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Full text of "Mathematical questions and solutions"

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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http : //books . google . com/ 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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-»)^ Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 * / ,% ^ Digitized by VjOOQ IC 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 Digitized by Google 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 Digitized by VjOOQ IC 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; Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by Google 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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^ Digitized by VjOOQ IC 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 Digitized by Google 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. Digitized by VjOOQ IC Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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). Digitized by VjOOQ IC 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- Digitized by VjOOQ IC 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, |.] Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 26 Solution iy Prof. Wolstbnholmb, Sc.D. ; J. W. Sharps, M.A. ; and others. Writing a^^p, b^^q, let U» - f*';^ ^ , . then Un -T (XJ„-i) ' T-i(^«-2)i "^d 80 on. JoJ» + ytan«« Jo(1+«')CP + ^2*) i'-^Jo \1+«' P + q^J ^^~g\'\p} ^^\P + Vpgl " V^\V>~v^MVi)' HenceU, = iir^-^-^-^^^-^^ -J-^^irJ^^Hf ^ ) ir 1.3.5... 2 (»-3) 2a^^'^d 2.4.6...2(»-2) (-1)" / 1 ±Y'' ( 1 \. 6...2(»-2) \a rfa/ \a + */ ' "*"2^ 2.4.6...2(»-2) where a^ ^ are positive. Thus * 4 o^A 4*(a + ft)«* 4 a5(a + A)2 4 a» (a2-*8)» ' IT =?5±-JL W 1 I ^ ^ = ir /g aM3g + 3) \ * 16a***16A a U2(a + A)« a (a + 3)3/ 16a*A \ (a + 3)» / " 16 a* (a + *)3 16 a» (a»- ^)» In the same way the value of the definite integral f>' tan^^a; ^ Jo (j» + (7tan2a:)« ' where m, n, and n—m are positive integers (or zero), will be found to be 2 «! \dp) \dql \p+Vpq)' (Obviously, iip^ qhe of opposite sign, the integral will be infinite.) Another method of evaluating Un is : — Put Vq tan x a Vp tan z, then p + qteai^x^ paet^z, VqBec^xdx ^ Vpsec'^zdz, and the limits 0, ^w are unchanged, so that '**I>«V? Jo (1+i?/? tan2«)(l + tan2«)"-i j^** Jq j^sin^z + jcos^* ' or XJ^^ V^^y !*'( P*" p*-(p-q)»cos^z \ ^^ P**(i'— ?)** Jo V^ sin"'' « + y cos' « p^{p^q) coai^z J — i»" ''(/>—?)' cos* «—... to « terms j dz Digitized by VjOOQ IC 27 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. Digitized by VjOOQ IC 28 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 '^ "" "^"^ * ^ (!-«')*• Digitized by VjOOQ IC 29 (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. Digitized by VjOOQ IC 30 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. Digitized by VjOOQ IC 31 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). Digitized by VjOOQ IC 32 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 Digitized by VjOOQ IC 33 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, Digitized by VjOOQ IC 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). Digitized by VjOOQ IC 35 [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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 37 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. Digitized by VjOOQ IC 38 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 Digitized by Google 39 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©. Digitized by VjOOQ IC 40 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. Digitized by Google 41 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/). Digitized by VjOOQ IC 42 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), Digitized by VjOOQ IC 43 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. Digitized by VjOOQ IC 44 • 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). Digitized by Google 45 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 Digitized by VjOOQ IC 46 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. Digitized by VjOOQ IC 47 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 Digitized by VjOOQ IC 48 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**; Digitized by Google 49 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- Digitized by VjOOQ IC 60 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 Digitized by VjOOQ IC 51 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). Digitized by VjOOQ IC 52 [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.] Digitized by VjOOQ IC 53 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. Digitized by VjOOQ IC 54 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. Digitized by VjOOQ IC 55 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. Digitized by VjOOQ IC 56 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. Digitized by VjOOQ IC 57 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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 69 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). Digitized by VjOOQ IC 60 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. Digitized by VjOOQ IC 61 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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 64 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' Digitized by VjOOQ IC 65 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 67 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. Digitized by VjOOQ IC 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^^ Digitized by VjOOQ IC 69 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 Digitized by VjOOQ IC 70 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), Digitized by Google 71 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), Digitized by VjOOQ IC 72 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 Digitized by VjOOQ IC 73 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. Digitized by VjOOQ IC 74 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. Digitized by VjOOQ IC 75 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) . Digitized by VjOOQ IC 76 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 Digitized by VjOOQ IC 77 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 Digitized by VjOOQ IC 78 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'. Digitized by VjOOQ IC 79 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.] Digitized by VjOOQ IC 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, $f Xi be any functions of ar, and u = 1 €*»*<*) }l/{z)dz + e (2) ; then, differentiating (2) with respect to (x)y (Todhuntbr's Int. Calc.,^. 198) dx j^ ax ax ax Digitized by VjOOQ IC 81 or S = S^t *"*^'^ • ^ W ^ W ^ + N (3), where N =» ^ €*.♦(•) 4,(0)- ^c'l^W.V'CiS) (4). Differentiating (3) with respect to x, we have where M - | . £ [^.« (•). + (,)]- g . £ [.-.♦'») . .^ {0)] (6) = gi{g •'■♦'•' ^ («) + (»)- g .»■♦ W. ^(J8) ♦(j8) j (6). Ffom (3) and {6) we obtain rfN .(7). Integrating the part affected hy the integral sign in (7) by parts in the usual way, we have r.'.*w . ^ (»)'+(«) <fe - r t^iM I [,-.♦(.)] & if }$ '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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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^, *'. «^ Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 87 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 89 [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, Digitized by VjOOQ IC 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}. Digitized by VjOOQ IC 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, Digitized by VjOOQ IC 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. Digitized by 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 Digitized by VjOOQ IC 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' Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 99 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 Digitized by VjOOQ IC 100 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)». Digitized by VjOOQ IC 101 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 Digitized by VjOOQ IC 102 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 Digitized by Google 103 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 Digitized by VjOOQ IC 104 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- Digitized by Google 105 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). Digitized by VjOOQ IC 106 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.] L Digitized by VjOOQ IC 107 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 Digitized by VjOOQ IC 108 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 Digitized byCjOOQlC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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°. Digitized by VjOOQ IC 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, Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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.] Digitized by VjOOQ IC 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). Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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. Digitized by Google 121 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. Digitized by VjOOQ IC 122 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/}, Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 124 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; Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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, Digitized by VjOOQ IC 127 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); Digitized by VjOOQ IC 128 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 \ Digitized by Google 129 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__^ Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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). Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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., Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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}; Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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) Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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''») — -. Digitized by VjOOQ IC 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. Digitized by Google 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. Digitized by VjOOQ IC 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, Digitized by VjOOQ IC 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- Digitized byCjOOQlC 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 150 (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. Digitized by LjOOQ IC 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 Digitized by VjOOQ IC 152 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 Digitized by Google 153 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 Digitized by VjOOQ IC 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 Digitized by Google 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 160 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). Digitized by VjOOQ IC 161 Now, equate the shortest sides 2« -i- 3 ^ 6< + 15, say. And let 2« + 3 » side of given square => 21, say, then a ^ 9, r — I. Then in (1) make r «- 1, and in (2) r = 9. Thus 20« + 21« = 29», or 21« - 293-202 (1). And 220» + 21« = 2212, or 21« » 2212-220' (2). Similarly for any other square whose side is given. 7. Draw a straight line cutting two eonemtric circles so that the part intercepted by thetn is divided into three equal portions. Let R, r be the radii of the outer and inner circles respectively. As the outer intercepts are always equal, let x be the middle intercept. Then we must have E'— r* « 2a^, 2mr The analysis gives us m3-2* Let r =« m^—2; then x « 2m, and R = fn« + 2 ; e.g., if m =» 4 inches, then r^ XyB,^ 7, 4, 9 inches. Therefore, measure ok 4 inches from the outer circumference cutting the inner one, and produce the straight line joining the points. The three intercepts are equal. 8. (Question 2814.) — To Jlnd three rational square integers in Arith' metical Progression having a common difference o/ 13. "We have {x^+{x + 1)2} {{x+ 1)2 + {x + 2)2} db {2 (a; + 1)}2 = {2x^ + 4a? + 3)2 or (2a;« + 4a?+l)2 (1). Also (w2 + «2)2±(,rt + «)(m-n)(4mf») = {m2±2in«-«2}2 (2). Now 13 = 22 + 32 ; therefore let a; = 2. Thus (1) gives us 13x26 + 36 = 192 and 13x26-36 = 172. Let i? « 13, j2 « 25, r2 ^ 36, and puti?j2 for #» and r2 for « in (2) ; then {pY + *'*)^±(pg^-^r'^{P9^—^^{^P9^r^) = two squares. But (^j^— r*)(4^2r2) = a square, therefore fi^—r^ = a square, and consequently Jo2(74 + r* 0^164568*241 . , — i^H -r^ ^ 3Q *^:»»^vu^:«x _ gq^are required, (l^q^^r*)^^ 376584400 ^ « -reHiurwu, and the other two = -_|V + »1^^13^ (P'q^-r^){^q^) Hence it appears that the question will always admit of a solution when the given number plus or minus a square are both squares. 9. To find a number which, being added to or subtracted from a sqtuire, the sum or remainder shall be a square. Use theorem 2 in (8), supra. Let m = 2, w = 1 ; then 62db24 = 72 or I2. N.B. — The three squares are inArithmetical Progression, thus 12 : 52 : 7'. Digitized by VjOOQ IC 162 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. Digitized by VjOOQ IC 163 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 . Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 166 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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 168 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. Digitized by VjOOQ IC 169 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. Digitized by VjOOQ IC 170 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. Digitized by VjOOQ IC 171 {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^ Digitized by VjOOQ IC 172 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». Digitized by VjOOQ IC 173 (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 Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 175 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. Digitized by VjOOQ IC 176 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). Digitized by VjOOQ IC 177 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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. Digitized by VjOOQ IC 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. Printed by C. F. Hodgson k Son, Gough Square, Fleet Street, E.C > Digitized by VjOOQ IC MATHEMATICAL WORKS PUBLISHED BY FEANCIS HODGSON, 89 FAREINGDON STREET, E.G. In Svo, clotli, lettered. PEOCEEDINGS of the LONDON MATHEMATICAL SOCIETY. Vol. I., from January 1865 to November 1866, price 10s. Vol. II., from November 1866 to November 1869, price 16s. Vol. III., from November 1869 to November 1871, price 208. Vol. rV., from November 1871 to November 1873, price 31s. 6(i. Vol. v., from November 1873 to November 1874, price 16s. Vol. VI., from November 1874 to November 1875, price 21s. 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