 0016,269,743,320,834,322,04 

0-012,459,138,219,882 0-000,088,170,737,846,122,04 

0-003,486,546,697,882 0-000,013,038,123,257,302,04 

0-001,617,538,344,682 0-000,001,311,427,089,610,04 

0-001,355,988,972,482 0-000,001,130,953,285,928,04 

0^001, 147,764,455, 162 0^000,000,093,452,567,496,24 

O^OOl, 133,244,891,498 0^000,000,002,215,359,935,12 

0-001,133,065,393,107 0-000,000,001,082,204,797,44 

The final remainder = 0-000,000,000,062,445,936,90 

Having thus found that Xi =« 1*2179819, we revert to (2), whence we oh- 

tain x^ + yx-z ^ 0, and x^-^tix + v ^0 (7,8). 

Expressing (7) in the terms given under (2), we have 

z^-(ax + x^ z'^''{cX''d)Z'^dx^ =^ (9), 



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or s*-6-8080339445656l23-26*93434047497581s 

-21-70986819346878 « (10), 

which has two real roots, namely «i « 9*7874906, zj = - 1-4846695. 

iVo«(7),(8). .-=.f..}("^)'j'-i^fi^) (.1). 

"-b('-^:y-~M'-^) (->• 

Taking «i, we obtain, from (11), ^i = 1-2179819 and r, = -8-0352616, 
and from (12) the two imaginary roots l-2227875±(- -0000075)'. 
Taking z^, we obtain from (II) two more imaginary roots, 

l-2l84609db(--0000225)*, 
and from (12), X2 — -803.32616 and ^3 « 1-2266901, which is quite dis- 
tinct from Xi. 

For the service of those who may wish to carry the investigation to a 
great degree of nicety, the logarithms of a, ^, <;, d are here given to 24 
placet of decimals, by aid of Peter Gray's admirable method : — 
log a - 0-640612401175296337879050, 
log* - 1-397318277386039907465102, 
loge; - 1-533126449938424905356617, 
logef - 1- 1653755 17215043330021209. 



9392. (Professor Gbnese, M.A.) — If the tangent at any point P of a 
folium of Descartes meet the tang nts at the node in X, Y, and the curve 

11 3 

again at Q, then prove that p^ + p y * pH- 



Solution by Professor Wolstenholme, M.A., Sc.D. 

The equation of the curve being a^* + y* « oay, any point P may be 

taken, x « — -, y « - — - ; and if any transversal px + gt/=^a meet this 
1 + ^ 1 + ^ 

in three points, the values of t at the three points will be given by 

pt-^qfi ^ 1 + t^i if t^t^^i be the three ^1^3^ = — 1. Hence, if ^, ^ be the 

values at P, U, iH' ■» — 1 ; also the equation of the tangent at P will be 

X {2i -'t*)-¥y (2^'- 1) — aflf and if be the origin, and PR the harmonic 

mean between PX, PY, the equation of OR will be y = — tx. Now, the 

coordinates of a point dividing PQ in the ratio m : / will be in the ratio 

m mn , It mt' It* mt^ , U mt* , 

l + <a 1 + <'8'1-H^3 i + t'i* ' i + ^'*"^«_i • 1 + ^3 ^_i' 

which = t {/(^-l) + m} : /(^»-l)-»t^»; if m = 2/', this ratio is -t : 1. 
Hence PR = ,PQ, and j^ . ^i^ = X . ^^. 

The pedal of the parabola y^ ^ iux with respect to the point ( — 3ff, 0) is 



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an orthogonal projection of the folium, and hence the property will be 
true for this pedal, hut I find on trial it is not true for any other pedal of 
the parabola. In this pedal, the circular points are inflexions ; as, in the 
folium, the three points at infinity are inflexions. 



8503. (N*Importe.)— A rod of length flr\/2 rests in equilibrium 
in a vertical plane within a rough sphere of radius a, one extremity 
of the rod being at the lowest point of the sphere ; show that the 
coefficient of friction is >v/2 — l. 



Solution by Gboboe Goldtuokpb 
Storr, M.A. 

Let AB be the rod, O the centre of the 
sphere, /* the coefficient of friction, and W 
the weight of the rod ; then, resolving ver- 
tically and horizontally, and taking moments 
about B, we have 

/iRa + Ila=Wja, 
whence /*=» v'2— 1 = -4141 nearly. 




9436. (W. Gallatly, M.A.) — AB is a mirror swinging on a hinge 
at A . At C is a candle flame, and at D an observer ; the line ACD being 
perpendicular to the axis of the mirror. Find geometrically the position 
of the mirror, when the observer at D sees the image of the flame on the 
point of disappearing. 



--.£ 



Solution by Professor Schoutb. 

Let the ray emitted by C, that would 
be observed at D when the mirror in the 
arbitrary position AB was long enough, fall 
on the produced mirror in E. Then AE is 
the external bisector of angle E of triangle 
CDE. Therefore EC/ED = AC/AD. This 
proves that the locus of E is the circum- 
circle of the triangle AEF, EF being the 
internal bisector of angle E. So the two 
limiting positions of the mirror are given by 
the joins of A with either of the points P, Q 
common to this circle and the circle with A as centre and AB as radius. 




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6911. (W. R. Westropp Roberts, M.A.)— Let H and H' be the 
Hessians of two binary cubics respectively, B their intermediate co- 
yariant ; then, using the notation ot Salmon, proye that 
9e'-36HH'-6PJ + H(6J). 



Solution by D. Edwardbs. 

"With the notation of Salmon's Higher Algebra^ Art. 216, the sources 
of'H, H', e, H (6 J) are respectively, 

flkJ-62, aV-*'2, ac'-\-ca'-2hb', Z&ia^^-^a^), 
Now 4 (ac-l^) (aV-*'2)-(a<?' + aV-2W)2 

=« 4 {be') {ab')^{eay « 4 K-JP) Oo-W ; 
hence, for the corresponding covariants, we have the stated result. 



9369. (W. J. C. Sharp, M.A.) — Prove, from the theory of com- 
binations, (1) that i+i?L.iL + '!Li!?iz:Il.!Ll«=ii) + ...=.fi^^ 
^ ' 11 1.2 1.2 mini 

must be true ; and (2) deduce that, if (m) be a prime greater than («), 
(m + «) ! — w! nl and ^ — ^ are respective multiples of (w^), (i/i). 



Solution by Professor Iqnacio Beyens. 

1. Designant par Cm le nombre des combinaisons de {m) lettres prises (») 

^ («), on aura d'abord ^^±^' == C«+» et aussi, 
ml nl 

Cm + » = Cm . C + Cm • C» + Cm . C/* + . . . + Cm CJi, 

mais Cr* = Cl, Cr* - Cn, 

done 1+ ^ . ^ + ^i!^i^.^i!!:i±) + ... «fei±^!, 

111.2 1.2 ml nl 

2. De cette Equation on deduira 

(m n . m(m—l) «(w— 1) . \ . i /_ . m t i 
-~ . — + —^ — -—!■ . — ^ — p-^+ ... ) m! ft! = (m + «)! — *n! «!, 
1 1 1.^ 1 , Z J 

et si (m) est un nombre premier plus grand que («) est Evident que le 
premier membre est toujours multiple de m ,m «= w^, done 

(m + «) !-m ! n ! « M . (m^) ; 
et de la mSme relation on a 

,uifl + ^ M_ »«(m--l) «_(n-l) \ (m + w)2 
■ V 1*1 1.2*1.2 •*; nl ' 

done -^ — J — * est aussi multiple de m. 



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9149. (Charlotte A. Scott, B.Sc.) — If ABCD be a quadrilateral, 
in which the sides BA, CD meet towards A and D in H, and the sides 
BC, AD meet towards C and D in K ; and if from a point L in HK, 
LAG, LFC be drawn meeting BC in G and AD in F, respectively ; show 
that BF and GD meet in HK. 



Solution by D. 0. S. Davies, M.A. ; 
G. G. MoRKiCE, M.A. ; and others. 

The triangles ABG, CDF are co- 
planar. Hence the diagonals of the 
quadrilateral ABCD, AGCF pass 
through same point. Therefore the 
triaugles AGD, CFB are copolar, and 
therefore coplanar. Therefore GD, 
BF intersect in HK. 




9414. (R- W. D. Christie.) — If 2^—1 is a prime, show that p is also 
prime. [Better thus : — What prime p will make 2'' — 1 a prime ?] 



Solution by Professor Ionacio Be yens ; D. Watson ; 

Si ( p) ne f usse pas nombre premier, supposons p = np' 
2P-1 = 2"»»'-l « multiple de (2»»-l) = multiple (2?'- 
serait pas de nombre premier, ce qui est contraire k V 
Be YEN 8 adds the following generalization : — * * Si (« + 1 )p — 
premier, Texposant {p) est aussi un nombre premier, parce 
pas de nombre premier, soit p = mp' ; alors (» + !)''—«'' = 
serait un multiple de (m + 1j»^-«p') et de (w+ l)"*— «"».'* 
remarks that the solution shows that p is a prime is a 
tion, but not that it is a sufficient one, which is the real 
feet numbers.] 



and others. 

; alors on aurait 
1), et 2P-1 ne 
enonce. [Prof. 
nP est un nombre 
que si {p) n'etait 
: (w+l)'"P'-w»«P' 
Mr. Christie 
necessary condi- 
problem in per- 



9164. (Professor Nilkantha Sarkar, M.A.)— Prove that 

— c« co«* Bin (e sin x) sin nx dx « — -. 
» Jo « ! 



Let 



Solution by D. Edwardes. 
Sf = 1 + c cos a: + — cos 2a: + —^ cos 2x + &c., 



c^ -:„ 



S, = csina;+ -— sin2a; + &c., and^* — — 1. 

Then So+^'S, = €'^(co8*+> .inx) „ ^ccos* |cos (c sina:) +ysin {c %\nx)} ; 
therefore S, = c*" «<>•' sin {c sin x) . 



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Hence, substituting the series for €*«»•' sin (tfsin ^), the integral reduces 
to — . — ^ sin' nxdx ^ — -. 






9427. (Professor Genbsb, M.A.)—If A, B, C, D be points in a plane, 

prove that HC.AD ^ CA /BD ^ AB.CD 

^ 8m(BAC-BD0) 8in(CBA-CDA) 8in(ACB-ADB)' 

where any angle BAC means the angle through which AC must be 
turned in the positive sense to coincide with AB. 



Solution by Professors W. P. Casbt, Matz ; and others. 

Describe a circle about ABD, produce ^^ "2:^^ 

AC to H, and join DH, BH. 

Then Z HDC - I BAC - I BDC, 
and Z CBH « Z ACB - Z ADB, 

but CD/CH = sin DHC/sin HDC, 
and CH/BC =« sin CBH / sin CHB ; 
thus 

CD/BC « sin DHC sin CBH/sin HDC . sin CHB, 
but sin DHC/sin CHB - AD/ AB, 
therefore CD . AB / BC . AD - sin CBH/sin HDC 

= sin (ACB- ADB)/sin (BAC -BDC). 
Again, Jiake Z CDR=» Z CBA, and therefore Z ADR= Z CBA- Z CDA. 
Produce BC, DC to V and S. Join VS, SB, and AR, and from the 
similar triangles of this figure, after a little reduction, we get 

BC . AD/AC . BD - AD . CD/CH . AR = sin HDC /sin ADR, 
or BC . AD/sin (BAC-BDC) = AC . BD/ sin (CBA - CDA). 




9391. (Professor Satis Chandra Ray, M.A.) — If the diagonals of a 
cyclic quadrilateral ABCD intersect in ; and if AB = «, BC — = *, 
CD = <?, DA « rf, Z AOD = ADB ; prove that 

{be + arf) {ed + ab)/ (ac + bd) ^ a*. 



Solution by J. Young, M.A. ; G. G. Storr, M.A. ; and others^ 

For all cyclic quadrilaterals — - — =» ■ « -— , 

AC Du AO 



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and in this case d^AO; hence (^<^+f^{f+'^^) « , 

AC . BD 
and AC.BD - ac + bd. 



9380. (Sarah Marks, B.Sc.) — Tangents are drawn to a parabola 
from a point T ; a third tangent meets these in MN ; prove that the polar 
of the mid -point of MN and the diameter through T meet on the 
parabola. 



Solution by C. E. Williams, M.A. ; 
R. Knowlbs, B.A. ; and others. 

The diameter through T is TQV, QO 
the tangent at Q, meeting MN at ; Mm, 
L/, Oo, Nn diameters; then w, «, o, V, 
are the mid-points of P/, P7, rV, PFj 
therefore o, are the mid-points of #w«, 
MN ; hence the polar of O, the mid-point 
of MN, is QL. 




8826. (Professor Sircom, M.A. Suggested by Question 2845.)- 



Show that l+-|.a;2+|^ 



6 3.6.7 



x{l-x-)^ 



Solution by the Proposer ; Professor Chakravabti, M. A. ; and others. 
The differential equation satisfied by the given series is found by the 

usual methods to be (jp— a:^) dy/dx + ( 1 - 2x^) y « 1 , 

of which the solution is ar (1 — a:^)* y = sin-^ a; + C, 

which is satisfied by the given series if C =» 0. 



9384, (Professor Bordaob.)— Show that the roots of the equation 
(a; + 2)«+2(x + 2)A/aj-2a;-3>v/ar-46 = 0are9, 4, | {-.13±3(-3^)}. 



Solution by Eleanor Bobinson ; G. G. Storr, M.A. ; and others. 
This is (a; + a;* + 2)2- 3 (a: + a;* + 2) « 40, whence a; + a;* + 2 — 8 or —5, 
a;* =-3, +2, i{-l ±3 (-3;*}, and ;p « 9, 4, J{-13 =F 3 (-3)*}. 

VOL. XLIX. K 



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9371. (J. Brill, M. A.)— Prove that in any triangle, n being a positive 
integer, a* cos «B + b** cos « A 

« c» - nabc^ -2 cos (A- B) + li|zi) aSJV -* cos 2 (A - B) 

-^(^-^H^-^)a3y^-6co83(A~B) 
3! 

H.^(^-^)^^7^'(^"^)a4^V-Scos4(A-B)-&c. 

Solution by Professor Sircom, M.A. ; H. Fortey, M.A. ; and others. 
We have a cos B + * cos A = <?, a sin B— i sin A « 0, whence 

Now the sum of the «*^ powers of the roots of x^^px + q = (Tod- 
hunter's Theory of Equations, p. 182) is 

and fle'B, be-'^ are roots of a?2—ca: + ai^-*(A^-®) = 0, a^-*^, ic*^ are roots 
of ar^— fa? + a3e^(^-B) «= 0, whence substituting for p and y from each of 
these equations, and adding, we obtain the required result. 

[The same method applies to Quest. 8290, which is otherwise solved on 
p. 96 of Vol. XLV.] 



9319 ft 9364. (Professor Bhattacharyya.)— (9319.) Show that 
(2w+l)(2m-t-3) ...(2m-t-2r-l) (2m + l)(2m-t-3) ... (2f» + 2r~3) 2/1-1 
r\ {r-'\)\ ' \ 

(2m+l)(2m4-3)... (2m + 2r-6) (2«~l)(2n + l) 
(r-2)! • 2! 

^ (m + « + r-2}_! 2r. 
(w + «— l)!r! 

(9364.) (W. J. Grebnstreet, B.A.) — If q is any positive integer, 
prove that -JL ^ i ^^^ to-J) ^^^ (g"l)(y-2)(y-3)^ ^ 
^ q-¥l ^ 2! ^ 4! 

Solution by R. F. Davis, M.A. ; W. J. Barton, M.A. ; and others, 

(9319.) This identity follows from equating the coefficients of a;** in 
the expansion of (1— a;) to the power — (m + «) and in the product of the 
expansions of the same binomial to the powers — \ (2m > 1) and 
-i(2«-l). 

(9364.) This identity follows from the fact that the sum of the odd 
coefficients in the expansion of (1 + x) to the power (g + 1) = i . 2« ♦> = 2'. 



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9325. (S. Tbbay, B.A.)— a, B, C are the dihedral angles at the base 
of a tetrahedron ; X, Y, Z the respective oppofijtes ; show that, if 

Tj « (1 - C082 B - cos^ C - C082 X - 2 COS B cos C cos X)*, 
with similar expressions (denoted by Tj, T3, T4) for the other solid angles, 

TaTaCosX + TaTiCosY + TjTjCosZ = 1 -cos^A-cos'B -cos^C 
— cos B cos C cos X — cos C cos A cos Y— cos AcosB cosZ + cos Xcos YcosZ. 



Solution by D. Edwardes ; Prince de Polignac ; and others. 
Denoting the areas of the faces by P, Q, R, S, we have by projection 

the equations P cos A + Q cos B + R cos C — S — 0, 

-P + QcosZ + RcosY + ScosA = 0, PcosZ— Q + RcosX + ScosB«0, 
PcosY + QcosX-R + ScosC = 0, 



therefore 



cos A, cos B, cos C 
cosZ, —1, cosX 
cosY, cosX, —1 



= S 



Therefore, squaring, P^ | &c. | ' = 
metrical determinant 

— 1, cos A, cosB, cosC 
cos A, —1, cosZ, cosY 
cos B, cos Z, — I , cos X 
cosO, cosY, cosX, —1 

— 1, cos B, cosC 
cosB, —1, cosX 
cos C, cos X, — 1 

P2 



. S2 I &c. 



1, cosB, cosC 
- cos B, — 1 , cos X 
— cosC, cosX, —1 
I 2. But we have the sym- 

cosA, cosB, cosC ' 
cosZ, —1, cosX 
cos Y, cos X, — 1 



therefore 



— 1, cosZ, cosY 
cosZ, —1, cosX 
cos Y, cos X, — 1 

S2 



— 1, C9sZ, cos Y 
cos Z, — 1 , cos X 
cos Y, cos X, — 1 



1, cosB, cosC 
-cosB, —1, cosX 

- cos C, cos X, — 1 

t.(f., P/Ti«S/T, where T-l-co82X-cos2Y-co8-Z-2co8Xco8YcosZ; 
therefore P/Tj = Q/ Tj = R/T3 = S/T. 

Now, from above, PQ cos Z + PR cos Y « P^ - SP cos A, 
and two similar equations. Adding these, 

2(PQcosZ + QRcosX + RPcosY)« P«+Q2 + R2-SS 

and substituting for P, Q, R, S the quantities Tj, T2, &c., to which they 
are proportional, and reducing on right side, we have the required result. 

[If we form similar relations for the other three faces, and add all four 
together, we obtain 

T2T3 cos X + T3T1 cos Y + TjTj cos Z + T1T4 cos A + T2T4 cos B + T3T4 cos 
« 2-cos"A— cos^B-cosSC— cos^X— cos^Y-cos^Z-cosB cosC cosX 

— cosC cos A cos Y— cos A cosB cosZ—cosX cosY cosZ.] 



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80 

9200. (Professor Nbubbbo.) — On casse, au hasard, une barre, de 
longueur Sa, en trois morceaux. Demontrer que la probabilite que le 
produit des longueurs de deux quelconques des morceaux soit moindre 
que a* est : 

I lofiTe tt (3 + ^6)] + 2 - -v/6 = 0123 (k tr^s-peu pr^). 



Solution by Professor F. X. De Wachtbr. 

Dans le triangle Equilateral de hauteur 3a, deter- 
minons le lieu des points dont les distances u deux 
des c6tes ont pour produit a'. Ce lieu se com- 
pose de trois arcs byperboliques ayant le centre du 
triangle pour sommet commun et les deux c6tes / 

correspondants pour asymptotes. II est aise de / 
voir que I'espace favorable h Pev^nement consid^re /^,/^ 
se compose de 3 quadrilat^res mixtilignes egaux, L.i.. 
situ^ dans les coins du triangle. Done la pro- 
babilite cberch^e a la valeur donn^. 



5440. (^' Rawson.) — ^Prove that the general solution of the equation 

dx^ i Xi \dx J dx^]dxdx^ x^ \dx ) 
t xi \dx I dx^ ) dx Xi \dxj 

where o, )8, Xi are given functions of x, and 

i^dx dx ) 

M = ^3 — ' f — €'i*(*)-^*(*) . <t> {ay^*^ip' (a) 
dx i^dx 



-^ €'.♦(- 



dx 



Solution by the Proposer. 

Let o, $f Xi be any functions of ar, and u = 1 €*»*<*) }l/{z)dz + e (2) ; 

then, differentiating (2) with respect to (x)y (Todhuntbr's Int. Calc.,^. 198) 

dx j^ ax ax ax 



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81 



or 



S = S^t *"*^'^ • ^ W ^ W ^ + N (3), 

where N =» ^ €*.♦(•) 4,(0)- ^c'l^W.V'CiS) (4). 

Differentiating (3) with respect to x, we have 



where M - | . £ [^.« (•). + (,)]- g . £ [.-.♦'») . .^ {0)] 



(6) 



= gi{g •'■♦'•' ^ («) + (»)- g .»■♦ W. ^(J8) ♦(j8) j (6). 



Ffom (3) and {6) we obtain 
rfN 






.(7). 

Integrating the part affected hy the integral sign in (7) by parts in the 
usual way, we have 

r.'.*w . ^ (»)'+(«) <fe - r t^iM I [,-.♦(.)] & 
if }$ 'f* («) *« 

. L- r*-^^ |2^(.) ♦ (.)+ 1^^ - f('y%if)fmdz 

Xidxildx dx Xi x^Jfi ^'(2) C 0' W j 

(8); 

where L - JL(^"^^'^» f (f f H - ^"^^^^ • f (f ^ (^) ] (9). 

Substitute the value of (8) in equation (7) ; then 

dh* I q dxi dx^ I dxi \ du \ 

dx^ \ x^dx dx^l dxJdx j 

^(2^1-^/f^ON^^ + ^ + M+f^VL I (10). 

^ \Xidx dx^ldx) dx dx \dx I f ^ ' 

x,\dx)}/ 1>'(z)V^^ 9'(z) i ) 

In equation (10) take 



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82 



then we have ^(«) = i^j e -<*»♦('>. 0(z)^>0'(») (12). 

Obsenring the conditional equation (11), equation (10) becomes 

dx^ \ x^ dx dx^l dxidx x^\dx I ^ 

\ Xi dx dx^ I dx I dx dx Xi\dx I \dx I J 

(13). 

The integ^l of equation (13) leads therefore, obserying the equation 
(12), to the results in the Question, the constants M, N being as there 
given, and 

L = £a. {€*>*(•) -<^>W .0 (o)^« **-€'»♦('») -'^'♦('') . 0()8)^t+2j. 

[By assuming the particular values, arj = — a*a?*», <? = 0, <?2 = 1, 
nc^ = - 1, e^n — — «— 1, ^ {z) = z-**, a == oc , jS = a*:c*, we obtain 
herefrom a solution of Question 5100 ; see Vol. xxviu, p. 76]. 



8020. (Asparagus.) — ^A conic circumscribes a given triangle ABC 
and one focus lies on BC ; prove that the envelop of the corresponding 
directrix is a conic with respect to which A is the pole of BC ; and, if A 
be a right angle, the envelop is the parabola whose focus is A and direc- 
trix BC. [If (0, 0), (a, 6), («, — <j) are the coordinates of A, B, C, the 
equation of the envelop will be 

Abc{be-a^ x^^^a (* + r) {be-a^) xy ■¥a'^[^a^-¥ {b^cy]y^ 

+ a2(i + <j)2(2aa;-a2) =0.] 



Solution by K. Lachlan, M.A. 

1. Let px + qy + rz = 0, be the equation in areal coordinates of the 
directrix of a conic which peisses through the angular points of the triangle 
of reference ABC. If S be the corresponding focus, and e the eccen- 
tricity, we have at once SA « ep, SB — eq, SC = er. 

2. If S lie on a circle, we have /. SA^ + w.Sfis + w .SC^ « k. 

. And, if/+m + » «• 0, then S lies on a straight line ; thus we shall have 
Ip^ + tnq^ + nr^^ kje^. 



3. 


Again, S, A 


,B,C 


being 


0, 


1, 1, 


1, 


1 


1, 


0, SA^, 


SB', 


SC^ 


1, 


SA', 0, 


AB', 


AC- 


1, 


SB', AB2, 


0, 


BC 


1, 


SC", ACS 


BC, 






-0; 



0, 


l/»», 1, 


1, 


l/«», 


0, P^ 


*». 


1, 


t^, 0, 


«', 


1, 


«', 0'. 


0, 


1, 


r\ i^ 


«', 



1 



*' 



= 0. 



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83 



4. If then the locus of S be /.SA^ + m. SBHti .SC^ = ;t, 
the tangential equation of the envelop of the corresponding directrix is 
0, Ip' + mq^ + nr^, k, A;, k =0; 

k, fi, 0, <^, *2 

k, q\ e\ 0, aa 

k, r2, ^, aS, 
which may also be written 



0, 
k^c^m — h'^ny 

k^m-a-m, 
where 



k-i^m^bhi, k-i^l-a^n, 



k-bn^ahn 




= 0, 



0, 

0, P\ q\ 

i^, 0, c\ 

q\ cs, 0, 

r», ^, «^ 

Thus, if the locus of a focus of a system of conies circumscribing a 
triangle be a circle or a straight line, the envelop of the corresponduig 
directrix is a curve of the fourth class. 

5. If the locus of the focus be the straight line BC, we have 
a.SA2-*co8C.SB2-<jcosB.SC2« a*<?cosA, 
and the envelop of the directrix is 

= 0; 



-8^ 


\ 0, 




0, 





0. 


0, 


P\ 


(l\ 


r2 


8a2 


i"', 


0, 


^y 


ft2 


0, 


«', 


^, 


0, 


a2 


0, 


r^. 


h\ 


««, 






which reduces to { -p'^a^ + qU^ + rV}a - 4iV^V2 cos^ A = . 

Thus, if the focus of a conic circumscribing the triangle ABO lie on 
BC, the envelop of the corresponding directrix consists of the two curves 
-j»2a2 + ^2^2 + y2c2 + 2*<?cosAgr=0, -j!?V + j2*2 + ^c2-2*ccosA^r = (1,2). 

If D, E, F be the mid-points of ABC, (1) is clearly touched by DE and 
DF, and (2) by EF and ijie line at oo . Thus (2) is a parabola. 

Again, it is clear that A is the pole of BC with respect to (1) and (2). 

If A be a right angle, Jl) and (2 J coincide, and the envelop is clearly 
the parabola whose focus is A and directrix BC. 

6. If the focus lies on the circum-circle of ABC, we shall have 
;.SA2 + m.SB» + «.SC«-A:, 
where m(^-\-nl^ = ky Ic^-^na^ = k, Ib^ + ma^ = k ; 
hence the envelope of the directrix is 

= 0; or pa ±.qb ■;^rc = Q, 



0, 


I^, 


«', 


r" 


p'. 


0, 


0% 


«2 


?'. 


«', 


0, 


a" 


r^, 


*'. 


«^ 






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84 

Thus, if the focus of a system of circum-conics lie on the circum-circle, 
the corresponding directrix passes through one of the centres of the 
circles which touch the sides of the triangle. 

7. More generally, if the locus of S be given by an equation of the 
form Wn + «»-!+ ... +Wo = ®» 

where w,» is a homogeneous function of SA, SB, SC of the «th degree, 
then, substituting SA ^ ep, &c., we shall have 

«- ./« (Pqr) + e^-Vn-i {pqr) + ... = 0. 
And if <; be eliminated from this and the equation in § 3, we obtain the 
tangential equation of the envelop of the directrix corresponding to 8. 

8. If the locus of S be given by an equation of the form 

/(SA, SB, SC) - 0, 
where /is homogeneous of the nth degree, then clearly the envelop of the 
directrix is / (/?, j, r) » 0, a curve of the wth class, and conversely. 



7949. (B. Knowlbs, B.A.) — Prove that the sum of the series 

-3-i^log (^-/t.^')% 3-i^(tan->g4fl^.cot-i3i]. 

Solution hy Rev. T. C. Simmons, M. A. ; J. O'Keoan ; and others. 
Putting X ^ y^, and dividing by y, the left-hand side becomes 
11^-1!^ + iy«-...sS. 

the constant being added in order to make S and y vanish together. 
Hence the stated sum of the original series follows. 



8668. (Alpha.) — The ellipse whose eccentricity is i 'v/2 is referred to 
the triangle formed by joining a focus to the extremities of the latus 
rectum tlurough the other focus : prove that its equation is 
'y^+9($y + ya + afi)=^ 0. 

Solution by A. Gordon ; Rev. T. Galliehs, M.A. ; and others. 
If x^Ja^ + y^/b^— 1 = be the ellipse referred to its axes, we have 

-x + y2^/2-'^p ^a, -x-y2V2-Zp ^ fi, x-Zp ^ y (1,2,3), 

and cosa - -J, p = a/3V2, x^ + 2y^ « 18^?* (4). 

Eliminating x^ y, p between (1), (2), (3), and (4), we obtain the result. 



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85 

9449. (Professor Sylvester, F.R.S.) — If there exist any perfect 
Dumber divisible by a prime number p of the form 2'*+ 1, show that it 
must be divisible by another prime number of the form^^ ±, 1. 



Solution by W. S. Foster. 

Let the number N ^p^ ,q^ .r^...; then, since N is a perfect number, 
we must have one of the Victors (say, ^*) such that ^* ♦ * — 1 is divisible 
by j7, therefore ^*** « M (^) + 1 ; and, since p and q are prime numbers, 
ji»-i = M(j7) + 1, therefore d + 1 is a divisor of 2'» » 2* suppose. Let 
q — xp±hj then k^—l = M(i?) ; hence h must be some power of the 
remainder after dividing (2»»)***~* by 2'* + 1 ; therefore h must equal 1, 
and q ^ arp±\f which is a prime divisor of N. 



9468. (R. W. D. Christie, M.A.)— Show that the tenth perfect 
number is P^o = 2« (2«- 1) = 2,417,851,639,228,168,837,784,676. 



Solution by Professor Ionacio Bbtens. 

Le dixi^me nombre parfait est donne par M. Carvallo dans Touvrage 
ThSorie dea Nombrea parfaits. 

[Every divisor of 2f — 1 is of form 2px + 1 when piaa, prime ; but 2** — 1 
is indivisible by ii2x + 1 ; hence 2^ {2*^— 1) = &c. is a perfect number.] 



3419. (Artbmas Martin.) — The point Ai is taken at random in the 
side £C of a triangle ABC, B^ in CA, and C^ in AB ; the point Aj is taken 
at random in the side Bfii of the triangle A^BjCi, B^ in CjA^, and C^ in 
AjBi, and so on ; find the average area of the triangle AnB^C^. 



Solution by D. Biddle. 

It is difficult to understand why this question has remained unanswered 
so long ; but the reason may be any one of three : — (1) its extreme sim- 
plicity, ^2) the fear of some concealed pitfall, or (3) a mere disinclination 
to consitter the matter. Unless (2) be well grounded, there can be no 
doubt that the correct answer is (J) "ABC. For, let Ai, A], ... A« repre- 
sent the successive triangles drawn at random upon the given ABC, in the 
way described. Then, the average area of An will be Ja»-i ; of An-i, 
iAn.2; aiid so on, until by retrogression we arrive at Aj, which on the 
average is JABC. The fact, in regard to each pair taken separately, is 
well known. But, if A^ be on the average J of Aj, which on the average 
is i ABC, it seems impossible to escape from the conclusion that on the 
average Aj = (i)* ABC. In being J, on the average, of any Aj, on which 
it may be drawn, A2 is on the average i of ^ ABC. It is a case of multiple 
integ^tion in which, as each variable is eliminated, the additional factor 
^ is yielded to the result. 

VOL. XLIX. L 



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86 

9402. (The Editok.) — If the radios of the in-circle of an isosceles 
triangle is one-»*** of the radius of the ex-circle to the base ; prove that 
the ratio of the base to each of the equal sides is 2 (n- 1) : n + 1. 



Solution by Professors Emmerich, Ph.D. ; Ionacio Bbybns; aftd others. 

Draw perpendiculars TD, T«D« from the centres T, Ta of the in-circle 
and the ex-circle to the base on AB. From similar triangles, we have 
AD : AD« = 1 : w. But AD = J (2b- a), AD. « J (2* + a), a denoting 
the base, and b the other sides ; Uieref ore 2b + a:2d — a=nll; hence 

2a : 4ft — «— 1 : «+l, and a : ft — 2 (n-l) : « + 1. 



9440. (Rev. T. C. Simmons, M.A.) — Prove geometrically that the per- 
pendicular from the Lcmoine-point of an harmonic polygon on the Lemoine- 
line is the harmonic mean of the perpendiculars drawn on^he same line 
from the vertices of the polygon. [A proof by tiigonometrical series is 
given in Lond, Math^ Soe. Proceedingsy Vol. xviii., p. 293.] 



Solution by R. F. Davis, M.A. 

If a polygon ABC ... L (n sides) inscribed in an ellipse is such that each 
side subtends the same angle (2ir/n) at the focus S ; then, projecting 
orthogonally, we get a harmonic polygon abe ...I {n sides) inscribed in a 
circle whose Lemoine-point a is the projection of S, and whose Lemoine- 
line yy' is the projection of the S-directrix YY'. This property is the 
basis of Mr. Simmons' theory of harmonic polygons, as set forth in the 
above paper. 

The properties of the polygon ABC ... L may be derived in turn by 
reciprocating with respect to any point S a regular polygon of n sides 
circumscribing a circle. Since (by a well-known theorem) the sum of 
the perpendiculars of 8 upon the sides of the latter ^lygon = n (radius), 
we have 2 (k^/SA) = w (Ar^/SD), where SD is the semi latus-rectum of the 
ellipse. Hence, if SX, AA', BB' ... be perpendiculars upon Y Y', SX is 
the harmonic mean of A A', BB' .. . ; and this relation is unaltered by pro- 
jection. 



(W. J. C. Sharp, M.A.)— If (x^, yj, z^, m>i), (a^s* ^i, «2> «^t)» 
(^3» !/zi «3» ^p) he any three points, and A, fi, v the areal coordinates of any 
point in their plane referred to the triangle of which they are vertices ; 
show that the equation to the section of any surface U = by the plane 
will be obtained by substituting for x, y, », ta from the equations 

(\ + ;*+y) Jf « KZi + fiX^ + yX^, (a + ^+i.) y » Ay j + ftyj + •'^j* 

(A + fi -I- v) « = ATj + ^j + yz^f {\ + fi + y)w*^ AtTj + fiw^ + viTj. 



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Solution by D. Edwaudes. 

Let A be the area of the triangle formed by the three points. Then 
any point in the plane of the triaogle may be expressed 

l + m + n * l-tm + n 

Also k — ' A, &c., therefore Ax = XXi + fix^ + yx^f Ay — &c. &c. ; 

l + m + n 

and A = \+fi + v, therefore, &c. 

fThat the point ( ^^l+>»^^-^>>^3 If/i^my^-^ny, ^ \ 
L \ l&m + n l + m + n J 

is a point in the plane follows at once, because if this point be (or, y ...) it 



I point ^ 

satisfies the equation 



Xi 



X2, 



= 0, which is the equation to 
the plane of the triangle.] 



9350. (Professor De Wachtbb.) — A point being taken within a tri- 
angle, prove that the chance that its distances from the sides (a), (6), (0), 
may form any possible triangle will be 2abc/ {{b + e) {c + a) [a + b)}. 



Solution by Professor Ionacio Beyens. 

Soient A], Bj, Cj les pieds des bissectrices 
du triangle ABC ; il est ais^ k d^montrer que 
la droite Bfii est le lieu g^om^trique des 
points tels que leur distance au c6t6 BC est 
egale k la somme des distances aux autres 
deux cdtes AB, AG, et que par suite pour 
tout autre point situ6 dans Pint^ieur 
du triangle AB^Ci la distance k BC est plus 
grande que la somme des autres distances k 
AB, AC, et que pour un point du quadrilat^re BCBjCi la distance k BC est 
plus petite que la somme des distances k AB, AC. La m^me chose 
arrivera aux droites BjA^, AjCi, et par suite tout point interieur k AiB^Cj 
aura la propriety que Pune quelconque de ses distances aux cAtes AB, AC, 
BC, sera plus petite que la somme des deux autres, et par consequence la 
probabilite demand^e sera AA^B^Ci : A ABC — 2abc : {b + e) (e + a) (a + b). 




. 8344. (K. Knowles, B. a.)— ad, be, CF are drawn from the angular 

points of a triangle ABC, so that the angles BAD, EBC, ACF are each 

equal to the Brocard-angle of the triangle ; show that their equations are 

bcy—jah = 0, b'^x—acz = 0, abx-e^ « 0. 



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88 

Solution by Gbobob (Joutthorpb Storb, M.A. 
It to "be the Brocard angle of the triangle, we have 
cot « s cot A + cot B -I- cot G. 
Now the equation to AD is 

Jl_ . «^(A-a,) ^ rinA cot— COS A - -J$^^ = f', or bcy-aH^ 0. 
« sin ft» sin B sin C *c 

Similarly for the equations to BE and CF. 



9376. (A. E. Thomas.) — Solve the equations 

x4+3y22i^a*+2d;(y* + a^ (1), 

y< + 3«%««4< + 2y(2' + :c^, 2*+3a;V = «* + 2« (a' + y') (2,3). 

Solution by Professor Sebastian Sircom, M. A. 
Adding, ^fl+y^+s^^xy—yz^zx^ (x + 9ty + uh){x + t^ + att) 

-(«^ + ^ + <r*)» (4), 

(2)x«, (3)x«i8 give (a; + y + «)(a: + «*y + «2) = (o^ + wi^ + ci'i?*)* (5), 

(x + y + »)(ar + «y-l-«%) = («< + ^^^ + »<?♦)* (6). 

(5)x(6) gives ^ + y + . = (a^-^<^^^-^c^^)* (a* ^ a>^b* + ^)\ 
(4) *" ' (a* + ^ + <?*)* 

with similar expressions for x + »y+§^f &c. ; adding, we ohtain a; in a 
form that can easily he rationalized, and then the valnes of y, z can he 
written down. 



9430. (Professor Wolstbnholmb, M.A., Sc.D.) — In a tetrahedron 
OABC, the plane angles of the triangular faces are denoted hy a, jB, or 7 ; 
all angles opposite to OA or BG heing a, those opposite OB or GA are 
i3, and those opposite OG or AB are 7; the angles at have the 
suffix 1, those at B, G, D the suffixes 2, 3, 4 respectively ; prove that, if 
«i + i^i + 71 = as + iSs -h 7s s T, then 

71 + 01-^1-74+04-^4; Oi + ^i-7i- 08 + ^8-78 

7s+aj-^- 74 + 09-^5; oj + ^-72 » 04 + ^4-74- 



Solution by Professor Swaminatha Aitab, B.A. 

^i> ^2> ^3 fti'o >^7 three points not in the same straight line. is the 
middle point of Gi, Gj, A of Gi, G3, and B any point in the plane equi- 
distant from Gs and G3. Now the tetrahedron of which the &ices are the 
triangles GAB, OAGi, OBG3, ABGjis of the sort described in the question, 
and, naming the angles as directed, we see at once from the figure, 
since OA is parallel to G2G3, that 

ai + 7i-^i= 04 + 74-^4; 02 + 32-72 = 04 + ^4-74- 
A similar proof is easily seen to hold for the other part. 



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[Professor Wolstbnholmb remarks that he had not attempted a de- 
ductive proof of this theorem ; the lengths of the edges of the tetrahe- 
dron, by means of which he noticed the property, are 

DA - 7069273, BC = 713973, 
DB - 7-376329, CA - 64644, 
DC - 7-316126, AB « 8* 13924 ; 

the half angles at D are 29*»4'40"-45, the half angles at A are 28*»36'42"-86, 
21**38'39"-03 28**46' 4"-73 

34**16'40"-62 32°39'12^'-42 



(Ti = 90° 0' 0" ffi - 90° 

those at B are 26° 68' 14"- 76, those at C are 30° 42' 8"-66 ; 
24° 43' 23"-96 30° 40' 19"-78 

30° 14' 59"-77 36° 40' 63"- 19 



0' 0" 



<ra«81°d6'38"-48 



<r4 « 98° 3' 21"-52 



the dihedral angle opposite DA is 66° 1' 3"-76^ 
BC is 66° 31' 60"-4 i 



opposite DB - 65° 68' 29"-34 7 
„ CA = 69°ll'49"-62j» 

.yj + ai-i3i«73"26'23"-88 7 
74 + «4- i84 = 73° 26' 23"-92 j 

«, + fi^^y^ = 42° 63' l7"-92 7 
«3 + /38-78-*2°63'l7"-96j 



opposite DC iB 79° 42' 30"-7 7 
„ ABis86°46'22"-96j» 

7i+ai-i52 = 64°69'41"-08 7 
78+«B-iBs - 64°59'41"-12 5' 

09 + ^^-7% = 49°23'10"-32 7 
' a4 + i34-74 - 49°23'10"-28)' 



the difference in each case being "'04 ; which is as near as can be expected 
with 7 figure logarithms.] 



9006. (H. L. Orchard, B.Sc., M.A.) — ^Inside a hemisphere (of radius 
p) a luminous point is placed, in the radius which is perpendicular to the 
base, at a distance from the base » ip \^3 ; show that the illumination of 
the surface (excluding the base) is = 3irC. 



Solution by Rev. T. Galliers, M.A. 

Let be the luminous point ; ABD the vertical section 
of the hemisphere (its centre being at C) through CO ; 
Z CPO = ; Z FOB = e ; radius of hemisphere = a ; 
CO=e,a^eVZ\ also let OP » r. 

Then the illumination of a band generated by the 
revolution of the elementary arc at P about CB 



Now 



= 2irC (r sin tf . cos 4) <fo)/r2 « ^ (say). 

cosd) =s^-J^ ^ "^ . also a^ = r2 + c2+ 2(T cose ; 
^ 2ar 



therefore 



(r + c cos 6)dr ss er sin . rfd, 




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90 



■^^ ^-?l±££2B», therefore *--?-; 

dr tfBine dr eam$ 

thus ^.I^J-W.lJ..; 

therefore illumination of hemi-spherical sur&ce 

- — f^"'*"*^ i^-^ + 1| rfr - 3»0, the result given. 



ap a^bq,fi^er,y, .-. ^ = *-2, ...M«N^^^^ 



9433. (G^. Hbppbl, M.A.)— If, within a triangle ABC, be a point 
where the sides subtend equal angles ;. then, putting OA >« p, OB » q, 
00 « r, show that the equation to the ellipse with focus O, touching the 
sides in D,- E, F, is in (1) rectangular coordinates, with O as orig^ and 
OA as axis of ^, and (2) trilinear coordinates, ABO triangle of reference, 

{x^+y^^'=^\(pq + qr-¥rp)'^[{pr+pq'2qr)if-p(q-.r)xy/Z + Zpqr]...{l)^ 

aYa^ + b^q^0^+A^-^2bcqr$y^2earpya-2abpqafi - '.(2). 

Solution by W. S. Foster. 

Since is the focus of the ellipse touching the sides of the triangle, the 
angle AOE = AOF, and AOB = AOC, therefore BOF = OOE, therefore 
BOD » COD, therefore OD bisects the angle BOC, therefore AOD is a 
straight Une. Let L<r» + M/32+N7«-2LMa/5-2LNo7-2MNi87 = be 
the equation to the ellipse ; then the line AD is MjS— N7 = 0, and since 
this passes through O, whose coordinates are given by the equations 

^^Ji 

N er* ' ' bq cr "^^^ ap' 

Substituting these values of L, M, N in the equation to the ellipse, we 
have the equation given in the question for trilinear coordinates. 
; Let hjp a 1 + « cos be the equation to the ellipse, referred to the major 
axis as initial line, and let this make an angle Bi with OA ; then, since 
AE touches the ellipse at E, 

k 

— = tfCOs0i + co8AOE = tfcos^i + i, 
P 

and — «tfCos(-^ + aij+i««(-icos0i — sin^jj+i, 

|.-fco8(^+a,)+i=i,^-4cos0i+^sin0i)+i, 

therefore 

k^—-ML -, .cosO,= ^g^-^-i^g , .sine,»^/^^(^'^) : 

^{pq-^qr-^rp) 2{pq + qr-tpry 2(pq + qr+pr) 

therefore the equation to the ellipse referred to a line perpendicular to 

OAis A;/p - l+« sin (01 + 0), 

.-. (a?' + y2)*= i{pq + qr'^pr)-^{(pr+pq-2qr)t/-^/3p{q^r)x + Zpqr}. 



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91 



8461. (F. R. J. Hervey.) — Find in how many ways n lines of 
verse can be rhymed, supposing that (I) no line be left unrhymed, and 
(2) the restriction as to unrhymed lines be removed ; and show that, in 
the case of the sonnet, the respective- numbers of ways are 24011157 and 
190899322. 



Solution by the Pkoposeb. 

Let the numbers of ways be denoted, (1) by 0», (2) by/« ; and first 
consider /n. Let line-endings which rhyme together be considered the 
samei and denoted by the same letter. Let the first euding be denoted by 
fl, the next distinct ending by b, the third by c, and so on. Two lines 
give two cases, aa and ab. Three lines give five, for aa may be followed 
by a or 6, and ab by a, b, or e. Genersdly, an arrangement of this sort 
containing only the first r letters of the alphabet may be followed by any 
one of the first r + l letters ; and the complete sets of arrangements for 
successive values of n may be exhibited in a table such as the following. 
Each letter gives rise to a ffrovp in the next line ; a g^oup of r letters, 
wherever found, gives invariably r—l groups of r followed by a g^oup of 
r + l; and the number of letters in the n^ line is fn. I shall show that 
fn a A**fl ; this is contained in the two following propositions. 



b 



d 



ab abc abc abc abed abe abe abed abc a\be\a\ bed abed a \ bed \ a \ bed \ a \ bede 

I. Suppose the above table continued indefinitely , and form a new table 
thus. Suppress the first line {a) ; suppress a of tbe second line, and all 
that it gives rise to in the following lines. Call this the first derived 
table. From this derive another table, and so on continually, the rule 
being : — To form the (m + 1)^ derived table, suppress the first line of the 
m^ ; then suppress every a of the second line, and all that they give rise 
to afterwards. The numbers of letters in the successive lines of any 
derived table are the differences of the numbers in the successive lines of 
the preceding table. 

Let A denote either the original, or anv derived table ; B the next, 
and S the table consisting of all that part of A, with the exception of its 
first line, which is suppressed to form B. To each letter {k) in the first 
line of A corresponds a letter {a) in the first line of S ; A; gives a group in 
the second line of A ; the corresponding a (the first of this group) gives 
a similar group in the second line of S ; and these two groups, being 
similar, g^ve similar sets of groups in the next lines of A and S respec- 
tively ; and so on continually. In fact, A and S are, with the exception 
of their first lines, identical, letter for letter. But the(j3+l)**» line of 
A is made up of the^?*^ lines of S and B ; whence the proposition. 

II. The first line of the m**^ derived table consists of the same number 
of letters, arranged in the same numerical groups, as the m**» line of the 
original table. Let the first lines of the m^ and {m -i- 1)*** derived tables 
be denoted by P and Q respectively ; and assume the truth of the pro- 
position for P. Consider a group of r letters in P ; this is a group of 
r+l with its first letter suppressed ; it gives rise to r— 1 groups ^f >• -t- 1, 



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n 


12 3 4 5 


T 


1 2 5 15 62 


A 


1 3 10 37 


A2 


2 7 27 


A' 


6 20 


A* 


15 



92 

and a group of r + 2 ; and these, with their first letters suppressed, are 
found in Q. Hence a group of r in P gives r - 1 groups of r, and a 
g^oup of r + 1 in Q. But this is just what arises in tie (m + 1)"» line of 
the original tahle from a group of r in the m^ line. In fact, the table 
consistmg of the first lines of the series of derived tables only is an exact 
copy of the original table, with this exception, that the letters a, 3, ^ . . . 
of the latter are everywhere replaced by *, <?, <f ... respectively, in the 
former. 

From this property fn can be calculated (see 
example). The oblique rows are formed in suc- 
cession ; each value of /obtained is transferred to 
the left-hand column, giving a new oblique row, 
ending in a new value of/; and so on to any 
extent. 

Another process, and an expression for /», are 
thus obtained. 

If (r, n) denote the number of groups of r in the n**» line of the original 
table, we have evidently (r+1, « + l)=(r, «)+r . (r+1, n). Forma 
table in which the r*** number of n*^ column is (r + 1, « + 1) . The first row 
is 1, 1, 1, ..., the first column 1, 0, 0, ..., the above relation gives all the 
rest, and the sum of the n^ column is fn. But the above conditions are 
precisely those for the formation of a table in which the r*^ number of 
«ti» colunm shall be A*'0"/r!; (compare the above relation with the 
following — aW 0" = A^**"*)©"-* +r . A^*^ 0"-^ in which A^*") 0*» stands for 
A*" 0»/r !) Hence (r + 1, « + 1) « A(*"^ 0", and 

fn = A0'»+A20»/2 + A30»/3! + ... + A»»0»*/«t; 
in which the r^ term, being (r + 1, « + 1), is evidently the number of cases 
with just r distinct endings. 

From this expression an algebraic proof may be given of the theorem 
already derived from elementary considerations; namely, that the 
«**» term of the 8eries/i,/2,/3 ... is the same as the »*^ difference of its 
first term. We have, A and D referring to x and y respectively, 

D'». A*"a:i' = A»» .D^arv = A»» .arv (a;-l)" ; 

whence D'».A"*a;<*— A*"(jr— 1)*». Now, the above expression may be 
written without error as an infinite series, and the general term A** 0^/r I 
is also A»-i !»•-'/ (r-1) ! Hence 

D» . A»*OVr! = D" . A»-i lo/(r-l) ! « A'-iO«/(r-l) ! ; 
and the theorem follows at once. 

It remains to determine the relation between ^ and /. A line which 
rhymes with no other may be called an odd line. I shall show that the 
number of arrangements of n lines containing at least one odd line is 
exactly equal to the number of arrangements of « + 1 lines which contain 
no odd line. Take any one of the first set, and add an odd line at the 
end ; then replace the odd lines by as many lines rhyming all together, 
but not with any other line. The result is one of the second set. Con- 
versely, take any one of the second set, and replace the last line and those 
which rhyme with it by as many odd lines ; then erase the last line. 
The result is one of the first set. Hence the two sets correspond 
definitely in pairs, and the number in each is the same; that is, 
/n — ^ =s 4) (« + 1), whenife the values of 4) are derived in succession from 
thoseof/. Also <^(« + l) =/»-/(«- 1) +/(fi-2)-...dr /I. 



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Google 



93 



Of the two relations, fn = A"/l and/«— ^« = ^ (« + 1), either may be 
made a coosequence of the other, as follows. It is clear that the number 
of arrangements of n lines containing k odd lines is, in general, 
^{jfi—h) .» («— 1)...(«— A;+l)/A;! \k ^n gives one; k ^ n—l gives none, 
but <pl = 0. Hence, if /O = 1, we have 

/«= 1+«4>1+J«(;j — l)«^2 + ...+«4>(n-l) + «^w, 
true forn = 0, 1 , 2, . . . J which givjes <f>n = A**/0, and </>» + 4) (w + 1)' « A*»/l . 

The question of finding /« may be approached somewhat differently, as 
follows. Of the complete permutations (r" in number) which can be 
formed from r given letters taken n (not < r) at a time, let the number 
of those in which all appear be deiv>ted by Ur ; let P be one of them ; and 
let ab .,., hk ...f represent any two of the r ! simple permutations of the r 
letters. If then the result of writing h for every a, k for every b, &c., in 
P, be called Q ; it is evident that P and Q represent exactly the Fame 
arrangement of line-endings. (The selection of one out of each such set 
of r ! equivalent permutations may be made by requiring that the r letters, 
so far as the^r«^ entrt/ of each is concerned, shall present one invariable 
permutation; compare the notation already used.) Hence (»'+l, w+1) 
= nr/r !, and «r = A** 0". But the last result may be obtained indepen- 
dently ; (it is in fact essentially the same as that contained in the first 
part of Question 8390 [Vol. xlv., p. ] ; as may be seen by putting letters 
for persons in that question, and regarding iheplaceSf 1st, 2nd, 3rd, &c., 
occupied by the letters as the things distributed among them) ; the formula 
for fn will then follow, by the above reasoning. 

The values of/ and 0, as f ar as « = 14, are given below. 



n 


/ 





n 


/ 


1> 


n 


/ 


1> 


3 


6 


1 


7 


877 


162 


11 


678670 


98253 


4 


16 


4 


8 


4140 


715 


12 


4213597 


680317 


6 


52 


11 


9 


21147 


3425 


13 


27644437 


3633280 


6 


203 


41 


10 


115975 


17722 


14 


190899322 


24011157 



9413. (J' O'Byrnb Croke, M. a.) — If D be the distance between the 
centre of the cii'cumcircle and the point of intersection of the perpendicu- 
lars of a triangle, prove that 2D/(1 — 8 cos A cos B cos C)* = «/ sin A. 



Solution by D. Thomas, M.A. ; R. Knowles, B.A. ; and others. 

Let o, )8, 7 be the vectors of A, B, 0, cooriginating from the circum- 
centre, so that To = T)8 = T7 = R, S$y = -R=2 cos 2 A, &c. The vector 
of the orthocentre = cot B cot C.0 + ... + ..., therefore 

-D2 = -R2 {5 cot2 B cot2 + 2 cot^ A cot B cot cos 2A + ... + ...}, 
.-. D» = R2 {(5cot B cot 0)2-2 cot2 A cot Boot (1- cos 2A) + ...} 
•= R2 .[l — 2 cot A cot B cot 2 sin 2A} =» the result required* 
[By Question 8872, we have D = R (1 - 8 cos A cos B cos C)*, 
hence 2D / (I - 8 cos A cos B cos O) = 2R = a/ain A, &c.] 

VOL. XLIX. M 



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94 

9478. (Hev. J. J. Milne, M.A.) — If p be the sum of the abscissae, q 
the sum of the ordinates of two points P, Q of an ellipse ; prove that (1) 
the equation of PQ is 2J^x-¥2a^y = i^ + a^q* ; and hence (2) if either 
(a) p or qhe constant, or (/3) if p and q be connected by the relation 
Ip-i-mq ^ If the envelope of the line is a parabola. 



Solution by R. Knowles, B.A. ; Prof. A. W. Scott, M.A. ; and others. 
If hk be the pole of PQ, its equation is 

l^hx-'ta^ky = a»*3 (1), 

h ^piffiJe^-^b^h^l^a^lfif k =r ^(a2;t« + A2A8)/2a2^, 

^ + aV = 4a**V {a^ + ^2^2) . 

making these substitutions in (1), we obtain 2ll^px + 2a^y = b'^p'^+a^q^, 

(a) If p is constant the equation to the envelope is the parabola 
aV = ^P (p-2a:) ; if q, b^x^ = a-q {q-2y), 
(/3) If Ip + mq = 1, the envelope is 

(aHy-bhnx)^ + 2a^b^lx + 2a^b^y^d^b^ - 0. 



9439. (A. Kahn, M.A.)— Show, by a general solution, that the roots 
of 4x* + 4txi+lZx^ + 6x + S = are i{-l±(-7)*}, i{-l±(-3)*}. 



Solution by Professor Cochez ; R. W. D. Christie ; and others, 
L' equation du 4« degr6 x* + ax^ + bx^ + ex + d = pent se mettre sous la 
forme {x^ + ^ax)^ + {b - ^a^ {x^ + exl{b-ia^} +d = 0, 

et pourra etre r^olue par les m^thodes du second degr^ dans le cas oh 

L' Equation propos^e ix* + 4^ + 1 3:r2 + 6j; + 8 = est dans ce cas ; on peut 
I'ecrire (x^ -i- ^x)^ +Z{x^ + ^x) + 2 = 0. 

Posant x^ + ^x = y, on a k r^soudre y^ + Zy + 2 =0, dont les racines 
Bont — 1 et — 2. Par suite les quatre valours de x seront donn6es par 
les equations x^ + ^x ^—l et a;' + Ja: = — 2. 

[It p^—ipq + Sr = 0, any biquadratic x^ + psfi + qx^ + rx + s « can be 
immediately reduced to quadratics ; hence the given equation 

= {x^ + tnx + 2){x^ + nx+l) = 0, ^(xi + \x + 2){x^ + ^x+l) = 0.] 



9437. (H. FoRTEY, M.A.) — Show that, if a, /3, &c. are the p roots 
(excluding unity ) of a^*^—mxP + m—l = 0, the number of ways in which 
m letters can be arranged « in a row, repetitions being allowed but not 
more than p consecutive letters being the same, is 

m 2 (a-l)^a* *-^P 

{m-iy aP''^-{p+l)a+p' 



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95 

Solution by Professor Swamin atha Aitar, B.A. 

Referring to my solution of Question 9293 (Vol. xlix., p. 26), let Q^ 
stand for the required number of ways ; of these Q,, ways let those that 
do not begin with the letter a be 9» in number. Then we have 

Q* - ?»» + ?«-i ...••- 7h-f and qn = (»»— l)(^»-I + ?i.-8...+^M-.p); 

therefore Qm = ——^ ?»• And y„ is the coefficient of x^ in the expansion 
tn — 1 

of {l-(m-l)(x + a;«...+a:^)}-»; 

therefore Qn^-gL-^ [-^)'-'''' . 



8177. (Professor Hanumanta Rau, M. A.) — The images of the circum- 
centre of a triangle ABO with respect to the sides are A', B', C ; prove that 
the triangles A'B'Cand ABC are (1) equal, (2) have the same nine-point 
circle ; also find (3) the equation of the i^kcum-circle of A'B'C and the 
angle at which the two circum-circles cut each other. 




Solution by A. Gordon ; R. Knowlbs, B.A. ; and othsrs, 

1. OA' = AI and is parallel to it, /^ 
OB' = BI and is parallel to it, 

therefore A'B' is parallel and equal AB, 
&c. ; therefore the triangle A'B'O' is 
equal triangle ABO. 

2. I is the circum-centre of A'B'O' 
(for lA'isparallelandequalOA^Rs &c.)^ 
also I A' « A'B (each = R), therefore 
B» = «I ; therefore the lines AI, BE, &c. 
are bisected by B'C, A'O', &c., at m, «, 

&c., and these points are on the nine-point circle of ABO. 

But n is also the middle point of A'O' ffor In is a perpendicular from 
centre I on the chord A'O'), and is thereiore a point on the nine-point 
circle of A'B'O'. These two circles have therefore three points m, «, &c. 
common, and are therefore coincident. 

3. If the mid-point of 01 is taken for origin of coordinates, and any 
rectangular axis for reference, and if x cos a + y sin a—Pi is the equation 
of BO, &c., then the circle about A'B'O' is 

5 sin {x cos o + y sin a +j?j) (x cos jS + y sin +pi) = 0. 
The equation can also be written in trilinears — 

R2 (a sin A + fi)^ = 5 (iSy -7i9')«- 25 cos A (70'- 07') (ajS'- a'jS), 
where o', jS', 7' are the coordinates of I the orthocentre. 
The angle at which the two circum-circles intersect is given by 
(01)2 = R2_2Rr = 2R8-2R2co8d>, or cosrf> = 2»-fR 



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96 

9217. (Major-General P. O'Connell.) — In using either the French 
or English Arithmometer, any two nnmhers each containing less that 
nine figures can he multiplied together, and the sum of a series each term 
of which is the product of two sudi numhers, whether positive or 
negative, can he ohtained without writing down any figures. It is 
required to find a formula for the product true to, say, thirteen figures on 
two numhers each of sixteen figures, so that the result may he ohtained 
by the use of the Arithmometer alone, i.e., without intermediate record. 



Solution by the Proposer. 

Let A and B he two large numbers, and let their digits, counting from 
left to right, he indicated by numbers written under A and B respectively. 
Let Ai_8 mean Uie first eight highest digits of A, and B9-12 the 9th, 10th, 
1 1th, and Tith digits of B ; then, if A and B each contain sixteen figures, 
the following formula will give a result true to thirteen or fourteen 
figures. 

AxB = Ai.8 xBi_g + Ai.4xB9_i2 + Bi_4X A9_i2 + Ai_2 x Bi3_h 
+ Bi-2xAi3-14+A6-lcxB9-10 + B5.«xA9-l0 

+ A, XB15 + B1XA16 + A3XB13 + B3X Aia + AsxBa + Bjx An 

+ A7XB9 + B7XA9 (I). 

If A =» B, we have 

A2= (Ai.8)^ + 2{Ai.4xA9.i2 + Ai.2xAlS-14 + A5-6xA9-10 

+ Aix A15 + A3X Ai84 A5X Aii + Ajx A9} (2). 

In using the second formula, first sum the series under the vinculum, 
add the result to itself, and finally add (Ai g)^ J ^7 this means A^ will be 
obtained true to 13 or 14 figures. 

The following formula, to be worked with pen or pencil and paper, will, 
when its total is added to the above, give a result true to all but the last 
figure, which may be looked upon as approximative. The dots are to be 
understood as decimal points. 

Remainder = 2 x {A, + -Au + Aj x -Ais-w + A3 x -Au-ie + A4 x 'Aw- is 

+ A5X •Ai2_i4 + A6X An-is + Ayx •A10-12 + A8X A9_ii} (3). 

By formula (2), if A= 3 99999 99999 99999, 
A2 = 15-99999 99999 9867. 
Byformula (3), remainder = 132, giving A' =. 15-99999 99999 9999 for 
the corrected value. 

If A = 3-16227 76601 68379 33 - >v/10, 

by formula (2), A2 =. 9-99999 99999 9972, 

by formula (3), remainder = 27-418, giving 

A2 = 9-99999 99999 999948 instead of 10. 



8333. (Professor Hanumanta Eau, M.A.) — Prove that the equations 
x^ 4 19a?- 140 = 0, and 7^'- 12.i;^ + 46*-2+ 12.i;+ 7 = 0, have two common 
roots. 



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97 



Solution by Profs. Aitab, B.A. ; Sulcom, M.A. ; and others. 
If a, /3 be two roots of the second equation, the other two roots are 
evidently and . Therefore the second equation is reducible to 

the form (s^—px + q) 1^^+ ^x+ ]»0, and comparing the co- 
efficients we have p » 2 and ^ — 7 ; and 3^—2x-k-l is a factor of 
«« + 19iB-140. 



9018. 0^* '^' Grbbnstbbet, B.A.) — If the Earth and Jupiter are 
in heliocentric conjunction at the same time as Jupiter and one of his 
satellites, show that the times when the satellite will appear to an 
observer to be stationary are the roots of the equation 

£. + il + £ + ^ (J + ,)co82, ( i-J-V- -('' + «)co.2, (I-IV 
a b c be \ b e I ae^ ' \a e I 

-J(a + *)co.2,(i--J-).-0; 

where e^ jj s are radii of the orbits of the Earth, Jupiter, and the satellite, 
a, b, e their periodic times, the orbits circular and in one plane. 



Solution by W, J. Gbbbnstreet, B.A. ; Sabah Marks, B.Sc. ; and others. 



St 



J 
E 

8 



St' 



N' N N" 



Consider motions in two directions perpen- 
dicular to each other. After a time t^ draw 
E'N, J'N', S^N" perpendicular to &r. 

llien SN = eBmwt. 

The position of satellite relative to Earth 
is given by 

y sin 0)1^ + « sin fl»2^— ^sin »n 

ycos«i^ + «cos»2^— ^cosaO 
Kelative velocities in same directions are 

wij cos «i ^ + t0^s cos wj^— we cos »^, — «iy sin a»| ^— a»2' sin «3< + we sin co t. 
M^en the satellite is stationary, 

jsmwit + ssmaf^—eBJnwt w^j cos oa^t 4- <a<^ cos w^t^we cos w t _ ^ . 
jcoswit-¥SGOBw.2t—eQOBwt wiJ^nwit-^w^&iiLw^t—oieBUiwt ' 
therefore, simplifying, and putting wi » l/b, w^ =» l/e, » « l/a, 
WiJ^ + w^ + <at^ + (»! + ft»2)y» cos (aj—aj) <—«(» + «2) cos(«— wj) t 
— (« + wi)je cos (« — aj < = 0, 

or ^ + £ + i.' + f' (j + ,)co82x(i— LV--(* + «)co82»(l— L V 
a b c be \b e I ea \ e a I 

- ^ (rt + *) cos 2ir ( w = 0, an equation for t, 

ab \ a b I 



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98 

9277. (Rev. T. C. Simmons, M.A.)— Prove that the Taylor-circle of 
a triangle is always greater than its cosine circle, and that in an equi- 
lateral triangle the respective areas are in the ratio of 21 to 16. 



Solution bp R. F. Davis, M.A. 
If R be the circum-radias and » the Brocard-angle, we require to show 
that R{sin«A8in«Bsin'C + cos»Aco8>Bcos20}*>Rtan». 

If tan <f> a — tan A tan B tan C, since 

cot A + cot B-i- cot C— cot A cot Boot C « cosec A cosec B cosec G 
— a positive quantity, 
cot» + cot^ is always positive, sin(^ + «) is always positive, and 
^ + » < 180^. The expression for the Taylor-circle may be written 

T> cosec <t> -n sin « . 

it — . — s t\ ; 

cot « + cot ^ Bin (^ + w) 

hence the above inequality reduces to R . ^'^^ — is always > tan w« 

' 8in(«^ + ft») ^ 

or cos a > sin(a + ^), [« > and < 30°], cos^a > sin'(« + <^). 
We may square, as both are positive ; therefore cos <j> . cos {<j> + 2w) > 0. 

For an ocM^-angled triangle cot <f> is negative, and ^ lies between ^t 
and ir, and, since ^ + 2« < v + w{< 210°), both cosines are negative, and 
the inequality holds. When the triangle is o*^«*«^-angled, cos <t> is positive, 
and (p lies between and ^n ; hence cos (p . cos (^ + 2(a) > 0. 

But when the triangle is right-angled <t> » }ir, and a has any value 
from up to cot-^2. The radius of Taylor-circle « radius of cosine 
circle in this case. Therefore it would appear, for a triangle having one 
angle a little greater than a right angle and the other two angles nearly 
equal, the above theorem is not true. 

When the triaugle is equilateral, a — 30°, cot « -■ -v/3 ; 

cot^— (-L)' = -3-L^ooBec»^=l^l-|. 

1 8 

cot w + cot <b = 'v/3 — - — — »« - — — ; 

and ratios of areas - — ^^^^ ^ ,, : tan'« = 21 : 16. 
(cot « + cot <t>)^ 



8781, (Professor Hanumanta Rau, M.A.) — If S be the sun, and A 
dnd B two planets that appear stationary to one another, show that 
tan SB A : tsm SAB = periodic time of A : periodic time of B. 



Saiuiion by W. J. Greenstreet, B.A. ; C. Bickebdike ; and others. 
By a well-known formula, we have 

tan SBA : tan SAB = \rr^V I (-7^)* » a* : ** 
=- (by Kepler* 8 Law) periodic time of A : periodic time of B. 



I 



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9044. (S. Tebay, B.A.)— If A be the area of one of the faces of a 
tetrahedron ; X, Y, Z the dihedral angles over A ; and 

M = (l-co8«X-co8'Y-cos2Z-2cosXcosYco8Z)*; 

show that A/M has the same value for all the solid angles. 



• Solution by the Proposer; Prof, Ignacio Betens; and others. 

Let Aj, Aj, A3, A4 be the areas of the faces ; «, J, c conterminous 
edges ; and a, /3, 7 the angles contained by bcy ca, ab. Then, from the 
polar triangle, we have cos X + cos Y cos Z •= sin Y sin Z cos a. 

Squaring and reducing, we find M s sin Y sin Z sin a. 

Now V - I . ^^ sinY = f . ^^ sinZ ; 

b 

therefore V^ = ± . ^^^^^ sin Y sinZ 

be 

^ VA5A5 sinYsinZsina = 4.-^l^;^M. 
® ^csmo ® 2Ai 

Therefore ^ = 4 • ^'^'f^\ 

which is the same for all the solid angles. 



9122. (Professor Hudson, M.A.)— Prove that the locus of the feet of 
I)erpendiculars from the vertex of y^ — ^ax on chords that subtend an 
angle of 46° at the vertex is r^^2^ar cos e+lSa^ cos 2fl =• 0. 



Solution by R. Knowles, B.A. ; Rev. T Galliers, M.A. ; and others. 

Let PQ be the chord, A the vertex, M the foot of the perpendicular, 
and Xiy-iy x^^, hk the coordinates of PQ and its pole respectively ; the 

equations to AP, AQ are 2^1^ » yi^;, x^^y^ (1> 2). 

By the condition, yiX^—x^2 — ^1^2 + y 1^2 ^^ ^ (k^—4tah) = (4a + A)2...(3). 

The equations to AM, PQ are 2ay + kx = 0, ky = 2a {x-^-h) (4, 6). 

From (4) and (5), k '^—^ayjx ^-'2aiSi:D.e, h ^^{a^ + y^)lx ^r^—r.Beoe, 
and substituting these values in (3) we have the result in the question. 



(D. Edwardes.) — Prove that (1) the squares of the lengths of 
the normals drawn from a point xy to the ellipse b^x^ + a*y^ « a^^, are 
given by the equation {p^-^-iV +pW ■\'9q*)f^ + JJY}^ 

= 4{r*-(2V + 3i?2)r2 + 3U + V2}{(^4-3^)r*-(2p2U-3^4V)r2 + U2}, 
where U = b^x^ + a^y^ - a^^, V= x^ + y^-a^-b^, p'^a^ + b^, an'd ^=a«*2; 
and (2) if on the normal at P, a length PQ be measured inwards, equal 
to the semi -conjugate diameter, the squares of the lengths of the other 



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three normals drawn from Q are given by the equation 

+ {4(a-*)2PQ2(2aS+2*2 + a*)-4a«*2(2a2 + 2*2-7a*)} r' 

-4 {(«-*)2pQ2-a2*2}2 = 0. 

Solution by Professor Sebastian SmcoM, M.A. 

1. To eliminate k from the equations 

A;2"(A; + a2)2 (A- + ^2» (^k + a^y (k + 1^)^ ^ 
Eliminating a^ and y2 alternately, we obtain 

k* + 2*2^:3 + ^.2 {^4 + (^2 _ ^) y2_ ^2^:} _ 2>fca2^2^- a2i4y2 =, 0, 
k* + 2a2^ + ^2 1^4 + (^2 _ ^2) a;2 - ^2^2 J _ 2ka'^l^^ a*l^r^ = 0, 
whence, introducing IJ, V, &c., we have 

2A:S + X:2(H«-V)-(?4r2 = 0, k^ +k {V-ph'^-2g^r^ ^ ...{I, 2), 
the eliminant of which is the required result. 

2. The coordinates of Q will be given by 

whence U = («-*)« PQ2«a2i2, V = -2«*, then (I) may be written 

2X;2(A; + ad) + r2(A;2-a2^) = 0, 2;t2 + ^2;t-a^ = 0, 
and (2) becomes k^ + k {{a'-b)^FQ^ -a^^ ^{a* + H^^} ^2a^h^ = 0, 
and the required result at once follows by elimination. 



9401. (J. Brill, M.A.)— Prove that, if « and r be positive integers, 
{a+l){a + 2)...{a + n) _ {b+l) {b + 2) ... (b + n) (e+^) {c+2) ... (e + n) 
«i (»-!)! («-2)!2! 

wherea=»«r, *-(n-l)r-l, tf = (»-2)r-2, rf=: («-3)r-3, &c. 

Solution by W. S. Foster ; Sarah Marks, B.Sc. ; and others. 
Let (a + 1) (a + 2)...(a + ») = a'»+i?ia'»-i+ ...+jt7„; 

then the given expression 
= 5|2- J a»-«-.n . ^n-,+ i^ziL) ^.,_ ....| from y - to 17 = n, 

-^^J^^^'^-^'n[a-{r^l)y-^^ ^-^ [«-2(r+l)]n «-...jp,. 
and »'^-n[a-(r + l)]'^+^^il^[«-2(r+l)]*-... 

=s A! . coefficient of a;* in ^«—7j^<»-(»- ♦!)]«+ ^i!iziile[a-2(r+i)]«_ 

1.2 *"' 

f.<?.,in^[l-«j-(r+i)x-]n^ =0,if Ais less than «, and =»:w!(r+l)«,ifA=«; 
therefore the g^ven expression — (r + 1)». 



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9505. (Professor Wolstbnholmb, M.A., Sc.D.) — Prove, without 
evolution, or the use of tables, that 3x2* -2* lies between 3-5022831... 
and 3-502282... ; the latter being nearer to the exact value. 



Solution by D. Biddle. 

Let a = 2*. Then Za^-a=^ x\ also a* = 2, and 3fl»— a* « ax, whence 
a2 « 6 - aa; ; thus 3 (6-*r) -a = a-, and a {Zx + 1) = 18-a?. Cubing both 

sides, we obtain a^+ l%x « 106 (1). 

"We can now find x approximately by a series of trials, correcting accord- 
iDg to the successively reduced errors. Let A = the portion of x already 
found, say 3, which is easily seen to be the first figure, and let A + e = a;. 

Then we have (A + «)'+ 18 (A + z) = 106, A3+18A= 106-A; (2, 3). 

Subtracting (3) from (2), we further have 

3A«z + 3Az2 + z3+18s = A:, 2 = A;/(3A'+ 18 + 3Az + ««) (4,5), 

or roughly, especially as z diminishes, z — A;/(3A'+ 18) (6), 

A, A3, 18A, k, 3A2+18. 

3- 27- 64- 25- 45- 

3-5 42-875 63- 0125 64-75 

3-50228 42-95784 63-04104 0-00112 64-7979 

3-5022821, 42-9589211, 63-0410778, 0-0000011, 54-79794 
3-50228213 nearly = x. 

[Otherwise : If x be the value, and y = 2a:- 7, we shall then have 
y+7 = 6x2t-2^ (y + 7)3 = 216 x 4 -16-18 y 4 (y + 7), 
or(y + 7>3 + 72(y + 7) = 848sy3^.2iy2+219y + 847, or y3 + 21y2 + 219y«l. 
This cubic has only one real root which is positive, and 

which are the given limits. Since 21y ^y^\A nearly = -^, it is clear that 
the value of x is nearer to the inferior Hmit.] 



" Something or Nothing ? " By Charles L. Dodqson, M.A. 

In the years 1885, 1886, there appeared in regard to a Solution of Quest. 
7695 (see Vol. xliii., p. 86, and xliv., p. 24) a discussion about a 
difficulty in the Theory of Chances, of which the following question was 
treated as a typical example : — *' A random point being taken on a given 
line, what is the chance of its coinciding with a previously assigned 
point ? '* On one side it was maintained that the chance is absolute zero : 
on the other side it was maintained," by myself and others, that it is some 
sort of infinitesimal, and not absolute zero. The arguments on both sides 
were fully stated, and my only excuse, for re-opening the discussion, is 

VOL. XLIX. N 



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that I have a n^w view of the dif&culty to offer to the supporters of the 
** absolute zero ** theory. 

I assume that both sides accept the following axioms: — (1) that no 
aggregate, however infinitely numerous, of absolute zeroes can constitute 
a magnitudey however infinitely small ; (2) (an example of the preceding) 
that no aggregate, however infinitely numerous, of points can constitute 
any portion, however infinitely short, of a line ; and hence (3) that, if 
the chance of a random point on a line coinciding with a single selected 
point be absolute zero, so also is its chance of coinciding with one or other 
of a selected aggregate of points ^ however infinitely numerous. 

I now propose two questions : — 

I. *' A random ^int being taken on a g^ven line, what is the chance of its 
dividing the line into two commensurable parts ? " It seems clear that we 
are here dealing with a selected aggregate of points ^ since it is impossible to 
mark off any portion of the lincy and to say " Wherever, in this portion, 
the random point shall fall, it will divide the whole line into two com- 
mensurable parts." I assume, then, that my opponents would answer 
** It is absolute zero,''* 

II. ** And what is its chance of dividing the line into two inewnmen- 
surable parts P " Here again they must answer " It is absolute zero,^* 

And yet one or other of these two events must happen ! Hence, the sum 
of the two chances must be mathematically represented by unity ; that is, 
one or other (though we cannot say which) must be — not only 
** something y^^ not only a certain infinitesimal y of some inconceivably 
high order — but must actually reach, if not exceed, the finite value 
of one- half ! 



9506. (Professor Hudson, M.A.) — Prove that (1) the parabola 
y3 = 2/ (a; + /) can be described by a force to the origin which varies as 
r/(« + 2/)3; and find (2) what ambiguity there is in the case of this law 
of force. 

Solution by Professor Wolstenholmb, M.A., Sc.D. 
The polar equation is sin^ — 2— cos 6 + 2— , 
or 2/i# « -cos + (1 + sin' 0)*, (« = 1/r), 

\ rf02/ ^ <i« V(l + sin»0)*/ 

= (l+sin20)» + -^5l26L._8in2^co8^e= 2 ^ 

(l+sin-e)* (l+sin'e)* (l+sin^e)* 
therefore the central force 

The centre of force is the centre of curvature at the vertex of the para- 
bola, and, when the moving point reaches the vertex, it can describe, with- 
out any extraneous pressure, either the circle of curvature or the other 



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half of the parahola, apparently at its own choice. Of course the Ques- 
tion is not one of physics at aU. 

[Otherwise .—Let A be the vertex, S the focus, C the centre of force, 
AS = SO « il; draw SY, CZ 
perpendiculars on the tangent 
at P, and SR perpendicular on 
CZ: then AY is perpendicular to 
AS, and the triangles CHS, ASY 
are similar ; 

hence CR : CS = AS : SY, 
or CR.SY = AS2, 

and CZ.SY = CR.SY + SY2 = AS' + AS.SP = AS(AS + SG) 

= AS (AC + CM + MG) = AS (aj + 21), 
where PM is the ordinate and PG the normal at P. But we have 
central force : normal force = CP : CZ, 

and normal force = ..^^^'^^y'' Z ■ ^^ 

radius of curvature CZ2 2SY^ 

central force = ^^'^1'r.tF = ^'- CP/ 2AS {x + 21)^ = ^/"^ ^, . 
2SY3CZ» ' ^ ' l{x + 2f)^ 

It would be easy to make this strictly geometrical, or rather to 
frame a geometrical proof on these lines: an expression involving 
A* cannot be made geometrical, but it could readily be proved that the 
central force must vary directly as CP, and inversely as the cube on AC] 



9087. (H. FoRTKY, M.A.) — Show that, when the cards are dealt out 
at whist, the probability that each player holds two or more cards of each 
suit is -2062806, &c. ; or the odds are about 4 to 1 against the event. 



Solution by the Proposer. 
If one suit be distributed anyhow amongst 4 persons o, jS, y, 8, the num- 
ber of possible distributions is the sum of the coefficients in the expansion 
of (a-i-/3 + 7 + 8)**. But, if each person must have at least 2 cards, the 
number of distributions is the sum of the coefficients of the terms divisible 
by a^BrY^ ; that is, the sum of the coefficients in 

13 ! o2i8V82( _M_ + — ^^ — + &c. \ , 

Call this expression M ; then, if all the 4 suits are distributed subject to 
the same restriction, the number of distributions will be the sum of the 
coefficients in the expansion of M'*. But at the game of whist 13 cards 
are dealt to each person. Therefore, with the above restriction, the 
number of possible deals is the coefficient of a^^0^^^d^^ in M*, or of 

2tt^ Sa^a Sa^fl^ 2a3j3y 

•(2!)3 6!3!(2!)2 6!4!(2!}2 6!(3I)22! 



{i3!r{^ 



^ Sa^fl^y ^ ^o?Byd 



(4 I;2 3 ! 2 » 4 ! (3 






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or (reducing) in 

86804 {18 (6) + 42 (41) +63 (32) + 84 (311) + 105 (221) + 140 (2111)}*, 
where for simplicity for 5a*, 2a*j3, &c., I write (6), (41), &c. 

Now, for 18, 42, &c. ... 140 write a, b, .../, and let a (5) = A, 
*(41) « B, .../(2111) = F. Then (omitting for the present the factor 
8580'*}, we have to find the coefficient of (5555) in the expansion of 

(A + B + C + D + E + F)^ 
or in 5A* + 45A3B + 65A-B2 + 125A2BC + 242ABCD. 

As a preliminary step I give the squares of the quantities A, B, &c., and 
the product of every two in terms of the symmetnc functions ; but as we 
want ultimately the coefficient of (5555), 1 omit from these values all 
functions involving an index greater than 5. 

A2 = a2.2(55), 

B2 = d* {2 (55) + 2 (541) +2 (442) +4 (4411)}, 

C« = c2 {2 (55) + 2 (532) + 2 (433) + 4 (3322) } , 

D2 = rf2 {2 (442) + 4 (4411) + 2 (4321) + 2 (3322)}, 

E2 =r tfS {(442) + 2 (4411) + 2 (433) + 2 (4321) + 6 (4222) + 4 (3322)}, 

F2-=/2{(4222) + 2(3322)}, 
AB =a*(54l), AC=a<;(532), AD = eK;(5311), AE»«^(5221), AF=0, 
BC = be {2 (442) + 2 (433) + (4321)}, 
BD = bd {(541) + (532) + 2 (5311) + 2 (4411) + (4321)}, 
BE = be {(532) + (5221) + (4321) + 3 (4222)}, 
BF = */{(5311) + 2(5221)}, 

CD = ed {(541) + (5311) + 2 (433) + (4321) +6 (3331)}, 
CE = «J {(541) + (532) + 2 (5221) + 2 (442) + (4321) + 3 (4222) + 2 (3322)}, 
CF =c/{ (5311) + 2 (4411) + (4321)}, 

DE^de {(532) + 2(5311) + 2 (5221) + (433) + (4321) + 3 (3331) +4 (3322)}, 
DF = df {(5221) + (4321) + 3 (4222) } , 
EF = <?/{(4321) + 3(333l) + 2(3322)}. 

We shall now have no difficulty in finding the coefficient of (5555) in 
every term of the expansion of (A + B -r C + &c.)*. Take, for instance, 
the term 12D2EF ; then, omitting for the present the factor 12 which 
multiplies all terms of that form, we have 
D2EF « DE . DF = d'-t/ {{5'S2) + 2 (5311) + 2 (5221) + (433) + (4321) 

+ 3 (3331) + 4 (3322)} x {(5221) + (4321) + 3 (4222)}. 
But (5555) can arise only from the products of the complementary func- 
tions (5221) X (433), (4321);4321), 3 (4222) x 3 (3331) (1, 2, 3), 

and the coefficient is 12 in (1), 24 in (2), and 9x4 in (3) ; therefore the 
coefficient of (5555) in D^EF = (t^efiVI + 24 + 36) = 72d^ef. 
It is convenient for the present to divide each coefficient by 24, which 
factor can be re-introduced after summation, and this being understood 
the coefficient of (5555) in D'^'EF is Zd'^ef, Determining the other co- 



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efficients in like manner, and collecting those corresponding to 2 A"*, 
2A-^B, &c., we get 

Group of 
terms. 
2A^ 
2A3B 



2A2B2 
2A2BC 

2ABCD 



Coefficient of (6556). 

a* + 9{b* + e* + d* + e*) +f* = P, suppose, 
^(2a + 6(i+/) + c3(2a + 6tf)+rf3(a + 6* + 3c + 6tf + 2/) 
+ e'^{a + Sb'^6e+l2d+7/)+P^ = Q, 

+ er^(8^+/2) + 2«2/2 = R, 
i^{ad+ Zed + 2ce + Scf+ 2de) 
+ c^{ae+2bd+ Zbe + 2bf+ Sde + df+ 2ef) 
+ rf- {2bc + 2be + b/+ Zee + 5cf+ Zef) 

+ e^iadi-2be+ibd + ib/+\^ed + y-i-Zdf)+P(ed + ee-^^)'S, 
abe (d + <?) + aede + be (ode + 2df+ef) + def{b + c) = T. 

Substituting for a, by e, &c., the numerical values 18, 42, 63, &c., and 
reducing, it will be found that 

P = 2,090,086,733, Q = 4,366,141,668, R = 1,676,987,172, 
S = 3,888,459,288, T = 366,476,292. 
Therefore the coefficient of (5555) in (A + B + C -r &c.)'*, or in 

^A-* + 42A^B + &c. = P + 4Q + 6R + 12S + 24T = 85,079,518,906. 

Now, remembering that every coefficient Was divided by 24, and that 
we have omitted the factor 8580^, we see that the nimiber of ways of 
dealing out the cards so that each player may hold 2 or more of each suit, 

= 24 X 8580-* X 85,079,518,906 = H suppose. 
Then log H = 28*0439855. 

Let K = number of ways of dealing out 4 hands, without any restric- 
tion. Then K = 52 ! -^ (13 I)-* and logK = 28 7295271, therefore 

logH/K = log H -log K = -3144584 = log -2062806, 
therefore H/K = -2062806 ... = the required chance, 

and the odds are about 4 to 1 against the event. 



9481. (W. S. McCay, M.A.) — AB is the diameter of a semicircle; 
show how to draw a chord XY in a given direction, so that the area of 
the quadrilateral AXYB may be a maximum. 



Solutions by [\) the Proposer ; (2) D. Biddlb and Prof. MacMahon. 

1. Drawing the chord XX' perpendicular to XY, the quadrilateral is 
equal to the triangle XX'B, and if the chord BC be drawn parallel to 
XX' the problem is reduced to construct the maximum rectangle standing 
on BC and having its vertices (XX') on the circle, the solution of which 
is well known ; X is the middle point of the intercept on the tangent at X 
between BC and its perpendicular bisector (Townsend, Vol. i., p. 47). 



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2. Let be the mid-point of AB, and centre of the semicircle. Draw 
AC parallel to the j?iven direction, and OD perpendicular to it, cutting 
AC in E. Bisect AE in F, through which, with centre A, draw the arc 
GFH. Also draw GI perpendicular to AB, and make IK = AG. Draw 
KL at right angles to AB, and make AM = AL. Bisect FM in V, and 
make EP = VF. Finally draw PX parallel to OD, and XY parallel to 
AC. The quadrilateral AXYB, being completed, is that required. 




For the conditions are fulfilled when, with infinitesimal bases tangential 
to the semicircle at X, Y, the pairs of triangles whose apices are respec- 
tively at A, Y, and at B, X, counterbalance each other, that is, when 
YT-YS = XR-XQ. Let AB = 1, then YS = XR = J(l-cosXOY) ; 
also XQ = i (1-cos AOX), and YT = J (1-cosBOY) ; 

whence i (cos AOX + cos BOY) = cos XO Y = 1-2 sin« DOX. 
But it is easy to see, by reference to p, q, r in the diagram, that 

i (cos AOX + cos BOY) : sin DOX = AE : AG. 
"We therefore have a quadratic, whence 

sin DOX = {(iAE)2+i}*-iAE. 
The construction follows this formula. 

[ Otherwise: — For the maximum area, the squares of the variable sides mus* 
be m Arithmetical Progression. For, if we take X'Y' a consecutive position 
of XY, then it is evident that the areas XX'Y + YY'X differ only by an 
infinitesimal of the second order from the areas AXX' + BYY' ; hence, in 
the limit. (AX)2 + (BY)2 = 2 (XY)^. Now take, further, Z the diametral 
j}uiiit ut X ; draw AZ, YZ ; draw the diameter GM parallel to the given 
direction and hence bisecting YZ at right angles, and draw AP' perpen- 
dicular to OF. Then AY^ + AZ2= 2YZ2= SM'Y^ ; hence AM'^ = 3M'Y2, 
AO^ + Oil's + 2PG. 9M' = 3(OY2-OM'2;, (P'G + 2GM0 (20M') = 20A2; 
and tbiiB the quadrilateral may be constructed by finding Q', so that 
l>(i\ i Hi'= 20A2, and through M', the mid-point of OQ', drawing M^ 
perpt'iidicular to GM' ; this determines the vertices Y and X.] 



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9459. (Professor Genese, M.A.) — If py 9 he the polar coordinates of 
a point whose coordinates referred to axes inclined at any angle w are 
«, y, then a:/p, y/p may be denoted by C (0), S {$), Prove that 

S(e-<^) = S(e).C(<^)-C(e).S(<^), 
c (e+<^) = c (e) . c (^)-s (e) . s (<f>). 



Solution by Profs. MacMahon, B. A., Ignacio Beyens ; and others. 

The following is a general proof by the method of projections. As cos B 
is the orthogonal projecting factor, so C (0) may be called the o^-gonal 
(om6gonal) projecting factor. Take any two angles e and <^, positive or 
negative, the initial line of ^ being the terminal line of ; then a unit 
distance on the terminal line of <b resolves into C (^) on its initial line and 
into S {<!>) in the w-gonal direction which makes angle « + d with a;-axis. 
Take the w-gonal projection of all three on this axis, then (whatever 
a.ud<p) C{0+<f>) =C{<t>) .C(0)+S(<p).C (« + e); but it is evident from a 
figure that C (co + 0) = -S {0), therefore 

C(0 + <t>)^C{0),C{<i>)-8{e),8(<p) (1). 

Now change into (a— 6), and observe that 

s («-e) = c (0), c («- 0) = s (e), 

therefore S{0-<t>) « 8 {0) .C {<!>)- C{0) ,S{<f>) (2); 

hence also 8(0 + <t>) ^ 8(0) .C (-<l>) + C (e) . S (<l>) (3), 

C{0-<l>) = C(0),C{^<l>) + 8{0).8{<p) (4). 

It is worthy of remark that the orthogonal formulas for sin (0 + <f>) and 
cos {0—<l>) are not so general in their nature as the other two formulas, 
since they require C (— ^) = C (<^), which is true only when tp = nir [n an 
integer). It may also be noted that, when <p ^ n ,^ir, (n an odd integer), 
C(-<^)«-C(^). 



9477. (Swift P. Johnson, M.A.) — A, B, C and a, h, c are two triads 
of points on a sphere ; show that, if the circumcircles of the triangles 
KbCf "Bca, Oab meet in a point, then the circumcircles of the triangles 
aBC, bCA, cAB will also meet in a point. 



Solution by Professor Schoute. 

A stereographic transformation and a transformation by reciprocal radii 
vectores, the centrum of which is the point common to the circles Abe, 
Bcttf Cabf lead to the following self-evident problem. When on the 
sides b'e\ c'a', a'V the points A', B', C are given, the circles a'B'C, 
h'Q'A!y c'A'B' meet in a point. 



9516. (I^- Biddle.) — Prove or disprove that (1) a circle B is not 
properly drawn at random within a given circle A, unless its centre be 
first taken at random on the surface of A, and its radius be subsequently 



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taken at random within the limits allowed hy the position of its centre ; 
(2) putting unity for the radius of A, r for the radius of B, and x for the 
aistance hetween the two centres, there are two things requisite in order 
that B may include the centre of A, namely, that x he less than |, and 
that r he hetween x and 1 — ir ; (3) from a favourahly placed centre, the 
chance of the radius of B heing such as to make it include the centre of 
A is (\-2x)l{\—x)\ (4) the chance is identical for 2ifX . dx positions, 
which form the circumference of a circle of radius x, around the centre of 
A ; (5) the prohahility that a circle B, drawn at random in a given circle 
A, shall include the centre of A, is not correctly foimd hy the formula 

14 rl-ar t\ r\-x 

xdxdy -f- 2ir xdxdy = i, 

ix Jo Jo 

since this assumes that the numher of circles capahle of heing drawn 
from anv centre is proportioned to the upper limit of the radius ; leaves 
out of account that one centre, (me radius, one circle B, are taken each 
time ; and gives a result which actually does not fall short of the chance 
that the centre alone shall he favourahly placed ; (6) the prohahility in 
the case referred to is correctly found as follows : — 

P = 2ir f*a: (^,^) dx-^ 2Tr T x , dx ^ H + 2lDgei 

+ 2-61370564 = 0-11370664, or less than |.. 



Solution hy W. S. B. Woolhouse, F.R.A.S. 

By " drawn at random,** as stated at the heginning of this question, it 
should he understood that there is not to he any conditiort whatever affect- 
ing the circle B, excepting that it must he wholly included within the 
circle A. In order that the circle B may be properly regarded as so drawn at 
random, the correct mode of procedure is to discuss all the cases that can 
arise from an indiscriminate admission of every position of the centre, and 
also, at the same time, of every possible magnitude of the radius consistent 
with the foregoing condition alone. This important principle is strictly 
carried out in (5), the first working stated in the question by which I con- 
sider the probability P = J to he correctly obtained. 

The probability otherwise deduced in (6) would be correct if the circle 
B were drawn at random in a very restricted sense, that is, assuming (I) 
that its centre is first taken at random within the circle A, and (2) that 
onlij one radius is allowable. This is undoubtedly special, and the circle 
B is not drawn at random in the free sense of the term, the cases taken 
into consideration being but an infinitesimal portion of the total cases. 

To further show how misleading partial considerations are in questions 
of this nature, suppose that the circle B is first taken at random from an 
infinite set of circles having radii from to 1, and then placed only once 
on the circle A, radius = 1 . Now, if the radius p of B fall between and |, 
it may or may not include the centre of A ; and if it should fall between 
\ and 1, it will, when placed on A, be certain to include the centre. The 
resulting probability according to this arrangement will therefore exceed 
\. Thus, when p = ... J, the probability will be 



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109 

and, when p «= ^ ... 1, it will be unity. The resulting probability is 
therefore f * -^— dp+Cdp^2(l- log, 2) = -61 21 . 

Jo (1-pr h 

If the gfeneral problem be considered absolutely, so as to include all 
possible cases of the circle B and its positions, then, when the radius p 
ranges from to \, the successful positions of the centre = 2irp2 ; and when 
p ranges from J to 1, the number of positions = 2ir (I— p)^, all of which 
are successful. Whence the total successful positions 

-2Tj'pVp + 2Tj\l-p)2^p=2ir(^ + V,) -iT. 

Also for every value of p the number of positions = 2ir (1 - p)', so that 
the total number of positions - 2ir [ {l-p^dp = Jjt. And therefore by 
division P « i, as before, which is the true result. 



9407. (W. J. Greenstreet, B.A.) — From a point outside a circle 
centre C, APQ is drawn cutting it in P and Q ; AT is a tangent at T : 
show that it is always possible to draw such a line that AP shall equal 
PQ, as long as AC < 3CT ; and that then 3 cos TAG =» 2 V2 cos PAC. 



Solution by R. Knowlbs, B.A. ; Sarah Marks, B.Sc. ; and others, 

PQ cannot be greater than the diameter of the circle ; it may be equal 
to it, and in that case AC » 3 times the radius. 

CosTAC = AT/AC, cos PAC - AN/AC (N the mid-point of PQ), 
therefore cos TAC / cos PAC = AT/ AN . AN = JAP ; 

and (Euc. iii. 36) AT =v/2 AP, therefore 3 cos TAC = 2^2 cos PAC. 



9425. (Professor Hanumanta Rau, B.A.) — Prove that the sum of the 
products of the first n natural numbers taken three at a time is 



Solution by Prof. Aiyau, B.A. ; Rosa Whapham ; and others. 
Let „P3 stand for the sum of the products taken three at a time. Then 

nP3= n-lPa + ♦»«-H 2 = n-lP3 + ♦»•!] ^ -f ^^ q J 

Therefore „P, - 2«„.,P, = 16 ^Ji±llf' + 2o'^^^6 (5±i)!i' 

6 ! ! 4 

^i^n^{n^\Y{n--\){n-2), 

VOL. XLIX. O 



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110 

Similarly, 

.p. - :S«->P, = 105 M!!V210 (?Lt |J!!! + 130 (l±l)!l' + 24 (l±l)^'. 

8 ! 7 ! o ! o ! 

And, calling the coefficients (106, 210 Ac), «, *, c, d, we have 
.P. = 9« (^^Jl^Vs (a.*) (!i±lll!l. 7 (*..) ^^' 

and 80 on for nT^, ^Py, &c. ; the first coefficient in J^r being 
1.3.6... (2r-l). 



9390. (N'Importb.) — In any triangle ABC, prove that 
a cos 2A cos (B-C) + &c. = - ?? - - ^. 



Solution by G. G. Storr, M.A. ; J. Young, M.A. ; and others, 
a cos 2 A cos (B — C) = 2R sin A cos 2A cos (B - C) 

= R (sin 3A- sin A) cos (B- C) ; 
2 8in3Acos(B-C) =0, 
2 sin A cos (B - C) = sin 2A + sin 2B + sin 2C =» 2A/R* ; hence, &c. 



9469. ("W". J. C. Sharp, M.A.") — If ji? be a prime number and r<p^l, 
prove that (1) r! (ji?— r — 1)! +(— l)** is a multiple otp; and hence (2), if 
p^ 2^-1, {(^-l)!}»+(-l)«-^ is a multiple of 2^-1. 

Solution byW.S. Foster. 

1. Since (jp-r- 1) ! = M (;?) + {^l)P-r-i (r + l)(r+ 2) ... (p-l)f 
therefore r! (jj-r-l)! ^^{p) + {-l)P-r-i (p-l)]; 

and, since j9 is a prime number, we have 

(i)-l)! + l =-M(j3), therefore r!(p-r-l)! = M(j5)-(-.l)i»-»-»; 
and p must be greater than 2, or r would be nothing, therefore p — l 
must be even, therefore (- 1)**"*""^ = (— 1)% 
therefore r! {p-r- 1)1 + (-ly = M (p). 

2. Let p ^ 2^-1, and r = ^-1, therefore {(^-l)!}2+(-l)«-i is a 
multiple of 2g— 1. 



9365. (W. J. Barton, M.A.)— In the expansion of (1-3« + 3j;3)-» 
ow that the coefficient of x*""^ is zero. 



show 



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Ill 

Solution by H. Fortbt, M.A. ; C. E. Williams, M. A. ; and other*. 
Let 
(l-3d? + 3a:2)->-.(l-«a:)(l-/B») = (1 +aar + a2x« + &c. + a«— ^x*— l + &c.) 

X (1 + i3x + ^2a;2 + &c. + i8^- 1 a*»- 1 + &c.). 
Here the coefficient of ««"-J = a)'*'-* + a**-* /S + &c. + a/S^"-* + )3*"-^ 

= (a'^-ZS^-j/Ca-iS). 
Now a + i8 = 3, a3 = 3; therefore a -i[3 + (-3)]*, 3 = i [3-(-3)]*, 
.«'-t[l + (-3)]*-3«, i8««i[l-(-3)]» = 3a>«, 
where « is a cube root of unity, therefore a* = 3^ « pfi, and a'"— /S*" = 0. 
[We know that, when ^ < 4atf, the coefficient of x^ in the expansion of 
(«r» + ^a? + <r)-iiB r;^ sin (n-h 1)7 ^ 

c sin 7 

where rco8 7 = — */2<?, and rsiny- (4ac— *')*/(2(j) ; hence, with the 
given values, we get for the coefficient of sfi'^- ^ 

r\ (6«-i) sin 6«7 ^ »i(«« - 1) sin 180£ ^ ^ , 
sin 7 sin 30° *-' 

The readiest way of obtaining the development is perhaps by Hornek's 
method of Synthetic Division ; and by observing the law of the series, we 
see that if A, D be two consecutive coefficients, these and the following 
coefficients will be A, 0, -3A, -9A, -18A, -27 A, -27A, 0, &c. ; 
hence, if occur in any term, it will occur every 6th term. But it occurs 
in the 6th term, and therefore occurs in the 6«th term, that is to say, as 
the coefficient of x^ *-i. 



9503. (Professor Bordage.) — Show that the roots of the equation 
22x+2 + 4i-«^ 17 are a? = ±1. 



Solution by A. M. Williams, M.A. ; A. H. Lewis ; and others. 
The equation becomes 2^* + -^ — -»/-. 
Solving in the usual way, we get 2^ • 2', or 2*'. 



9389. (Professor Hanumanta Rau, M.A.) — Prove (1) that sin 6° is a 
root of the equation IQjc* ^^x^-l^x^-^x-^-l =» ; 
and (2) express the remaining roots in terms of trigonometrical functions. 



Solution by J. Young, M.A. ; Professor Nash ; and others. 
The equation 2 sin 6a— 1 = has » = 6° for one solution, and 9 = 30** 
for another ; expanding and removing the factor 2 sin d— 1, we have the 
equation given in the question, on writing x for sin 0. The remaining 
roots will be found to be sin 78°, sin 222°, sin 246° ; or, in terms of functions 
of acute angles, the four roots are sin 6°, cos 12°, -cos 24^, — C08 4b°. 



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112 

9410. (A. E. Thomas.) — If n and r are positive integers, and n>r, 
then {e being the Naperian base) 

l+!i±l^.i . (!i±^Ji!L±2) ^l(n^ l)(n ^ 2){n ^ 3) ^^^ ^ 
ri-l 2! (r+l)(r+2) 3! (r+l)(r+2)(r + 3) '' 

1 h**-^t ^ {n-r)(n-r-l) 1 (n-r)(n~r- l)(n-r-2) . •) 

r + 1 2! (r^l)(r + 2) 3! (>•+ l)(r + 2Xr+ 3) '*• *j 



'[ 



Solution by R. F. Davis, M.A. 
This identity follows from equating the coefficients of «♦•-•• in the de- 
velopment of (l+a:)'»^+* = ^{(l+a;)»»^}, and dividing each side by 
,»Cr ; and the necessity for f» >r is easily seen. 



9499. (Professor Ath Bijau Bhut.) — Prove that the orthocentre of a 
triangle is the centroid of three weights, proportional to tan A, tan B 
tan C, placed at the comers A, B, C. 

Solution by W. J. Greenstreet, B.A. ; Col. H. W. L.Hihb ; and others. 

With trilinear notation, o. /8, 7 being the centroid of masses oaj bfi, ey 
or a sin A, /B sin B, 7 sin C, at the angular points, we have 

a = aabBinc / {aa + bfi + cy) ■■ o . 2 A/2a = o. 
Similarly, 3 = 3 and 7 = 7. 

Now, the orthocentre is sec A, sec B, sec C, therefore it is the centroid 
of masses sin A, sec A, etc., or tan A, tan B, tan C. 

[The in-centre is 1, 1, 1, therefore is centroid of masses sin A, 
sin B, sin C ; the circum-centre is cos A, cos B, cos C, therefore is cen- 
troid of masses sin 2 A, sin 2B, sin 2C ; the ex-centre (A) is —1, 1, 1, 
therefore is centroid of masses —sin A, sin B, sin C; the nine -point centre 
is cos (B — C), etc., therefore is centroid of masses sin A cos (B— C), etc., 
i.e.y of masses sin 2B + sin 2C, etc. ; the symmedian-point is sin A, sin B, 
sin C, therefore is centroid of masses sin^ A, sin' B, sin- C ; the cen- 
troid of the triangle is cosec A, etc., therefore is centroid of masses 1,1,1.] 



9482. (S. Tebay, B.A.)— AB, AC, AD are edges of a tetrahedron ; 
BE, CF, DG perpendiculars on the opposite faces ; P, Q, R their areas ; 
jp, q, r the areas CED, Di B, BGC ; and S the area of the base BCD ; 
prove that Pj» + Q^ + iir = S^. 

Solution by W. S. Foster; Professor Beyens ; and others. 

Let vectors AB, AC, AD be o, P, 7 respectively ; and let xYafi be the 
vector DG ; then, since G is in the plane ABC, 

SajS (7 + xVa/8) = 0, therefore x (Vo/3)- = - Sa/8y, 



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113 

Now r : R = tetrahedron DBCG : tetrahedron DBCA 

= S(a-7)(i3-7)^Vo/8 : S {a-y){&^y)(-y) 
= xS{a^-ay-yfi)Yafi I —80/87, 

and R- = - i {Ya6)^ 

therefore Rr : - i ( Voi8)2 = ( Vai3) - 2 S (Va& + V70 + V/37) Va/3 ; 

therefore Rr = - JS ( Vaj8 +- VJ87 + V7a) VajS. 

Similarly, Q^ = - iS ( VajS + V/87 + V70) Yya, 

and P^ « - iS ( Va/3 + Vi87 + ¥70) Y$y, 

therefore Pi> + Q^ + Rr = - ^ ( VajS + VJ87 -i- Yya)^ 

= i ( ^AH^ )^ ^^ ^^ ^® *^® perpendicular on BCD 
6 vol. ABCD ^ 



1; 



-(' 



2AH 



I \2 / 3 vol. ABCD \2 ,, . . ^_,^. , 
■ ] = ^ -^ ) = (triangle BCD)'. 



[The theorem has heen otherwise proved by ordinary methods.] 



8941. (W. J. C. Sharp, M. A.)— Prove that the cond-tions that th© 
binai y quaaiic (a, b, c . . . ^ .t-, y)" bhould be reducible to a binomial form, 



ff, b, Cf 
b, c, d, 



=■ 0. 



[This is a generalisation of the catalecticant of the quartic ; those of 
quantics of higher order admit of similar extension.] 



Solution by D. Edwabdes. 

Let the factors of the binomial form be given by Aj-' ^ giry + Cy^. Then, 
substituting differentiating symbols for the variables and operatiug on 
the quantic, the result must vanish identically. We thus get 

Ac — Bi + Ca = 0^ or a, b, c, d ... =-0. 



M-^c-^Cb = 
Ae-Bd^Cc = 

A/-B«+C<? = 
&c. 



a, 


i. 


0, 


*, 


0, 


d, 


1 c, 


d, 


«, 



/. 



9511. (E. B. Elliott, M.A.)— Of inhabitants of towns p per cent, 
have votes, and of country people q per cent. Also of voters r per cent, 
live in towns, and of non-voters * per cent. Find the proportion of the 
whole pojjulation who have votes ; and show that jo, ^, r, « arc connected 
by the one relation 100 {qr—pb) = {p + f) qr— {q + r) ps. 



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114 

Solution by E. F. Elton, M.A. ; Rev. J. L. Kitchin, M.A. ; and others. 

Let V and N = total numbers of voters and non-voters ; then 

100 — « : p = town non-voters : town voters = * N : -^ V, 

100 100 * 

100 — tf : q = country non- voters ; country voters = - ^^ N ; V, 

* •" -^ 100 100 ' 

therefore ^-^^^ . ^ = iL = L?^^ . 12?J1-'- 

p 8 \ q 100-* 

H^nce lOO(^r-/?») = (j» + »)^r — (^ + r)i?*, 

«» V 

xind = • — number of voters : whole population. 

lOOr-^r+i?* K + V ^ ^ 



9403. (RusTicus.) — Baby Tom of baby Hugh the nephew is and 
uncle too. In how many ways can this be true ? 

Solutions by {\) Professor Macfaklane, LL.D. ; (2) D. Biddle. 

1 , Using the method of the analysis of relationships described by me 
in Vol. XXXVI., p. 78, let T denote Tom, H Hugh, m male, / female, e 
child, c* parent; then the data are 

T = mccc-^inR, T = mcc-^c-^ mH (1, 2). 

By combining the two, we obtain ^ — mc c c-^ mc c c-^- mH (3) . 

Now, if the sex symbol before the third and fifth symbols of relationship 
ftxo the same, the equation reduces to 

H = mcccc-^mH, i.e., c-^ mH = c c c^ niE. (4). 

Now the sex of c-^ on each side cannot be the same, for then cc would be 
1 ; thkit is, a person could bt» identical with his or her own grandchild. 
Nor can the sex be different, for then one parent would be the grandchild 
of the other parent, which is against the law of marriage. Hence the 
Bpx isymbol in the third and fifth places cannot be the same. 

Kor can the sex symbol in the second and sixth places be the same, 

for (3J is equivalent to c-^niK = e c-^ mc c c~^ niK (5); 

and, Tinder the hypothesis, the sex of the parent of H is the same ; then 
cc-^mcc =ly or c~^mcc=c-^ (6). 

Now the sex of the first and third symbols cannot be the same, for 
then c = e-^, i.e., the child of a person could be the parent of the person ; 
nor ean the symbols be different, for then the consort of the child of a 
person could be a parent of the person, which is against the law of 
mairiJigo. The only cases left are 

mcincmc~^mcfefc-^m = 1, mcfcfc^mcmcmc-^m = 1 ...(7, 8), 
f)icmcfc-^mcfemc-^m= 1, mcfcme-^mcmcfc-^m = l...(9, 10). 
JTt>w (7) means that Hugh's father and Tom's mother have married each 
thtj Eippropriate parent of the other; (8) that Hugh's mother and Tom's 
fiif,h(jr have married each the appropriate parent of the other ; {9) that 
the fiithors of the two babies have married each the mother of the other ; 
And ( I Oj that the mothers of the babies have married each the father of 



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115 



the other. Hence there are four, and only four, possihle ways in which 
the phenomenon may he true. 

2. Otherwise: — In relation to an uncle, the species "nephew" com- 
prises the following varieties : (1) a brother's son, (2) a sister's sen, (3; a 
half-hrother's son, (4) a half-sister's son, (5) a wife's brother's son, (6) a 
wife's sister's son, (7) a wife's half-brother's son, (8) a wife's half- 
sister's son. This list comprehends all the varieties. 

Of these eight varieties, the four last are excluded from our present 
consideration by the epithet **baby," attached to both individuals, 
which puts the existence of a wife of either out of the question. 

Again, although we may suppose a grandfather to marry a grand- 
daughter, and that she and her mother bears sons about the same time, 
one named Tom, and the other Hugh ; and, although we may suppose, 
what is still more preposterous, that a grandmother marries her grandson, 
and bears a son about the same time his mother does ; such marriages are 
contrary to law, so that we must exclude also varieties (1) and (2), and 
restrict our attention to (3\ and (4). 

Let A, B represent husbands, and A', B' their roRpective wives, 
whilst figures at the foot signify first or second. We then obtain the 
following eight cases : — 



Case 1. 
A'l ^ A = A'j 

B'=pBi 



T 



Hugh Tom 
Case 3. 
A'i=pA = A'2 
J _ ^_ 

^2 T B' T Bi 

I 

A'2=^A 

Tom Hugh 
Case 5. 

AiyA'rsAs 

A2=FA' 



■gh Tom 



Hui 



Case 7. 
Ai =F A' = A, 

AajA' 
Tom Hugh 



Case 2. 
A'l =r A = A'a 

B'2 nr B "T B'l 

A'2=f A 

I 
Hugh Tom 

Case 4. 

A',=T=A = A'2 

B'2 =F B =j= B'l 

A'2=r A 

Tom Hugh 

Case 6. 
Ai=f=A' = Aj 

A2=fA' 

Hugh Tom 
Case 8. 
Aj q= A' = Aj 

I — — , 
B 2 "T~ B -J- B'l 

A2=r A' 

Tom Hugh 



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116 

It is quite possible, in these cases, that no disparity of age shall exist be- 
tween A and A'j, or between A' and Ag. 



Note on a Rectangular Hyperbola. By R. Tucker, M.A. 

The equations to the hyperbola with respect to which the triangle of 
reference ABC is self- conjugate, and to a tangent (or polar) thereto, are 

4, = a2(/>2-c2)+/3'(er2_a^)+y(«2-^S) = (1), 

aa'{b'2--(^) + fifi'(c^-a-)+rY'{a^-b^ = (2). 

From (2) we see that the polar of either of a pair of inverse points, 
i.e., foci of an in-conic, passes through the other point. The same property 
holds for the related points 

a=* ^ C * 6' + ^ C' + a^ a2 + A2 ^ ^* 

From (1) we see that the curve passes through the circumcentre fO)*, 
the Symmedian p')int (K), and the in- and ex-ceutres. Its centre, which, 
of course, lies upon the circumcirele, is 

a(*--c2)/a = fi{c-^-a^)lb = y{a^-b^)lc (4), 

the ** inverse'* of which is the **isotomic" of the Steiner point 

The tangent at the in -centre is a (i'-c^) ■»-... + ...= 0, a line which 
passes through {a\l^yC% {b^-\-e\ c^-\-a\ a^-k-l^), 

(cot A, cot B, cot C), and (l/*4-<?, 1/c + a, \la-\-b) (5). 

The tangent at K is oa (**— c^) + ... + ... =0, which passes through the 
centroid (1/a (A + c), ...), (cot A/rt, cot B/*, cot C/c), 

the last point in (5) and other points through which Kw passes (6). 

(See Note on ** Symmedian-point Axis, &c.'* Q. J., Vol. xx.. No. 78.) 

The tangent at O is a cos A (*2_^) + .,. + ... = q (7), 

which passes through the orthocentre and centroid. 

The polar of the centroid is o (^ — c^) /« + ... + ...= 0,. 
which is the circum-Brocardal axis. 

The polar of the Steiner point is aja + fijb + yje = 0, that is, the axis of 
porspective of ABC and of the triangle formed by tangents at A, B, C to 
the circumcirele. The "isotomic" of is the point (1/a-cosA, ..., ...), 
i.(?., the point P of my paper on " Isoscelians '* ?cf. Proc. L. Math. Soc.y 
Vol. XIX., No. 316), hence the inverse of P is (a^ cos A, ..., ...), the polar 

of which is a^ocos A (^2-c2) + ... + ... = (8), 

i.e., the line Pit of the above cited Note (Q. J., Vol. xx.). 

The polar of the centre of the Brocard ellipse is 

aa(*4-c*)f... + ... = (9), 

[♦ This follows also from the fact that O is the orthocentre of the 
ox-centric triangle.] 



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117 



which passes through the centroid and the inverse of Kiepert's point 

Similarly we can ohtain results from the polars of the two Brocard 
points. 

The asymptotes are given hy P^j + ^— c^. <?2— a', a'— i» (aa + .. . + .. .)*=0. 
where P « a^^c^ (1-8 cos A cosB cos C). 



9479. (A. Kahn, M. a.)— Solve the equations xyz = 24, 

«(y-2)2 + y(«-dr)2 + «(:f-j^)2 - 18, x^ {y-z) ^y^{z-x) -k- z^ {x -y) --2. 

Solution by R. Knowles, B.A. ; H. L. Orchard, M. A., B.Sc. ; andolhes. 
Let y — mxy z ^ nx\ then 

ars- 80/(m + m2« + «2) =. 82/(n f iw«« + m^) - 2^1 mn. 
Eliminating m from these equations, there results 

12«»-41n2 + 40/i-12 = 0, or («-2) (4«-3) (3«-2) = 0, 
whence » =« 2, f , J, and m = f , 4, J ; and, since x^ = 2i/mn, therefore, 
when m = 5, II — 2, a: = 2, y = 3, « «- 4. 

[By inspection we have iP = 2,y=»3,2 = 4; and the symmetry of the 
equations shows that congruent values are 

a: = 3, y = 4, 8 = 2 ; a? == 4, y = 2, « = 3.] 



9183. (A. R. Johnson, M. A.) — Investigate the induced magnetisation 
of an elHpsoidal shell composed of any number of strata hounded by con- 
focal surfaces. 



Solution by the Proposer. 

Let there be m strata bounded by confocal surfaces which may be con- 
veniently indicated by the suflfixes 0, 1, 2, ... m, counting outwaids. 

Let fin% V„ be the magnetic inductive capacity and the total potential 
in the stratum between the (w — 1)*^ and n^^ surfaces. Then, if the in- 
ducing potential be ES, the proper assumption is Vn = (A,»E + BhF) S. 
The conditions at the «*^ surface give 

An>iEn + B„+iFn = A„E,» + B„F„, 



whence 
where 



/t„^i(A„,iE,. + B„,iP,.) = fin (A„E,.+ B„Fn) ; 
fl„A»+i = -<7«Art — CrtB,,, a„B,,*i = inA„+/»B,, 

, E„ Eh a — / '*'» — 1 \ ^'*^»* /. _ Jf!L __ 1 

„ ^ — — i On — \ 1 I —t Cn — 1 

^n F„ ^M#»*l 'F..F„ /*«♦! 



F„ 

f iln E„ E„ 

/t« + l *M Frt 



M» En E„ 



gn = 



VOL. XLIX. 



M«.l F„ 

P 



(1). 



(2). 



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118 

From(l) ?^A„j- lii -"-^Xa^^i^ (^-^ -^^ A„ = (3). 

The internal potential Vq cannot contain F. Therefore Bq = 0, and 
therefore, from (1), ^qAi = — ^o-^* 

From (3) and from the relation just obtained, there results, after some 
reductions, 

*A,. . 1 ^ ( fln(Hn.l-fl>^ [" /*/.- !-/*» 1 M»(^»-^»»-l)-Mti-l(^»-^n-l) 1 
A„ 1 Mm t 1 (/*/» - /*«-l) L H-n^l — ^n On J 

fin- fin- 1 On )/ C M« (/*«-! — /^n-2) 

M'l-M'i-l 0«_l J /t»-I-/*'»-2 0M-l3' 

(M3(Mi-Mo) L'^i-Mi «i J Mi-Mooii' 

f Ml A) — Mo3i) 7 /^\ 

I oo i 



F 



wlierp /8„ = ^^', /S; =^; On = fin (fin- fi'„). 



Hunoe An,\IA^ « product of continued fraction (4) and those derived 
frriiii it by writing? «-l, w-2, etc., instead of n; the last being 
{;ti$ii ^^Bo)lao' That is to say, A„ + i/Ao = N„ + i, where N„+i is the 
numerator of the continued fraction (4) expressed as a proper fraction by 
the method of converarents. 

Now the potential due to the induced magnetisation can have no term 
in E in external space ; i.e., A,„ + i = 1. Therefore A^ = l/N„,+i, and 

A„ = N„/N,„u (5). 

Wy Diny in the same way find B„ from the equation 

0«+l \ On On Ml \ On Onhnl 

or proceed as follows. 

Frrjin (1) whenM;;f>w, B,. = ^JL^-X-^9n^n jgj^ 

a,n a'» 
so thnt (Mm+i-M»») B"»*i = ^^^-^0/» + M»n + l )8'«— MwiSw* (7). 

[* When the fraction is written in the customary way, the last term 
within each each pair of parentheses is to be placed over all that follows 



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119 

In the case of a composite spherical shell, E = c*, F = r-'-', and ' 

». = >•;'*>, 0:, = -:-^ '^», «» = ^4^ '-^r •, 

* " » + 1 '* t + 1 " 

SO that (4) becomes 

( fln{flu*l-fln) r /*nM-Mn-l _ {>'/*n4l + (* + 1). Mnj { 1 ~ (^n-l/^-'"' '} "[ 

_ /i,..i(M».i-M.O ('Viiy'^^^/etc. / ipi+{i + l)fio 
Mi.(m«— /u«-i) \ r„ / )/ '. / (2i + l)/ao * 

Fri^m the comp'ete potential subtract the inducing potential, and the 
result is the potential of the induced magnetisation. 



8853. (A. Russell, B. A.)— Prove that 



V = 
Jo 



J -^ V 4a%'^' U-p^' ic^q^J ^ ^ 



is a solution of the differential equation 






Solution by the Proposer. 
Since 



r r f / (^- _£L, t^ -^„-„, ^- — ^ e-i'^*'P'^'i')dHdpdq, 
Jo Jo Jo \ 4«%2' 4^2^2' 4^^.; ^ ^» 






therefore 3^ = 3 ( f \ f{...)e-i'^'-P'^9')dndpdq, 

at JQ Jo Jo 

also 



therefore 
a' 



e^^^ Jo Jo Jo V 4a2;*2' 4Z»V 4(:V/ 

Thus we see v satisfies the given differential equation. This solution of 
the equation of the motion of heat in an eolotropic solid is suitable for the 
case of a time-periodic source. 



9524-. (Rev. J. J. Milne, MA.)— If yi, J/a* 2^3 ^^^ *^® ordinates of 
three points P, Q, R on the paiabola y^ = 4«.r, such that the circle ou 
PQ as diameter touches the parabola at R, prove that 



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120 

Solution hy A. E. Thomas, M.A. ; Rev. T. Gallibrs, M.A. ; and others, 

(1) is derived from the well-known theorem, *' The algebraic sum of the 
ordinates of the points of intersection of a circle and parabola is zero," by 
supposing two of the points to coincide. 

(2) The points where the circle on mj, mj, as diameter meets the para- 
bola again are given by 

(am*— awi') [am^—amj) -k- {2am i — 2am) {^am^^2am) = 0, 
or, discarding the factor a' (m— /Mj) {in — m^jj 

nfi + (wi + iwj) m t »»! *»3 + 4 =0, 
which has equal roots if iwi <*» mj = 4, i,t. if yi '*' ya = 8«. 



9267. (Professor Han VMANTA Rau, M.A.)— Given the base and the 
vertical angle of a triangle, prove that the envelope of the nine-points 
circle is itself a circle. 

Solution by R. F. Davis, M.A. ; D. O. S. Davies, M.A. ; and others. 

Since the circura -circle is fixed, the radius of the nine-points circle is 
also fixed. The nine-points circle therefore touches, at the other extremity 
of its diameter through the mid -point of the base, a circle having this 
latter point lor centre and radius equal to the circum-radius. 



9314. (Professor Beni Madhav Sarkar, B. A.) — Solve the equation 
4- + yz =a a = 384, y + «x»»* = 237, « + a:y = <? = 192. 



Solution by R. F. Davis, M.A. ; D. Watson, M.A. ; and others. 

Since x — a-yz = (b^y)/z - {c—z)/y, we find y = (az— ft)/ (a^— 1) 

and X = (fta-«)/ (22-1) ; hence («-c) («»- l)^ + (az-b) (bz-a) = 0. 

There are, therefore, five solutions as each value of z determines a single 
value for both x and y. In the given example, a; = 10, y = 17, « = 22 by 
trial. The equation determining z is the subjoined quiutic, 

2*- 1922^-2s3 + 9l392s2_ 203624s + 90816 « 0, 

which is reducible to the following quartic by rejecting the root 22, 

24-170.»-37422« + 9068s-4128 = 0, or 

(22-852 + 40-728)2-=11048-46z2-15992z + 5786-77=- (10511z-7607)2&c., 

whence z = 189-6, 1-62, -63, -21-9 roughly; so that (-11*7, -18-1, 
-21-9), (--8,238, 1-62), (1*24, 2-02, 189-6), (380, '5, -63; are approxi- 
mately the other four solutions. 



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9092. (A. E. JoLLiFFE, M.A.) — Prove that 

Ml (2n-l)! i2n-2)l ... to (« + 1) terms = 1. 

nlnl l!(«-l)!(n-l)! 2!(«-2)!(«-2)! ^ ^ 

Solution by K. E. Thomas, M.A. ; Prof. Matz ; and others. 

3^ (1 +a:)- ^ (I ArxY [(1 +a:) -l]*- 

= (l+ar)'»*'--r(l+a:)«**-i+ ^-^^ (1 +ar)'»+'-2-... 

1 < ^ 
Equating coefficients of sf on both sides we get 

1 « (jL±r)l ., ('Ll^jdll + Li!^^ {n.r-^y. 

r\n\ r\ («-l)! 2! r! («-2)! ^ ^ ^ '^ 

r!«! l!(r-l)! («-l)! 2! (r-2)! («-2)! * »*" ^ -^ ^ ^ 
If r =» «, we have 
, {Iny. ('2M-1)! ^ (2M-2)! . , ... 



8782. (A. Russell, H.A.) — Prove that, if 

a3 (* + <y) + ^3 (c + a) + c» (a + *) = lahc (/? + * + <?), then 

<'>(*T^'--)/(.*-)-r-r=-«)/(.^-*) 

(3) (*2-(;2)fa-^V{3a' + a(i + c) + 3r} 
+ (c2-a«)(*--^y {352 + *((. + «)+,.«} 
+ («3-*2)fc--?^)^{3c2 + c(a+^) + a*} =0. 

Solution by the Pkoposer ; Professor Sarkau, M.A. ; and others. 
1, 2. The given rehition may be written 
{a^—bc)(b-{-c)a t ... + ... = 0, and since {a^-bc){b-\-c) + ... + ... = 0, 

therefore (a^-hc) ^ = {b'^-ca)^-^^ {iP-ab) 'i^\ (a). 

b — c c—a a — b 

c.. ., ^ ab-i- ac—2'>c bc + ba-2ca ca + cb — 2ub ... 

S""'l«rfy -iJZ? ^-i ^^ZiT- w. 

a{b-c) b(c-a) c{a-b) ^^^' 

From {a) we deduce ('i), and from (b) and {d) (1) follows. 



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3. The given relation may also T)e written 

a— ^■b +c =- ; 

-r c c -t a a -t o 

and putting a: = a — - — , y = &c., z = &c., x + i/ + z = 0, 
■*■ c 

and therefore . a* (y - r) + v** (z - a-) + r^ (x - »/) ^ . 
Substituting for x, v, &c. their values, after a little reduction, the re- 
quired result follows. 



9416. (J 0*Byrne Croke, M.A. Suggested by Question 9360.)— 
The sides of a polyhedron are of areas inversely as the pnrpendiculars on 
them from a po'nt O, and 00' mtiels them in Pj, P^, I'a ... Pm, respec- 

tively ; prove that ^^^4 ^^-^ . — ^ ... ^ -^,-; = n. 



Solution by Professors Curtis, M.A. ; Beyens ; and others. 

T\\^^ volumes of the pyramids which are subtended by the sides of the 

Eolyhudron at O are equal each to n-'V, where V = vol. of polyhedron; 
unrt.', ciiUiug the vols, subtended by these sides at O', t'j, r.^, t'a, &c., we have 

OPi OP2 •• ' w-iy u-^Y 



SBSG. (E. ViGARiE.)— Dans un triangle ABC si (a) est le pied snr BC 
do lu jivmediane issue du sommet A, et si (a') est le point conjugue 
hanniHiiquo de (a) ; deniontrer que Aa' est egale au rayon du cercle 
d'AiKiilonius coirespondant a BC. 



Solution by Professor Ignacio Beyens. 

Be k propriete de la symediaue bien connuj — = — on deduit 

aC b^ 

c/B _ oB ^ c^ J, . ,-D _ gg" ,Q _ f^^*^ 
a'C aC " Z»2' * - ^.2_^2' * ^' - ^2_^ ' 

Ertaia m [m) et (;«') sont les pieds des bissectrices qui partent du sommet 

A, on Liura mC = - — et Qm' = ,, d'ou le rayon du cercle d'Apollonius 

o-\-c c — b 

*^ ^ «^'"' niC-\-Q)n' abe ... 

ftera = =» = n (1). 

Jlais, par le theorerae de Slewart, dans le triangle ABa' nous aurons 
Ka'- . BC + AB^ . Ca' = B/ (AC^ + BC . C/}, 



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123 

et rempla9ant Ca', Bo', AB, AC, BC par leurs valeurs on aura 
rayon du cercle d'Apullonius d'apr^s (1). 



9412. (A. B. Johnson, M.A.)— Show that, if 1, 2, 3, 4, 5, 6 be six 
points on a conic, then = 2 (023) (031) (012) (156), 
2 denoting summation with respect to all terms obtained from the one 
presented by cyclic interchanges ; O denoting any point in the plane of 
the conic, and (456), etc. the areas of the triangles 456, etc., described in 
the order named. 

Solution by Professor Curtis, M.A. ; G. G. Storr, M.A. ; and others. 

If the six points (j'i, yi), (j^o, ya)* &c., all lie on the conic 
(flf, ^ ^/, ^» ^'2^, y, 1) = 0, 
we must have six linear equations in rr, ^, c, /, ^, ^, the condition for 
whose being simultaneous is the determinant 






= 0, or 2 



^1% ^L'/i* i/i" 



X 


^4« y4» 1 




^5» Vb. 1 




^6» y6» 1 



« 0. 



V> ^eye* ye'* arg, yg, 1 
Now, the first of the two determinants here written down is equivalent to 

- (^2^/3 - ^3^2^ (^3^1 -^\yz) (^1^2- ^2^1) » 

or to eight times the product of the triangles (023), (031), (012), and the 
other determinant is double the triangle (456). Hence the theorem 
stated. 

If we make O coincide with the point (jTg, yg), we see that the condition 
that six points may lie on one conic may be written 

2 (023) (031) (012) X (045) = 0, 

there being ten such products to be taken each with its proper sign. Of 
course, each when developed would express the property of Pascal's 
Hexagram. 



8766. (S. Tebay, B.A.)— If AX, BY, CZ be opposite dihedral angles 
of a tetrahedron, show how to construct the solid in order that 

{tan i (B - Y) " tan i (C - Z) } tan i ( A + X) 

+ {tan i (C-Z)-tan \ (A-X)} tanHB + Y) 

+ {tun i (A -X) - tani (B- Y)}tan i (C + Z) =0. 



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Solution by the Phoposer. 

Let a, 3, <; be conterminous edges ; x^y^z the oppositcs, and A}, A], A3, A4 
the areas of the faces bex^ eay^ abz, xyz ; then 

volume = I ^^^sin A = J ^^sin X ; 
a X 

therefore ^i^:^ - i*!"? A.» = *^ A,» = ^^^ Aj'. 

A4 ar sm A y sin B z sm G 

Tj sin A sin B sin O , 
If . ^ = -i—^r — -r— Ti» wo have 
sm X sm Y sm Z 

tan \ (A + X) = Mtan ^ (A-X), tan i (B + Y) = Mtan \ (B- Y), 

tan|(C+Z) =Mtani(C-Z); 

M being a common factor. Whence the proposed relation. We also have 

xyz 
These conditions furnish the required construction. 



9354. (Professor Mahendra Nath Ray, M.A., LL.B.) — A pencil of 
four rays radiates from the middle point of the base of a triangle, and is 
terminated by the sides. If the segments of the rays measured from the 
prigin be x^^ yi, aig* V^y ^3» Vzt and 3-4, y^^ show that the identical relation 
connectiog these lengths is 

= 0. 



*:'. 


'i'' 


*.-'. 


*;' 


y:^ 


yi\ 


yi•^ 


y;' 


('.V,)-' 


, (^-jyj)"' 


. (^ay.)-' 


, (^4^4)-' 


1, 


1, 


1. 


1 



Solution by J. O'Byrnb Crokb, M.A. 
Obviously the relation which is to be established must be independent 
of the position of the origin. Let » = vertical angle of the A ; and let 
c, e' be the parts into which it is divided by a line of length k drawn to 
the origin of rays. Then, we easily find that 

1/A;2 8in2 » = a- -2 sin^ € + y{^ sin' €'-2 (j-i^i) "* sin € sin c' cos « ; 
which may be written 

«i ^r H 'C2 yf ^ + 'ca (j-iy 1) - * + ^4 = 0. 
And, eliminating the constants between this and the three other similar 



equations, we have 



r-2 



x-2, 

^4"'» ^4"'* (^4y4)-^ 

a relation identical with that in the Question. 



y,-^ (^iyi)-S I 
y-2, (x2v^-\ 1 

2. (T.yA-K 1 



= 0; 



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APPENDIX I. 



SOLUTIONS OF SOME UNSOLVED QUESTIONS. 
By W. J. CuRRAN Sharp, M.A. 

2144. (Professor Wolstenholme, Sc.D.) — If from the highest point 
of a sphere an infinite number of chords be drawn to points uniformly- 
distributed over the surface, and heavy particles be let fall down these 
chords simultaneously, their centre of inertia will descend with accele- 
ration Iff, 

Solution, 

Let A be the highest point, AB the vertical diameter of the sphere, 
and let P bo a point such that BAP = ; the acceleration on a particle 
falling down AP is ff cos 0, and its depth at time t is ^ff cos^ . ^. 

From symmetry it is evident that the centre of inertia lies in AB. 
Now the area of a belt bounded by horizontal small circles such that 
the lines to them from A makes angles and + b0 ^ 27rrsin20 x 2r86, 
and the depth of the centre of inertia is therefore 

f *' ^Sf cos2 0fix 4ir>'2 sin 20 d0 f *' cos^ sin d0 
«Ji>__- «|^^Jj- = i^^, 

4irr2 sin 20^0 cos sin 0^0 

Jo Jo 

and the centre of inertia descends AB, as a particle would which started 
from rest at A, under the action of a uniform acceleration \g. 

I have, of course, assumed that the particles were equal, as the data 
would be insufficient without some rule as to their mass, and this simple 
supposition gives the correct result. All question might be avoided by 
inserting the word equal — thus, ** and equal heavy particles be let fall." 

The result may be confirmed by the fact that all the particles reach the 
sphere at the same time, viz., that in which the one which falls along AB 
reaches B, and that then the centre of inertia is at the centre, as it 
should be. 



2146. (Professor Nash, M.A.) — D, E, F are the points where the 
bisectors of the angles of the triangle ABC meet the opposite sides. If 
a:, y, z are the perpendiculars drawn from A, B, C respectively to the 



VOL. XLIX. 



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126 

opposite sides of the triangle DEF ; p^^ p^, p^ those drawn from A, B, C 
respectively to the opposite sides of ABC : prove that 

^ + ^' ^.^-ll + SsinAsin-S sin-^. 
x^ y* ^ 2 2 2 

Solution, 

AE - -*^- , AF - — - (EucKd, vi. 3) ; 
a+e a+b 

therefore the triangle AFE - i -— sin A = lil^^ 

and FE2 - *«(j2 i-A- + -J_ - . -? ^^ cos a) 

= , ^r— 7^ {2a2+2aft + 2a<? + A2 + <52-2(a2 + a* + fl<j + *c)cosA| 

(a + r)2(a + i)5 «^ > ' j 

A2-2 

— — V {2a2(l-cosA) + 2a(*-<JC08A) + 2a(tf-*cosA) + an 

(a f cy {a -k- b)^ ^ ^ ' "* 

= ^fjf^ {j-cosA + cosB + cosC} ; 

,.. ,, ^'^'f-^' ,, {i~cosA-hcosB-hcosC}«i- ^^^l__«in«A; 

therefore ki^^a^ {{-cosA + cosB + cosC} - Ji^cSginS^ ^ iPi^a^; 
therefore 3 - 2 cos A + 2 cos B + 2 cos C — Pi^/x'. 

Similarly, 3 + 2 cos A - 2 cos B + 2 cos C « P2^li/"f 

3 + 2cosA + 2co8B-2cosC =p^lz^\ 

therefore ^ + -^ + -^ = 9 + 2(cosA + cosB + cosC) 

«= 11 + 8 sin JA sin |B sin \G, 



2173. (Professor Wolstenholmb, Sc.D.) — The qnadric 
aa:2 + V + C22« 1 

is turned ahout its centre until it touches a'x^ + b'y^ + c'z^ = 1 along a 
plane section. Find the equation to this plane section referred to the 
axes of either of the quadrics, and show that its area is 

Tr{a-^b^-c-a'-b'-c')^l{abc-a'b'c^)^. 

Solution. 

Let kx^ + By2 + Cz^ + 2Fys + 2Gzx + 2Yixy = 1 he the equation to the 
quadric, the onginal equation of which was ax'^ + by^-¥cz^ «» 0, after it has 
heen turned about its centre. Then, by question, 

Ar2 + By- + 0^2 + 2Yyz + 2Qzx + 2Ha:y s a'x^ + h'y^ + cfz^ 

' +R(a:co8o + ycos/3 + «cos7)2, 



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therefore A = a' + R cos'o, B -= *' + R cos^ /3, C « (?* + R cos^^, 
F = Rco8/3co87, G = Rco8 7C08o, H — RcosaoosiS. 
Now, by a Paper in the Froe. Zond. Math, Soc, Vol. xin., pp. 193 — 94, 
A + B + C = « + * + c, 
AB + BC + CA-F*-G2-Hi»a* + ^ + M, 
ABC + 2FGH-AF2-BG2-CH2 = abc; 
therefore a + b + e^ a' + b' + e' + R, because co8'o + co8?/3 + cos'7 = 1, 
ab + bc + ca => a'V + b'c' + /a' + R {/»' (cos' /3 + cos' 7) 

+ V (cos' 7 + cos^ o) + c' (cos' a + cos' /3) } 
s= «'*' + b'e' + <^a' + R {rt' + *' + c'- a' cos' a-b' cos' /3 - c' C08'7}, 
abc = a'^<^ + R {^V cos' a + tfV cos' /3 + «'A' cos' 7} ; 
therefore a^b^c^a' -V -^ _ 1 



aAc — of b'e' b'(f cos' a + cfa' cos' 3 + a'b' cos* 7' 

which is equal to the product of the squares of the axes of the section of 

a'«' + yy' + tf'z' =s 1 by arco8o + yco8i8 + «cos7 =» 0, 
(Salmon's Oewn, of Three Dim., Art. 97). 

R and cos' a, cos' /3, cos' 7 are easily found from the above equations 
which are linear in those quantities, and so the equation to the plane 
j;cosa + yco8/3 + 2COS7 » 0. 



4721. (Professor Sylvester.)— Prove that every point in the plane 
carried round by the connecting-rod in Watta' or any olher kiud what- 
ever of three-bar motion has in general three nodes, and that its inverse in 
respect to each of them is a unicircular quartic. 



Solution, 

Let AB, BC, CD be the three bars of a three-bar system, where 
AB = a, BC « bf CD = c, DA = d, and P be a point rigidly connected 
with BC, and h and A;, respectively, the perpendiculars from P upon BO 
and the portion of BC intercepted between B and this perpendicular. 
Then, if BAD = <p, and be the angle which BC makes with AD, taking 
A as origin of rectangular coordinates, and AD as axis of ^, the coor- 
dinates of B are a cos <p and a sin ^, and those of C, 

a cos <p+b cos 0, and a sin ^ + ^ sin ; 
therefore c' = (a cos <^ + * cos d — rf)' + {a sin <^ + 3 sin 6)', 
or 2acos<^ (ftcosa— rf) + 2asin<^.isine = c^ + 2bdcoB0—tfi^b^—{P,., (1). 
And, if (iP, y) denote the point P, 

X ^ a cos (p-k-k cos 6 — A sin 6, and y = a sin <p + k sin 6 + A cos 0... (2), 
therefore 2(Arj? + Ay) cos0 + 2 (%-Ad;) sinO = ir2 + y'+ A' + ^-'-a' ... (3), 
and from (1) and (2), 

2(*a; + rfA?-W)cosd4-2(^y-rfA)sin0 = c2 + 2i^ + 2dii;-a'-A2-<i'... (4); 



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from (3) and (4), 4 {{bi/^dh) {kx + hi/)-(bx + dk-bd) (ky^hx)Y 
= {(*y-rfA)(a:2 + y2 + A2 + ^s_a2)-(Ary-Ax)(c2 + 2W- + 2d:r-a2-^-rf2)}« 
+ {{bx + dk-bd)(x^ + y^ + h^ + ki-a') 

^{kx + hfj){c^ + 2bk + 2dx-a^^b'-d^y (5), 

which is the equation to the locus of P. This evidently has double 

points, where 

by-dh ^b -^ dk-bd ^ <?-^2bk-^2dx--a^-b^-'d^ ^ sunnose • 
icif-hx kx-^hy a;2+ya + A2 + A-2_«3 ^' PP ♦ 

therefore p = dhx^dy{h-k)^h(c^^2bk-^a^-i;^^d^) 

__ by — dh __bx + dk-^bd ,^y , 

" ky-hx kx-ithy ^ ^' 

and, eliminating x and y from the equations (6), a cubic is obtained to 
determine /?, and each value of jp g^ves one node, and there are therefore 
three. 
The equation (5) to the locus of P is easily reduced to the form 

^{bh{x^^-y'^-bdhx-d{h'^^y^-bk)yY 

-= {*2(;r5 + y2)-2WAy-2W(*-Ar)-^ + rf2(A2 + *2 + ;t2_2*Ar)} 

X {a;2 + y2+A2 + A-2_aa} \f^2bk^-2dX''a^-V^^d^'\ 

which meets any circle at its intersections with a cubic, and therefore six 
times at infinity, i.e.^ it is a tricircular sextic. 

And, if the origin be at one of the nodes (and one must be real, since 
the cubic in p must have a real root), the equation will be 

Uo (a:2 + y2)3 + XJi (x2 + y2)2 + XJ3 (a;2 + t/VU3 + V2 = 0, 
where Uq, Ui, Uo, U3 are homogeneous functions of x and y, of order 
0, 1, 2, 3, respectively, and Vj one of order 2 ; therefore the equation to 
the inverse will be Uq + Ui + Uj + U3 + Vj {x^ + y^ = 0, 
a circular quartic. 



4828. (The Editor.) — If the comer of a page of breadth a is turned 
down in every possible way, so as just to reach the opposite side; 
(1) show that the mean value of the lengths of the crease is 

j{7A/2 + log(l + A/2)}fl, 

and (2) the mean area of the part turned down is \\c^. 

Solution , 

1 . Let ABHK be the page, and ADE the comer tumed down into th© 
position DCE, meeting the edge BH in C, and let ar, y be the lengths 
interctptud upon the edges AB, AK from the corjier to the crease ; then 



\ 



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we have AB = a, AE =» ar, AD = y, and (C being the point where the • 
comer A rests after folding) if AC cut DE in F, the angles at F are 
right angles ; therefore 

a2 = AB3 := AC3- BC2 « 4AF2- CE^ + BE^ 

therefore = 42:V- 2aa; (x* + y^), 
and 2a;y2 = a(a;2 + y3), 



and the crease : 



'<-*»=)'- (-^j' 
-•(^.)' 



Now ar may have any value from x ^ iatox — a; 
hence the required mean value 




^ 1 p / 2^\ efa; = i- r(c2 + «2)Ww, if c« = «,and2a;-a = w2, 
\ajia ylx—al a Jo 

= f^{7V2 + log(l-hV2)} = |- {7V2 + log(l+V2)}. 
2. Again, the area turned down = Ja-y, hence the required mean value 

iajja \2ar-a/ 4>v/a Jo Vz ' ' 16 



6391. (J- J* Walker, M.A.) — If O, A, B, C, D are any five points in 
space, prove that lines drawn from the middle points of BC, CA, AB 
respectively parallel to the connectors of D with the middle points of OA, 
OB; OC, meet in one point E, such that DE passes through, and is bisected 
by, the e^ntroid of the tetrahedron OABC. [^Quest. 6220 is a special case, 
in two dimensions, of the foregoing theorem in three dimensions.] 



Solution, 

This may be generalised into — If ABCD... be a simplicissimum in 
space of n dimensions, and Bi, Cj, D, ... the mid-points of the lines drawn 
from any point P to B, C, D ... (all the vertices but A), and if parallels to 
AB„ ACi, AD, ... be drawn through the centroids of CDE ..., BDE..., 
BCE..., &c., (the faces of BCDE ...) respectively ; these will all meet in 
the same point Q, AQ will pass through the centroid of PBCD ..., where 
it will be divided in the ratio of «— 1 : 2. 

Any straight line through (A.', /*', v ...) may be represented by 



a 



where a + b-\-c. 



b e 



And the parallel through (a.", /ia", v" ...) is 



\-x" 



_/t-M__^ 



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130 

If, then, ABCD ... be the simplicissiiiium of reference, and P (\', /*', i^ . . .), 
Bi,Ci,Di... are {K, i(M' + V), */...} {K, i/*', i(»'' + V) ...}, &c., 
and the equations to ABj, AC], ... are 
X— y __ ji ¥ jr __ A.— V ji^ V __ jr_ 

&c. &c. &c., 

and those to the parallels through the centroids of (C, D, E ...) 
(B, D, E...), *.<?., through 



are 



X « M ^ p-y/jn- 1) ^ ir~V/(«-l) ^ 
X'-2V"/i' + V 1^' "* ir' 

;i~V/(w-l) ^ _p_ ^ y~V/(n~l) ^ 



X'-2V /*' iZ + V ^^ 

which all meet where 

^ _ M~V/(n-l) _y~V/(n-l)^ ^ / ^\ 

x'-2v / »^ "* V«-ir 

80 that, if (x'', /a", p" ...) be the point of intersection Q, 

X" = - — 2--, /'«-'•-+ -, w = +- -, &c., 

fi— 1 n-r «— 1 «-l n— 1 n-1 

X — V it V 

and the equations to AQ are -T/ -rr "--=#/-=*••• » 

X —V fJL" V 

x-V ._. M. p _ 



x'-(» + l)V /+V / + V **•' 

and the Hne passes through (-^, '^^t-Y, ^;^j •..) or (X,^,7...), 

the centroid of (PBCD ...) ; and, since 

(«+l)X = (fi-l)x" + 2V, (» + 1)m= («-1)/*" + 2x0, 

(« + l)ir« («-l)v" + 2x0, &c., 

this centroid divides AB in the ratio ft— 1 : 2. 

[Solutions by Professors Minchin and Gbnbsb are given in Vol. xxxiv., 
p. 40.] 



7131. (W". J. C. Shaup, M.A.) — Prove that the vector equations to 
the centrodes of a three-bar motion, which are easily derived from one 
another by a linear substitution, are of the third degree in the vectors, and 
reduce to the second where the algebraical perimeter of the figure is zero. 



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131 

Solution. 

If AB, BC, CD be the bars, a, b, e their lengths, and AD = d the 
distance between the fixed ends, the instantaneous centre is O, the inter- 
section of the lines AB and CD. (Clifford, Di/namie, p. 166.) 

It then follows at once that, if AO = r and BO = r', 

2 cos AOD = tf±r^ = ^f)!±irl-_£)!=:^. 

®' rr" (r''a){r'^c) ' 

which is the yector equation to the fixed centrode, and is, in general, of 
the third order. If, however, r + r'±<?=s(r + r'— a— <?)±*, i.e., 
a+c ^ ±(b ±d), a factor will divide out, and the equation is of the 
second order. 

If BO =» R, and CO = R' the relation between R and R' will determine 
the movable centrode, and it is easily seen that this is derived from the 
above by putting r = R + a, and r' « R' + c ; 

and the equation to this centrode is of the pame order as that to the other, 

(R + R^ -g-g)^-rf3 ^ (R.|.R72_^2 

,^** (R-a)(R'-<j) R.R' ' 

which may therefore be deduced from that to the fixed centrode by writing 

b for d and vice verad. 



8177. (B. Hanumanta Rau, B.A.) — The images of the circum- 
centre of a triangle ABC with respect to the sides are A', B', C ; prove 
(1 ) that the triangles A'B'C and ABC are equal ; (2) that they have the 
same nine-point circle. Find the equation of the circumcircle of A'B'C 
and the angle at which the two circumcircles cut each other. 



Solution, 

If a, b, c be the middle points of the sides, evidently B'C is parallel to 
and double of be (Euclid, vi., 2), and therefore parallel and equal to BC, 
and so for the other sides of A'B'C, and this triangle is equal to ABC in 
all respects. Also O, the circumcentre of ABC, is the orthocentre of 
A'B'C, since A'O, B'O, and CO are perpendicular to BC, CA, and AB, 
and therefore to B'C, C'A', and A'B', respectively ; and a, i, e are the 
middle points of the portions of the perpendiculars intercepted between 
the orthocentre and the vertices, therefore the circle through abcy the 
nine-point circle of ABC, is also the nine-point circle of A'B'C. The 
relation of the triangles AliC and A'B'C is mutual, for, since 
BC s= BO = BA', the perpendicular from B on AC or A'C bisects A'C, 
and the orthocentre of ABC is the circumcentre of A'B'C. 

If 0' be this point, the angle at which circumcircles cut is 

where R is the circumradius «= cos-^ i (1 - 8 cos A cos B cos C). 



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132 

If {jr\ /, r') be the trilinear coordinates of O, the points A', B', C are 
-x', p,— y*, ;»3-s'; p^-x', — y', ^-r'; Pi-r', ;>j-y', - c' respectively, 
where Pi, p^ p^ are the perpendiculars of the triimgle ABC, and by 
sabstituting these in 

yz sin A + ex sin B + jry sin C — (jr sin A<f y sin B + z sin C) (Ax + Ary + fa) = 0, 
hj k, and / are easily determined, and the equation to the circumcircle of 
A'B'C The line hx + Ay + fc = 0, the radical axis of the circumcircles is 
the line bisecting 00' at right angles. 

The equality of the triangles ABC and A'B'C may be otherwise demon- 
strated in a way which explains the geometrical meaning of the equation 
to a conic circumscribed to a triangle, and demonstrates an interesting 
property of such figures. 

If P (y, y', z') be any point, A", B", C" its projections on the sides of 
the triangle of reference, and A', B', C points on PA", PB", PC", such that 
PA' = /. PA" = /x', PB' = y . PB" = y/, and PC = A . PC" = hz' ; 
the area of the triangle A'B'C 

= i{PB'.PC'sinA + PC.PA'sinB + PA'.PB'8inC} 

= i {yA sinAy'c' + A/sinB z'x'+^sinCx'y'} (1). 

Now, if/ = y=A = 2, and x', y', / be the circumcentre, the points 
A', B', C are those in the Question, and 

A A'B'C = 2 (// sin A + c'x' sin B + x'/ sin C) 

= 4 (A*Or+ AeOa+ AaOb) = aABC. 
Now, returning to equation (1). If P lie upon the circumscribed conic 
yA sin Ayz + A/ sin B zx +fy sin C xy = 0, 
the triangle A'B'C vanishes, i.^.. A', B', C lie on a straight line. Or, 
in other words, if the perpendiculars upon the sides of the triangle 
of reference from any point on the circumscribed conic 

Ays + fizx + vxy = 
be produced to points A', B', C, such that 

PA'=/.PA", PB' = y.PB", PC = A. PC, 
the points A', B', and C will lie in a line, if 

/: y : A :: sin A/ a. : sinB//i : sinC/v. 
The well-known property of the Simpson lines is a particular case 
of this. 



8592. (Professor Mathews, M.A.) — Through a point P are drawn 
three planes, each parallel to a pair of opposite edges of a tetrahedron 
ABCD. Prove that the 12 finite intersections of these planes with the 
edges of the tetrahedron lie on the same quadric surface ; and that, if 
BC'« + AD2= CA2 + BD2 = AB2 + CD2 {i.e., if each edge of the tetra- 
hedron is perpendicular to the opposite edge), there is one position of P 
for which the quadric surface is a sphere. 

Solution, 

If Ai, fij, vi, TTi be the tetrahedral coordinates of P referred to the 
tetrahedron as tetrahedron of reference, the equations to the three planes 



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133 

through P parallel to opposite edges of the tetrahedron are 

\-\-fi __ y + ir \ + y __ /i-t-ir , X-t-ir __ /i-t-y 

Now, if Aii\2 + Aj2fi2+ ... +2Ai2\fJL + 2A^X»+ ... - 

be any quadric, this meets x=:/a=0 at points at which 

A85i'2 + 2A34i/ir + A44*^=0 (1); 

but, by Question, these points lie on 

A- + y ^ M + ir ^n^i A + ir ^ t^ + p ^ 

and therefore the equation (1) is equivalent to 

(-1 ^U-!^ =^)-0; 

vAj + i'i Mi + ^'l' ^Ml + ''l A.i + iri/ 
therefore Ajs : — 2A34 : A44 

.. 1 . 1 , 1 . 1 

" (^1 + "1) (Ml + yi) ' (Ml + iri) (Ml + "1) Ui + "1) (^1 + iTi) ' (Xi + iTi) (/ii + Ti)' 
and so on. And the quadric 

X2 f^ 

(^1 + Mi)i(^i + ''i) (^1 + »i) (Ml + "1) (Ml + »i) (Ml + A^i) 

-^^^( I X 1 ]...«0 

(^1 + Ml) l(>^i + ''i) (Ml + ^i) (Ml + ''i) (^1 + »i) 3 

meets the edges at their finite intersections with the three planes. 

If Aii\2 + Aj2/i'+ ... +2Ai2\fi-" = be a sphere, 

•°-ii •** -^22 ~ 2-^12 _ Aji 4 Aa3 -- 2Aj3 _ A32 -j- A33 — 2 Ag __ « 
(1.2)2 (i.3)a (2.3)8 •' 

where (1 . 2), &c. are the edges of the tetrahedron of reference {Proceedings 
o/Lond, Math. Soe,, Vol. xvm., p. 341). 

And {ui + iTi) (Ai + Ml + 2vi) (\, + Ml + 2iri) « r (1 . 2)«, &c., 

or (^i + iTi) (O— (w 1 - I'l)'} = r (1 . 2)2, &c., where X + M + v + ir=0, 

and therefore, if Aj = /*i "■ •'1 = 'i> ♦•*•> if P be the centroid, 

(,., + iri)C-=r (1.2)2, (^j + »j)C2=r(1.3)2, (mi + i'OC^ = r (1 .4)', 
(^i + Mi)C« = r(3.4)«, (Ai + v,)C'=r(2.4)3, (a^ + ^,) C* = r (2 . 3)», 
and (1 . 2)2 + (3 . 4)« = (1 . 3)2+ (2 . 4)2 = (1 . 4)2+ (2 . 3)2 ; 

and the tetrahedron is rectangular (see Solution of Question 3228, 
Appendix III., Vol. xlviii., p. 168). 



8940. (W. J. C. Sharp, M.A.)— If 
S = a;r2 + by^ + cz^ + dw' + 2/ys + 2mzx + 2nxt/ + 2pxw + 2gyto + 2rzw, 
and Fi. 2 ^(1X1X2 + by i^j + WjZj + <?m?i«72 + ^ (yiSj + yjjc J + &c. ; 

VOL. XLIX. R 



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134 



show that Si88S,+ 2Pi.8P2.3P8.i-SiPj.3-S2l^.l-S3P? 2 



yi» i^2» ^3 M «1» «2> «3 W'l, ^fiy ^Z 

«i» «2i «3 + ... + 2L t/;i, tt'a* "'s h^i> «2> ^3 

t^l, «'2» ^^3 I I ^1» ^2» ^3 I I yi» t/%» VZ 

where A, m are the first minors of the discriminant of S. 



+ &c.y 



^ Solution, 

If X, y, Zf w be eliminated from S by means of the relations 
l\ + fi + y)x = XXi + fix^ + px^, 
(\ + /i + >/)y= Xyi + zuya + yyai 
(A. + jU + v) « = A«i + /*22 + •'«3» 
(A. + fl + v) M> « Aw?! + ^"^2 + ''"^3> 

the result is 

,,^^^ ,2 {^'Sl + M'S2 + I.SS3+2\MPl2 + 2/il'P23 + 2l'XP3i}. 

Now, as I have shown in a paper read before the London Mathematical 
Society in December, 1883, and in a note, JReprint,Yol. xliii., p. 47, A, fx, v 
are areal coordiDates of any point (a:, y, «, w) on the plane through three 
points (:r„ yj, «i, «?i), (a-g, yg, Zj' «'2»)» (^3» S^s* ^» *^3) referred to the tri- 
angle of which those points are the vertices. And the plane equation to 
the section of the quadric by the plane is 

A2Si + /i2S2 + I^Ss + 2X/iPi2 + 2/U>/P23 + 2«'AP31 =0 (1). 

If the plane satisfy the tangential equation to the surface, the section 
must have a double point. Now the equation to the plane is 

= (2), 



^1 


y, 


2, 


W 


^1» 


yi» 


«ii 


iTi 


a-2, 


y2i 


«2, 


w^ 


Hy 


^3, 


53, 


W3 



and therefore SiS883 + 2P23P8iPi2-S.P28-SaP3i-S3Pj2 = 0, 
the condition that (1) should have a double point, and 



yii 


!^2» y^ 


«i» 


Hy «3 


m;„ 


m;2, w^ 



. + 2L I 



M?i, iTa, t^al +...= 0, 

^i> ^2» H 

y\y y%y yz I 



«1> «2> ^3 
I ^11 ^2» ^3 

the condition that (2) should touch the surface, hold simultaneously. 
The sinisters are both of the same order in the coefficients and variables, 
BO they can only differ by a factor which is easily found to be unity. 

[Mr. Edwardes sends the following solution: — ^As in the solution 
by 8970, wg have 



^1 
^3 



rfSj 



^1, 


y\y 


«i» 


w^ 


« 16 


Si, 


P12. 


Y,3 


^2» 


y<i^ 


«2» 


w^ 




P12. 


S2, 


P23 


Hi 


yai 


Hy 


w^ 




Pl3, 


P23, 


S3 



1 



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135 



and writing S = aj!^ + by* + ez* + dtt^ + 2fi^+2ffMX'\-2hxi/-¥2lxia 

+ 2myw + 2mw, 
the same is 2 dSj ^ dOi 

dxi difi dzi 

rfSg dS« rfSg 

dx^* di/2 dz^ 

rfSa rfSa rfSa 

dx^ dy^ dz^ 






2 


«, 


A, ^, M 




^i» yij 




A, ft, /, m 




^2, y^y 




9y /, ^ » 




^3» ya. 


= 2JD 


iPi, yi> «i 




yi, 


x^ y^y «2 


+ L 


y2» 




^. 


*3» y3» H 






ysi 



«2> 
^3, 



«'2 
W?3 



yi, 
y2» 



^8. ysi 



+ N 



= result J 



t^8» 



2i» 

«2» 
«3» 

^3» 



^2 +M 22> «^2i ^2 

^^3 I I «3» ^'Sl ^3 

yiK |^i» yi» 

y2 ? ^ U2» y2, 

ysK Us, ys, 



8969. (W. J. C. Sharp, M. A.) — ^If the ternary n-ic be written 

1 1 • ^ 

and <m; + ft|y + b^ be written for a, 

hix + Cjy + Cjg be written for ftj, 
b^ + c^ + e^ be written for ftg, and so on, 

in any invariant or covariant ; the result will be a covariant of the 

(m + l)-ic <m;»»*i + "-ii (bji/ + ftjz) x** + &c. 



For any covariant of the (n + l)-ic, the operation y — - must be equivalent 

dx 



to 



"T.^2*.J-.*,^ + 3..|^ + 2..^^,3^^,&c. 



^ft| 



Writing the (« + l)-ic. 



aV» + Y (^'i y + ^'a *) ^**"* + &C'> where a' =^ax + biy + ftj?, &c., 



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Then 



136 

I in question, we have 

y ^ • y f (n + 1) <w» + n + 1 . n (*,y + ij«) «*-* 
dx (^ 

+ ftj {Mf^-^yz + ff«t"- V + « . n— 1 y« z*-*) 

— (»+ 1) aa^y + « + 1 . fi (*iy + *j«) a^-^ 

+ ^L+LllliLli (ciys + 2<?2y« + <y«) a:"- V + &0. 

— yj(» + l)ajc* + n + l.fi (*,y + b^) a?"-* 

+ ^:tiili±li(cjy8 + 2<?5y8 + c3a;3)a^-2 + &c.| = y^ from above. 

Similarly, « -— is equivalent to 
aar 

and therefore, &c. 

[Any invariant or covariant of a quantic maybe expressed as a function 
of the differential coefficients of the quantic, and this same function of its 
differential coefficients will be a concomitant of any other quantic. 
Now, if Un and «n +1 denote the gfiven quantic and the one derived from it by 
making the proposed substitutions, dP*<i*' Un^ij dsiP , dy^ , d^ is the result 
of making the same substitutions in df^t*^ Uh/ dxP . dy^ , dsi^, and the 
symbolical form of the transformed concomitant is the same as that of the 
original one ; therefore, &c. This may, of course, be extended to A;-ary 
quantics, and to substitutions of a higher order in the variables.] 



8970. (W. J. C. Shaep, M.A.)— If X, Y...U denote the deter- 



minants 



^l> t/lf «i» ^u ««i 

^2» t/3y H* ^2> «2 

^8, t/3y Hy «^8» «3 

^4» y4» «4» ^iy ««4 



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137 



and Vi, Vj, Vj, V4 be the values of the quinary quadratic V when 
(^1, y\y 2i» «^ii ««i), (^2»'yj. «2i w's* «*2)> &«. are put for {x, y, «, «;, «), and 

S1.2 , &c. stand for J ( a>i -- + yi — + ... J Vj, &c., 
\ 00:3 ay2 / 

«AX» + BY« + &c., 



v„ 


8, .a, 


8,.,, 


S,.4 


s,.„ 


v„ 


8,.,, 


ai.4 


Si. J. 


Sa.j, 


v„ 


8, .4 


81. 4> 


S,.4, 


8,.«, 


V4 



where A, B, &c. are the first minors of the discriminant of V. 



Solution. 

This question is a generalisation of 8960, which is itself one of the 
property (otherwise) proved in Salmon's Conica^ Ed. 6, Art. 294, and 
which is the analytical ground of the theory of reciprocal ^olars. 

The proof in this case is the same as that in S940 ; for, if or, y, s, w, u 
be eliminated from V = 0, by means of the equations 

(\ + t^ + y + ir)x — \xi + fir^+yx^ + nx^ 

(x + fi + i' + ir)y « Ay, + fiy2 + i/y8 + Ty4, 
&c. &c. &c. 

the result will be the equation to the section of V — by the linear 
locus through (ari, yi, «i, «;„ w,), (a^a, yg, ar„ t(?2, «,), (arj...), (3:4...), 
and, if this locus have a double point, the determinant in the question, the 
discriminant of the locus of section, will vanish, and also the result of 
substituting X, Y, ...U, in the reciprocal equation; and those two 
quantities being each of the same order in the coefficients of Y, and in 
X, .V, ... &c., can only differ by a factor which will be found to be unity. 
The property may easily be extended to space of any dimensions for 
A;-ary quadratics) and proves that the theory of reciprocal polars holds for 
space of all dimensions. 

[Mr. Edwardbs sends the following solution : — 

Since Wi —-* +yi -—1 +&c. - 2Vi, we have 

dVj dVj rfVj 
dtoi du^ 

dw2 ^^ 

dj^^ 

dwJ 



dxi 

rfVj 
dx^' 

dVj 
dx^ 



dx, 
dyi 

dVj 

dyi 
dy^ 



dzi 
dz2 



dV^ 
dzA ' 



2Vi, 
2Si2, 
2Si3, 
2Su, 



2S21, 2S3,, 

2V2, 2S30, 

2523, 2V3, 

2524, 2S34, 



3 

rfV4 

dwl 

2S41 

2S4, 
2S43 
2V4 



or,, 
^2> 
^3> 
a^4» 



yii 
y2> 
yzy 

^4. 



*2» 
«3» 
«4, 






du^ 
dV, 
du^ 

and obviously S12 = S21, &c., therefore 
this determinant becomes 



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16 






138 



V2, 823, S24 



Sii> Sjj, Vj, 834 

Sl4» S34, S34, V4 

But expanding in another way, we have 
cfV, rfVj ^i dVj 

tfo-j dt/2 



rfV4 

and this = 162 



<^3 



&c. 



^4» y4> 



IT, 



«, A, ff, I 

If frt, w, <f 



a?i, yit «!, wi 

^2> yj» «2> <^2 

^3» y3> ^3' <^3 

a?4» y4> «4> «^3 



viz., 16XA.X^, the quadratic being written 

(abedefghlmnpqrs IS^xyzwu)^,^ 



9561. (W. J. C. Sharp, M.A.)— If (1.2), (2.3), &c. denote the 
edges of a tetrahedron, and D}, D^, D^ the shortest distances, and &i, ^ji h 
the angles between (2 . 3) and (1 . 4), (3 . 1) and (2 . 4), and (1 . 2) and 
(3 . 4), respectively ; prove that 

^^^ ''''' ^' ^ 2(2.3K1.4) ^^^ • ^^'^ ^^ • '^^'"^^ • ^^"^^ • ^^'^' *''" ^''•' 
and (2) the square of the volume 

= ^{4(2-3)Ml-4)'-[(1.2)» + (3.4)»-(2.4)»-(1.3)»]»} = &c.,&c. 



Solution. 

Let ABC!D be a tetrahedron. From D draw 
DE parallel and equal to BC ; join BE and AE. 

BCDE is a parallelogram and its diagonals 
bisect each other (in F). 

Now 61 = ADE ; 

therefore cos ^i =» cos ADE 

= ^ (AD» + DE2-AE2> 

2AD.de ^ ^ 



1 
'2AD.BC 



{AD2 + BC2-AE2}; 




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139 

but AB2 + AE2 = 2BF2 + 2AF», 

D A8 + AC? - 2DF2 + 2AF2, DB^ + BC? = 2BF2 + 2DF2, 

therefore DA^ + DB^ + AC^ + BO^ = 2BF2 + 2AF2 + 4DF2 

= AB2 + AE2 + CD2, 
therefore ADs + BCa-AE^ = AB^ + CDS-DBS-AC^, 

and therefore cosei = _^ {AB^ + CD^-DB^-ACaJ 



2AD.B0 

1 
'2(1. 4) (2. 3) 



{(1.2)2+(3.4)2-(2.4)2-(1.3)2}, 



From these values it follows that the opposite edges are at right angles, 
if (1 . 2)» + (3 . 4)2 - (1 . 3)2+ (2 . 4)2 « (1 . 4)2 + (2 . 3)2, 

and conversely. If HK be the shortest distance Dj, it is at right angles 
to AD and BC, and therefore to AD and DE, and so to the plane ADE ; 
and since BC is parallel to that plane, it is equal to the perpendicular 
from C upon the plane ADE. 

Now Vs tetrahedron ABCD = tetrahedron ACDE 

— sin 01 X perpendicular from C to ADE 



6 
AD. BO 



sin 01 . Di ; 



therefore V2 = 42!.i^ D^a (i « co82 0{) 

36 

== ~ {4 (1 . 4)2 (2 . 3)2- [(1 . 2)2+ (3 . 4)2- (2 . 4)2- (1 . 3)2]2} « &c. 

From the above (2.3) (1 . 4) Di sin dj = (3 . 1) (2 . 4) Dj sin 63 

-(1.2)(3.4)D3sin03. 

[Mr. Edwardes sends the following solution : — Denote the sides of the 
tetrahedron (1 . 4), (3 . 4), (2 . 4), (2 . 3), (1 . 2), (1 . 3), by «, *, e, d, e,f, 
respectively. Through the point (c, d, d), on the plane of abe, draw a 
parallel to a, and let angle (a, d) — 0i, &c., also, let angles (^, c), (e, a), 
(c, a), (c, d), {ed«y abe) be X, /ia, if, o, <j), respectively. Then we have 

—cos 61 = cos a cos 1^- sin a sini^ cos d), and cos d> = cos /^~<*08^ cosy 

sin A sini' ' 

therefore -cosOi - cosy (cosa + ?^5L?L22i^\ -^Iffcos/* 
\ sinA / smX 

2ac \ 2dc d lee I d 2ae 

2ad 2ad 2ad * 

hence we have cos 0, = —^—P 

2ad 

" 2(2.3K1.4) ^^^ • ^^'^ (3 .4)2- (2 . 4)2-(l . 3)2}, &c., &c. 



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140 

Again, Dj « — — ; therefore, V = -J arfsin e, ; 

odBmBi 6 

therefore, V»^gLWBin»(>,»gl%«^[l-( ^^^^"f-/ Mn 
36 * 36 i V 2a^ / j 

= ^{4(2.3)M1.4)2-[(1.2)«+(3.4)»-(2.4)»-.(1.3)2p} = &c.,&c.] 



7384. (Professor S. RfiAus.) — Etant donn^e la s^rie ilKmit^ 7, 1 3, 25, 
43, 67, 97, 133, 137, ..., dont le tenne general, celui qui en a « avant lui, 
est An — 3 (f»^ + f»j + 7 : demontrer les propositions suivantes : — (1) sur cinq 
termes consecutiis, pris k volonte dans la serie, un terme est divisible par 
6 ; (2) sur sept termes coDs6cutifs, deux sent divisibles par 7 ; (3) but 
treize termes consecutifs, deux sont divisibles par 13 ; (4) aucun terme 
de la s^rie n'est egal k nn cube; (5) une infinite de termes, tels que 
Aj = 26, Aj* = 4225, etc., sont aes carr^s divisibles par 25 ; (6) la 
deuxi^me et la troisi^me proposition sont comprises, comme cas particu- 
liers, dans la suivante : si N est un nombre premier, de la forme 6m + 1, 
sur N termes consecutifs de la serie, deux sont divisibles par N ; (7) on 
pent affirmer aussi que, k I'exccption de 5, aucun nombre premier de la 
forme 6in— 1 no peut diviser aucun terme de la s6rie. 

Solution, 

Out of any five consecutive numbers one must be of the form 5p + 2, 
and the corresponding term of the series will be 

3(5i? + 2)(5p + 3) + 7= 25(3p2 + 3p + l), 
which proves (I) — indeed, that one term in five is divisible by S^— and 
(5), since an infinite series of squares of the form Zp^ + Sp-k-l can be 
found [see Solution of Question 4535, vol. xxiii., pp. 30, 991. Similarly, 
out of seven consecutive values of n two are of the forms ip and 7p + 6, 
and in each case 3 (w^ + n) + 7 is divisible by 7 (2) ; and out of thirteen, 
two are of the forms 13;?+ 1 and 13p+ 11, which each of them makes 
3 (w^ + fj) + 7 divisible by 13 (3). No term is an exact cube, for if 

3 (n« + «) +7 = 3 (n« + « + 2) + 1 
be a cube at all, it is the cube of a number of the form Zp + l, and 
n''^ + n + 2 must be divisible by 3, which it never is (4). 

Again, 4A„ = 3 (4«2 + 4» + 1) + 25 s 3j»2 + 25 

if p — 2n + lf and if 6w + I be a prime, 3/>2 + 25 = (6m + l)y has real 
solutions. If 6m— 1 be a prime, Sp^ + 25 = (6m- 1) y has not, except in 
the special case m = 1 (7) ; and if, when n — q, 3 (w^ + «) + 7 is divisible 
by 6m + 1, itis 80 whenu = 6m — ^ or (6m + l)p + q ot {6m + l)p + 6m-qf 
which proves (6). 

It may be interesting to point out that 

2oA«ar« = (7~8a:-7rr2)(l-ar)-3. 



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APPENDIX i^l. 



NEW QUESTIONS. By "W. J. Curran Sharp, M.A. 

9776. If perpendiculars jp, q, r he drawn from the vertices of a 
triangle upon any tangent to the circumcircle, these are connected by 



the relation 



9, ^\ 



c2, *2 



-0. 



r, l^y a\ 

Prove this, and show that a similar relation holds in space of n 
dimensions. 

9777. If /A + W/A + WV+... « be the equation to a linear locus in 
space of n dimensions, in terms of the simplicissimum content coordinates 
(areal, tetrahedral, &c.), (see Question 8242) ; show that /, m, «, &c. are 
proportional to the perpendiculars drawn from the vertices of the simpli- 
cissimum of reference upon the locus. 

9778. If the variables a, jS, 7, 8 be removed from the tangential 
equation to a surface, by substitution from 

(X + M + >')a = Xai + Ma2 + i'a3, (A + /i + 1') jS = AjSi + /i/Sj + v^j, 
(X + /i + i^)7= A7i + /i78 + i0'3, (A. + M + >')8=^ ASi + fiSj + irJa; 

show that the resulting equation in (x, /*, y) is a tangential equation to 

the tangent cone whose vertex is at the intersection of the three planes 
(«!» ^1, 7i» 81), (o,, /32, 72, 83), and (oj, ^3, 73, 83). 

Hence determine the number of tangent lines of different classes 

which can be drawn to the surface from any point. 

9779. Prove the following identities 

0, 1, 1, 1, ... 1 

1, 0, iPl + ^8> ^1+^a ••• *l + ^n + l 
1, jri + a?3, 0, X^-¥X^ ... X^^-Xn^l 
1, ^1 + ^8) fl's + 'Si ••• ^8 + ^»» + l 

1, Xi+Xn^li *i + *n + li a's + ^n*l ... 

^^{^2Yx^X^,.,Xn^\\^ + — +... + —), 
VOL. XLIX. S 



(1) 



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U2 

^1 + ^8, 0, r^ + x, 

a-j + a-j, a-j + a-,, 



^1 + <F«* 

Xz + Xn^l 



l[(«^l- 



0, 1, 1 ... to n columns 

1, 0, 1 ... 
1, 1, ... „ 






(3) 



= {-!)-' (»-I), 



the last to be proved independently, and then shown to be consistent 
with the first two. 

9780. In ppace of n dimensions, the ppherical loci (hyper-spheres) 
described ahout the n + 2 simplicissima (see Question 8242), eadi oi which 
is bounded by m + 1 out of n + 2 given linear loci, all pass through the 
same point, when n is even ; but not, in general, when n is odd. In 
space of two dimensions this is Micquel's Theorem. 

978 1. If a.v + bt/ + cz -^ dw = 0, and a'x + b\i/ + c'z + d'w = 0, where 
fl, i, c^ df a', b'y (?', d' are functions of a parameter 6, be the equations 
to a line, this line will generate a ruled surface of order m + «, where m 
and n are the orders of the two given equations as functions of e. 
Especially examine the case when m = « = I, and show that a second set 
of lines exists in this case. 

9782. If Pnr denote the coefficient of a:*" in the expansion of 
(1 +iF)'*, &c., C„r denote the number of combinations of n things taken r 
together; form the equations of differences which determine P„r, and 
Cnn and hence show that these are equal. 

9783. If ABC be a triangle, in which AC > AB, AD the perpen- 
dicular from A upon BC, and if DC be taken upon CD produced, so that 
DC = CD ; the circle through A, B, C will also pass through the ortho- 
centre of ABC, and will be equal to the circumcircle of tiiat triangle, 
which will pass through the orthocentre of ABC. 

9784. If the sums of the squares of the opposite edges of a tetra- 
hedron be equal to one another, show that the nine-point circles 
inscribed in the triangular faces are all sections of the same sphere ; show 
also that this is the condition that the perpendiculars from the vertices on 
the opposite faces should meet in a point. Also show in space of n 
dimensions, that if (r«) denote the edge joining the Hh and «th vertices 
of a simplicissimum (Question 8242), and (r.«)2 = Ay + A, (where 
Ai, A2 ... An .1 are w + 1 arc^l magnitudes) for all values of r and «, the 
perpendiculars from the vertices upon the opposite faces will all meet in 
a point, and the nine-point circles of all the triangles formed by joining 
the vertices are sections of the same spheric (hyper- sphere). 

9785. li m *^ m', there are in general (m— 1)2»» («+ 1) points which 
have the same linear polar with respect to each of two loci, of orders m 
and m', in space of n dimensions. Hence deduce the conditions that the 
loci may touch. 



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143 

9786. If a circle cut the sides of the triangle of reference at the feet 
of concurrent lines from the vertices, the line joining the isogonal con- 
jugates of the points of intersection passes through the centroid. 
Enunciate the corresponding proposition in the Geometi^ of Higher 
Space. 

9787. If the sums of the opposite edges of a tetrahedron he equal to 
one another, show that the circles inscribed in the triangular faces are 
all sections of the same sphere. Also show, in space of n dimensions, that 
if for all values of r and s [{r.s) denoting the edge joining the rth and «th 
vertices of a simplicissimum (Question 8242)], {r ,s) ^ dr + dg, where 
<?,» d.2f &c., dn^i are « + 1 linear magnitudes, the circles inscribed in the 
triangles formed by joining the vertices are all sectioijs of the same spheric 
(hyper-sphere) . 

9788. In space of n dimensions, the Jacobian of n + 1 quadratic loci 
(which is a locus of the n + l^^ order) is the locus of the points which are 
conjugate with respect to each of the loci, or the locus of the points whuse 
first polars with respect to all the quadratic loci meet in a point. 

9789. Show that 

(fi-y) (?-«) (»-iB) (a;-a)2+(8-y) (?-«) (a- 8) (x-^fi)^ 

+ («-«) (a-iS) (i3-8) (;r-7)«+ (i3-a) (7-/8) («-.7) (:p-«)« 
vanishes identically, and hence deduce that the sextic covariant J of the 
binary quantic {x — o) (aj - 0) {x -y){x- 5) 

= {{a''fi){x^y){x^n)^{y-B){x-a){x^0)}{(a-y){x--fi)[x-9) 
-(8-i8)(a:-a)(2:-7)}x{(«-8)(a.-/3)(a:-7)-(i8-7)(2:-a)(a;-8)} 
s {(a-iS) (^-7) (^-8) 4- (7-8) (a:-a) (a;- i3)} {(«-7) (.^-i8) (^-8) 
+ (8-i8) (2:-a) (a;-7)} X {(a-8) (a:-i8) (a;-7) + (/3-7) (a;-«) (a;-8)} ; 
and confirm this by showing that 

(a-/3) (ar-7) (ir-8)-(7-8) (a:-a) (a^-iS) 

= (a-7) (^-/3) (^-8) + (8-/3) (a?-a) {x^y). 
Also explain the geometrical relation between the points determined by 
the roots of the covariant, and those determined by the roots of the 
quantic. 

9790. If u be a rational and integral symmetrical function of 

Xi, X2 ... Xny show that x? -^'- x^ — and x^ ^ — x^i ^ 
dXf. dxg dxr dxg 

are divisible by Xr — Xg for all positive integral values of p, r, and », 

9791. Show that the secondary form of Maclaurin's Theorem, given 
in Boole's Finite Liffereticea^ p. 23, Ed. 1., viz., 

0W = ^(O)^0(|)o.^ + ^«(|)o«.j^+&c.. 
leads at once to the equation ---^ = log (1 + A) ^ ^. ,^ 

€r — \ A 

wheye the A*8 only act on the zeros. Also from this form deduce 
(^~1)»= 2 (A". 0''») — -. 



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144 

9792. Show that, if (A) «*-i»i «•-»■»•/?, «•-«...)"• 

and that, if m - 2, ^^ - PrPo+Pr^i -Pi + Pr-i -i^ ... +J?oPr. 

Also deduce the expansion of ^ (/r) in powers of x, where (^ (z) is an 
integral and rational function of x, and 

^ 9793. If XJ =» be the equation to an tn-ic locus in space of n dimen- 
sions, referred to simplicissimum content coordinates, and 

Ssx>i(1.2)« + /ii'(2.3)« + ..., 
where (1.2), &c. are the edges of the simplicissimum of reference, the 
equations to the normal to U at the point (X/i ...) are 

dhf d\* dfi''^ dfi ' 

dV dV -0, 

dk' dti 

1, 1 

where xV, &c. are the current coordinates. Hence show that m" normals 
can, in general, be drawn from any point to U «*0. 

9794. If (1 . 2), &c. denote the edges of the tetrahedron of reference, 
and X, /A, Iff » be tetrahedml coordioatos, show that the circle at intinity 
is represented by the equations 

X+/i + v + xs=0, and X/i(1.2)2 + /ij*(2.3)2+ ... - 0; 

and that in space of n dimensions, if (1 . 2), &c. be the edges of the 
simplicissimum of reference, and X, /i, v.., content coordinates (see 
Question 8242), the Lypersplicre at infinity (in space of n dimensions) is 
represented by X + /i + y+ ... « 0, and X/a(1 . 2)* + fir (2 .3)*+ ... — 0. 

HencQ determine the conditions that an equation of the second degree in 
tetrahedral or higher content coordinates may represent a sphere or 
hypersphere. 

9795. (Suggested by Question 6420.) If S,,^ denote the coeflScient 
of tj in the developed product of (1 + /) (I + 2<) ... (1 +tO> 8how that 

Si.i« Si.l., + iS.-8j-l+»(»-l)S<_3.y-2+... 

+ i(»-l) ... (»-y+ 1) Si-j-i.o. 

and that the product itself is the coefficient of x**^ in the expansion of 
(l-tx)- 1/< multipUed by (» + 1) ! 

9796. Show that the transformation from rectangular to areal coor- 
dinates, or vice versd, may be eflfected by substitution from the equations 

{\ + fi + y)x = \Zi + fix.2 + yX3, (X + /it + v)y == Kt/i + fiy^+^Jsy 

where (j*i, yj), (a-j, yj), {x^, y,) are the vertices of the triangle of reference. 



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145 

And similarly, that the rectangular and tetrahedral coordinates of a point 
in space of throe dimensions are connected hy the equations 

(X + /i + K-i-»)a; = ^Xl + /UTj + yx^-\-wx^f 

(X + /4 + y + x)y-Xyi + /*y2 + yy8 + iry4, 
and (X + /i + y + x) « =• A«i + fiZ^ + W3 + irr^, 

or more gf nerally that, in space of n dimeneions, the connection between 
orthogonal and simplicissimum content coordinates (see Question 8242) is 
given by the equations 

(A + /* + !'... +t) z « Xi*i + ^2+ ••• +Tir„+i, 

{\ + /i + y... f T)y « Xyi + /iiy2+ ••• +Ty«,i, 

&c. &c. 

9797. If P ^ + 2Q ^ -h % = X, where P, Q, R; X are functions 

UX' ax d / V \ PR 
of X only, and are subject to the condition — ( — ] + — ^ 1=0, 

show that y - f-J^'^*^ [[ J- cK'^ dxK 

9798. A quadratic locus in space of n dimensions, has n principal 
axes {i.e.f axes which are at right angles to the linear loci which bisect 
chords parallel to the axis). 

9799. Deduce the solution of ^ — a^ -t « from the expansion of 

dx' dy^ *^ 

^ (x, y) in ascending powers of x and y. 

9800. If «, A, A be given in a spherical triangle, deduce the con- 
ditions that the triangle should be impossible, unique, or ambiguous, from 
the discussion of the equation 

cos a =: cos ^ cos e + sin * sin ^ cos A, 
where there are two triangles ; show that, c and (/ being the third sides, 

tan i (c -I- c') ^ tan b cos A, 
and confirm this by the case when the radius of the sphere is infinite. 

9801. If Pr denote the Legendre's coefficient of the r**» order of 
J(A; + 1/A:), showthat 



j: 






{(1-2:2) (1-^2^2;}* '3 '6 2r + l 

9802. If there be two series of functions of x, P©, Pi, Pj..., and 
Qo, Qi> Qa •••» and one of operations, Rq, Ri, Rj, &c., each of which gives 
a result independent of x : then, if K,» . Pn . Qp = 0, whenever t/i, n, 
and JO are not all equal, but not when they are, any function /{x) may 
be developed in the forms 2 A»P„, or 2 BnQn- Apply this to some known 
expansions. 

9803. Show that for any proper cubic the Cayleyan of the Hessian is 
the Hessian of the Cayleyan ; and that the discriminant of the polar 
conic of any line vanishes doubly when the line touches the Cayleyan. 

9804. Show that the nodes on the locus of a point rigidly connected 
with the middle bar of a three bar system lie upon the fixed centrode. 



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146 

9805. If PBC be a small circle of a sphere, B and C fixed points, and 
P any other point upon it, then, if the arcs BC, PC, and PB be bisected 
in D, E, and F respectively, and if the arc DE meet BP in B' and B", 
and DF meet CP in C and C" ; show that (1), for all positions of P, B', 
C, B", C" lie on the same great circle, that of which O the pole of the 
small circle is the pole, and that B'C = B"C" = w-B'O" = x-B"C' ; 
(2) that these arcs are of constant length for every position of P on one 
side of BC ; (3) that the values of those arcs corresponding to positions of 
P on opposite sides of BO are supplementary ; (4) that the points 3', B'', 
and C, C" are the poles of the arcs OF aud OE which bisect the sides 
BP and CP of the triangle BPC at right angles ; (5) that the six points 
in which the sides of the triangle D£F meet the corresponding sides of 
the triangle PBC lie on the gi-eat circle of which O is the pole. 

9806. If ABC be an acute-angled triangle, o, /S, y the circular 
measures of its angles; show that P being a random point in the 
triangle, the chance that the angle BPC is obtuse, is 

^ .^"". {8in2i8 + sin27 + »-2a}. 
2 8mi8 8m7 »■ •* 

U the angle at 3 (i3) be obtuse, the chance is 

— ^iBA_/27 + 6in27}. 
2sin)88in7 ^ '^ 

9807. The perpendiculars from the vertices of a triq,ngle upon the 
central axis (the line which passes through the circumcentre, the ortho- 
centre, the nine-point centre, and the centroid) are proportional to 

cosAsin^B — C), cos B sin (0— A), and cpsOsin(A— 3), 

those on one side of the line being reckoned positive, and those on the 
other negative. 

9808. The mean value of the pedal triangle of a random point in a 
triangle ABC is ^W (1 + cos A cos P cos 0), where A, B, and C are the 
augles of the triangle, and B the circumradius. 

9809. Given forces act along the sides of a tjiangle, in the same 
sense. The value of the mean sum of their moments about a random 
point in the triangle is the mean of their moments about the vertices t)f 
the triangle. If the forces be such that, if applied at a point, there would 
be equilibrium, the sum of the moments is the same for all points. 

9810. A, B, are fixed points, P another point, construct the resul- 
tant of forces acting along PA, PB, PC, when they are proportional to 

(I) PA»,PB»,andPC2; (2) ±, -L, and ^; 

(8) ^. pgi. and ±^. 

9811. If masses P, Q, and It be placed at the vertices A, B, and 
respectively of the triangle of reference, show that the trilinear equations 
to the principal axes at the C. of G. of the masses will be 

lx-tmi/ + nz =s 0, and tx -f m*y + n'z == 0, 



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147 

where I : m : n and f : m' : n' are determined by the equations 

a e a b e 

pi?l + Q^ + R«4'„0,and 

fl- ^ c* 

//' + mm' + «ff'— (iwn' + #»'«) cos A— (w/' + n'l) cos B — {hnf -f Z'm) cos C = 0, 
and that at any point the principal axes are conjugate {i.e.^ each passes 
through the pole of the other) with respect to the conies 

— .- + -2^: — -h — ■■ 

fl^ b'^ c'^ 

and ^+m2 + n2— 2m« cos A— 2n/ cos B — 2/m cos C = 

(the circular points at infinity). 



9812. If p. denote j;^^.. and Q„.£- 

show that P„ = -— — P„_i, 

2» (1— a^^)" 2n 



•a;)'* 



and that 



_ 1 k'^'lr2»'l 2n-l r, 
dx 



Jo4(l-; 



){(l-a;2)(l-A:2a:2)}* 

-'•-i(?)'-H(?)'-■-^^M"^--• 

9813. Show that 

J(l-a;-)* 2» ^ ^ 2n J(l-a;2)* ' 

-JL — .^«f!i_if _? dx. 

0(1 -a;2)* 2» Jo(l-a;2)* 

And hence show that 

^■' 2«! i2» + l 1 ■ 2 ■2« + 3 1.2 2»'2« + 6 

Jo {(!-*') (1-AV)}» 2i IS Vi/ 1».22 V2; 
is.22.32 V 2 ^ ^"■'•j- 



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148 

9814. Show that, for all integer yalaes of n, 

Ji J_ J_ 1 1.3 J^ _l 1.3.5 J^ 1 , 

2/1 + 1 1 ' 2 * 2w + 3 "'' 1.2* 22 *2n + 5 1 . 2 . 3 * 2^ " 2« + 7 ^' 



« i ( 1- J!lri + (2n-l)(2n-3) ^^^ | 
2'»i (2»-l) 2(»-l).2(»-2) ) 



■^(-')"2^iW^^«^('^^2). 

9816. If dashes ahove the line denote differentiation with respect to ar, 
and dashes helow with respect to y, and ^ and tp stand for any functions 
of X and s/ ; show that the equation 

+ 4>y — <^'Y + «<^''4'' + b\i^'^ = 

may always be transformed into a linear differential equation with con- 
stant coefdcients. 

9816. If the perpendiculars PA", PB", PC" from any point P on the 
conic \i/z + /xzx + yxp^K {z Bin A + y BinB + z Bin. C)^ = 

be produced to A', B', and C respectively, and if 

PA'=« ?i^.PA", PB'=?^.PB", PC=!^.PC", 

show that the area of the triangle A'B'C is constant. 

9817. If A', B', C be the reflections of any point P, on the circum- 
circle of the triangle ABC, with respect to the sides ; show, by Euclid, 
that A', B', C lie in a straight line which passes through the orthocentre of 
ABC. Hence deduce the theorem (Quest. 2145) that ** the feet of the 
perpendiculars let fall on the sides of a triangle from any point on the 
circumscribing circle lie in a straight line. Show that this straight line 
is equidistant from the point and from the centre of perpendiculars of the 
triangle." 

9818. If ABC be a triangle. A', B', C the feet of the perpendiculars 
from any point upon the sides, and A", B", C" points in PA', PB', PC 
(produced if necessary) such that PA" =/. PA', PB" =^ . PB', PC'^= h . PC, 
respectively ; show that, when A", B", C" lie in a straight line, the 
locus of P is a conic circumscribed to ABC ; and that, when the area of 
A"B"C' is constant, the locus of P is a conic having double contact with 
first at infinity. Also, when the conic is given, determine flgih, 

9819. If the sides AB and AC of a spherical triangle ABC be divided 
in F and E respectively, so that sin AF : sin BF : : sin AE ; sin CE, the 
great circle FE will cut the great circle BC in a point Q such that 
BQ + CQ = », and the great circles through all such divisions meet in 
the same points, and conversely. 



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149 

9820. Prove the following— (i.j If abc and A'B'C be the pedal tri- 
angles of the circumcentre of the triangle ABC and of any other 
point P, A'B'C = iAABC x (R2^ OP'^/R^ where R is the circumradius. 
(ii.) If and K be the centres of two circles whose radii are R and r, P 
any point on the second circle, and PL the perpendicular from P to the 
radical axis of the circles, 20K . PL = R=« *- 0P=^. (iii.) The area of the 
pedal triangle of any point P on a circle, the centre of which is K, with 
respect to a triangle ABC, of which is the circumcentre and R the 
circumradius, is ^aABC (OK . PL)/R2, where PL is the perpendicular 
from P upon the radical axis of the two circles. 

9821 . Show that, if a point be taken at random in the circumscribed 
circle of a triangle, the mean area of the pedal triangle is ^ of the tri- 
angle. 

9822. If y =« a:*»-i log oj, prove that, when r is not > w, 
g=(»-l)(«-2)...(»-)^— [log.+ (^^^,4... + j^J|; 
and henee show that 

(«-li(«-2H«-3Jf_l_ _1_ _1_^ 
1.2.3 i»-l n-i »-3j 

in-1 «-2 «-3 1> 

+ £lfl - li± + Jl:!^ 1-2.3^ ^^^7 

n \ «+l (n+l)(n + 2) (»+I)(» + 2)(» + 3) j 

9823. Prove (1) 

f ^ ^JL±J,tan-ifflV.tane]-^Il-^ ^"^ ; 

J(acos20 + ^8in-^6)2 2a»A* \\ a I ) 2aba + bt&ii^d 

(2) that Xs« + *ar», 

f a;"»-i loga:.X^^.^"^^^^^^""^)x'' - *!??fa;"»*«->(mloga:- l)XP'^dx; 
J m- m^ J 

(3) that (*' cos- <psinn<p.d<p = i m^^^I^Mmil^ , 

9824. If the diagonals of a quadrilateral are at right angles, the sums 
of the squares of the opposite sides are equal. Hence, if «, A, c, d be the 
sides in order, a, c, i, d will be the sides in order of a quadrilateral in a 
circle the area of which is ^ [ac + bd). 

9825. If particles be projected along lines meeting in a point, with 
velocities proportional to the projections of the same v< rtiral line upon 
each line ; at any time the particles will lie upon the sphere descrihtMl 
upon the space traversed by the vertical particle as diameter. Hrnco find 
the line of quickest transit when an inelastic particle is dropped from () 
to A (a vertical distance /*), and after impact upon an inelas<tic line AB 

VOL. XLIX. T 



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(the line to be detennined) proceeds along it to B, a point on a given 
circle in a vertical plane through OA. 

9826. If particles be projected from a point A, with the same velocity 
V, along lines meeting at A ; at any time (t) they will all lie upon the 
surface generated by the revolution, about the vertical line through A, 
of the bicircular quartic, which is the inverse about A of a conic whose 

focus is at A, its directrix horizontal at a distance — — upwards from A 

I where k is the radius of inversion, and its eccentricity -7^ ] . The 
conies corresponding to different values of t envelop a circle of which the 
highest point is at A and the radixis ^. 

9827. If XJ — be the homogeneous equation to a curve of order n, 
and ^3X,|.+y,A+^|.; 

show that the discriminant of 

X»Ui + X«-»/iAUi + \\** - V'AlJi + &c., 
only differs by a factor from the result of substituting 
for a, fiy and y in the tangential equation to the curve. 






9828. If U » be the homogeneous equation to a surface of order n, 

A J. ^ ^ d , d' 

and As.,- +y,- +«,-+«-,-, 

- . d d d d 

dxi dj/i dzi dwi 

show that the- discriminant of 

X»»Ui + X~ - » (/tA + vA') Ui + iX"- « 0*A + kA^s Ui + &c. 
can only differ by a factor from the result of substituting 

^i> Pv «i> ^i 

^2» t/2* hf ^3 ^^^ <h fif 7> <uid 8 in the tangential equation to 

^31 P3f «8t ^3 

the surface. 
Show that this may be extended to higher space. 



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APPENDIX III. 



UNSOLVED QUESTIONS. 



1087. (The Editor.) — ABCD is a conic whose centre is 0. If the 
radii vectores OA, OB, 00, OD represent in magnitude and direction four 
forces, show that the direction of uie resultant passes through the centre 
of a second conic which is parallel to the first, and passed through the 
points A, B, 0, D. 

1196. (The Editor.)— Given the vertical angle and one oi the con- 
taining sides, construct the triangle, when the ratio of the base to the sum 
of the other side and a given line is given, or a minimum. 

1207. (Alpha.) — To determine the position of a rock (R), the angles 
subtended at it by the distances between three headlands (A, B, C) were 
observed, viz., 

BRC = 161° 2' 42", CRA = 133° 25' 67", ARB = 75° 31' 21" ; 
and it was known, from a previous survey, that AB = 8883, BO = 9870, 
and OA — 10857 yards. Find the distance of the rock from each of the 
headlands. 

1228. (Alpha.) — A messenger M starts from A towards B (distance 
a) at a rate of v miles per hour ; but before he arrives at B, a shower 
of rain commences at A and at all places occupying a certain distance z 
towards, but not reaching beyond, B, and moves at the rate of u miles an 
hour towards A ; if M be caught in this shower, he will be obliged to stop 
until it is over ; he is also to receive for his errand a number of shillings 
inversely proportional to the time occupied in it, at the rate of n shillings 
for one hour. Supposing the distance z to be unknown, as also the time 
at which the shower commenced, but all events to be equally probable, 
show that the value of M's expectation is, in shillings, 



a X z V V* u )' 



1274. (The Editor.) — If an indefinite number of parallel equidistant 
lines is drawn on a plane, and a reguleir polygon, the diameter of whose 
circumscribed circle is le8sthan.the distance between consecutive parallels, 
is thrown at random on the plane ; prove that the probability that the 
polygon will fall on one of the lines is 1/ /', where I is the perimeter of the 
polygon, and /' the circumference of the greatest circle that can be placed 
between the parallels. 

1913. (The late Rev. R. H. Wright, M. A.)— Find the condition in 
order that a straight line passing through an angular point of the triangle 



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of reference ohall bo a normal to the conic whose equation is 
(/a)*+(mi8}* + (ti7)* = 0. 

1016. (Sir R. Ball, LL.D., F.R.S.) — Show that the equation of 
squares of difFerences of the biquadratic (a, A, c, rf, e) (a?, 1)* = has for 
its discriminant (H being ■» b^—aCf &c., as in Quest. 1876) 

(27 J2- 13)2 (4H»- aSiH - a^J)^ (65296H3J + 2Z0iaWP- 1 6632a2HIJ 

-625a3P-9261«'J2)«. 

1919. (The late Professor Townsend, F.R.S.) — If a system of quadrics 
touch a common system of eight, seven, or six planes, their director 
spheres (that is, the spheres which are the loci of the intersections of their 
rectangular triads of tangent planes) have a common radical plane, axis, 
or centre. 

Prove the three general properties involved in this statement; and show 
from them, respectively, that— 

(1) The director spheres of all quadrics passing through the four sides 
of any skew quadrilateral have a common radical plane with the two 
spheres of which the two diagonals are diameters. 

(2) The director spheres of all quadrics passing through a common line 
and touching four common planes, have a common radical axis with the 
four spheres of which the four connectors of the intersection of three planes 
with that of the line and fourth are diameters. 

(3) The diameter spheres of all quadrics touching six common planes, 
have a common radical centre with those of the fifteen quadrics determined 
by the fifteen different triads of intersections of the planes taken in paiis. 

1928. (N'Importe.) — Given the four cones -ci/^+bz'^-'fu^ ^ 0, 
ex^ - az^-gvt^ = 0, - hx^ + ay2- hw^ = 0, fx^ + gy^ + A«2 = o, and the four 
conies which are the sections of these by the planes a; = 0, y = 0, « = 0, 
«> = 0, respectively; prove that (1) any line touching three of the four 
cones touches the fourth cone, and (2) any line meeting three of the four 
conies meets the fourth conic. 

1936. (N'Importr.) — (1) Three points are marked at random on a 
given straight line ; find the chance that, of the four parts into which 
it is thus divided, any three will be together greater than the fourth. 

(2) Again, a rod of given length has a piece cut off at random ; from 
the remainder a piece is again cut off at random, and the piece then left is 
divided into four parts at random : find the chance that any three of the 
parts will be together greater than the fourth. 

1989. (Professor Cremona.) — On donne un tetra^dre abed, Con- 
siderons tous les cones quadriques S qui ont leur sommet en e et sent tan- 
gents aux plans cad^ cbd le long des droites ca^ cb ; et tous les cones 
quadriques S' qui ont leur sommet en a et touchent les plans ode, abc 
suivant les droites ad^ ab, Un cone S et un cone S' etant choisis arbitraire- 
ment, se coupent suivant ime courbe gauche G du 4^ ordre ; et toutes ces 
courbes C ont un rebroussement en a et sent osculees en b par un meme 
plan stationnaire ebdy etc. (voir Comptes Rendm 17 mars 1862, p. 604). 
Demontrer geom^triquement les proprietes qui suivent : — (1) Si d'un point 
quelconquejo de I'espace on m^ne les plans osculateurs h, toutes les courbes 
C, les points de contact formeront une surface P de 3« ordre et 4« classe, 
qui passe par le point donne p et par les six aretes du tetra^dre abed ; 
(2) Les plans osculateurs des courbes C, aux points oil celles-ci sont 



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couples par tm plan quelconque donne x, enveloppent une surfece n de 
3^ classe et 4^ oridre qui touche le plan -k et passe par les six aretes du 
tetrafedre ; (3) II y a une correlation de figures, dans laquelle k un point 
Pf donne arbitrairement dans I'espace, correspond le plan «■ qui oscule en 
p la courbe C qui passe par ce point ; et» vice versdy ^ un plan donn^ x 
correspond lepoint j? de contact entre ce plan et la courbe C qui est osculee 
par ce meme plan. Si le point p decrit un plan x', le plan x enveloppe la - 
smface n' ; si le plan x toume autour d'un point fixe p% le lieu du point 
p est la surface P', etc., etc. 

1999. (R. Tucker, M.A.) — (1) P is a given point on the side of a 
triangle, and Q another given point in the same plane : it is required to 
inscribe in the triangle a maximum triangle having P for a vertex and its 
base passing through Q. Again (2), P is a given point on a circle, and 
Q a point in the same plane with it : it is required to inscribe in the circle 
a maximum triangle having P for its vertex and its base passing through 
Q. Discuss the several cases fully. 

2233. (The late T. Cotterill, M.A.) — 1. If two points are given, 
there are two others in the same plane such that the distances of points 
taken from each pair vanish. In orthogonal diameters of orthogonal 
circles, the diameter of each circle cuts the other in such pairs of points. 

The products of the distances of a point in the same plane from the 
pairs are equal, and the cosine of the angle subtended at the point is 
real. 

2. If a triangle is given by a point on a circle and the intersections of 
a line and the circle, give . a construction for the centres of the circles 
touching the sides of the triangle. 

2287. (W. B. Davis, B.A.) — Find what algebraical curves can be 
expressed by arcs of the circle. 

2293. (Professor Whitworth.)— If B, B' be two real points and 0, 
O' the circular points at infinity, prove that (IJ the fourteen-points conic 
of the quadrilateral BBOO' is the rectangular hyperbola whose conjugate 
axis is BB' ; (2) the critical circumscribing conic of the same quadrilateral 
is the circle on BB' as diameter, and the critical inscribed conic is the 
ellipse whose foci are B, B', and whose excentricity is ^ a/2 ; (3) these 
three conies have double contact at the vertices of the hyperbola — the 
minor vertices of the ellipse ; (4) if EE' be the third diagonal of the 
quadrilateral BB'OO', then the critical circumscribing conic of the quadri- 
lateral EE'OO' is an imaginary circle, concentric with the circle on BB', 
their radii being in the ratio a/(— 1) ; I, and the critical inscribed conic 
is the imaginary ellipse whose real foci are B, B', and whose excentricity 
is i a/2 ; and (5) the critical circumscribing conic of the quadrilateral 
BB'EE' is the rectangular hyperbola conjugate to the former one, and 
the critical inscribed conic is another rectangular hyperbola, similarly 
situated to the last, and having B, B' as foci. 

2314. (The late T. Cotterill, M.A.) — Taking points in a plane; 
prove that (1) the sum of the squares of the distances of a fixed point 
from the four centres of the circles touching the sides of any triangle in- 
scribed in a fixed circle is invariable, and this holds, if two of the points, 
or even if all three, coincide on the circle ; (2) more generally, if A, B, C, D 
be four such centres, and we denote the square of the length of the tan- 
gent drawn from a point — to the fixed circle by (M) ; to the circle on the 



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154 

diatncfter AB, by (AB), &c. ; to the circles circumscribing BCD and copolar 
to it, by (A) and (a), &c. : then we shall have the following system of 
identities : — 

2(M) « (BC) + (AD) ^ (OA) + (BD) = (AB) + (CD) 
«(A) + (a) = (B)+(3) - (C + 7) = (D) + (5) 
«i{(A^) + (B) + (C) + P)}=i{(«) + (i8) + (7) + («)}. 
Hence AP^ + BP^ + CP^ + D^* « a constant is the equation to a circle 
concentric with (M) ; and (3) if A, B, C, D are any points in the plane, state 
the nature of a pair of points, corresponding to the circular points at 
infinity, such that a similar system of equations must exist between the 
corresponding conies through the two points. 

2367. (The late T. CoTTBRiLL, M.A.)— * 
afe 

1. If the determinant fbd yanish, the equations 

e de 
o(3«« + <jy«-2rfy2) = fi(ex^ + az^-2ezx) » y{ay^ + bx'*-'2fxy) 
are satisfied by two conjugate pairs of values 'of the variables, one pair 
being independent of the constants a, i9, 7. Find the equation to the 
quadric satisfying the values y = « = 0;« = a?=sO; a;=»y = 0, and the 
last pair of values ; and the linear equations satisfying each conjugate 
pair. 

2. Show that such a system of equations is the analytical representation 
of the projection of the three circles described on the lines connecting the 
opposite points of intersection of four lines in a plane as diameters, the 
circle circumscribing their diagonal triangle, and the line at infinity. 

2366. (Professor Burnsidb, M.A., F.R.S.)— Determine (1) the locus 
of points such that the polar conies with reference to the curve IT shall 
be equilateral hyperbolas, where 

U - 2 ^^^^^ ^ ; 

Ayz + j::* (_ <p cog ^ ^ y cos B + « cos C) 

and show (2) that this locus passes through the vertices of the triangle 
ABC and through the feet of perpendiculars of the same triangle. 
2390. (The late G. C. Db Moboan, M.A.)— Prove that 

f** 1/ {xtp (x-alx)} dx ^ 0, 

« being anything positive ; and 

J+co - r + oo 

1 / {o? {x + alx) jdx ^2\ ^x/x dx, 

a being infinitely small and positive, and <p being such a function that 
the subject of integration is finite for all finite values of a;, however small 
a may be. In the second case, if ^/a? be infinite when a? = 0, the 
integral on the r%ht must be replaced by 



J -00 



2322. (The late Professor Townsend, F.R.S.)— Express the radius 
(B) of a circle orthogonal to three others in a plane, in terms of their 



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155 

three radii {p, q^ r) and the three sides (a, b, e) of the triangle determined 
by their three centres. Investigate the corresponding formula for the 
radius of the circle orthogonal to three others on the surface of a sphere. 

2402. (R. TucKBB, M.A.) — Prove that the locus of a point whose 
distance from its polar with reference to a given conic is equal to its dis- 
tance from a given point is a quartic curve, which, when the conic be- 
comes a circle, degenerates into a cubic curve. 

2419. (The late T. Oottbrill, M.A.)— 1. If A A', BB', CC are the 
opposite intersections of a complete quadrilateral, an infinite number of 
cubics can be drawn through these points and another point D, touching 
DA, DA' at A and A'. Amongst these cubics, there are two triads cS 
straight lines and four cubics having respectively a point of inflexion at 
B, B', C, C 

2. The locus of the intersection of tangents at B, B' is the conic 
DAA'BB' ; and of tangents at C, C is the conic DAA'CC. 

Give the reciprocal results when the class cubic degenerates. 

2436. (Professor Crofton, F.R.S.)— If pi, f>2, pg, P4 are the distances 
of a point from four concyciic points 1, 2, 3, 4 ; and if a, ^, 7, 5 are the 
triangles formed by joining 1, 2, 3, 4 ; then the equation 

Pi V W - f^ -/O) + Pz Viy)' P4 '\/(«) » 0, or p,p3 ^(ay) « p^ ^/(3»), 

represents two circles cutting each other and the circle 1234 orthogonally. 

2439. (The late T. ColrrBRiLL, M.A. ) — If a, *, c, d are coUinear points as 
well as a;, y, t.t and«ln the same plane ; then of the 16 intersections of ax^ 
aj/y bxf by with ez^ ety dz, dt, 8 lie on one conic and 8 on another. The 
4 points of intersection of these conies lie on a third conic through abxy, 
and a fourth through cdzt^ and these conies are respectively harmonics to 
the lengfths (a*, xy) and {cd, zt). Also, the tangents to the four conies at 
any point of intersection are harmonic. 

2440. (S. Roberts, M.A.) — 8how that the centre of a curve of thefith 
degree (t\^., the centre of mean distances of the points of contact, if 
parallel tangents) is the 0. M. D. of the poles of the line at infinity. 

2442. (The late Professor Townsbnd, F.R.S.) — Through a given 
point in a plane draw a line the simi of the squares of whose distances 
from any number of given points in the plane shall be a maximum, a 
minimum, or given. 

2443. (J. Griffiths, M.A.) — Prove (1) that the Jacobian of the three 
conies represented by the trilinear equations 

8 = sin2 A . a2 + &c. - 2 sin B sin C .*)37- &c. = 0, 
S'=co82A.o2 + &c. -2cosBco8C.)37-&c. = 0, 
F r= sin 2A .a* + &c. - 2 sin A . fiy- &o. = 0, 
breaks up into the three right lines 

^ , y ,0 y . « - 

8in(C-A) 8in(A-B) ' sin (A - B) sin (B - C) ^ * 

+ —- ^ «0. 

8in(B-C) sin(C-A) 

Hence show (2) how to construct geometrically the common self-con- 
jugate triangle of the three conies in question. 



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156 

2467. (Professor Cropton, F.R.S.) — A Cartesian oval, having a 
given focus F, is made to pass through three fixed points on a straight line. 
Show that the fourth point in which it meets the line is also fixed ; and 
that the locus of the points of contact of its douhle tangent is a circle with 
F as centre. 

2468. (W. B. Davis, B.A.)— Prove that a curve of the fifth order 
and fifth class has three double points, three points of rebroussement, 
three double tangents, and three points of inflexion. 

2487. (The late Rev. R. H. Wright, M.A.) — If a conic be circum- 
scribed about a triangle ABC, and tangents be drawn at A,*B, C, and 
produced to meet so as to form respectively three triangles having the 
sides of the triangle ABC for their bases ; find forms for the bisectors of 
the angles of the external triaugles, and the equations to their circum- 
scribing circles in trilinear coordinates. 

2488. (The late Professor ToWnsend, F.R.S.)— Apply the method of 
homographic division to draw the two right lines which intersect four 
given right lines in spoce. 

2496. (The late Dr. Booth, F.R.S.)— Let U s <p (|, v) be the tan- 
gential equation of any plane curve, t the portion of the tangent between 
the point of contact and the foot of the perpendicular p from the origin. 
Then generally 

i being the angle between the perpendicular and the radius vector. 
Now a weU-known formula for the rectification of plane curves being 



1 « ^± (pd\, 



V • V COS A. > J Sin \ 
in which =» I and ■■ 

P P 



we shall find the rectification of a plane curve of which the tangential 
equation is given generally as easy as to find the quadrature of that whose 
projective equation is given. [To apply these principles, take the Caustic 
of the circle for parallel rays, discussed in the Editor's Note to Question 
1509 (Vol. II., p. 21).] 

2601. (N*Importb.) — Find the equation whose roots are the differ- 
ences of the roots of the equation («, A, c, rf, e) (a;, 1)* = 0. 

2508. (Professor Ceofton, F.R.S.) — 1. A point being denoted by 
(p, tr, t), its tri-polar coordinates or distances from three poles R, S, T,' 
show that all circles represented by Ap- + Ba-^ + Cr^ = are orthogonal to 
the circle RST. 2. If O be the centre of the circle Ap2 + Btr^ + Cr^ = D, 
show that A, B, C are proportional to the triangles SOT, TOR, ROS. 

2538. (The late T. Cottbrill, M.A.) — Prove that, if the equation to 
a curve is of the form x^y^isT = k (where p-\-q-¥r = 0), the order and 
class are the same, and the singularities are reciprocal ; (2) if the variables 
denote point coordinates, the locus of a point on a tangent to the curve, 
which with the point of contact is harmonic to the lines ar^ + y^ + 2axy=\)^ 
is of the form Pp . Q^ . R»* = jwg*", P and Q being linear and R of two 
dimensions in a? and y ; and find (3) what is the reciprocal theorem, if the 
circular points at infinity are the pair of points. 

2555. (The late Professor De Morgan.) — The following is a theorem 
of which an elementary proof is desired. It was known before I gave it 



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157 

in a totally different fonn in a communication (April, 1867) to the 
Mathematical Society on the eonic octogram ; and the present form is as 
distinct from the other two as they are from one another. If I, II, III, 
lY be the consecutive chord-lines of one tetragon inscribed in a conic, 
and 1, 2, 3, 4 of another; the eight points of intersection of I with 2 
and 4, II with 1 and 3, III with 2 and 4, lY with 1 and 3, lie in one 
conic section. A proof is especially asked for when the first conic is a 
pair of straight lines. There is, of course, another set of eight points in 
another conic, when the pairs 13, 24 are interchanged in the enimciation. 

2660. (J. J. Walker, F.R.S.)— Given that either of one pair of 
impossible roots of the equation 3a:* — 1 6a:^ + Z^x^ + 8a: + 09 = gives a real 
result when substituted for x in bs^^l^^—lx^ it is required to find the 
four (impossible) roots of the biquadratic. 

2564. (The late M. Collins, B.A.) — A being a curve whose equation 
is given in the usual Cartesian rectangular coordinates, B the evolute of 
A, and C the evolute of B ; required a general differential expression for 
the radius of curvature of C, on the usual supposition of dan being taken 
constant, and likewise on the supposition of dx^ + dy^ {— dz^) being taken 
constant. 

2666. (Professor Crofton, F.R.S.) — 1. A convex boundary of any 
form of length L, encloses an area A. If two straight lines are drawn at 
random to intersect it, the probability of their intersection lying within 
itUp^ 2irflL-2. 

2. The probability of their intersection lying within any given area », 
which is enclosed within A, is p « 2ir(itfL~^. [An interesting but much 
more difficult problem is to find the chance of their intersection lying on 
a given area », extei-nal to A.] 

3. If an infinity of random lines are drawn across the given area A, 
their intersections form an assemblage of points covering the plane, the 
density of which is clearly uniform within fl. Show that at any external 
point P the density varies as 0— sinO, where is the angle A subtends 
at P. 

4. If A be any plane area, enclosed by a convex boundary of length L, 
and e be the angle it subtends at any external point P {x, y), prove that 

|J(0-sinO)</j;dy « JL«-fl, 

the integral extending over the whole external surface of the plane. 

2678. (Professor Crofton, F.R.S.) — If A be the difference between 
the whole length of a complete hyperbola and that of its asymptotes, and 
if be the angle between the tangents to the curve from any external 
point (ar, y), then jj {$— sine) dxdy ^ J a', the integral extending over 
the whole surface of the plane outside the hyperbola. [When the two 
tangents touch the same branch, 6 is the exterior angle which they make.] 

2680. (A. W. Panton, B.A.)— 1. If F and F' are the foci of an oval 
of Cassini, and C its centre ; prove the following construction for the 
second pair of foci. The circle through F or F' and the two points where 
any line through C meets either oval (the curve consisting of two distinct 
ovals) cuts the axis in one of the required points. 

2. If S and S' be the foci thus found, and P any point on the curve, 
prove that PS . PS' is proportional to PC. 

VOL. XLIX. ' V 



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158 

2602. (Professor Crofton, F.B.S.)^1. If be the angle between the 
tangents to an ellipse from an external point {x, y), then jjedxdy ^ tA, 
the integ^ extending over the annular space between the curve and any 
outer similar coaxal ellipse ; A being the difference of the parts into 
which that space is divided by any tangent to the inner ellipse. 

2. Show toat this theorem holds for any two convex boundaries, so 
related that any tangent to the inner cuts off a constant area from the 
outer. 

3. Show that, if the same integral, with regard to any convex boundary, 
be extended over the annulus between it and any outer convex boundary, 
jj ddxdi/ ^ IT (e— 2A), e being the area of the annulus, and A the average 
area cut from it by a tangent to the inner boundary (the tangent being 
supposed to alter by constant angular variations). 

2629. (The late Professor Townsbnd, F.R.S.) — For a system of 
quadrics inscribed in the same pair of cones, real or imaginary, and having 
. consequently double contact with each other at tiie extremities of the chord 
common to the two planes of intersection of the cones, show that a 
variable line touching in every position the two cones determines (a) two 
systems of pointe inversely homographic to each other on every quadric of 
the system, {b) four systems of points, all homographic with each other on 
every two quadrics of the system, (c) pairs of variable chords cutting each 
other in constant anharmonic ratios in every pair of quadrics of the system, 
{d) triads of variable chords in involution with each other in every triad of 
quadrics of the system. 

2632. (The Editor.)— Prove (1) that 1 . 2 .3 ...n < 2*'»(»-»); and 

(2) that < 4a or 4^, when a and b are both positive. 

2664. (The late T. Cotterill, M.A.J — 1. Five points (no three in 
the same line, and no four in the same plane) determine, by the lines and 
planes through them on a plane, a system of ten points and ten lines, the 
points lying m threes on the lines, and the lines passing in threes through 
the points (Caylby). Show that the figure is its own polar reciprocal to a 
conic ; and that, if a conic and triangle in its plane are given, the rest of 
the figure can be constructed. 

2. A quadric through the five points cute the plane in a conic contein- 
ing triangles conjugate to the fixed conic. There is a quadric passing 
through the fixed conic, to which one of the five points and the plane 
are pole and polar, the remaining four pointe forming a self -conjugate 
tetrahedron. 

2672. (R. Tucker, M.A.) — From a point on an ellipse chords are 
drawn parallel to fixed straight lines ; find the maximum triangle formed 
by the three chords of section. 

2683. (R. Tucker, M.A.) — To each point on the circumscribing 
circle of a triangle corresponds a foot -perpendicular line ; this cute the 
circle in two pointe ; required the locus of the intersection of the feet- 
perpendicular lines corresponding to these pointe of section. 



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APPENDIX IV. 

NOl'ES, SOLUTIONS, AND QUESTIONS. 
By R. W. D. Chkistie. 

(A.) DIOPHANTINE ANALYSIS. 



This branch of Algebra derives its name from its inventor, Diophantus 
of Alexandria, in Egypt, who flourished in or about the third century. 

'* No person has ever surpassed Diophantus in the solution of these 
problems, and few have equalled him." 

1. To find three square integers whose sutn is a square integer. 
Let a?2, y', «' be the three required squares, and let x'^ + y^ ^ o^. 
Assume a^+z^ = (na^z)^, then a « {2nz)l{n^~ 1). 

Let z = n*— 1, then a =• 2«. 

Similarly, y « m^— 1, a: = 2m, a = m2+ 1 « 2fi. 

Therefore the squares are (2w)2, (w«-l)2, {(m'-l) (m2 + 3)/4}2, where 
m may be anything, but, if integers are desired, then m must be an odd 
number. Otherwise, let (2«y)2, {y(«'— 1)}^ and x^ be the required 
squares. Assume (2«y)3+ {y {n^-l)}^ + 3F^ = {(»2+ l)y}2 + a;2 « (ny + x)^, 

say. Then a: - ^(!^±illl^| y. Let y = 2«; thena; = n* + «8+l, 

and the three squares are (2«)*, {2f»(«3-i)}2^ {«* + w2+l}«, 
where « is anything. 

It is clear the process may be extended to 4, 5, 6, &c. squares, but 
the following method is simpler. 

2. Find four integers in Arithmetical Progression, the sum of whose squares 
is a square integer. 

Let a?— 1, X, a?+l, a; + 2be the required numbers. 

Assume (a;- 1)2 + {sp) + (a; + 1)2 + (a; + 2)^ = 4ar2 + 4a> + 6 = (2a;-y)2, say. 

Then x = -sL , where y is anything integral or fractional. 

4(y + l) 

Let y = 6, then a: = ^, and the four squares are (i^)', (H )2, (f|)2, 

(11)2, or,.rejecting denominator, 12 + 16- + 292 + 432 == 542, 

It is plain the process may be extended to any (m2) squares, e.g,, 

22 + 62 + 82 + 112+142+172 + 202+232 + 262 = 482. 



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3. To find three square numbers in Arithmetical Progressum, 
Let a> be the first square, and 2az -¥ ^ the progression. 
Then the three squares are a^; a'+ 2<wf + «* ; a2 + 4<»j? + 2r^ 
The first two are already squares. It remains to make 

«' + ^ax + 2x^ » a square ^ {nx- a)\ say. 

Then ,«?1^. 

Let a «n2-2; thenar- 2(« + 2); 

2ax + x*^ com. diff. = {in) (« +!)(« + 2), 
and the required squares are {««-2)2; («2 + 2ii + 2)2; («5+4« + 2)*, 
when n is anything. 

4. If N — a3 + ^, prove that it also equals 

where a, 5, m, and n may be anything integral or fractional. 
Let nx^a, and mx-^b » sides of squares sought. 
Then (#M:-a)2 +(»«?-*)« - N. Therefore {n^ + m^x ^2{mb + na), 

Let a be any of the following expressions, viz., 

2n, 4(« + l), 4(2»-l), 3(2» + 3), 12 («+!), 8(2« + 3), 6(2» + 5), 
7(2« + 9), 10(« + 6), {m(2w + m)}, &c. &c. 

Similarly, b the corresponding expressions 

(««-l), {(2#f + l)(2» + 3)}, (4«2-4«-3), 2f»(#f + 3), (4n2+8f»-6), 

(4n8+12#f-7), 2f»(« + 6), {2(«+ l){ii + 8)}, «(» + 10), {2«(« + m)}, 
&c. &c. 

Then N becomes a square, and we shall have divided the sum of two 
squares into two other squares, N also being a square. 

6. We know, by Eulbr*8 theorem, that a sjrstem of any number of 
binomial factors being multiplied together, their product is the sum of 
two squares, e.g,, {a^ -i- 6^ {e^ -^ d^ i^+P) (y^+A^ «i?^ + ^. 
If, now, we make ad ^ be andfff '^ eh, q vanishes, and thus we can divide 
a square into four different pairs of squares, as e.g,, 

1230« = 12002+ 270« = 7382+9842 - 11222+6042 = 7982 + 9362. 

6. To divide a given square number which equals the difference of two 
square numbers, into the difference of two other squares. 
Generalising the expressions given above, we easily obtain 
{2r2+2r(2« + l)+4#f}2+{(2fi-l)(2r + 2n + l)}2 

= {2r2+(2»+l)2r + 4i»2+il8, 
where n and r are anything. 
In this equation make « — 2, « « 1 respectively, and we get 

(2f«+10r + 8)2+ (6r+ 16)2. {2r2+ 10r+17}2 (1), 

and (2r2 + 6r + 4)2 + (2;- + 3)2= {2r2 + 6r + 6}2 (2). 



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Now, equate the shortest sides 2« -i- 3 ^ 6< + 15, say. 
And let 2« + 3 » side of given square => 21, say, then a ^ 9, r — I. 

Then in (1) make r «- 1, and in (2) r = 9. 

Thus 20« + 21« = 29», or 21« - 293-202 (1). 

And 220» + 21« = 2212, or 21« » 2212-220' (2). 

Similarly for any other square whose side is given. 

7. Draw a straight line cutting two eonemtric circles so that the part 
intercepted by thetn is divided into three equal portions. 

Let R, r be the radii of the outer and inner circles respectively. As 
the outer intercepts are always equal, let x be the middle intercept. Then 

we must have E'— r* « 2a^, 

2mr 



The analysis gives us 



m3-2* 



Let r =« m^—2; then x « 2m, and R = fn« + 2 ; e.g., if m =» 4 inches, 
then r^ XyB,^ 7, 4, 9 inches. Therefore, measure ok 4 inches from the 
outer circumference cutting the inner one, and produce the straight 
line joining the points. The three intercepts are equal. 

8. (Question 2814.) — To Jlnd three rational square integers in Arith' 
metical Progression having a common difference o/ 13. 

"We have 

{x^+{x + 1)2} {{x+ 1)2 + {x + 2)2} db {2 (a; + 1)}2 = {2x^ + 4a? + 3)2 

or (2a;« + 4a?+l)2 (1). 

Also (w2 + «2)2±(,rt + «)(m-n)(4mf») = {m2±2in«-«2}2 (2). 

Now 13 = 22 + 32 ; therefore let a; = 2. Thus (1) gives us 
13x26 + 36 = 192 and 13x26-36 = 172. 

Let i? « 13, j2 « 25, r2 ^ 36, and puti?j2 for #» and r2 for « in (2) ; 
then {pY + *'*)^±(pg^-^r'^{P9^—^^{^P9^r^) = two squares. 
But (^j^— r*)(4^2r2) = a square, 

therefore fi^—r^ = a square, and consequently 

Jo2(74 + r* 0^164568*241 . , 

— i^H -r^ ^ 3Q *^:»»^vu^:«x _ gq^are required, 

(l^q^^r*)^^ 376584400 ^ « -reHiurwu, 

and the other two = -_|V + »1^^13^ 

(P'q^-r^){^q^) 
Hence it appears that the question will always admit of a solution 
when the given number plus or minus a square are both squares. 

9. To find a number which, being added to or subtracted from a sqtuire, the 
sum or remainder shall be a square. 

Use theorem 2 in (8), supra. 

Let m = 2, w = 1 ; then 62db24 = 72 or I2. 

N.B. — The three squares are inArithmetical Progression, thus 12 : 52 : 7'. 



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10. Comtruet a parallelogram toho$e sides and diagonals may he represented 
by integers. 

Let a, d be the two contiguous sides, and 0, d the diagonals ; then 

(« + 3)«+(o-3)2-c« + <P. 

Assume p, q ^ any integer, and a, b, e, d are now easily found. 

11. Tioo chords within or tvithout a circle intersect at right angles. Con* 
struct it so that the four segments as well as the diameter mag be represented 
by integers. 

Resolve any square into four other squares ; e,g., 
D«-a« + A2 + c2 + ^ (V, infra). 

Then D — the diameter ; a^b^c^d^ the four segments of the intersecting 
chords. 

12. Construct a regular decagon and a regular pentagon in a circle so that 
the sides of each, as well as the radius, mag be represented by integers. 

Let P «= side of pentagon, and D ^ side of decagon, B = radius. 
Then P^ « R^ + D'. Therefore assume any of the expressions given in 
A . 4 for R and D, making n » any integer. 

13. Construct a series the terms of which may be taken to represent the 
three sides of a right-angled triangle, and find the sum, 

Wehave (2fi-l)2+(2it.ft-l)« = (2«2--2it+l)«. 

Therefore assume the n^ term = (2n-l)(2f».fi-l)(2M2— 2fi + 1) ; 
and the series becomes 1.0.1 + 3 . 4 . 6 + 5 . 12 . 13 + 7 . 24 . 26, &c., 
and the sum = Jw' {(4f»2_i)(nS~l)}. 

14. To find a series of biquadrates equal to a series of squares. 
We have 2i.„..«M)/sJ(„.«,„ = (3«' + 3«-l)/6. 

If M :b 6, we have 

l* + 2^ + 3H4* + 5* + 6* « 62 (12 + 22 ■|.33 + 4« + 6' + 6«). 
From this value of n others may be obtained thus : — 
Suppose that re — /, y = g is a solution of the equation x^—'^y'^ « a, 
and let a: = A, y = A; be any solution of the equation x^ -Ny2 » 1 ; then 

a?2-Ny2 - t/^-N^2)(A2-NA:2) = (/4 + Ny*)2-N {fk^ghf. 
By putting x=^fh± ^gk, y =fk ±gh, 

and ascribing to A, A; their values found by convergents, &c. 

15. For X^ = 1* + 2' + 3* ... w*, see an interesting solution in Vol. xlix. 



(B.) RESOLUTION OF SQUARES. 

1 . If any number N can be resolved into the sum of n squares, then 
2 (n — 1) N can be resolved into the sum of « (»— 1) squares. 



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Let N = fl2 + A2, then 4N » (2o)« + (2*)2. 
Again, if M = o2 + ^2^.yi^ |;hen 

4M = (a + i3)2 + (a-^)2+(a + 7)2+(o-7)2 + (i8 + 7)'+(i8-y)2. 
H a2 = o* + ^2, by any of the formulaa in (4), we easily get 

-[i(a + )3)P + [J(a-«P+[i(a + 7)?+[i(a-7)P + [i(^ + 7)? 

+ [i(i3-7)P. 
Ex.gr.: 102 + 22-62 + 82 + 22- l2 + 2« + 32 + 42 + 52 + 72. 

Again, let N » a2 + ^ + c2 

«[i(a + *)]2+[i(a-*)]2+[i(« + .)p + [i(a-.)]2 + [J(* + ^)P + [i(*-^)?. 

Ex.gr.: N=* 122 + 62 + 22-102 + 82 + 42 + 22= 92 + 32 + 72 + 52 + 42 + 22. 

2. If any number N can be resolved into the sum of n squares, 
\ {n— 1 . ft— 2) . N can be resolved into the sum of 4n squares if n > 3. 

LetN = a2 + ^2 + ^ + ^^ 

Then3N«[i(a + 3 + c)]2 + [J(o + *-tf)]2 + [J(a-* + c)]2+[J(* + tf-a)]2 
+ [i(a + ^ + rf)P + [i(a + 3-rf)]2 + [i(a-* + rf)p + [i(5 + rf-a)]2 
+ [i(a + <? + rf)]2 + [i(a + c-rf)]2 + [i(a-<j + rf)p + [J((, + d_a)]2 

+ [i(* + (J + rf)]2 + [i(* + c-rf)? + [4(*-^ + ^? + [i(^ + ^-W. 

Similarly, 6N = sum of 20 squares, ION => 24 (or 30 squares), &c. &c. 

And, since the sum of the first n natural members is a perfect square, 
if w is = A;2 or A:'*— 1, where k is the numerator of an odd, and k' the 
numerator of an even convergent of ^^2, if either (1) J(#i— l.fi— 2)— N; 
or (2) N and i (»— 1 . « — 2) (a sequence from unity} are both squares, w© 
can thus resolve a square into 16, 20, 24 ... 4 (» + 3) squares. 

3. Weknowthat(a + *)2 + (* + c)2 + (r + a)2 = (a + i + c)2 + a2 + i2 + c2«M. 
If a - i (Sb-e), we have also [i(7*-<?)]2 + K(* + c)]2 « M. 

If also we make 5{b + e) : 7b— e ::»»:«, where »»2 + «2 „ ^2^ ^q g^^ 
M2 - [i (7b-e)y + [$(* + c)]2 - (a + 3)2 + (* + e)^ + {e + af 
= (a + * + <j)2 + a2 + 62 + c2. 

Or again, M2 - [| (7a + 17*)]2 + [J^ (a + *)]2=3 squares=4 squares, &c, 
Ex. gr. : 1302 „ 6O2+1202 - 602 + 962 + 722-1092 + 132 + 372 + 692 
— five squares, &c. 

= 782 + 1042 = 402 + 782 + 962 =. 1072 + 1 12 + 292 + 672, &c. 
Generally, to get N2 - a;2 + y2 ^^ (a + 3)2 + (a + c)2+(i + c)2 

« (a + 3 + c)2 + a2 + ^ + c2, 
let (« + *):(* + <?)::#»:«, where »»2+«2 « pi. 

Then ^^m^^^^j)-nb^ ^^ (a + 3)2 + (a + c)2+(3 + c)« 

n 
(m2 + w2)f3 + g)2 ( (w-n)3 + (m + >»)g y 
" «2 +[ ,n j 

^ p(3 + g) |2 ^ ( (^~n)^ + (m + »a)g |2 ^ ^2^^3 . 



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164 

and to make x^-^y* ^ N^y assume 

p{b-k-e) : (m-fi)ft + (»t + fi}c ::?:>•» where y' + r^ « «2^ 
and we get e ^ pr^{m - «) ^, * = (w + «) ?—!>*•, « « (m— «) y +/?r. 

Thus finaUy, N^ = (2i»«)2 = (2^?^' ^ (2pr)3 = (2i«(7)2 + (2/>r)2 + (2n^)2 
- {jt?r + (m + f»)y}^ + {(m-fi)^ ••'■i'''}' + {(♦» r«)y-i?r}2 ^. {jt?r- (w-n) j}2. 

And since >n, » ; gt fl Pt * &re interchangeahle, we can always secui-e 
four different resolutions of N^ 

Ex. gr. : 1302 = 1202 vSO' . 96«'r 502+722 -.- IO92 + 372+I32 + 692 
-502+1 202 « 302 .;. 1 203 .> 402 « 952 ^ 552 + 252 + 652 
-1042+782 = 402 + 782 + 962=. IO72 + II2 + 292 + 672 
- 782+1042 « 302+ 1042 + 72 - 1032+312+12 + 732. 

The following transformations of squares may here he noticed : — 

where N and r may he iuf,egers > unity. Or, again 

N2.N2f'^-^|%N2f-f'\V, 
Cm2 + r25 (jn^ + f^y 

where m and r ntay he assumed at pleasure m > r. 

(2) (a2 + ^ ((j2 + rf3) - (a<? ± &rf)2 + (arf :p ^C)2 =r (flk?)2 + (Atf)2 .?- (a4)2 + {irf)2, 
(a2.;.62)(^ + rf2)(^+/») „(A3 + B2)(u2+/2) « (A^±B/)2+(B^=FA/)2, 

(a2 + i2)(c2 + rf2)(u2 +/2)(^2 + ^2) « (^2 + B2)(C2 + D^) 

= (AC±BD)2+ (BC=rAD)2 ^p^-i-q^ 
where /) =• AC + BD = (a<j + W) (^^ +J%) + (*c-a(/) {fg-eh), 
q - BC-AD = (itf-flr^ («^ +/A) - (a<? + i<0 (/y— *A)> 
is one of eight solutions. 

(3) a2 + *2j.^= [j(a + i + <j)]3 + [|(a + J-<j)]-i+[|(a-*+<j)]2 

+ [i(3 + .-a)]2 (1), 

and a2 + ^ + ^ + ^«[j(a + j + tf + rf)]2 + [^(a-i_c + rf)]2 

+ [i(«-* + <?-^? + [i(« + *-«-<^P... (2); 
also (a + * + £;)2 + a2 + ^2 + ^= (a + 3j2+(^ + ^)2+(^ + rtj2. 

Thus (1), (2), ife2 + /2 + ^2 + „2«a;2 + y2 + 22 + ^ + ^. 
(1), (3), i?2 + ^ + ^«a;2^.y2 + 5;2 + ^+(^ + 5 + ^)2j 

(4) (fl* + 3«2+l)2« {2«(f»2+l)}8+{„4 + „2+l|2 

«(2«)*+{2n(«2-l)}2+(„4 + „2+l)2; 

(«) (y-«)2 + (2-a;)2+(a;-y)2+(a; + y + 2)2« 3(a:2 + y2 + j,2). 

Thus 

a? + J2 + c2={t(y-«)}2+{J(i5-a:)}2+{Har-y)}2+{j(^ + y + «)}*, 
where a;, y, z are the sides, and a, 6, « the medians of any triangle. In 
this way other identities may he made practically useful. 



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165 

(6) (ey-bz)2+(az^cx)^-¥(bx-ayy+(ax + bi/ + czy 

= (ax)^ + (ay)2 + {azy + (bx)^ + (by)^ + (^2)2 + {cz)^ + ((?y)2 + (ez)^ 
= (fl2 + i2 + c2)(a;2 + y« + a2). 

4 . By Pollock's Theorems we can resolve any number N into one, two, 
three, or four (not more) squares as foUows : — 

Take any trigonal number (or sequence from unity), sayi? =« 10. 
We have 2;?+ 1 = 21 = m^ + m+l = (2m ± 1)^, and the squares are 
(n ± 1)2, «2, «2, »2; and 21 = 32 + 22 + 22 + 22. 

Again, if ^ = sum of two trigonal numbers, say 15 ; we have 
2i?+l = 15=:2a2 + 2a + 2&2 + l = (^5 + i)2^ ^2^ ^2^ ^2^ 

whose roots are (*+!)> (~^)» (+^)> (~^) = !• 

Thus 15 - (3)2 + (-2)2 + ( + 1)2+ (-1)2 (roots - 1). 

Lastly, if ^ = sum of three trigonal numbers, say 47 ; 
we have 2i?+l = 47 = 2a2+2a + 2*2 + 4it2±2« + 1, 

and the roots are ^ a+n ± I, a—rif b + n, b—n, (—1). 
Thus 47 = (-3)2 + ( + 2)2+( + 5)2+(-3)2 (= 1). 

Therefore, since every number must either be a trigonal number or 
composed of two or three trigonal numbers (Lbgendre, Theorie des Nombres), , 
every odd number may be resolved into integral square nimibers (not 

exceeding four) whose algebraic sum will be 1, 3, 5 2»— 1, up to the 

maximum. Even numbers may also be resolved into square numbers 
(not exceeding four) the algebraic sum of whose roots may always equal 2. 

It may be interesting to reproduce here Pollock's method of resolving 
odd numbers into squares. 

Eesolve 57 into four squares the sum of whose roots may equal 9. 

Then2«j + 1 = 9; thus w =» 4, andw2 + «j+l = 42 + 4 + 1,= 21. 

Now 
57-21 = 36, and 36 + 1 = 37=(- 1)2 + ( + 2)2 + ( + 4)2 + (-4)2 (roots = 1). 

2 +2 +2 +2 
Therefore 21 + 36 = 57 = l2 + 42+62+(-2)2 (roots = 9). 

By this theorem we can instantly transform one square into two, three, 
or four squares, ad libitum ; 

e.g., 181 = 12 + 42+82+102. 

Then 181-42= 155 « 12 + 82+102 = 42+62 + 72+82; 

thus 12 + 102 = 101 = 42 + 62 + 72= 12 + 62 + 82. 

Then 42+72 = 12+32 = 65=22+32 + 42 + 62; 

thus 72 = 22 + 32 + 62, &c. , &c. 

5. We know, if N = a2 + ^2 + ^ + ^2^ there are two resolutions of 4N into 

VOL. XLIX. X 



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uneven squares, viz., 

= A2 + B-'+C« + D2, say (1), 

«(A-2rf)2+(B + 2rf)2 + (C + 2tf)2+p + 2tf)2 (2). 

£.ff., let N = 92 - 23 + 42 + 62 + 62 - &c., by Pollock ; then 
4^-^182=42 + 82+ 102 + 122= 172 + 62+ 12+ 32 = 52 + 72 + 92+ 132, by Gauss. 

6. In order to resolve M2 into N integral squares, we may make use of 
the following principle : — 

(« + 1)2_ (^^2 _ 2» + 1 = any odd number > 1, 

therefore take any N squares whose sum = 2« + 1, 
Bay 22+32= 13 = 2«+l; 

thus #f = 6, then 72 = 62 + 22 + 32, or generally 

(«■+«- 1)2 = (w.« + l)2+(«+l)2 + w2. 
Or again, we have 12 + 22 + 32 + 42 + 52 + 62 = 91 = 2« + l and » = 45 ; 
thus 452 = 442 + 12 + 22 + 32 + 42 + 52 + 62, or generally 
(3«2+l5« + 28)2= (3n2+15w + 27)2 + 

n2+ (» + 1)2+ (« + 2)2+ (» + 3)2+ (« + 4)»+ (« + 6)2, 
where n is arbitrary, and so on. 

7. The same object may be effected thus : — We know by easy analysis 
that if a; = a^-b^, y = 2ab, 

then «2 = a;2 + y2. 

also, if a; = a2+ ^2-^2, y = ^acy z = 2*r, 

then «2 = a;2 + y2 + j;2j 

again if x = a2 + ^3 + ^2_^^ y = 2<w?, z = 2bd, w = 2crf, 

then «2 ^ u;2 + fl;2 + y2 + ;52 . 

nnd generally, if A, ^ a^-k-a^^a^ ... — «J, 

and Aa = 2«, a,„ A3 = 2,a^ a^ An = 2a„-i<i„, 

then U2= A,2 + A22 + A32 +Ai. 

E.g., let «i = 1, ^2 = 2 flj = ^ » 

thea 112= 12 + 22 + 42+62 + 32, 

and flo on for any number of squares. 

H . Numerous resolutions may be obtained thus : — Take the first n^ 
iiaturttl nimibers and arrange so. Let n = 4: ; then 

1, 15, 3, 13, 6, 11, 7, 9, 
16, 2, 14, 4, 12, 6, 10, 8. 

Ntitc! the odd numbers in the top row and the even numbers in the bottom 
TH^f also the sum of the numbers at the top and bottom = 17 = »2+ 1. 



I 



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167 

Take any four odd numbers whose sum =32 from the top row, and the 
remaining four even numbers from the bottom row ; e.g.^ 

6,7,9,11, 2,4,14,16, 
8, 7, 9, 13, 2, 6, 12, 16. 

The sum of the odd numbers =32, and the sum of the even numbers = 36, 
and the total sum = 68 = m . »'*+ 1. 

Then we have at once : — 

(a; + 5)2+(a? + 7)H(a; + 9)2+(a?+ll)3+(a? + 2)2 

+ (a; + 4)2+(a;+14)2+(ar+16)« 
= (a: + 3)«+(a; + 7)2 + (x + 9)2 + (iF+ 13)2+ (a? + 2)2 

+ (ar+6)2+(a:+12)2 + (;r+16)2, 
= (a: + 1)2+ (a?+ 16)2+ (a:+ 14)2+ (ar + 4)2+(a:+ 12)2 

+ (a; + 6)2+(ar + 7)2 + (a? + 9)2, 
= (a?+8)2 + (a? + 10)2+(a:+ 11)2 + (a: + 5)2+(a?+ 13)2 

+ (a: + 3)2+ (a? + 2)2+ (a: +16)2, 

- (a:+ 1)2+ (a;+ 12)2 + (ar + 8)2+ (:c+ 13)*+ (a?+ 15)2 

+ (a: + 6)2+(a;+10)2+(a? + 3)2, 
= (« + 1)2+ (a; + 6)2+(a; + 11)2+ (a?+ 16)2 + (a: + 4)2 

+ (a: + 7)2+ (a: + 10)2+ (a: +13)2, 
= (ar+16)2+(a;+14)2+(a;+12)2+(ar + 9)2+(a; + 3)2 

+ (a; + 2)2+(:F + 5)2+(a: + 8)2, 

- (a:+ 11)2+ (a: + 16)2 + (a?+ 10)2 + (a?+ 13)2+ (ar + 1)2 

+ (a; + 4)2 + (a: + 6)2 + (a? + 7)2, 

- (a?+ 11)2 + (a; + 16)2+ (ar + 4)2 + (a: + 7)2+(a:+ 14)2 

+ (a: + 2)2+ (a: + 9)2+ (a? + 6)2, 
&c., &c. 

But as these squares are not all different, we get, after elimination of 
the identical squares, — 

(1) (a: +14)2+ (a; + 4)2+ (a: + 7)2+ (a; + 9)2 

-. (a: + 8)2+(a; + 13)2 + (a;+10)2+(a? + 3)2r=4a:2 + 68a: + 342, 

(2) (a; + 1)2 + (a: + 15)2 + (a; + 12)2 + (a- + 6)2 

= (a: + ll)2+(a? + 2)2+(x + 6)2+(ar+16)2 = 4a:2+68a; + 406, 

(3) (a: + 15)2 + (a?+14)2+(aj + 12)2+(jr + 9)2 

= (a;+ 11)2+ (a;+ 16)2+ (a;+ 10)2+ (a?+ 13)2 « 4^2+ I00a?+ 646, 

(4) (a;+3)2 + (a: + 2)2+(ar + 5)2+(a; + 8)2 

= (j;+l)2 + (a: + 4)2+(a: + 6)2+(ay + 7)2-4a;2+36a:+102, 

(5) {X + 12)2 + (X + 8)2 + (a; + 15)2 + (a; + 3)2 

= (a:+ll)2 + (a; + 16)2+(a. + 4)2+(a?+72) = 4a;2 + 76:c + 442, 

(6) (a: + 14)2+ (a; + 2)2 + (a: + 9)2+ (a; + 5)2 

= (oj + 1)2 + (x + 6)2 + (a; + 10)2 + (a? + 13)2 = 4a:2 + 60a? + 306, 
&c., &c. 



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It is now easy to make any of these » a square. 
We have (1) 4a;2 + 68a: + 342 = (2a:+ 17)2 + 63 = y« + 63 = z\ say; 
and, since 53 » 27 + 26, we have at once 2 — 27, ^ = 26, and ^ » f . 
From this value of x we easily obtain 

642= 172 + 232 + 272 + 372= 162 + 262+292 + 362. 
Similarly (2), 4a?2 + 68a; + 406 ; if a: = V» ^® establish 

1182 = 432+712 + 662 + 632=632 + 452 + 612 + 732. 

(3) givesus 222 = 52 + 72+112+172=32 + 92+132 + 152, 

and (4) the same. In order, therefore, to find another value of x we have 
by Eulbb's metiiod a: = y + i. Thus 

4a;2 + 36a?+ 102 = 4y2 + 40y + 121 = {py-Uf, say. 
Then y = V, and a; = V^. 
Neglecting the denominator, we have 

1902= 792 + 862 + 972+1152 = 732 + 912+1032+1092. 
In this way endless resolutions may be obtained. 

n = 6, by a similar process, will give us a still larger variety of 
equations, of which one is 

(a: + 1)2 + (a: + 33)2 + (a? + 7)2 + (ar + 27)2 + (a: + 22)2 + (a; + 21)2 
= (ar + 36)2 + (i; + 4)2 + (a; + 30)2 + (a;+10)2 + (ir+15)2+(ar+16)2, &c., &c. 
If ft = 10, we may obtain any number of resolutions such as 
(a; + 100)2 + (a; + 68)2 + (a; + 92)2 + (ar + 86)2 + (a; + 4)2 + (a; + 6)2 + (a? + 6)2 

+ (a: + 7)2 + (a: + 8)2+(a? + 99)2 
-.(aJ + l)2 + (a? + 3)2 + (a? + 9)2+(a;+ 15)2 + (a; + 97)2 + (a: + 96)2+(a; + 96)2 

+ (a: + 94)2 +(-,; + 93)2 + (a; + 2)2, 
or, if we prefer it, 
(2)2 + (2«2 _ «)2 + (2» + 2)s + (2»3- 3f»)2 + (4« + 2)- + (2«2- 5«)2 + {fin + 2)2 

+ (2«2- 7„)2 + (13« + 2)2 + (2«2_ 9«)2 
= (2n2)2 + („ + 2)2 + (2w2 - 2»)2 + (3« + 2)2 + (2w2^ 4w)2 + (5» + 2)2 
+ (2»2-6«)2 + {In + 2)2 + (2«2_ 13«)2 + (9« + 2)2, 

the principle of which will readily suggest itself on a comparison of the 
upper and lower squares ; and generally, for any unevenly even number of 
squares, say 18, we have 

(2)2 + (2«2_ w)2 + (2« + 2)2 + (2n2_ 3«)2 + (4f» + 2)2 + (2«2_ 5n)» + (fin + 2)2 
+ (2it2_7„)2 + (8» + 2)2+(2f»2-9„)2+(i0n + 2)2+(2»2-ll«)2+(i2n + 2)« 
+ (2«2- 13n)2 + (14» + 2)2 + (2«2_ 15«)2 + (25» + 2)2 + (2«2_ I7n)2 

»(2«2)2+ („ + 2)2+ (2«2_2«)2+ (3« + 2)2+ (2»2_4«)2+ (5» + 2)2+ (2«2_6«)2 
+ (7w + 2)2+(2«2-8«)2+(9» + 2)2+(2;»2_io«)2+(ll« + 2)2+(2w2_l2»)2 
+ (13« + 2)2+ (2w2- 14»)2+ (15/* + 2)2+ (2W--25/02+ (17» + 2)2. 



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9. To make s'/^i = a square, where 

2; = !'• + (« + !)♦• + (« + 2)** («+n-l)^ 

and a the common difference, we have 

2i/2i = l+a(«-l)+«2[i(ft.w-l)]=M2. 

(a = 1.) Then »« + « = 2M2 ; thus SM^ + 1 =« a square = N2. 

Thus 8M^— N^ = — 1, an equation easily solved by the even convergents 
of ^/2. 

(a = 2.) Then 2«*- 1 = M^, solved by the odd convergents of ^/2 ; 

^ Tw 'T ^x. 13 + 33 + 53+73 + 98 -o 
..^.,n = 5,M = 7;thus ^^3^,^,^, =7- 

Consequently, 13 + 33 + 53 . . . (2w - 1)3 = (nM)^, where n is the denominator 
and M the numerator of any odd convergent of //2. 

(a=3.) Wehave 9w2_3»-4 = 2M2, or (3«-J)2 = J (8M2+ 17), 
M = 1 obvious. 

Therefore, let M=N+1 ; thus 8M*+17=8N2+ 16N + 25-(i;N-5)2, say. 

Thus N = ^-^1^;^; i? = 4, &c. ; N = 7, &c. ; M = 8, &c. ; « = 4, &c. ; 

13 + 43+73+103 Qj 

'•^" 1 + 4 + 7 + 10 =^' 

and so on for higher values of a. 

When any one value of n is obtained, others are readily got by substi- 
tution. 

In a similar way we can convert 2^ / si into a square, where 

2? = a'' + (a+l)'' + (a + 2)»* (a + n-l)'*, 

a being the first term ; e.ff., a = 3 gives us, iin^—n-^ 2a (?e — 1) + 2a2= 2M2, 

w2 + 6» + 12 = 2M2 or (« + f )2 = i (8M2 - 23), M = 2 obvious. 

Therefore, let M=N + 2 ; then 8M2-23=8N2 + 32N + 9 «(i?N-3)2, say. 

Then N - '^{^P-^^^\ ^ == 4, N = 7, M = 9, « = 10. 

p — 8 

Thus 33 + 43 + 58 + 63 +73 •»• 83 + 93+ 103+ 113+ 128 ^ 

3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 ' 

and so on for any value of a. 

10. To resolve a square into two, three , four , or more squares. 
Applying the results given above, we easily establish 
(2w4 + 4;>3 + 4„s + 2» + l)2= (2«2 + 2« + l)2+ {(2» . «+ l)(«2 + «+ i)p 
= (2;j + l)2+(2».«+l)2+{(2».«+l)(n2 + »+l)}2 
= four squares, by Pollock, = &c. 
Ex,gr., (w= 1) 13'-^= 5^+ 122 = 32 + 42 + 122 = 42 + 62 + 62 + 92 « &c. 



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11. Again, to resolve three squares into three others, the sums of wh6se 
roots are equal, we may make use of the following identity, 

(b + ecCf + {hcf + rf2 = {be + rf)2 + ja + (^^2^ where *, Cy d are arhitrary, 

the sum of the roots heing (* + i^(c+l); and, if * « 2», c = « + l, 
<? = 2« + 1, we easily get 

(2«2 + 5» 4- ly + (2» . « + 1)3 + (2« + If =x (2w2 + 5» + 1)2 + (2n2 + 2n + 1)2 
-(2»2 + 4n+l)2+(2«)2+{(n + l)(2» + l)}2. 

Ex. gr., 82 + 42+32 = 82 + 52 = 72 + 22 + 62, 

192 + 122 + 52= 192+132= 172 + 42+152, &c. 
Similarly {m (« + fH)}2+ {n («+#«) }2+ (mw)2 = {m^^mn-^-n^Y. 

12. To resolve an even square into eight or sixteen squares. 
Primes are of two classes, viz., 4«+ 1 and 4» — 1 form. 

Now it has been demonstrated by Db la Grange (Afemoirs of Berlin, 
1768), and others, that primes of the former class may be resolved into 
the sum of two squares ; and, since it may be shown that every even square 
> 5 equals the sum of four 4« + 1 primes, it follows that every even 
square > 6 equals the sum of sixteen squares, since Pollock has shown 
that every odd number may be resolved into four squares ; thus 

62= 1 + 5 + 13 + 17 = sixteen squares (by Pollock) 
« eight squares (by Lagbanoe), 

82 = 6+13 + 17 + 29- &c., 

102= 13 + 17 + 29 + 41 = &c., 

&c. &c. 

182= 37 + 41 + 97 + 149 =41+61 + 73 + 149 = 41 + 63 + 73 + 167 

= 41 + 73 + 97 + 113 = 41 + 73 + 101 + 109 = &c., &c., 
= sixteen squares in six ways at least. 



(C.) RESOLUTION OF CUBES. 



EuLER has demonstrated that it is impossible to find any two cubes 
whose sum or difEerence is a cube ; also that the formula a^:ht/^ = 2s^ ia 
impossible. 

There are two methods of resolving a cube into three cubes. 

(a) ff two of the cubes are given, and a third be required to make the 
three cubes equal a cube, we may use the following formula : — 

{a(b^-a^)Y+{b{b3-(^)y+{a{2b^ + <^)Y= {*(^ + 2a3)}3. 
Thus any cube may be resolved into three cubes ; thus, 

lb (b' + 2a' ji lb (// + 2a«)) {b (L' + 2a»)) * 

and a similar principle applies throughout. 



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{b) Again, if it be required in general to find three cubes whose sum may 
equal a cube, Euleb gives us 

X =p + q=^ {fi + ^ffu) + (fft-'fu), 
y=.p-q = {ft + Sffu)-{fft-^fu), 
t> =. r + « = {kt—hu) + {ht + dku), 
z = r— « = {kt~'hu)^(ht + dku), 

where u =f{P + Ss^^^h{h^ + 3k^), t = Zk{hf + Zk^)^Zff{P + 3ff^), 
and ffSl^hjk are arbitrary. 

.-. x^ + f/^^f^-2^ = {{ft + Zffu) + {^t-fu)Y+{{ft + 3ffu)-(s/t-fu)Y 
= {{kt-hu) + {ht + Uu)Y-{{kt-hu) - (fit + Uu)Y. 

(c) To obtain four cubes equal to a cube, we may use the following equa- 
tion:— («-a)3+(»-*)3+(*-.c)3 + 3a*<? = «3. 

Assume Zahc = (3mn)', thus abc = dm^^. Now 9m^n^ = abc may be 
assimied in any combination ; e.g., let a = 9m, b = m^n, c = n^. Thus, 
e.ff.y if M = 2, n = 3, we easily get 13 + 63 + 73 + 123 = 133. 
Other combinations are a = 9m\ b = n, e ^ n^ ; 
a — 9m', b = wn, <? = f»2 ; 
a = 9»j, b = m3, c = n^, &c., &c. 
To secure the desired result make m<n, 

(e?) If m>n, we may secure the result that the sum of two cubes equals 
the sum of three cubes thus : w = 3, » = 2 gives us 
133 + 413 + 363= 493 + 53. 

{e) "We may secure four cubes equal to two cubes by the use of the 
following identity : — 

(a; + y + 2)3+(a; + y-2)3 + (a;-y + «)3+(a;-y-«)3«4a;(a?3 + 3y2 + 3552) 

= 4a:3+12a:y2+i2a;22 = a:3 4.3^(a;2 + 4y2 + 4z2) = ipS + 3;c (M2), 
if a; = a2 + ^-c2, y == «<?, and z = Jc, making M = a^+i^ + ga. 
If, also, we assume 3a; = M, i.e. if a^ + i^ _ 2^2, by making 

a^^-\-2qp-q\ b =^ p'^-lqp-q^, c=^^^-q\ 
we establish (a: + y + «)3 + (a; + y — 2)' + (a;— y + 2)' + (a;— y — «)3 = a;3 ^ (3^j3 . 

(/) Also, if a:, y, and 2 are so chosen that any two are > the third, we 
have the sum of three cubes resolved into three other cubes, e.g., 

i? = 2or3, ^=1, gives us 113+133=13 + 33 + 53+153, 

p = i, q^l, gives us 7993 + 5613+173 = 2893 + 8673 + 2213. 

{ff) To find n cubes whose sum may equal a square. 
Let 2;|= l'' + 2'' + 3'-... +«^ 

It is easy to prove that 2* + (22')' =* (322)2, 

(13 + 23 + 33... +^.3)^(^2^^)8^^ ^•^^^^l■^2;^+l j2^ 



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e,g,f if n » 5, we have 

l» + 2» + 3» + 4» + 6»+30»= 1662- {3 (12 + 2* + 3' + 48 + 52) p. 
[si - (si)' is well known.] 

(A) Tojind two cubes whose sum may equal the difference of two squares. 
Since 2* = (S^', we get from (^) 

(22')'=(32,)»-(2')», 

and, since we know that «' = ( ^lli! \'_ ( ?!i:5 V, 

we have at once ««+ (»'+«)• - (" " ^ Z"''^)'- (^)*- 

C/) To find a number of odd cubes whose sum may equal a number of even 
cubes. 

We have i»-2» + 3*-4»... =-(4«» + 3«2) ; 

therefore l» + 3» + 5»... +(2n-l)» + 4N» + 3N2 = 2» + 4» + 6»... + (2N)», 
or U + 3» + 5»...+N» + 3N2(N + l) -2» + 4» + 6»... + (2N)». 

To make 3N2 (N+ 1) = a cube, let N - 8 ; thus 

l» + 3»+6» + 7» + 9» + ll»+13»+15»»2»+4»+6»+10»+14»+16». 

To obtain other values of N, we may proceed by Eulbr's mode to sub- 
stitute N = M + 8 in ZW (N -i- 1), and equate it to (pm + 12)>, and so on. 

(k) To resolve a cube into a number of squares. 

We know by the Analysis that x^ + y* can be made « (/?2 + y2)*». 
Let X = y _ zpq\ y = Zp^q-q* ; 

then a?2 + y'=CP« + ^» = {i»(i^ + ^')}2+{^(i>' + «^}*; 

similarly 

(^2 + j3 + r2)>={p(i;2+^ + r2)}2+{y(j,2^.^ + ^p+|^(^ + y3 + r2)}2, 

and generally 

{il« + r + r2...««}»= {p (1^ + 52+. ..;5J)p+{^(pS + ^8+...;52)}+&C., 

C>n.fw + 1.2m + l')» r,«/m.fn-^l .2m + l\72 

1 — 6 — j-ri — 1 — )i 

(/) Every cube may be resolved into an Arithmetical Progression. 

We have S = a+(a + *) + (a + 2*) ...a + (w-l)d = «|a+ **-! . j| . 

Let f> = 2m + 1, then 2 == (2m + l)(a + ^). 

If a + ^» = (2m + 1)2, then sum of n terms '= «'. Thus let 

a = 2w-l, * = 2(fi-l), «=»6; 
tf.^., 9 + 17 + 26t33 + 41 = 5». 



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(m) To resolve a number into three cubes. 

Take 6 for example. 

Assume 6 = (2 - *ar)» + (ex- 1)8 + (ete- !)», 

and make ^ » (0 + d) / 4 ; 

then x^ {66(c8 + eP)-16crf}/{2l(c8 + d8)_tfrf((j + rf)}. 

If tf « 7, rf = 6, 3 = 3, then 6 = ( J)8 + (j)s + (j^)8. 

Again, take 8 =« 2^ = {i-bx^+iex -f 1)«+ (dx^ l)i, 
and make 3 — {c + d)/4; 

then « » {3 (dS-u«)-6*3} / {c» + rf»-.A»}. 

If d^ 7, e — 5f then ^ » 3 and a: » ;^, and we have 
36» + 98'« 92» + 69», 

or the sum of two cubes resolved into two other cubes. 

If rf = 6, tf = 7, * = 3, we get 3; = - f and 20* - 7* + 14» ^- 17», or a 
cube equal to three cubes. 

Similarly, for any number if we take care to make the given number in 
the original equation vanish and to equate the coefficient of a; to zero. 

(n) To find a cube equal to the difference of two squares. 

Assume a;3— ^'= a^ » (ar-fty)', say; 

then 2»a? = y (y+n'). 

If y s 2n, then x = ft'+2» and a — 2n—n^; thus 

(2»)' - (»2+2ft)2-(»2«2n)«, 
where n is arbitrary. 

(0) To resolve the sum of four cubes or biquadrates into four other cubes or 
biquadrates. 

We have (a + 3+<?)'» + a* + ft* + <J» — (a + 3)* + (* + <?)'• +(c + »)•• + X", 

(«=3). (a + 3 + (;)» + a» + *» + (j"=: (a + b)9 + (b + c)t + {e + a)t + 6abc. 

To make 60^0 = X', proceed as in (C. e), and assume 6abe = (6mn)^ 
in any order, as, tf.y., a = 36m, ^ = m%, c = nK 

Thus m« 2, « = 3 give us 3' + 4' + 24»+ 31» « 7' + 12»+ 27» -1-288. 

(fi — 4). Again, for 

(a+b + c)* + a!^+b* + c*==^ {a + by + {b + e)*+(e + a)*+l2abe{a + b + e), 

Tomake I2abc{a + b+c) - X* == (nb)* b&j ; 

let c = 1, a = 3^*^ thus 36 (3»2* + 3 + 1) = (w*)2. 
Now assume Zn^ + J + 1 = (2«*— 1)' say ; 

thus b — -^-^ , where n is arbitrary ; e,ff., « • 1 gives us 

1^+2^ + 94 = 344.74 + 34. 

VOL. XLIX. Y 



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174 

(D.) SOLUTIONS OF OLD QUESTIONS. 



1010* (The Editor. ) — ^Pind integral yalues of oj, y , z which will make 
«'-2y', y*—Zz\ aj^— 6«3 all squares. 

Solution. 
An obvious solution \ax ^Zz, y=2z\ but this makes two of the squares 
identical. Therefore, let y - 2«, then «8-8«5 = a\ a^-6z^^^\ 
thus a«-»-3«2 = J8. 

Let a = jf^—Zq\ z « 2py ; then b —p^ + Zq*. 
To find X, we have a^ = a« + 8«2 »« ^ + 6«3 = j?* + 26/?3^2 + 9^ 
— (jp2— 6^)^, say therefore, 3p = 2q, 
Thus a? «j93-5^2, y = ipq, z = 2p^, where 3i? = 2y ; 
<f.^., 1» » 2, J = 3, a? = 41, y = 24, « « 12 ; 

a2 = 23«, ^ == 31', c2 « 12. 



1014. (The Editor.) — Find the least integral value of x which will 
make the expression 927a;^— 12362; + 413 = a square. 

Solution. 

We have 103 (33?- 2)2+ 1 = p^yji.e., y2-103r' = 1. 

The twelfth convergent of V^103 gives us y = 227628, z = 22419, and 

the 24th y = 1035379»1667, z = 10201900464 ; but neither of these values 

of z will give us an integral value of a? ; so that either we must go fiirther 

or else the method of convergents does not ensure us integiul values. 

If a; = I, we have 927a;2_ 1236a; + 413 = 1. 
Suppose the quantity a-¥bx + ex^ ^ ff\ where a; » /, so that we have 
a + bf+e/^^ff^y 

., . . /m^— 3fn--927\ 

then we get ,-,.^ ____), 

where m is arbitrary, an equation giving all other possible values of x ; 
whence it is clear x can never be an integer. 



1042. (The Editor.) — Find values of x which will make each of the 
expressions 3a;3 + 1, a;^ + 1^ 2ar*— 3a;2+ 2 a square number. 

Solution, 

EuLBR has shown that a;^ + 1 can never become a square except in three 
cases, viz., a; — 0, —1, and 2 ; so that, although we may readily obtain 
values for x in the other two equations, it would appear that no value of 
X will simultaneously satisfy the three equations. 



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2814. (The late Matthew Collins, B.A.) — Can the common differ- 
ence of three rational square integers in Arithmetical Progression be ever 
equal to 17? 

Solution. 

Square numbers are of two forms, 4N and 4N + 1. 

Let the three required squares be x'^y y'^y t?. 

Suppose (1) a;2 = 4N ; 
then y'^ = a:^ + 17 »= 4N + 17 = 4M + 1, which is possible, 

«2 =a?2-17 = 4N-17 = 4M-1, „ impossible. 

Again(2),leta:2=4N + l; 
then y' = 4N + 18 = 4M + 2, which is impossible. 

Thus, by either supposition it is impossible for a;-, y', «2 to be squares 
simultaneously. 



8930. (R- W. D. Christie.)— Prove that, whether (n) be odd or even, 
sin»fl= sine { (2cos0)»-i-(»-2),2co8a)'»-3+ &L=:^^;^IlD (2 cos »)»•-» 

>("-^K^~^H^-6)(2cose)n-7^...j'. 

Solution,* 
Cette formule est certaine pour les cas n = 2, n = 3, parce que I'on a 
d'abord 

sen 26 = 2 sen 6 cos 6 = sen a {(2 cos 6)2-1 1^ 

sen 30 = sen 2a cos <; + sen cos 20 = 2 sen cos^ + sen (cos^ - sen' 0) 
= sen0(4co82 0-l) -= sen {{2 cos 0)2 -(3 -2) (2 cos 0)o} 
= sen0{(2co8 0)«-i-(3-2)(2co8 0)3-8}. 

Cela pose, nous aliens prouver que la formule qui est verifi6e pour le cas 
(w— 1), lo sera pour le cas \n). 
Par supposition nous avons 

sen (»- 1) « sen J (2 cos 0)»»-2- (n - 3)(2 cos 0)»-* 

■, («-;K«-s) (2coB6r-'...| (1). 

Mais sen w0 = sen . cos («— 1) + cos sen (n— 1) ; 

mais (voyez Carr, Synopsis of Fure Mathematics, page 177) Ton sait que 

2cos(»-l)0 = (2cos0)^-'-(n-l)(2cos0)"-8 + ^^'~|^^^~'*^ (2cos0)»-^.., 

1 . iS 
d'od cos (» - 1) 

= 2»-2(cos0)— i-(«-l)2«-*(cos0)«-s+ (^IllK|:=i)(cos0)«-«.2~-«..., 

et substituant la valeur de sen («— 1) en (1) nous avons tr^ facilement 
la formule demandee. 

* This solution is due to Professors Betens and Catalan. 



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9444. (R- W. D. Chwstie.)— Solve (1) in integers «* + irV* + y* =«* ; 
and (2) note the result when a b 3. 

Solution, 

We have {a^+xi/ + ip){x^-'xy + y*) = oA. 

Assume a?±xy + y^ — a or * » ar^i ary/w ; 

then ay«±fiy/»-l. Lety-ii-1; then 3; = ±n; also a«n'— n + 1. 
n may now he assumed any integer at leisure. 

II ** + «V +5^ - «' = («*- «V)', sayi 

then ««(2ii? + l)-(n*-l)y2. 

then jc'=tn*— 1. 



9621. (R. W. D. Christib.)— Prove that (j;**Mir*)/6 is an integer 
where p is any perfect number and ir any prime number except 5. 

Perfect numbers end in the digits 6 or 8 ; therefore p^ ends in 6. 
Prime numbers end in 1, 3, 7, or 9 (except 2 and 5) ; therefore it^ ends 
in unity, also 2^ ends in 6. 
Therefore CP^-ir^)/5 is an integer, except ir — 6. 



9608. (Sbftimus Tbbat, B.A.) — Find the least heptagonal number 
which when increased by a given square shall be a square number. 

Solution, 
The general form of heptagonal numbers is \ (6x'— Zx), Let a> be the 
given square, and h the number sought. 

Assume a^ + A = (a- w)' ; 

then h s=s n3— 2af>. 

But 40A + 9 is always a square ; therefore assume 
40 (n3~2an) + 9 « (6fi + 3)3. 
Thus n a 20a + 9, and a may be assumed at pleasure. Let a = 1, 
then ft = 29, therefore h =. 293-58 =. 783. Thus 783 + 1 = 283. 



9629. (Professor Gbrondal.) — Partager 90° en deux parties ar, y 
telles que la tangente de I'une soit le quadruple de la tangente de Tautre, 
et prouver que tan |a? = 2 sin 18®. 

Solution, 

We have a? + y = 90" (1), 

and tana; = 4tany (2). 



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Then 
therefore 



therefore 



tanrtany ^s 1 

4tan2y - 1 

tan y = ±h 

t&nx =±2 = (2tan Ja;)/(l-tan2ix) ; 

tania: = J(<v/6-.l) « 2siIll8^ 



(1); 

(1), (2), 



9643. (R. W. D. Christie.) — If 2n^V +2'' + Z'' ... »!»•, prove that 
2n is exactly divisible by 2n when r is odd. 



Solution. 

Obtain 2n by the formula (n+1) 2»i« {duldx)'XH*i and separate o^^ 
values of r from the evens, and we shall find that 2n is a constant factor 
of the odd and (2n + 1) / 2 (r + 1) of the even. Thus, let « ~ 2n ; then 
r = 1 gives us », 

«^, 

**'(4*-l), 

i««/l6«3-20«« + 3(4*-l)}, 
4«2|i6^_32«» + 34«3-6(4a-l)}, 

--L- «» {960*»-2800»< + 4692«»-4720«- + 691 (4* - 1)}. 

iA;.[4(6*-l)], 

|A:a[4(12»2-6«+l)]. 

>A:« [^ (40«3-40»2 + 18«-3)]. 



r= 3 
r- 6 
r- 7 
r- 9 
r = ll 

r = 13 

r=. 2 
r== 4 
r« 6 
r= 8 
r = 10 

r = 12 






;j{48«<-80»« + 68«2_5(6«-l)}], 
-L- . {3360a« - 8400«4 + 1 1480»»- 9440«2 



+ 691 (6« 



r= 14 



-1)}], 



^ks [J {l»2««-672«5+ 1344»<-1760«» 

+ 1436«3-106(6»-1)}], 
&c., &c. 

Thus the theorem appears to be true for odd values of r only ; that.it 
is not true for even values of r may easily be tested by making r ^ 4, 
n = 2, 3, or 4, for example. 



9668. (Professor Vuibert.) — Si Ton designe d'une mani^re generalo 
ir Sm la somme dos puissances do degrc m dcs n premiers nombres entiers, 
jmontrer qu'on a (3JS6 + 2Si^) /0S4 = Sj/Sj. 



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178 

9643. (R W. D. Chmstib.)— If 2: - !»• + 2'' + 3** .. . n»- ; prove that 
21 is divisible by 2j». 

9226. (J. White.)— Prove that 

13 + 23 + 33 ... M3 is a factor of (l»+2« + 3« ... M«) x 3. 
8784. (R. W. D. Christib, MjV.)— Prove that, if 
«« 1 + 2 + 3+.. . + », 82 = 12 + 22 + 32 + ... +w2, S3=P + 2S+3»+...+n», 

2 - l* + 2* + 3<+...+», <r- l« + 2» + 3« + . .. + »«, 
then {3<r + 2«») / 62 - Ss/S^. 

9883. (R. W. D. Christie, M.A.)— If 2r = l»* + 2»*...+«»-, prove 
that 726 + 624 = 1222 2s. 

9042. (H. L. Orchard, B.Sc., M.A.)— Prove that 1' + 2^ + 3' + ... + a:^ 
is a factor of the expression Zsfi + 12x7 + 1^0^—7 x^* + 2jt^. 

9102. (H. L. Orchard, B.Sc., M.A.) — Show that the series 
V + 27 + 3' + 47+ ... + 97 is divisible by 27. 

8647. (R. W. D. Christie, M.A.) —If «« 18 + 23 + 33+ ... +«>, 
8 = l* + 2* + 3» + . ..+«*, 2 = 17 + 27 + 37+. ..+n7; prove that 2 + 8 = 2»2. 

9142. (R. W. D. Christie, M.A. See Quest. 8700.)— If 

2r = l'' + 2»' + 3'' «^ 

prove that (92ii + 302, + 92;) / 23 - (112,0+3028+ 726) /2j. 

Solution.* 

In 1834 Jacobi proved that Sgn^i contained 8,2 as a factor, and that 
S'in contained Si as a factor. Another proof was given by Prouhet in 
I80I ; and agaiu, an a priori proof by Caylby in 1867 ; and there are pro- 
bably others. The simplest of those mentioned is Prouhet's; viz., 

writing (l+A)*" in the form 1 +riA + r2A + ..., 

it is easily shown that (Sr = l*" + 2*" + 3'' + . . . + n**) ; 

(r « odd), riSr-i + r3Sr-3 + ...+»*r-4S4+rr-2S, = J[(» + l)'- + fi'--2n-l] 

(1); 

(r «even), riSr^i +r,Sr-3 + ... +rr.8S3 + r,..i8i := i[(n + l)'' + n*"-l] 

(2). 

In (2 ), (« + 1)*" + «•' — 1 vanishes when n = and when « + 1 = ; there- 
fore, by putting r = 2, 4, 6, ..., we prove that Sj, and therefore S3, and 
therefore 85, and so on, are successively divisible by i»(« + l). But 
Prouhet proved the full theorem. Let 

K « {n + \Y+n^-'l-2rSi = (w + l)'-+W-l-m(« + 1) ; 
then both K and dK/dn vanish when « =« and when « + 1 = 0, there- 
fore K contains n^ {n + 1)2 as a factor. Hence, by putting r = 4, 6, 8 ..., 
we prove successively that S3, 85, S; . . . contain Si'*^ (or 83) as a factor. 
Similarly, the secoud part of the theorem is proved. (No. 9643.) 

* This solution is due to Mr. J. D. H. Dickson. 



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179 

The following theorem is capable of proof — 

(2n— l)S2n-2 = 38, (aoS2»-6 — «2S2n-7 + ^4S2»i-9— .••±<»2n-6SiT«2n-fi} 

-(3). 

2llS2i»-l = 483(^0^211-5— *2S.n-7 + *4S2H-9—...±^n-6Si=Fd2n-5} 

.(4). 

where ^q == 2«— 4, ip = 2n— 4, 

oj = i (2ii-4)(2n-6), i, - i (2«-4)(2«-6), 

the remaining a's and d's being somewhat complicated functions of 
Bbrnoulli's numbers. The first few cases are appended — 

654 = 3S,{28,-i} (5), 

655 = 483 {28i-i}... (6), 

786 =3S3{483-28i + i} (7), 

8S7 =483 {483-48, + !} (8), 

988 =883 {685-683 + ^1^1-1} ....(9), 

108, =483 {685- 883 + 681-1} (10), 

ll8iof= 382 {887- 1285 +1683-1081+4} (11), 

12811- 483 {887-I685 + 2683-2OS1 + 6} (12), 

138i2- 3S2{l0S9-20S7 + 44S5-^FS3 + i||^Si-f§i} (13), 

148i3 = 483{l0S9-¥87 + i^pS5-HFS3 + Hi^i-W}... (14), 
&c. 
No. 9226 follows from equation (6). 
No. 8784 (the same as 9668) may be written in the form 
(3S5 + 28,83)/S3 = 684/82; 
and by equations (5) and (6) each side equals 6S1— 1. 

No. 9683 comes from (5) and (7) by simple addition. 

No. 9042 is "prove that 248; = multiple of 83." And No. 9102 is 
nearly the same question, with 9 written for x. 

No. 8647 is 85 + 87 = 283', and, like No. 9683, follows from equations 
(6) and (8). 

No. 9142, by equations (8), (10), (12), and (7), (9), (11), shows that 
each side of the given relation is equal to 24 (87 + 85). 

The number of relations like the above maybe indefinitely extended by 
the theorems (3) and (4). 



9767. (R- W. D. Christie.) — Prove that «"» is the sum of n con- 
secutive odd numbers. 

Solution. 
«»»»-i— « is always even = 2p, suppose. 
Then w"» = 2iw + »2 

= the sum of 2p + 1, 2p + 3, 2j» + 6, ... to « terms. 
Thus also 

5n= (w.m+1 .2m+l)/6 = n + 3 (w-1) + 5 (n-2) ... 2m-l. 



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180 



9876. (R- W. D. Christib.)— Prove that 

2tan-»— ±tan-* « iir, 

where a is the coefficient of a?*» and d of «»♦ * in the expansion of - — -. 



Solution. 
If 2a* ±1 -*3_a2^ then 

2tan-i 4- ±tan-» -— i- = iir 

becomes tan**— -^, itan-^-— ; — - = 1. 

2ab±l 4ab±l 

Now tho coefficient of a?* in the expansion of is the sum of 

(u+1) terms of l + 2a:-a?* 

f 2>' + (n~l)2''-gH- ^~^;^^"'^ 2'«-* »+l or l](-l)% 

and those coefficients bear the assumed relation ; the sign depending on b. 
Examples, — If n = 2, then a » 5 and 3 => 12 ; thus we have 

2tan-»-^-tan-i-^ « Jir, 
12 239 * 

whioh lA Machin's formula. 

If » = 5, then a ~ 70 and b » 169 ; thus we obtain 

2tan-i — + tan-i-^— = ix. 
13-^ 47321 ^ 



(E.) NEW QUESTIONS. 



9877. DioPHANTUs' Epitaph. 

Hie Diophantus habet tumulum, qui tempera Titae 

mius mira denotat arte tibi. 
Egit sextantem juvenis ; lanugine malas 

Vestire hinc coepit parte duodecimo. 

Septante uxori post haec sociatur, et anno 
Formosus qninto nascitur inde puer. 

Semissem aetatis postquam attigit ille patemae 

Infelix subit^ morte peremptus obit. 
Quatuor aestates genitor lugere superstes 

Gogitur : hinc annos illius assequere. 

DST8. Every number contains an even number of factors, and tftre- 
l^rc tho numbers of odd and of even factors are either both odd or both 



Digitized by 



Google 



181 

even, except when the original number is a square, and then the reverse 
is the case {i.e.y it contains an odd number of odd factors, and an even 
number of even factors, and consequently an odd number of factors). 

9879. Prove that a'^h'^tf.., (where r = 2, 3, or 4) is of the same form 
as the squares, cubes, and biquadrates themselves, viz., 4N and 4N + 1, 
9N and 9N±1, 16N and 16N + 1. 

9880. 1. If N = a2 + ^2^ prove that it also equals 

{2m»*+ (»2-m2) a}2/(w2 + «2)2+ {2i»w«+ (w2-«2) i}2/(ws + n2)2, 
where a, *, w, «, are any integers whatever. 

9881 . Show that the sum of » terms of the following n series 

lr+2»- + 3»- n\ 

lr + 3r + 5r (2n- 1)% 

lr + 4r + 7r (3»-2)% 



l'' + (n+l)'' + (2« + l)»- («2-«+l)»*, 

where 2|[ = sum of l** + 2*' + S** to n terms. 

9882. Let »= 12 + 22 + 32 n^, 8 = 13 + 23 + 33 n\ 

5= 1^ + 26 + 35 w5, then S + 22= 3*2. 

9883. Prove the following property of prime numbers. Distribute 
the primes from unity together with their multiples as in Quest. 9225, 
into groups oifour (having however ^^ in i\\Q first group) then 

^„ = ^ + w = 6(^7-1-2), 
where g means the group, t the tens, and u the units in any prime. 
Ex.gr., • ^1 = (1+2 + 3 + 5 + 7) =6x3, 

, (11, 13, 17, 19) = 2+ 4 +8+10 = 6x4, 
(23, 29, 31, 37) = 5 + 11+4 + 10 = 6x5, 
[41, 43, 47, (49)] = 5 + 7 + 11 + 13 = 6x6, and so on. 

9884. Prove that the sum of the factors of any number is 

S/= (2''*i-l) f ^^-^ j, a = any prime > 2, 

where/ (a prime) the number of odd factors is Xjn^ of the number of 
even factors. Show also that, if n is any even number, N is a square 
(except when /= 2). [jEa;. — Let N have 7 odd and 14 even factors. 

Then 87 



vhen /= 2). [Ex.—LQi N have 7 od 
= (23-1) I ^^y) » and N = a square.] 



98 .5. Let <r= 12 + 22 + 32 n\ «= 13 + 28 + 33 n^, 

S = 14 + 24 + 3* «S 2= l8+2» + 36 n\ 

Then 7S + 5S = 4« x 3<r. 



VOL. XLIX. 



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182 

9886. Every square number is divisible into two sequences from m 
any integer). 

9887. Prove the following equation 

«;♦' = «»(«!:)+«:-', 

where a means any prime number, and a!^ — sum of factors of a^ ; hence 

show, if we could solve 2 = ^ ( >)+<»> j^ integers, an odd perfect 
would be found. *''**' 

9388. Divide the sum of two cubes into two other cubes. 

9889. Take any number of my digits (1, 2, or 3 together) , and I am equal 
tf] a sequence from unity. Cast out the nines from my dozen divisors 
arid you'll find the factors of each of my digits. I am a famous number, 
but not a perfect number, and both myself and the sum of my digits are 
ilhni^ible by a perfect number. 

9890. Find a number from the remainders after dividing it by a number 
of primes, say 3, 5, 7, 11, and 13. 

9891. Draw a straight line cutting two concentric circles, so that the 
jiurt intercepted by them is divided into three equal portions. 



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