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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/ MATHEMATICAL QUK&^rtONfci SOLUTIONS. I VOL. LXXIV. Digitized by Google MATHEMATICAL QUESTIONS AND SOLUTIONS. FROM THE "EDUCATIONAL TIMES." WITH MANY PAPEES AIJTD SOLUTIOlJrS IN ADDITION TO THOSE PUBLISHED IN THE *' EDUCATIONAL TIMES.*' EDITED BY D. BIDDLE, M.R.C.S., PELLOW OP THE ROYAL STATISTICAL SOCIETY; PORMERLY STilTISTICIAK TO THE GENERAL MEDICAL COUNCIL. ^OL. XjXIX:!^. LONDON: FRANCIS HODGSON, 89 FARRINGDON STREET, E.C. 1901. Digitized by VjOOQIC iroa/yL B^r.S"^ LIS W C^^/-Ov^ i-AAA^c( TORSV b'k Abbott, R. C, B.A. ; Marl! Kdaxr, B. W. ; Guernsey. Alsxaitdeb, J. J., M.A.; 1 Allev, Bev. A. J. C, M.A. Afdeesof, Prof., M.A. ; Qu^« » x^« Afthofy, EDwnr ,M. a. ; Newton Ho. Appleton, Miss E. ; LiverpooL Abchibald, E. C, M.A. ; Strassburg. Abmenavtb, Professor ; Pesaro. Abnold, I. ; Institution, Belfast. Atetoit, Mrs. ; Hyde Park, W. Ball, Sir Robeet Stawell, LL.D., F.E.S., Pro- fessor of Astronomy and Geometry, Camb. Ball, W. W. Bouse ; Cambridge. Babniyille, John J.,M.A.; Krishnagar,Bengal. BabeetTjB. p., B.A. ; Stroud. Babtob, W. J., M.A. ; Highgate, London. Bates, G.N., B.A. ; Bamsgate. Battaolini, Prof. Giuseppe : Univ. di Boma. Bebcav, Prof. W. W. ; Univ. Michigan. Besaitt, W. H., D.Sc., F.B.S. ; Cambridge. Betens, Prof. lONACio, Lt.-Col. du G6nie; Cadiz. BHATTACHAETTA,Prof. MUNINDBANATH, M.A. Bhut, Professor Ath Bijah, M.A. ; Dacca. BiCKEBDiKE, C. ; AUerton, near Castleford. BiCKMOBE, C. E. M.A. ; Oxford. BiDDLE, D. (Editob) ; Charlton Lodge, Kmg- ston-on-Thames. BiOGS, B. ; Portora, Enniskillen. BiLLUPS, H. B. ; Sandown, I. of W. BiBCH, Bev. J. G., M.A. : Bunralty, co. Clare. BiBTWiSTLE, G., B. A., B.Sc.; Pemb. Coll., Camb. T5T-r---^=:. J„M,A.; Nnr^v^K."]. JilAULJi, JnEJEPJI ■ Budt'ii iniut II. liLYTWE, W, H.. M.A,; iSt. Mark'^J Li^iigt*, ^>vnl- BOLAJSD, J,; j*and HOl^ Gibt-altar. Wmm. Booth. Prof. AV. ; Hmfcl.v Crillcp\ Hs'npnL BORUAGB, ProL Edmond; Co)L iU« Xllllluzi. BouBJTB, Eev. C, T\'., M.A.; Kirii^^K i v.31. hjdi. Boston, V. i„ H,Wc. ; Huinpttm Uk-k. BOWSFR, Prof. E. A- ; New .ti*rse>% U.S.A. BOTa^, Ptxjf. ; Hfljal ColJ^p' nf Science. Bbjll, J., M.A. 3 rruTibrsclcTj., BEocAnp, H.,€ln'fU4' l?:iriislllnTi: Bwr-k^-Buc. BR<HJKa, Praf. E.t Milks^xilK Ppuiiaylvimia, Brows. A. Ceu^, D.-St. : EtiinburFli^ Brows, Prof,- Providence . It,s,Ap BfircF:, Rtv. IUjbebt, B.T>. ; Hurtrtersflphl. BBt?IS¥ATE. W. E, ] Trinity CollJlliLntbriagf. BBTATf. G. H.. P.R.S.; Ban^r. BHYASfT. H., iJ.Sf,, B,A.: l^ndon tT9Htihitiaii. Buck. Edwarc, M.A.; ("hikbill, \k\t\\. BtXBEE, J„ M.A,: TriuitV CnUs^pi?, Diililm. BuBNStDB, Professor. M/A. i Vniv. or Dulstin, BrRStAI-L. H. J. W,: Kinir's ColK. Utndoik Ca]IIa,B. N.. M.A, ; Ht. JoLiii's Coll, t'tujihridp'. OjVMtbeli.. J. E., M.A. s Hortfonl Cull., OsfoKi. CAJi:Kl,, M. N,, Lf„ B. ; Beclfoid !>qiiniv, I/hhIpil Cafuam. D. : Nopwic'h. CaBmodt. W* P,. B.A,: Ch^ntiipl Rraiii. Scliml. Care. G. S.. M.A. : Nottiii^ Hill. W. CAVAU4N, Prfjl,, M^A. . rnivprHJiy of Fpsnla. Cave, A. W., E.A. : MjucdJilfn Collem\ OxIorLl. Oavb, F. E. ; Sti>^i[tl»ini, CHAKBlVAItTI, Pn^fessor Hvo^WKHA : Culcntfn. CBArDBA BaSc. Pioftwsor, M.A. e Ciilciittn. CflAPMAJf, VmL H. \V., M. A.; Vm\\ nf Atblairjo, S. Austral in. Chabtkjjs. R*i CarU^Sttwt.^ljMidirstpr. CBRISTTE. B. W. D.i 8taiik\v Cdll.. Eivi-rixiril. Claire, A. : AlM*r>Mtwytli, Clabitk, Colonel A. R., C.B., F.Ji.S, : U^stlliill Ci.ATTON, Frof.t M.A. : i'mrkville. MiUnnurrje. COC13ES5, Profesiaoi'; Paj-is, CoHEX.ABTnuR.M,A..Q.C..M.P.; Tfijlknrl Pk. CoLffON, C. G*. M. A, : t^niversity f>f St. AiKlr* wsi. Doubtable. Rev. W. J., M.A,: Fppini.^lkiiriK Coos Doo . Prof . M A H E ^- ] m A N A T 1 1 : EiTuii'es. Coos BY. W, a ; Hiffb ^lnwL iHlblfll. CoopRR, T. M. A., M.A,; Clirlt^ nhmiL ""'iTT^BrLUiJ, H,. >1.A,; K. N. Coll . Giwnwich. WFORD. G. B., M.xi.i Clifton, HristoS. «. Luioi ; Bome. ressor M. W., P.B.S. ; Bognor. ITENE, M.A. ; Dublin. J. ; St. Beuno's Coll., St. Asaph. ^.^ . , Prof., M. A.; St.David'sCoU., Lampeter. KNiNOHAM,ALLAN,Lt.-Col.,B.£.;Ken8inffton CUEJEL, H. W., M.A. ; Southport. Curtis, Prof., M.A., S.J.; Univ. Coll., Dublin. CzuBEB, Dr. EMANUEL ; Prag, Bohemia. Dallas, B. J. ; Pemberton G^dens, N. Daniel, Vincent ; Cardiff. Daniels, A. E., B.A. ; Nottingham. Dabboux, Professor ; Paris. Data, Prof. Pbomathanath, M.A. ; Delhi. Da VIES, Prof., Brown Univ., Providence, U.S. A. Davis, Bev. B., M.A. ; Principal of Jai Narayan College, Benares. Davis, B. P., M.A. : Chiswick High Boad. Dawson, H. G.,M A. ; Christ Coll., Cambridge. De Lonochamps, Professor ; Paris. De Wachteb, Professor, M.A. ; Schaarbeck. De Vries, Dr. Jan ; Utrecht. Dey, Prof. Nabendba Lal, M.A. ; Calcutta. DiBB, J. H., B.Sc. ; Hendon. Dick, Hon. G. B,; Begistrar Gen., Mauritius. Dickson, J. D. H., M.A. ; Peterhouse, Camb. Dickson, W.; Edinburgh. DOBBS, W. J., M.A. ; Earl's Court. D'Ocagne, Maurice ; Paris. Droz-Parnt, Prof. A. ; College de Porrentruy. Drurt, H. D., M.A. ; Marlborough College. DuPAiN, J. C. ; Professor au Lyc6e d'Angoulfeme. Easton, B£lle, B.Sc. ; Lockport, New York. Edmondson, Prof. T. w. ; New York Univ. Edwardes, David, B.A.; Bangor. Elliott, Prof. E. B., F.B.S. ; F. Mag. C. Oxon. Elton, E. P., M.A.; Wellington College. Emmerich, Prof., Ph.D. ; Miilheim-am-Buhr. Eve, a. E., M.A. ; Marlborough College. Everett, Prof. J. D., D.C.L. EwiN, Cecil ; Drumcondra, Dublin. FiNCKEL, B. F. : N. Lewisburgh, Ontario, U.S.A. Flood, P. W. ; Drumcondra, Dublin. Forsyth, Prof., ScD.: Cambridge. FoRTEY, H., M.A. ; Clifton, Bristol. Foster, F. W., B.A.; Chelsea. Foster, Prof. G. Carey. F.B.S. Foster, W. S. ; Hoddesdon, Herts. Franklin, Prof. Christine Ladd.M.A., B.Sc. Freeth, Bev. Dr. ; Fotherby, Louth, Lines. GALLiERS.Bev.T. , M.A. ; KirsteadRect. .Norwich Gaskin, C. H. ; Birkenhead. Genese, Pi-of., M.A.; Univ. Coll., Aberystwyth. Georoi, Maurice ; Paris. GERRANS,H.T.,M.A.;Fel.&Tut.Wor.Coll.,Oxon. Gibson, T. Paul ; Hendon. Ghosh, Srish Chandra, M.A. ; Calcutta. Ghosh, Prof. U. Chandra : Allahabad. Glaisher.J.W.L., F.B.S.: Fel.Trin.Coll.,Camb. Goldenbero, Professor, M.A. ; Moscow. GoPALACHARi, Prof . ; Presidency Coll., Madras. Gordon, A., B.A.; Bamwood Ho., Gloucester. Gould, S. C. ; New Hampshire, U.S.A. Grace, G. M., B.A., B.Sc. ; Hammersmith. Grace, J. H. ; Peterhouse, Cambridge. Graham, B. A., M.A. : Trinity College, Dublin. Graham, Thomas ; Kilkeel, Co. Down. Greenfield, Bev.W. J., M.A.;Dulwich College. GREENHiLL,A.G.,M.A.,F.R.S.;ArtiI.C.,Wwich. Green8TREET,W. J.,B.A.; Marling Sc, Stroud. Greenwood, James M. ; Kansas City. Griffith, Gertrude, B.A. ; Southampton. Griffiths, D. T., M.A. ; Caermarthen. Griffiths, G. J., M.A. ; Fellow Ch. Coll., Camb. Griffiths, J.. M.A. ; Fellow of Jesus Coll., Oxon. Gross, W., LL.D. ; Bournemouth. Grove, W. B., B.A. ; Perry Bar, Birmingham. Hadamard, Professor, M.A. ; Paris. Haigh, E., B.A., B.Sc. ; King's Sch., Warwick. Hall, Professor Asaph, M.A. ; Washington. Hamilton, J. P. ; Penang. Hammond, J., M.A. ; i l^iftW^&©§l^-^- Hanumawta Rau, Professor, B. A. ; Madras. Hardy, G. H., B.A.; Trinity Ck)ll., Cambridge. Habkema, C. ; University of St. Petersburg. Harlet, Rev. R., M A., F.R.8. ; Forest Hill. Heax, W. E., M.A. ; Marion, Ind.. U.S.A. Hendebsov, R., B.A., Queen's Coll., Belfast. Heppel, Geobge, M.A. ; Baling. Hervet, F. R. J., M.A.; Worthing. Hill, Rev. E., M.A.; Cookfleld, Suffolk. HiLL,Prof.M.J.M.,M.A.,D.Sc.,F.R.S.;Univ.Col. HiLLHOUSE, Dr. W. ; Newhaven, Conn., U.S.A. HiLLYER, C. E., M. A. ; Cambridge. HiME, Lt.-CoI. ; Alexandria. HiNTOir, C. H., M.A. ; Cheltenham College. HiPPiSLEY, R. L., Major, R.E.; Famborough. HOBSON, Dr. ; Cambridge. Holt, J. R., M.A. ; Dublin. Hooker, J. H., M.A. ; 184 Evering Road, N.E. Hopkins, Rev. G. H., M.A.; Stratton, Cornwall. HOROBIN, C. J., B.A. : M.L.S.C. H0W8E, G. F. ; Balliol College. Oxford. Hudson, C. T., LL.D. ; Manilla Hall, Clifton. Hudson, J. F. ; Gunnersbury House. Acton. Hudson, W. H. H., M.A. ; Prof, in King's Coll., Hughes, Edith K. A. ; Stroud. [Lond. Hughes, W. R ; Jesus College, Cambridge. Ikin, a. E., B.Sc.; Kettering. Jackson, F. H., M.A.; Hull. Jackson, Miss F. H. ; Towson, Baltimore. Jacobs, H. J. ; London Institution, Finsbury. Jago, G. ; Plymouth Public School. Jenkins, Morgan, M. A. ; Tunbridge Wells. Johnson, Prof., M.A. ; Annapolis, Maryland. Johnstone, W.J.,M.A.;Univ.Col.,Aberystwyth. JoLLiPFE, A. E., M.A. ; St. David's College. Kahn, a., B.A. ; Mildmay Grove, N. Kalipadu BasO, Prof., M.A.; Dacca College. Kathern. R. p., M.A.; Fleetwood, Bath. Kelvin, Lord, F.R.S. : Largs, Ayrshire. Kennedy, D., M.A. ; Catholic Univ., Dublin. KiTCHiN, F. L., B.A., R.N. ; Bxmouth. KiTCHiN, Rev. J. L., M.A. ; Exmouth. KiTTUDGE, Lizzie A. ; Boston, United States. Klein, Professor Felix ; Univ. Gdttinpen. Kniseley, Alex. ; Columbia City, Indiana. Knowles, R., L.C.P. ; Tottenham. KoEHLER, J. ; Rue St. Jacques, Paris. KoLBE, Rev. Dr. ; St. Mary's Coll., Cape Town. Krishmachandra De, Prof. ; Pudukkottai. KRi8HNAMACHARRY,Prof.,M. A.; Tiupati, India. Lachlan, R., M.A. : Newnham Terrace, Camb. Lampe, Prof., Ed. of Jahrb. der Math. ; Berlin. Langley, E. M., B.A.; Adelaide Sg., Bedford. LAVEBTY,W.H.,M.A.;lateExam.inUniv. Oxford. Lawrence, E. J. ; Ex- Fell. Trin. Coll., Camb. Lawson, J. ; Newcastle-on-Tyne. Layng, A. E. ; Grammar School, Stafford. Legabeke, Professor ; Delft. Leidhold, R., M.A. ; Finsbury Park. Lemoine, E. ; Avenue du Maine, Paris. Leudesdobp, C, M.A.; Fellow of Pembroke College, Oxon. Lewis, tlOHN ; Cardiff. Lewis, Wm., M.A.; Headmaster, Llandyssul. LlAGRE, General ; Rue Carotz, Brussels. London, Rev. H., M.A. ; Wimbledon. LoNGCHAMPS, Professor de : Paris, L0NGHUR8T, Rev. E. S., B.A.; Newark. LowRY, W. H., M.A. ; Blackrock, Dublin. LoxTON, C. A. ; Cannock. LuDWiG, Professor ; North Carolina, U.S.A. McALiSTER, Donald, M.A.,D.Sc.; Cambridge. Macaulay, F. S., M.A. ; St. Paul's School. McCay, W. S., M.A. : Fell. Trin. Coll., Dublin. McClelland, W. J. ,B. A. ; Prin. of Santry School. MacColl, Hugh, B.Ay; Boulogne. McCoy, Rev. R. ; Stonyhurst. McCubbin, J.. B.A. ; Burgh Academy, Kilsyth. Macdonald, W. J., M.A. ; Edinburgh. Macfarlane, Prof. A., D.Sc. ; Univ. of Texas. MclNTOSH, Alex., B.A. ; Bedford Row, London. Mackay, J. S. ; Edinburgh. MACLEOD, J., M.A. : Elgin. MAcMAH07r,Mftjfipp. A„ F.R.S. i R.M,Aciul^iny, McMaHOS, Prnf., M.A i Cmiii?:!! Uriiv„ Ithacflk, MACMtTRCHT, A., H.A.: Uni\^ Coll., Tomntfl. McVrcKKR, C, E.. M.A. : KMiasiiil^ton, MADriAVATiAo, pTnf,, M*A. i ViduniurrBin, Maddibo?s, [gABRl.B.A. ; KinffsHuM Reading. Mainprise, B. W, j R. Niiviiil SchnoL Elthami. Ma LET, .I.e., M.A.. F.E.B.: Dublin. Mann, M, J, J,; t'Kirnwdl Ittiad, Fj^iidon. M A \?i II Ki M^ 51. 3 Vn^l* i r Ecole Fol.vti^tih, PttriK. Marks. Mjhm C. L, B.A.; KFn.MinKton. Makttn, Artkmas, M.A.tPliD.; Wa^shinglon. Mah3Et^ W, H.; Twvfnnl, H«rks*. M A T n E WB , Prof i'ssor G . B. . M . A . ; llsioKor . Matz, FnTif., M.A. ; Ki fig's Mnunlfiih, CtiroliiiiL, Mrtlur, ir. W., M.Sc. : Oh til!- f ull.. M'chtSiter, Mhrrif[ELIj, J„ LL,D„ FJl.A.S.' Plymouth. M EfiittMAN, Masbfielu. M.A. ; Viile Collepe. Mkyeb. Marv S. j Girtou Collogt\ C«mbritl|ei*. Miij.EK, W. J. Claejek. B.A. (Kmekjtlb Editor) , Bournemouth. Milne, Rev. J. J., M.A. MiNCHiN, G. M., M. A. ; Prof, in Ciooper's Hill Ck)l. Mitchell, Hugh, M.A. ; Edinburgh. MiTCHESON, Rev. T., B.A. ; Marazion. Mittao-Leppler, Professor ; Stockholm. MoNCK, H. St., M.A.; Trin. Coll., Dublin. Moore, H. K., B.A. ; Trin. Coll., Dublin. Morel, Professor ; Paris. Morgan, C, M.A. ; R. Naval Coll., Greenwich. MoRLEY, Prof., Sc. D. ; Haverford Coll., Pennsyl. MoRRiCE, G. G., M.A., M.D.; Salisbury. MuGGERiDGE, G. D., B.A. ; Pemb. Coll., Camb. MuiR, Thomas, M.A., F.R.S.E. ; Cape Town. MuiRHEAD, R. F. : Glasgow. MUKHOPADHYAY,Prof.A8UT08H,M.A.,F.R.S.E. MuKHOPADHYAY, Prof. Syamadas; Chinsurah. MuLCASTER, J. W., M.A.; Allendale Town. MuNiNDRA, Nath Ray, Prof. M.A., LL.B. Nanson, Prof. E. J. : Melbourne. Neale, C. M. ; Middle Temple, E.C. Neuberg, Professor ; University of Li6ge. Newcomb, Prof. Simon, M.A. ; Washington. NiSNER, I. ; Weatboume Grove, W. NixoN,R. C. J., M.A. ; Royal Aciad. Inst., Belfast. North, Francis C. ; Ashton-on-Mersey. O'CoNNELL, Major-General P. ; Cheltenham. Oldham. C. H., B.A., B.L. ; Dublin. Openshaw, Rev. T. W., M.A. ; Clifton, Bristol. Orchard, Professor, M.A., B.Sc. ; Hampstead. O'Regan, John ; New Street, Limerick. Orpeur, Herbert ; East Dulwich. Owen, J. A., B.Sc.; Tennyson St., Liverpool. Panchapagesa Aiyar, K.G. ; Madras. Panton, A.W., M.A. : Fell, of Trin. Coll., Dublin. Paranjpye. R. p., B.A. ; Cambridge. Paroling, Professor ; Cornell University. Peachell, F. H., B.A. ; High Wycombe. Peirce, Prof. B. O. : Harvard College. Peiris, M. R. ; Moratuwa, Ceylon. Perrin, Emily, B.Sc.; St. John's Wood Paik. Phillips. F. B. W. : Balliol College, Oxford. PiLLAi, Professeur, M.A. Trichy., Madras. Plamenewbki, H., M.A.; Dahgestan. PocKLiNGTON, H. C., M.A. ; Yorks Coll., Leeds. PoLiGNAC, Prince Camille de ; Paris. Pollexfen, H., B.A. ; Windermere College. Poole, Gertrude, B.A. ; Cheltenham. PoRTELA, Seftor Francisco de V. ; Fregenal de la Sierra, Spain. Pressland, a. J., M.A. : Academy, Edinburgh. Preston, G. W., B.A. ; Wheatley, near Halifax. Price, H., M.A. ; Penarth. Prudden, Frances E. ; Lockport, New York. Puckle, G. Hale. M.A. : Windermere. Purser, Fredk.,M. A.: Fell. Trin.Coll., Dublin. Purser, Prof., M.A. ; Queen's Coll., Belfast. Putnam, K. S.. M.A. ; Rome, New York. Pyddoke, H. W.. M.A. ; Oxford. QuiN, A. G. B. ; Streatham. Quint, Dr.; Gravenhage, Netherlands. Radhakrishnan, Prof., M.A. ; Pudtikkottai. Rama Aiy an gar. Prof. ; TrkjhinopolyxM^dras. digitized by V^XjOvtc Bamachandra Row, Prot.M. A. ; Trichinopoly. Ramsey, A. S., M.A. ; Fettes Coll., Edinburgh. Ranoachart, V, Srtntvasa^ B.A.; Gk)vero- iti'".! ' "!■-:■, K ■ '■ ' iHiiii, InrllaH RAtf, V. RiiJi"SAfU : Tin u pat i, Indbi. Bawsow, Robert; Havana . Hnnttj. Heat, Lionel E., U.A.; Wnttojiundar-EflKe, Ekes. E. W,; Fetiartli, CartiilT. REliVEj}, Q, M., M. A. ; Lee. Kmt. E K v>''o r jj a , H . , M.A,? Cumiiri dgi?, RicK^ Jaaieb; Cra.iffit^ Rymid, BinfoHt. KicriARD^, DAnii, B.A. ; Oliriat'aColL, Btocod. RiciU^DBUN, Rev. G.. M.A.s Wmt^hMk^r Coll. Roach. Rev. T.. MhA.; Twyfurd. Hants. Ro»H, A. A.; R. Aefid. In«r., Bt^Ufwt. Roberts. A. D., B.A.; Jt-iiiiK ColL, Oironl, Roberta, R, A.^ M.A.- Ni^w York, Roberts. W. R.W,.M. A,-, Ft-U, nf Trin.OolU.Dub. Roe3t>N,HAL, B,A, ; Sidney SusHex ColL. CnTTib. ROU^L. Professor; Mittwpidn, Gi^rmauy. RoE^KRS, li. J,, M+A^s Yorkshire Coll.. ij>eds. Ro wTOJs , Rev. S. J. . M , A . . M uw. l>fJo. ; Folksaitone, Roy, l*it)fetisoi' KALrpRASASNA. M.A. ; Aiptk. Rr^^GJKRo, SiiioTiKLLi ; UnivpraiU di U»?jsm. RuiaiaKLL. AJ.KSANDEK, B.A.5 Oxford. RimaELL, iJ, ; Pembroke hudg^, biiiilimfiTid. Ru^SELJ.p J. W.» M.A, ; Mellon Col leaf, Otford, Rrsa^LL, R., M.A. ? Fell, of Trin, Coll.. Dubliu. RcsaELL, T. ; C'fthis* Colle^, Cambridge. RiTTTEa, K, ; Stmdffrlimd. BaHLER^ JDIL^ \ Bi!-kt*Tihf^d, St* CtJiTB, i.e., M.A.; Rvili r Strret, London. SAUtoir, Rev. GEoFLiiK, n H., F,E.fi. ; Provost of TriJilty Colle^p, Dublin. Sal^uon, \V. n., B.A.; HftiHlJiil^d. Sa y D EBiJO^ . Eev . t.\ M . , M . A . : Brj Il^1on Rectorj. Sasjana^ Piiol; BhnvnjiJ^ur, tk^inljay. S.*EADARA>'JAV Rat. Fi'of,^ M.A. ; IJfifcn. Sajucar, Pifif, Heni Maluiav, M.A.: Agm, BAfi£AR. YnrL NlLKA>TnA, M,A.5 Calentta. BArAOE, Prof, T, ; 8t. Patri<!k'B ColL. Armasli- BCff ARAB, C. H. ; Scrantoii, IJ^S^k, BCBZV¥MB.t Ptfllessor; MeixMrsbury Coif.. 1^. SCHOTTTB. Prof. P. H* ; irijivi^rHit^vof GniDtrijire'n. ScxJTT, AAV». M, a. ; St. Diivid s Cnll. . Luraptittfr. Scott, ProfeaBorCHABTjOTiE A.. D.St, Scott. Qeob(je. BI.A.3 (;iv\>i<'T'ti6 s, Scott, R F.,ll.A.; Fdi, St ,luhnNColL,Cflmb. SBrtAR, Hl'GH W.; Tmut.v loll., i'auibridKi*. SE3i,Pnir. Raj.Moijaw J ilajshiiti,vK'rtU.,Bfapib Sewell, Rev. H.t B.A. ; Bury Grammar School. Sbtmqvr, W. R.p M.A. 3 St- Innards. SUARPE, J. W,, M.A*^ BonrtiPUKnith. SiiRrnKBn, E4'v. A. J. P., B.A,; Fellow of QiiOf n ' \i Col leKti . Oxford . Bhielub. Prof.. M.A,; Co<iiJwood, Mississippi. BiMMo^H, Rev. T. C. M,A.! Grtiinthorpe VicuTO^*, North Thort^Nhv S. O,, Tji nrs. SlRCOM, Plfir. SEllAeTIAN. \I,A.; StCHl.vtllirst, SlTAR^tMAlYAR, PmL, Iljtj^lii i nil., Tjiiiu vclly. SiVERLRY, WalTEH; Oil Ctly, IViiiisyhiitiia. Bkrf u SHIRE, Eli V, E,,M.A. : Lbndftff. Smith, C*, M.A.j SiiirKy SuH.sejL Collfpf, Camb. Km t Til, Proft^sBor D* B.: Brorkport, h.Y. SJaiTH, J. G. ; Portora, Enmskillen. 8MYLY, J. G. : Mi^rrinn Square, Dublin, SOTER. H* E., B.A.; Hfrhu^ti'. Sl'Allit. Rtw. F. M.: I.i ^vi^liiim, I^mdtHi. St A > IT A « , W. C, , li A . : 11 1 w:] 1 bur v P Utc*.*. ^TKEQE. B,H., M.A. ; Tiiibly tViille^Kt, Dublin. Steggali., Prtif. J» K. A., m!a.; Diuidt^. aTePHBN,ST.Jo«Jf.B.A.^Csiu8Coll.,Caiiibndge. Stewart, H., M*A. ; Islington. Stoops. J. M*, B.A. ; Victoria Collt^e. Bellast* Stoops, W. ; Newry. Stormonth, James ; Edinburgh. Storr, G. G, , M.A. : Clerk of thfTMedical CotUiciL '•' >' ■'• *: ■ ' -. C-i>., LiviTpL >ii.:....'.\, .1. \:., Anvil. J. L....a..[i, S.E. J^WAMiNATiiA AlVAK, l-mSf'.swor. M,A-; Madma. SVMON«. E. \Y.. M,A.; Givjvt" IbiUise. Walsall, TAiT, Pt^f, P. G.. M.A.; Univ., Kdinburgh. TASsEH.Prof.lLW.L.,M.A.;aWalt?aUuiv.Coll. TAaLETO.v, F. A.. M.A.i Fdl. THn. Coll.. Dublin, Tatk, JAsiKs, M.A,; Kedltiijton Rmd, Derby, Tavi,oh, JU'v. CirARLEd. D.D. : Maattr of SL John's College, CAmbritlwe, Tayi,ob. F. G., M,A,, B.Sc; Xottin^ham. Ta ¥ LOR, H , M . , .M , A . , F. E S, ; Fello * Trio, CoU , , Cftinbriiijirt\ Tatlok, J. H*. M,A, f Littlt Trinity. Owubridge^ Tatlob. \y. W.. M.A. i Oifnrd. Tkery^ Eev. 1\ E„ M,A,; llsley fteciory. Tjuri^ J. A., M.A.! Beitb. Ayi-sMre. Thomas. A, E., M,A.; Memjti CoJIoffe, Oxford. T < 1 o ^1 A H , lie V , D„M , A . 3 Gara ington R*dct , , Oi f ord. TjioMPSfjs. C- H.: St. David sCflll.. Lampi!t€r. Tm i>Mau5 JU V. F. D. .M.A.; E£-Fel.St,J.Col.,Oani, THOM^rjN, W. [., ; KdinbHrKh, TiiYA<iAfiAOAiVAs, V. B.,M,A,3 Bwigolore. TtRE(4,i, Dr. FRANCEacos Univ. di Roma. Toun, E. Walter; Belfast, Thaii-l. Anthony. M.A., M.D.; Fell. T.C.D. T u € K E li, R, , M , A . i H i llmar Eon Eoad , N. V\y DEB Heyi^es. a. F., B..\,; G^ttiugBtt, Venkata, Prof. K, Ramawatya; MudmB, ViOABlfe. EM11.E; IjftiiJSllC, Av6?TtT0n. ViiCEiszo, jAConi^fi; tintvptsiift di Romn^ Vt:»aE, PrrtfysBOF G. B. ; Washington. Wai^hnf, W. H. ; Mt-m, Pbys. Society. London, Walker, G, F. r Queens' CoUe^te, Gimbridge. W^ALMjiLEY, Ji, B.A,; Bi..clt^a, MiinchoidcT. W AFt U t? RTO S - Wil IT E J R . , B, A, ; Sal iJibnJ->^ AV A a [ > , B E A T R 1 c B A. , B.Se. ; Cheltenham, Waki>, F. L,, B,A. ; HartlepiKjl W arbe.v, a, T.. M.A,; Usk Tomuce, Brecon, Warren, R.. M.A. ; Trinity Culle^t^, Dublin. WAsaoK. J, C.5 Kells. Menth. WATHERSToXf Eev. A. I.., M.A. ; Bowlon. WATS05. D., M,A.; Folkestone. \yAT30s%Rev, H. W,; Hxr ell. Triaity Coll., Camb. Watts, J. O.? .Midillesbrough, Wbbh, G. H,: Hiilifpj. Webb. H. JL; Trinity CoHgitb, Qimbridtte, WfiLLACOT. Kev, AV. T,. M,A.; Newton Abbot, W^BBT^cii. Yrahz; Weiniar. West, J. WooU'COTE ; Kt'T*' Crn^s, Wu ALLEY, L. J.. B.So.; Ley tn] 1st nut , Whapiiam. R. H. W,, M.A.: M^nir h, st-L WntTK. E. A.. M,A.E Eekinjitojj, Ptrsihisi-e. WniTEJ^iKE, G. M.A.; EiMjlejfton, Ijinoishir^. Widi^voKTH, Rt>v, W. A., M.A.; Louiion. W J rmi s a . Rov. T, , B. A . ; Rnelmnmton. Wilcox, Rev. A, M.^ M,A,; Branhope^ Leeds. WiLKi^E^ON, J. F,; Bficup. Williams, A, E,. .M.A.; Texa*, WlLLiAMB, D, J., M.A. : WoiTt^stor Coll., Ox€tn. WlLLlAMBO^f , B,, M,A,5 F, k T, Trtn. OjU,, Diib, Wilmot, Rev, W. T.. M.A.s Newton Abbot WiLHOft-. VeQ, Art'bdwit'on J, M,, M.A., F.G.S. WiLSfJN, G. D.. B.A. I Qimbridf?t\ MiLftox, Ri^v. J„ M.A,; Bannockburo. WiT^ON, Eev. J. R., M..4.; Royston, Cambs, WOLT E RB , J . B, ; n rrmin^en , WooiiALL. H. J.. A.R.C.S.i Stockport, WooncocR, T., B.A. ; Twic-kenham. Wbiout, W. J. ; New Jersev. U.e,A, You^NS, John. B,A.; l\»rtsicIown. Ireland, 3SEUe.p Professor, M,A. ; Staunton, Virinnia, %♦ Of this series of Mathematics there have now been published seventy-four Volumes, each of which contains, in addition to the papers and solutions that have appeared in the Educational Times, an equal quantity of new matter, consisting of further solutions, and of completions of, or additions to, solutions that have been paxily published in the pages of that journal. These volumes contain contributions, in all branches of Mathematics, from most of the leading Mathematicians in this and other countries. *^^ew Subscribers may have any of these Volumes at Subscription-prices^. ..^ . ^OOQ IC CONTENTS. iWatftematital papers!, $tt^ Page Editorial Note on Quests. 14520 and 14670 74 Alternative Proof of Ptolemy's Theorem. (R. F. Davis, M.A,) 82 Note on Quest. 14447. (G. H. Hardy, B.A.) 99 Farther Editorial Note on Quests. 14620 and 14670 105 Editorial Note on Quests. 2810, 3260 107 Euclidean Proof of Pascal's Theorem. (R. F. Davis, M.A.) 112 4953. (W. J. 0. Miller, B.A.) — A king is placed at random on a clear chess board, and then, similarly, (1) a bishop, or (2) a rook. Find, in each case, the chance that the king is in check so as to be unable to take the attacking piece ; and find also (3) the chance of cheeky with or without the power of taking, for any combination of two or three of the pieces. [If we estimate the powers of the pieces (a) by the chances of simple check, as investigated in the solution of Quest. 3314, Eeprinty Vol. XV., pp. 60, 61, in January 1871 ; (6) by the chances of safe check, as shown in an interesting paper by H. M. Taylor in the Fhilosophical Magazine for March, 1876 ; (7) by the results given in the Berliner Schach- zeitung, we have the relative values of the knight, bishop, rook, queen as (a) 3 : 6 : 8 : 13; (jS) 3 : 3J : 6 : 9J; (7) 3 : 3J : 4J : 9i] 128 6703. (Rev. W. G. Wright, Ph.D.)— A chord is drawn through the point (2, 0) in the ellipse whose semi-axes are 3 and 2. Find the locus of the intersection of normals from the ends of this chord 34 5731. (R. A. Roberts, M.A.)— If from (1) a:2/a2 + y2/^-4 «0, (2) a%2 + %2_(a2 + ^2^2 ^ 0, tangents be drawn to x^/a^ + y^/b'^^l = S = 0, prove that they form, wit£ their chord of contact, a triangle whose (1) centre of gravity, (2) intersection of perpendiculars, lies on S = 134 6419. (The late J, J. Walker, M.A., F.R.S.)— Three lines in space are determined each by a pair of planes Wi = Biy + Ci2?+1 = 0, a; + «i =» 0, (wj = Diy + Eiz)... . Prove that the equation to the pair of planes through the axis y » 0. z =s 0, and one of the two lines meeting it and each of those three lines, = 6T 1 1 1 «h Wg nh «1 «2 «3 VOL. LXXIV. a Digitized by Google VI CONTENTS. 6498. (J. W. Russell, M. A.)— Show that is divisible by {a + b + e)^ + abc 117 6514. (W. J. 0. Miller, B.A.)— Find, to 4 decimals, the value of JoJo Jo ^ (l-^)(l-y)(l-.)(;. + y + .) 6687. (W. E. Wright, B.A.) — From a point on a curve of the second degree tangents are drawn to another curve of the second degree. Find the envelope of their chord of contact 63 6630. (Professor Nash, M.A.)— If three tangents OP, OQ, OR be drawn to a semi- cubical parabola from any point O, prove that (1) the circle through P, Q, R meets the curve in three other points P', Q', R', the tangents at which will meet in another point 0' ; (2) the middle point of 00' always lies on a fixed straight line ; and (3) the lines joining: O, O' to the cusp make equal angles with the axis 51 6632. (Professor Genese, M.A.)— Find the envelope of the asymptotes of conies inscribed in or circumscribed about a given quadrilateral .. 119 6642. (J. R. Harris, M.A.) — When two or more spherical soap- bubbles, blown from the same mixture, are allowed to coalesce into a single bubble, prove (1) that we obtain for the radius of the bubble an equation of the form «3— ri>— rj^— ... = a(ri2 + rj2+...— ir»), I'n r^, ... being the radii of the bubbles, and a some positive quantity; and (2) verify (what one would infer also from physical considerations) that this equation implies a reduction of the total surface 121 6648. (H. Fortey, M.A.) — In a plane triangle ABC, bisect AB in D, and take DBA' opposite to DA, or DAB' opposite to DB, each equal to half the sum of AC and BC ; and prove that the semi-perimeter AA' or BB' will subtend an acute angle at C if the base AB does not exceed half the sum of AC and BC 97 6679. (Rev. T. R. Terry, M.A., F.R.A.S.)— Show that the value of the continued fraction — — *- —^ — ^ .... where N = fw? and 1+ 1+ 1+ 1 + N.^ ^'-'' x\ is (l + ^r-(l-^)" 135 (2r-l)(2r+l) (1 +a;}«+ (1-j;)" 6701. (Professor Crofton, F.R.S.) — Find a function X such that, whatever F be, F(tf'D)X = F(«)X 119 10358. (R. W. D. Christie.) — If a^ and 013 are irrational cnbe roots of unity, prove that, if w + 2 is a prime number, ^w-l.>»^n+j^^^.e^^n-6) + &c. „o 61 10364. (L. B6n6zech.) — Si b est inferieur k Va^ a ei b etant des nombres entien positifs, ^expression a (1 + a + 2d) ne sera jamais un carr^parfait 87 Digitized by Google CONTENTS. Vll 11069. (J. J. BamivUle.)— Prove that 15 + (15+ 25) 2"^ (l* + 25 + 36) 2""+ ... = 2744, l»+(l» + 33)2-^ + (13 + 33 + 63)2-V(l3 + 33 + 63+108)2-'+... » 6416, ia + (l2 + 42)2"^ + (12+42 + 102)2"'^ + (12 + 42+102+202)2-^+... = 2016; #i.l3 + («-l)23+... + 2(w-l)3+l.«3=,^iy„(„+lj(„ + 2)(3»2 + 6« = 1); in thefigurate series 1, 7, 28, ..., 66u„-k-26{Un+i + Un.i) + Un+2 + Un..2 = a sum of consecutive fifth powers ; the ultimate term of the series 1, i, f, -|, f, ^y ... is 2 sin-j^ir 80 11075. (R. P. Kathem, M. A.)— Solve the equation tan^e + asecd+itand + sec'd = 86 11427. (R. Lachlan, M.A..) — If the points of contact of the three tangents which can be drawn from the point P to the cardioid r = a (1 + cos 0) be coUinear, prove that (1) the locus of P is a circle r + acoBd = ; and (2), if the feet of the three normals which can be drawn from P be collinear, the locus of P is the circle 3r » a cos ... 78 11762. (Professor Lucas.)— D^montrer Tidentite X-XP xP - gP" XP^^^XP"**^ {I^X)(1-XP) {l-XP)(l--XP^) "' (l-xf^){i-x^^') x--xf^^^ {\-'XP){\-XP^''^) 87 11830. (Professor Lebon.) — Deux plans parall^les k un cercle , commun h. plusieurs spheres et 6quidistantd de ce cercle determinent dans les spheres qu'ils coupent des segments equivalents 93 11852. (Professor Zerr.) — If S represent the length of a quadrant of the curve r»» = «, cos w6, p the radius of curvature of the path of the pole when the curve rolls along a straight line, Si the length of a quadrant of the first pedal of the curve, prove that (1) SSi = piray2r ; also (2) p/r = (m+ l)/«» = a constant ... 86 11864. (Professor Lucas.) — Mettre le nombre x^^—b^t/^^, oh x et y sont des nombres entiers positifs, sous la forme d'un produit de trois facteurs rationnels 86 11888. (Professor Zerr.) — A paraboloid floats in a liquid which fills a fixed paraboloidal shell; both the paraboloid and the shell have their axes vertical and their vertices downwards : the latus rectum of the psura- boloid and shell are equal, and the axis of the shell is m times that of the paraboloid. If the paraboloid be pressed down until its vertex reaches the vertex of the shell, so that some of the liquid overflows, and then released, it is found that the paraboloid rises until it is just whoUy out of the liquid, and then begins to fall. Prove that (1) the densities of the paraboloid and the liquid are in th e ratio 2{w2 + m + l-(w+l)\/w2-l} : 3V'(m+l)/(m-l), the free surface of the liquid being supposed to remain horizontal through- out the motion ; and (2), if cone and conical b e read for pa raboloi d and paraboloidal, the ratio is 3 {m*—l-'{m^-l)\/m^-l} : 4Jv/w»3— 1, sup- posing the vertical angles of both equal 88 Digitized by Google Vlll CONTENTS. 12055. (Professor Clayton.)— ABCD is a quadrilateral figure formed by four lines of carvature on an ellipsoid. If p^ g, r, a be the central perpendiculars on the tangent planes at A, B, C, D, respectively, then pr^qs 72 12561. (C. E. Hillyer, M.A.)— (1) A, B are two fixed points on a circle whose centre is O, and C is any third point on the circumference ; BC meets the tani^^nt at A in D, CA the tangent at B in E, and AB the tangent at C in F. Prove that DEF is a straight line which envelops a conic of eccentricity c, where €* = SOA^/AB^. (2) Generalize the above and Quest. 12462 (solved in Vol. lxii., p. 89) by reciprocation 67 13339. (R. W. D. Christie.) — Prove that in general any integral square may be resolved into three integral squares in six ways at least 72 13391. (B. W. D. Christie.) — Prove that the solutions of X«-5y»«-4 are X«a»-i2 + 4a^ Y = a^-i-b\ where a and 6 are any two successive terms of a eontinuarU 71 13480. (R. W. D. Christie.)— Find a series which gives all the odd primes in order as factors , 84 13496. (Professor Sanjkna, M.A.) —If pr denote the cofficient of sf in (l-«)'*, prove that log4 =Pi + 4Ps + ijP8+---» log(v'2 + l) -i?i + ii?8 + iP6+... = l-iPi + iAHP8 85 13663. (The late Professor Wolstenholme, M.A., D.Sc.) — Normals to the parabola y^ « iax at the points P, Q, Q' meet in a point, and QQ' passes through the fixed point {—Cy 0) : the envelope of the circle PQQ' will be the pedal, with respect to the origin, of the parabola 4ay* = (^(x—ia + e). The origin (A) is a double focus of the envelope (which is a lima9on),and there are two single foci Si, Sj at the points (— <?, 0), (e—i^/ia, 0) ; the vector equation being, for the infinite branch, (<j-4a) . SiP + 4a . SjP =« <J . AP. When e » 4a, the envelope is the pedal of the parabola itself with respect to the vertex (a cissoid), which is Uien a triple focus, ahdthe vector equa> tion becomes nugatory. When c > 4a, there is no oval branch. When e = 8a, the envelope becomes the straight line 2; + 4a — 0, Si, S, coincide, and the vector equation is SiP = AP 89 13668. (D. Biddle.) — It is known that the opposite edges of a tetra> hedron are equal, and the trilinoar coordinates of the projection of the apex on the base are given. Find the cubic contents 69 13746. (Lionel E. Beay, B.A.)— Find the area of the triangle with sides equal to the medians of a given triangle. Show whether such a triangle is always possible 109 13764. (Professor Umes Chandra G-hosh.)— Aa, Bi, and Ce are the perpendiculars from the vertices of the triangle ABC on the opposite sides. O is the orthocentre of the triangle ABC. Oi, O2, and Os are the points of intersection of the opposite connectors AO, be ; BO, ea ; CO, ab of the tetrastigms AbOe, BaOc, and CbOa. A0„ AOj cut BC at ai and a.; BOi, BOs cut AC at ^i and 6,; and COi, COs cut AB at e^ and Ci. Prove that Digitized by Google CONTENTS. IX (i.) Cc {l/eci + l/eci) + Bb {l/bbi + l/bb^ + Ka (l/aai + l/aa^ = 6Ctf/A<? Bb/Cb AafBa; (ii.) Ce [lleci-llcci) + Bb {Ijbbi^llbbi) + Aa(\laai~llaa^ «0 133 13978. (J. J. Barniville, M.A.) — Having formed the series 1, 1, 3, 7, 17, 41, 99, ... {un = 2«,._i + t«,._2) and ' 1, 1, 3, 11, 41, 153, ... (w„ - 4«„_i-«„.a), prove that 1 — + — —+ — ? ... » — , ^ 1.3^3.7 7.17 17.41 V2' — !— + ■^+— L_+_L_+ =_L 1+V2 3+^/2 17+V/2 99+V2 '** ^2* and _L.+_:L.+ _1_,+ _J_ + ...= ^ 39 12 + 2 32 + 2 ir^ + 2 41« + 2 4 14027. (J. J. BamiviUe, M.A.)— Prove that 24-1 54 _i 54_1 84-1 ^ 14028. (G. H. Hardy.)— Reduce the evaluation of (' co8(;?/f)<ftrf» •^ ' Jo 1 + 2^0084) + ^2 where ^, ^ are integers, p<q and ^< 1, to the integration of a rational fraction. Prove, in particular, that t" cos i4) d<t> __ 2 tanh-^ \^t . 1 + 2^ cos <^ + «2 ~ 1 4. ^ ^i ♦ and deduce (and also prove independently) that rtan-if2^^)-^=8tan-iV^tanh-i^/^ 122 Jo \ l-t^ /sin 14) 14109. (Professor Cochez.)— On donne les deux paraholes y^ -, 2px. x^ sr 2qi/, Lieu des points M tels que les tangentes issues de ces points a chaque parahole soient rectangulaires 119 14132. (I. Arnold.) — Describe a square in a given sector, having two angular points on the arc and the other two on the radii 116 14158. (Lt.-Gol. Allan Cunningham, K.E.) — Express (3.2»+l) in one or more of the forms {c^7!h2cP), {A^ripZB^; or show that it does not admit of this expression 93 14173. (D. Riddle.) — The sides of a triangle being given, a> b> e^ draw a line parallel to one of them, such that the qua&latanl formed shall have the maximum area possible in proportion to its perimeter, and find both area and perimeter 125 14184. (J. J. Bamiville, B.A.)— Prove that 1,1,23 5 _17 23 "^33-1 58 + 1 88-2 13» + 3 *" 90' -i- + -l- + -^+-??^+ -1 2.3 5.8 13.21 34.55 '* ' 7 31 183 835 ^ Z_ ^^ 3.5 8.13 21.34 55.89 *" 2 Digitized by Google X CONTENTS. 14188. (Salutation.)~Bi8dCt AB (= unity) in C, and AC inD; on AB describe a semicircle ; from A, D draw parallel lines intersecting the semicircle in P, Q respectively ; S, T being the projections of P on AB, and of S on DQ, prove that 4ST is the sine of an angle — 3PAB 46 14201. (R. Tucker, M.A.) — P, Q, R are points on a parabola, such that PQ, QR are normals to the curve. If SP, SQ, SR are denoted by n» *-j» *••» prove (^2 -n)' = (n +*-2)M2»*2-»-i-^3) ; hence they cannot be in A. P. Show that the circle PQR is given by m2 (m3 + 2)2 (a;2 + y») - 2pqax-ipmay + m- {m^ + 2) {m^ + 4) qa^ = 0, where P is the point (am^, 2am) and jo, q stand respectively for m* + 4m* + 2, m* + 6»»2 + 4. Find also when the :ircle passes through the focus 29 14203. (V. R. Thyagaragaiyar, M.A.)— Show that the roots of the equation 32a;5+ 16a:*-32a:3-12a:2 + 6j:+ 1 = are cos-^ir, cos-j^^* cos-ft^ir, cos-j^^, and cos-J^ir 70 14213. (Robert W. D. Christie.)— If An « m»» — «.fn*»-2+ m"-* m"-*+ ... . 2 ! 3 ! for all integral values of m and n, then X*'^AHX"4l-(X2-mX+l)(aiX^-Va2X^"' + «3X^'*"*+...+«8X+l), where On = a series allied to A«. E.ff, — If m ^ 5y ♦» « 3, then a;«-110a:3+l = (x^^dx+l) (x* + 5x^ + 24x^-hbx+l). There are two other allied theorems for positive values of An and m : it is required to establish them 130 14219. (I. Arnold.) — If a and b be the two parallel sides of a trapezoid, and h the line which bisects those sides, the centre of gravity G of the trapezoid is in this line. It is required to find the distance of G from a in the line h in terms of «, b, and h 68 14222. (Professor Elliott, F.R.S.)-If P + «oQ, in which P and Q are free from Aq* is annihilated by «o;r" +2ai — +303-^+ ... to oo, dai doj odz show that — Q « 0, and that, when m > 1, ;) r-m-l ■^'i 5^U--i 2 r-| P 43 14236. (R. Tucker, M.A.)— A.BCD is a square. P, Q, R are points on AB, AD, BC, respectively, such that PQR is an equilateral triangle. Find maximum value of triangle. Show also that locus of intersection of AR, BQ, as P moves along AB, is a parabola 135 14238. (Rev. W. Allen Whitworth, M.A.) — If a straight line be divided at random into any number of parts, the expectation of the square on any part taken at random is double of the expectation of the rectangle contained by any two of the parts taken at random. [This can be proved by algebra without the integral calculus] 115 Digitized by Google CONTENTS. XI theorem:- xAO^'^" = ___£^1±^__- (modP), 14250. (Bobert W. D. OhriBtie).— Prove the following very general rt+* P^+ a? {T^(XP+l)}*modP^ where x, n, k are any integers, P any odd prime, p the period of 1/P, m any integer required to make the remainder an int^er (always possible) . Ex, ^.—(1) a; =» 3, Ar - 5, P » 7, X « 1, 3, 7, 9, when P ends in 9, 3, 7, 1, respectively. Therefore 3. lO**** - ^J^^(°i<xl7) = ^^ - 1 mod7. 5* mod 7 3 Thus 3.10**-^ « lmod7. (2) n-7, A;=l, P - 19. 7.10*®"**- (19m+7)/2*» 13 mod 19. Thus 7.10^***^ ^ 13modl9 76 14251. (K. Enowles, B.A.) — Prove that the sum of the first r coefficients in the expansion of (1 — «) "** is «[r (r + 1) . . . (r + «— 1) }/« ! 76 14262. (Professor Sanj&na.)— Prove that J_ J_.JL _L + ± 1.2 1 1.2.3 _ 6ir»— 7» . 1*4 2*4.6 3*4.6.6 4*4.6.6.7 '" 36 ' and show how to find the value of i i- + i. — i— + i- __Ll2__ 1 n 2 *»(« + l) 3 •»(»+l)(» + 2) ***' where n is any positive integer 71 14266. (B. F. Davis, M.A.)~If O be the centre of inversion (con- stant — K^, investigate the formula of transformation tangent from point P to the circle a X (tongent from inverse point P' to inverse circle 0^, and show that X - OP (or k^/OP')/ tangent from to C. Apply this to Quest. 13801. (See Vol. lxx., p. 73) 56 14270. (H. MacColl, B.A.) — If A; be a positive constant, and the variables x and y be each taken at random between and 1, show that the chance that the fraction k{\^x—y)l{l~'y^ky) will also lie between 0andlis(Ar2+l)/{2A;(A; + l)} or (1 + 2A;-A;2)/{2(A;+1)} according as Aris greater or less than 1 49 14278. (I. Arnold.) — Two non -concentric spheres intersect, forming a shell. Find the centre of gravity of the larger shell and its distance from the centre of the larger sphere, the distance between the centres of the spheres being d, and the radii of the spheres being B and r 28 14284. (Professor Neuberg.)— Soient O, I, la, Ifr, !« les centres des oerdes circonscrit, inscrit et exinscrits au triangle ABC ; soient D, E, F les pieds des hauteurs et Aj, B^, C^ les p61es de BC, CA, AB par rapport au cercle 0. Les quatri^mes tangentes communes auz cercles ^I, Ia)» (I, It), (I, le) ferment un triangle a^y homoth^tique aux triangles ^i^i^i) I>EF. Le centre d*homoth^tie des triangles a/Sy, AiBiCi partage la drolte 01 dans le rapport B : r, et est le conjugu6 isogonal du point de Digitized by Google Xll CONTENTS. Gbrgonne de ABC ; sea coordoxmees normales par rapport au triangle a/37 sont 1/a, 1/^, ijc. Le centre d*homothetie des triangles aBy, DEF a pour coordonnees normales, dans cos triangles, tan }A, tan }B, tan }C 66 14299. (Rev. T. Mitcheson, B.A.) — Let PiQiRi be an equilateral triangle such that P^ is on one side of a square, Q^ and Bi on the adjacent sides, Qi'Ri parallel to the other side, and the mid -point of QjRi ; and let PQR be any other equilateral triangle, whose angular points aro the same sides, QR passing through O, and let PiQi meet PQ in S, PjRi meet PR in T. Then the circle passing through P, Pj, S, T touches QR in O, and circles passing through O, T, R, Rj and O, 8, Q, Qi, respectively, are each one third of the first circle (An echo of Quest. 14236) 14301. (J. J. Bamiville, B.A.)- -Sum the series 1 + 3-^ "^ 1 2 + 42 ■*"3 + 62 + ..., 2 + 42"^ 1 5 + 72 ■*" 8 + 102 + ..., 22+3"^ 1 32 + 7 ■^ 42 + 11 + ..., 1 + 1 + ' + 92 52+7 ' 92+I6 ' 132+23 14309. (Professor Oochez.) — Lieu des points d'oil I'on pent mener a r ellipse quatre normales dont la somme des carr6s soit constante ... 48 14312. (Professor N. L. Bhattacharyya.)— A parabola slides between the two foci of an ellipse, such that the focus of the parabola always lies on the ellipse. Find the envelope of (1) the directrix, (2) the axis, of the parabola 113 14315. (B. N. Cama, M.A.) — If parabolas be described cutting an equiangular spiral orthogonally, and having their axes in the direction of the polar subtangent, the loci of the focus and the vertex are copolar spirals whose linear dimensions bear a constant ratio 63 14329. (J. A. Third, M.A., D.Sc.)--L, L', M, M', N, N' are points en a conic. LL', MM', NN' form the triangle ABC ; MN', NL', LM' the triangle A'B'C ; and M'N, N'L, L'M the triangle A"B"C". The straight line AA'A" meets BC, B'C, B"C" in X, X', X" respectively ; the straight line BB'B" meets CA, C'A', C"A" in Y, Y', Y" respectively ; and the straight line CO'C" meets AB, A'B', A"B" in Z, Z', Z" respec- tively. Show that the following are triads of concurrent lines : — YZ, Z'X', X"Y" ; ZX, X'Y', Y"Z" ; XY, Y'Z', Z"X" : YZ, Z"X", X'Y' : ZX, X"Y", Y'Z' ; XY, Y"Z", Z'X' ; and that the points of concurrence lie on a conic 124 14338. (Professor Sanj^ina, M.A.) — In Quest. 14110 denote (2tf— 1) «2 + y2 + ^2 i,y fl5j . take ^i, c^ similarly; call e the ratio of the Tucker circle, and let A = (l — tf)tan«. Then prove that (1) the equation of the Tucker circle (whose ratio is e) XX'YY'ZZ' is $y/be + ya/ea + afi/ab — (a/a + $/b + y/e) ( I — tf) tan o> + ( 1 — tf)2 tan^ « = , Digitized by Google CONTENTS. XIU or (o— «rX) 08— *A.) {y—cK) =» 0^7 ; (2) the envelope of this circle, as its ratio varies, is aVa'i + fiyi^ + '^le^-2a$/ab^2fiylbe-2yalca = 0. or V(a/a) + '^(iS/*) + ^^^(7/*^) = 0, the Brocard ellipse ; (3) the radical axis of two Tucker circles of radius/and ff is a/a + $lb + y/e = (2 — /— ^) tan u, so that the radical axis of the Tucker circle e with itself is a/a + fi/b + yje s= 2X, which is also its chord of douhle contact with the envelope ; (4) the radical axis of the Tucker circle e and the circumcircle is a/a + fi/b + y/c « A, and the chord of contact of the circumcircle with the Brocard ellipse is the Lemoine line ; (6) if f+g = constant « 1 +*, the varying Tucker circles / and g have a fixed radical axis, which is the radical axis of the circumcircle and the fixed Tucker circle whose ratio is e ; (6) if f+ff — 2 sin^w, the varying Tucker circles / and ff are of equal area; (7) the polar of the symmedian point with regard to the circle e is a/a + fi/b + y/e « (4^ — 1 )/ (2^) X, and, if fy * J, the varying Tucker circles /and ff have the same polar for this point ; (8) the radical centre of the circles round AY'Z, ABC, and the Tucker circle, lies on BC, the radical axis of the first two being ficb^ + ybci = ; so also for BZ'X, CXT ; and these three radical centres, on BC, CA, AB respectively, are situated on the line (aa|)/a + {&bi)/b + {yei)le = ; and (9) the radical centre of the circles round AY'Z, BZ'X, CY is the point 0/(^1^1 cos A) = $l{ciai cos jS) « 7/(«i*i cosC), which lies on the curve fiy sin 2A sin (B — C) + 70 sin 2B sin (0 — A) + a$ sin 2C sin ( A — B) « 0, that circum-hyperbola of ABC which is the isogonal transformation of Euler's line. [The last result has been obtained by Rev. J. Cullen in Quest. 13921] 82 14344. (J. J. Bamiville, B.A.) — Having Un^i + Un+i = 4w«, prove that J- +-^ +-^ +-i- +...=-^, 1 + 1 3 + 1 11 + 1 41 + 1 2 * 1 _1^ _^ 1 ^ 3^3 1 + 3 "•■ 3 + 3 U+3 41+3 '** 10 ' -!- + -A_ + -^^ + -J_+ ^^^ 1 + 11 3 + 11 ll + U 41 + 11 *•* 38 ' 111 ^ x/2-1 + - T- •*- -. 7:; +...=- 2+ a/6 4+^6 It + a/6 ^3-r _1 !_ + J L_ + «1 1 + 2 2 + 2 7 + 2 26 + 2 "* 6* -1 L_ +_1 L_ +...= ! 102 1 + 1 5-1 19+1 91-1 2a/3 14367. (Professor N. Bhattacharyya.) — Show that the product of three numbers representing the sides of a right-angled Mangle is divisible by 60 94 14372. (B. C. Archibald, M.A.) — Parabolas with a common focus pass through a fixed point. Show (1) that the locus of their vertices is a cardioid whose cusp is at the common focus and whose vertex is the fixed point ; (2) that the locus of the points of intersection with the parabolas of the lines through the focus making a constant angle with their axes is a cardioid 62 Digitized by Google XIV CONTENTS. 14384. (W. H. Salmon, B.A.)— If a chord of a circle 8 subtend n right angle at a fixed point O, show that its envelope is a conic S' ; and that of the common tangents S and S' two pairs intersect on the polar of O, one pair at the centre of S, and the other on a fixed line. Show also that O has the same polar for S and S' 107 14394. (Professor Thomas Savage.)— Discuss, n being integral and positive, (l + l/ar)"<2, but (l + l/x)"*^>2 59 14400. (B. F. Davis, M.A.) — Find positive integral values for N, ar, y which will render N»-3z«, N« - 3y«, N«-3 (« + y)% N5-3 (x-y)' pcorfect squares. [A special Christmas puzzle.] 93 14402. (R. C. Archibald, M. A.)— Show that (1) the locus of the fourth harmonic point to P, S, P', where PSP is any cuspidal chord of the cardioid r a 2a(l— cosd), is the Cissoid of Diocles r = 2asiaetaxL0; (2) if r and r^ are the radii vectores respectively of the cardioid and cissoid for a given 6, r : r' « tan Jd : tand ; (3) referred to (—a, 0) as origin, the equation of the cissoid becomes r/a « ( 1 + tan' ^0) / (1 — tan' 1$) 5S 14407. (Rev. T. Mitcheson, B.A.)— If o, /3, 7 be the distances of the incentre from the angular points of a triangle, the diameter of the incircle ^ ^^(g-'j^iA + fi-^oo^^B + y-^coBiC) g^ a cos^A + /3 cos^B + 7 cos ^C 14410. (Rev. T. Wiggins, B.A.)— Inscribe in a given triangle the triangle of least perimeter 26 14412. (H. A. Webb.)— Three equilateral triangles are described outwards on the sides of any triangle as bases. Prove geometrically that the centres of these three equilateral triangles form the vertices of a fourth equilateral triangle 77 14413. (Robert W. D. Christie.) — Find integral values of « to satisfy the equation T^ » »Ty, and give general values for a; and y. (T a Iriang^ar) 87 14419. (Lt.-Col. Allan Cunningham, R.E.) — Find three sums of successive cubes which shall be in arithmetical progression 65 14424. (Professor Neuberg.)— Trouver le lieu des centres des hyper- boles 6quilat^res qui out une corde normale commune MN 25 14^25. (Professor U. C. Ghosh.)- Prove that \ Xil> (sin x)dx — \irr ^ (sin x) dx, and hence evaluate f . ^ 7" -dx 62 Jo 1— sma; 14430. (J. A. Third, D.Sc.)— A conic, whose centre is O, touches the sides BC, CA, AB of a triangle at X, Y, Z, and C is the point of concurrence of AX, BY, CZ. Show that O bears to ABC the same rela- tion that the isotomic conjugate of C bears to the anticomplementary triangle of ABC (the' triangle formed by parallels through A, B, C to the opposite sides) 76 Digitized by Google CONTENTS. XV 14432. (R. Tucker, M.A.)~PSQ is a focal chord of a parabola, and PQB is the maximum triangle in the segment cut off by PQ. Prove that the equation to the circle PQR is 8 (a;» + y»)-2 (7^ + 20) ax+p (3/?3-4) ai/ + 6p^a^ = 0, where p = m—l/m (P is aw', 2am). The locus of the centre is a cubic, and, if O is the fourth point of section, the locus of the mid-point of OR is a parabola, and the envelope of the chord ORis another parabola 64 14434. (Edward Y. Huntington, A.M.) — An astroid, two nephroids, and four cardioids are drawn on the same fixed circle of radius a, their cusps lying at the quadrantal points of the circle. Prove : a line of length 2a sliding between either pair of opposite cardioids envelops that nephroid which has the same cusps ; and a line of leng^ 30 sliding be- tween the two nephroids envelops the astroid. (Nephroid = two-cusped epicycloid ; astroid » f our-cusped hypocycloid) 34 14436. (Rev. T. Roach, M.A. Suggested by 14376.)— If I, Ij, Ij, I3 be in- and ex-centres of a triangle ABC, and o^, 0^, 0^ circumcentres or II3I3, Ilsli, II1T2 respectively, prove that o^liO^JsjOil^ is an equilateral hexagon, and find the value of its angles 51 14437. (R. P. Paranjpye, B.A.) — Show that there are six conies passing through three g^ven points and having contact of the second order with a g^ven conic ; and, further, that these six conies all touch a quartic having the three points as nodes 64 14439. (H. MacColl, B.A.)— There are five possible hypotheses, Hi, H3, Sec, of which one must be, and only one can be, true ; Uie chance of each being one-Jifth. Each of the three H^, Hj, H3 implies that the chance that a statement A is true is '52 ; whereas H4 and H5 lead each to the conclusion that this chance is *06. From these data prove the paradoxical (but not absurd or impossible) conclusion that it is probable but not true that A is probable ; and show that the chances that A is probable and true^ probable but not true, true but not probable, neither probable nor true, are respectively -312, -288, -024, -376 36 14441. (Rev. T. Mitcheson, B.A.) — A regular polygon of an even number of sides is inscribed in a circle, and lines are drawn from one of the angular points to each of the others. Show that the sum of these lines = (acotT/2n)/(sinir/t}) {a being a side of the polygon), and if the lines be hi, A3, A3, &c., then KAn-i + A„_2 + A„_3+ ...) — {Aj(„_2) + Aj(„_4) + Aj(,t_e) + ...) — R.. 40 14443. (R. Knowles.) — F, 8 are the foci of a rectangular hyperbola ; from a point T on the circle whose diameter is FS, tangents TP, TQ are drawn to meet the curve in PQ ; the circle TPQ cuts the curve again in CD ; prove that (1) the diagonals of the quadrilateral PQCD intersect in the axis; (2) two of its sides are parallel 31 14444. (P. Milnes.) — A conic cuts the sides of triangle ABC in D, D', E, E', F, F' respectively ; AD, AD' intersect the conic again in d, dl ; BE, BE' in «, <?' ; CF, CF' in/,/. Show that the intersections of ^% ^ t ff "^th the polars of A, B, respectively are coUinear 27 Digitized by Google XVI CONTENTS. 14445. (Rev. J. Cullen.)— Prove that (Z^^'^-lsO [mod(y.2'+l)], if y.2'+l be a piimu 27 14447. (H. W. Curjel, M.A.) — If / (x) is finite and continuous for all positive finite values of x except a finite number of values, then I B\n{/{x)}dx and T cos {/(a:) } rfa? are convergent or divergent according as limit "^^ is infinite or finite ; except in the case where limit f{x) « or «ir, when Tsin {/{x)} dx may be convergent, and the case where limit f[x) = Jir or (2«+ 1) Jir, when COS {/(a?)} dir may be convergent 98 E 14449. (Paul Gibson.) — Given that, in reducing 1/N (N prime) in scale 17 to a pure circulator, five consecutive remainders formed are 1<?, 4, 9, 21, 5 (in scale 17), to find N 35 14452. (Professor Umes Chandra Ghosh.) — If n and n' are the Brocard points of a triangle ABC, Ai, As, A3 and A'^, A\y A'3 are the areas of the triangles nBC, nAC, nAB and n'BO, n'AO, n'AB, show that .. V sin^M _ _Ai^ _ A2 _ ^3 _ ^'\ _ 4j_ = ^^ . ^ '' 4A «V- «2^2 ^2^ ,^2^2 ^2^2 ^2^.2 ' fii \ ?ilL!!! ^ HA ^ QB ^ nC ^ a' A. ^ Cl'B ^ a'C , ^* *' 2 A * 6^c ac^ an h(? ^ a^c "" a¥ ' where w is the Brocard angle and A the area of the triangle ABO ... 38 14453. (Professor A. Droz-Famy.) — Oonstruire un triangle, dont on connait la base, la hauteur correspondante et sachant que sa droite d'Euler est parall^le au c6te donne 114 14454. (Professor Sanj&.na, M.A.) — Solve, in rational numbers, the equation M^ — 2a;N*^ = x^—\, where x stands for any one of the natural numbers 2, 3, 4, 5, 6, [The solution gives N*+ 1 as the difference of two squares. I have reason to believe that 5 is the only small value of a; admissible. For the method see Ohrystal, xxxiii., §§ 15-19] ... 44 14455 . (Professor Cochez. )— Oourbe p3_3ptan« + 2 = 56 14466. (Professor N. Bhattacharyya.) — There are n smooth rings fixed to a horizontal plane, and a string, the ends of which are fastened to two of the rings, passes in order through them. In the loops formed by the successive portions of the string are placed a number of pulleys whose masses are m, ^m, \m, \in, -^m, &c. If, in the subsequent motion, all the portions of the string not in contact with the pulleys are vertical, show that the acceleration of the rth pulley is {(n — 2r)/«}^. Discuss the case when n is even 70 Digitized by Google CONTENTS. XVll 14458. (J. A. Third, D.Sc.)— XYZ is a triangle inscribed in ABC and having its sides proportional to the medians of ABC. Show that the envelope of the circumcircle of XTZ is the Lemoine ellipse of ABC 31 14460. (R. F. D.ivis, M.A.) — Given the base of a triangle in magni- tude (a 2a) and position, and also the length {= I) of the line bisecting the vertical angle (vertex to base^, prove that the locus of the vertex referred to the base as axis of x and a perpendicular to the base through its middle point as axis of p is (ar2 + y3 + a2)« = 4:rV+/*a:V(/2-y3) 52 14461. (Rev. W. Alien Whitworth, M.A.)— If a straight line be divided at random into three parts x^ y, z, show that the expectation of the volume (y + 2) {z + x) (x + y) is 14 times the expectation of the volume xyz 63 14463. (R. C. Archibald, M.A.)— Express the coordinates of any point on the cardioid as rational functions of a variable parameter, and show that the locus of a point which moves such that the triangle formed by joining the points of contact of the tangents drawn therefrom to the cardioid is of constant area and in general a curve of the eighth degree. [This theorem is due to Professor Zahraduik] 78 14464. (Edward V. Huntington, A.M.)— The angle between the principal axes of two given concentric ellipses is 90°, and a + b » a' -vV . Show that a line of length a—h' (or a' — h) sliding between these curves envelops an astroid ; and that any line rigidly connected with this sliding line envelops an involute of an astroid. (Astroid » hypocycloid of four cusps) 64 14467. (G. H. Hardy, B. A.)— Prove that j^^{4>(«-a)-^(a:-3)}^=(A-rt){</>(oo)-^(- 00)}, provided each side of the equation represents a determinate quantity. Deduce the values of f ^ , pf ^ ... Ill J _« cosh (a:— a) cosh (a:— A) * J _„ sinh («— a) sinh {x—h) 14473. (W. S. Cooney.) — Construct the triangle, being given miy ^Ar^« of the following six points: — the centres of the squares described externally and internally on the sides 124 14474. (R. Knowles.) — Tangents from a point T meet a parabola in P, Q ; the circle TPQ cuts the parabola again in C, D ; the sides PC, QD of the quadrilateral PQCD meet in E ; the diagonals in G ; M is the mid-point of EG; MN^, EN'2, GN3 are drawn at right angles to the axis; MNi meets the parabola in K. Prove that ENi^ » ENj . GN3 33 14476. (Professor E. J. Nanson.)— If + — - — + « 0, then — — p + 75 -^-z r «0... 30 a b e a^—be o'—ea e'—ab 14478. (Rev. T. Mitcheson, B.A.) — P, Q are the ends of conjugate semi-diameters of an ellipse, and a straight line drawn from the intersec- Digitized by Google XVlll CONTENTS. tion of the normals at P and Q, through the centre C, meets PQ in S, whilst the tangents meet at the point (A, k); show that CS« ^^^ 81 14479. (Salutation.) — I is the incentre of the triangle ABO, of which A is the greatest angle. P is a point on the incircle, and through P lines are drawn parallel to the three sides of the triangle, and meeting the incircle again in Q, R, S, respectively. QB, RS being joined, prove that the quadrilateral PQRS is a maximum when AIP is a right angle, and find its mean area 60 14481. (H. M. Taylor, M.A., F.R.S. Suggested by Quest. 14382.) —On the sides of a triangle A'B'C, triangles B'C'A, C'A'B, A'B'O are constructed similar to three given triangles. Having g^ven Uie triangle ABC and the three triangles, reconstruct the triangle A'B'C 41 14482. (Professor Neuberg.) — Soient a, b, e, d les cdt^ AB, BO, CD, DA d*un quadrilat^re spherique ABGD circonscrit k un petit cercle. D6montrer la relation sin a sin ^ sin' ^B » sin ^ sin ^ sin' ^D 96 14484. (Professor A. Droz-Famy.) — On joint un point A de la directrice d'une parabole au sommet S de cette demidre. AS coupe la ( ourbe en un second point B. La tangente en B rencontre en JP le diam^tre de la parabole mene par A. Tirons la deuzi^me tangente PC. La droite CB est normale en B ^ la parabole ' 132 14491. (R. Tucker, M.A.) — Squares are described externally on the sides of the triangle ABC, and tangents are drawn from their centres to the incircle of the triangle. Prove that 22(tangent8)2«2A(2 + 3cot«)-2(*<j) 64 14493. (J. H. Taylor, M.A.) — If A', B', C are vertices of similar isosceles triangles described all externally, or all internally, on the sides of any plane triangle BOA, the straight lines AA', 6B', CC are concurrent 28 14494. (Rev. T. Roach, M.A.) — Along the hedge of a circular field of radius r are placed 2n heavy posts at equal distances. A man brings the posts together, one at a time, to one post. Show that the product of the2n-l walks multiplied together - 2'^'*. r^'*"\ n 36 14495. (R. C. Archibald, M.A., Ph.D.) — The points i?i, p^, p^, where any three parallel tangents to a cardioid cut the double tangent, are joined to the centre O of the fixed circle. Prove that the angles Pi^Pit PfiPz *re each equal to 60° 131 14496. (G. H. Hardy, B.A.)— Prove that 2<r2(«)cri'(M)<r2M«')<r8M«') = l{P(M + <')+P("-«')}<rM« + v)<rM«-f); the notation being that of Weierstrass*s theory of elliptic functions, and the summation applying to the six possible divisions into pairs of the functions <r, ctj, (Tj, 0-3 37 14499. (Lt.-Ool. Allan Cunningham, R.E.)— Prove that the continued product P = {a^b){a^^h^){a^^b^)..,{a^''-^^h^-^) is divisible by {(2«)!-i-(2".«!)}"*^ if {a^b) is divisible by (2n)!-t-(2*'.>t!) 32 Digitized by Google 00J5TENTS. XIX 14501. (Rev. T. Wiggins, B. A.) — Given a triangle ABC, find a point D within it such that D A^ + DB^ + DO^ is a minimum 26 14603. (Robert W. D. Christie.) — Show that the primitive roots of 331 are connected with the associated roots by the modular equations r"*= « mod 331, rf =-«'mod331, where r is a primitive, and r^ an associated, root ; also » signifies one of the roots of 0:3+ 1 « 0, namely, i {l + a/C-S)} or i {I- a/(-3)} ; and generalize the result 114 14604. (R. Knowles.) — The circle of curvature is drawn at a point P of a parabola ; PQ is the common chord ; an ordinate from P to the diameter through the focus meets the parabola in R, and a diameter through Q in O. If T be the pole of PQ with respect to the parabola, prove that TO, PQ, and the tajigent at R are parallel 60 14608. (W. H. Salmon, B.A.)— The frustum of a pyramid with quadrilateral base is such that the intersections of the opposite fiices are coplanar (A) ; prove that (1) the diagonals of the frustum are concurrent (0) ; (2) each diagonal of the frustum is divided harmonically by O and its point of intersection with A ; (3) the diagonals of each face are divided harmonically by their point of intersection and the plane A 96 14609. (I. Arnold.) — Given two circles, one within the other, a point can be found such that the extreme portions of any right line cutting both circles shall subtend equal angles at the point lOU 14511. (John C. Malet, M.A., F.R.S.) — If, in the sextic algebraic equation afi — jp, x^ +P2X* —PzX^ +P4^^^P6^ +P6 = 0, the sum of three roots is equal to the sum of the other three, (1) prove ^P6Q4'-Qa<V'-P6Q^^+PbQ^Q3-P5'' = 0, where Q«=i?2-iM Qz^p^-kPiPi + ^Pi^ Qi^Pi-^PiPz + kPi^P^-^Pi^'y (2) solve the equation 42 14612. (Professor Neuberg.) — ^Trouver dans le plan du triangle ABC un point M qui soit le centre de gravity de ses projections A', B', C sur BC, CA, AB, pour les poids donnes a, fi, y 28 14516. (J. A. Third, M.A., D.Sc.)— X, Y, Z are three points in the plane of a triangle ABC, such that the pairs AY and AZ, BZ and BX, CX and CY are equally inclined to the bisectors of the angles A, B, C respectively. Y moves on the straight line «6, and Z on the straight line ue. Prove the following statements : — (1) the locus of X is a straight line Ua ; (2) if u^ pass through B, and m« through C, Ua passes through A ; (3) if Ub be perpendicular to CA, and Ug to AB, «« is per- pendicular to BC ; (4) if L, M, N be the points where «o, «6i w* meet BC, CA, AB respectively, AL, BJM, CN meet in a point P ; (6) AX, BY, CZ are concurrent in a point whose locus is, in general, a conic circumscribed to the triangle and passing through P ; (6) if Ub, Ue meet on the cubic circumscribed to the triangle, and passing through every pair of isogonal points whose join passes through P, viz., / (y^-z^jyz + m (z^-x^)lzx + n {x^-y^)lxy = 0, where /, m, n are the coordinates of P, Ua^ Ub, u^ are concurrent 94 Digitized by Google XX CONTENTS. 14516. (Professor Jan de Vries, Ph D.) — For each conic of a g^ven pencil the orthoptical circle (circle of Monge) is constructed. How many of these circles will pass by a given point ? 32 14618. (Professor A. Gk>ldenberg.) — Resoudre le syst^me (a; + 2y)(« + 2a) =a», (y + 2^) (y + 2a;) = ft», (z + 2a?) (a + 2y) = c» 96 14619. (Professor XT. 0. Ghosh.)— Find the sum of the products of the terms of the geometric series a, d^, a^, a*i ...» a*^, taken r at a time, r being less than n 126 14620. (Professor N. Bhattacharyya) and 14670 (E. W. Adair).— Required a direct proof of the old problem : — If the bisectors of the base angles of a triangle, being terminated at the opposite sides, be equal, show that the triangle is an isosceles one. (See Todhunter*8 Euclid) 73 14522. (J. H. Taylor, M.A.)— If A, B, C are vertices of equilateral triangles described all externally, or all internally, on the sides of a triangle A'B'C, and Aa, Bb, Ce are diameters of circles circumscribing tuose equilateral triangles, then AA', BB', CC are equal and concurrent, and a, b, e form an equilateral triangle and are middle points, each of a pair of arcs, on sides of the triangles ABC, A'B'C 76 14624. (R. F. Davis, M.A.) — If A, B, C, D be the angles of any convex quadrilateral, sin A {sinC + sinB— sin(A + D)} : sinC {sin A + sinB— sin(C + D)} « sin A + sinD-sin(A + D) : sinC + sinD — sin (C + D) 115 14625. (J. Macleod, M.A.) — KL is a diameter of the circle KML. From L any two chords LM, LN on the same side of KL are drawn and produced to meet the tangent at E in Q' and O. Through O a line is drawn parallel to MN, and LQ' is produced to meet it in Q. QQ' is bisected in Y, and the straight line OY in P; through P a tangent is drawn to the parabola which is touched by OQ, OU' in the points Q, Q', and meeting OQ, OQ' in R, R'. Prove that the angle EOL is equal to the angle of the focal distances of P and R 84 14526. (R. C. Archibald, M.A.) — With reference to the centre of the fixed circle, the corresponding tangent and normal pedal curves (positive or negative) of the cardioid are similar 101 14528. (R. P. Paranjpye, B.A.) — Show that any triangle can be projected into an equilateral triangle whose centre of gravity is the projection of a given point 96 14529. (Lt.-Col. Allan Cunningham, R.E.)~Show that ^= l(mod p) where a; = J.2104Q, Q, ^ q\ p = Q.210*<^+1 = pHme 115 14632. (Rev. J. Cullen.) — Let A be any conic in the plane of a given triangle ABC. A point P is taken on A, and parallels through P to BC, CA, AB meet A again in A', B', C. Prove that AP, BP, OP intersect B'C, CA', A'B' in three coUinear points L, M, N. (A parti- cular ease is that the intersections of the symmedian lines with the corresponding sides of Brocard's triangle are collinear.) Prove also that, if A be the circumcircle, then LMN is at right angles to the Simson-line of P 46 Digitized by Google CONTENTS. XXI 14634. (W. 8. Gooney.) — Let Oi, O2, O3 be the centres of squares described externally, and wj, dtfg* ^ ^^^ centres of squares described internally, on the sides «, by c, respectively, of triangle ABC. Join Oi to A)3 and »s, meeting side BC in P, P' ; O3 to wj and o»i, meeting CA in Q, Q'; Os to «i and 013, meeting AB in R, R'. Prove that A', B', C, the intersections of P'R, Q'P, R'Q are the centres of the insquares of ABC, and that, if AA', BB', CC meet sides of A'B'C in a, 3, 7, then triangle a/Sy is similar to ABC 45 14536. (I. Arnold.) — In any triangle the radius of the circumscribed circle is to the radius of the circle which is the locus of the vertex, when the base and the ratio of the sides are griven, as the difference of the squares of those sides is to four times the area » Ill 14538. (Salutation.) — Arrange in one plane two triangles of g^ven dimensions in such manner that two specified vertices may coincide, and the other four be concyclic 118 14640. (Professor G. B. Mathews, F.R.S.)— Prove that, if Q= 5*iyi(3n>i)-+*5^s«"^ then Q} «^^„ .00 -00 lOITK where X, A.' are the moduli into which k, nf are transformed by the change oiq into ^ 108 14541. (John C. Malet, F.R.S.) — If the roots a?„ x^^ x^, a?4, a^j, a;g, x-jf x^ of the equation afl—pix^+p^^p^+p^x^—p^+poz'^pjx+pg = are connected by the relations ^1 + ^ + ^8 + ^4 = ^d + *6 + ^7 + ^ and XiX^^x^ = x^x^jx^, (a) prove pj = ^/ps (p^-^PiPi + ^pfl, {QiPr-^Pi'^PB+^PiPB? = (02^-404) {PT'-ipsCU), where Q^^P^-iPi^ Q^= Pa-PiPiK^'^PbI-'^^P^* Q^=P%-Q^^Ps\ (h) solve the equation 106 14543. (Professor Morley.) — The greatest number of regions into which » spheres can divide space is 2n + -^»(n—l)(n — 2) 110 14546. (Professor Neuberg.) — Si les angles des triangles ABC, A'B'C verifient les egalites A + A' « ir, B » B', les cdtes sont lies par la relation €iaf ^ bl/ ■¥ e(/ 110 14547. Professor Langhome Orchard, M. A., B.Sc.)— Show that, if m be any positive integer greater than unity, l».f 28 + 38 + 48+. ..-^»^8-(l«^-^2«> + 3S + 4S + ...+nS) ^ ^ . (1 + 2 + 3 + 4 + .. . + n)a-(l* + 26 + 3* + 46+...+«6) * 14649. (J. A. Third, M.A., D.Sc.) — K is a conic circumscribed to a triangle ABC ; P is a point on it ; Q is the isogonal conjugate of P with respect to the triangle ; R is the point where PQ meets K again ; L, M, N are the points where AR, BR, CR meet BC, CA, AB respec- tively ; X, Y, Z are variable points, Y lying on QM and Z on QN, such that the pairs AY and AZ, BZ and 'BX., CX and CY are equally inclined to the bisectors of the angles A, B, C respectively. Prove that the locus of X is QL, and that the locus of the point of concurrence of AX, BY, CZ is K. VOL. LXXIV. I Digitized by Google XXll CONTENTS. The constmction usually given for Kiepert*8 hyperbola (see Casey's Anah/tieal Oeometryy p. 442) is a particular case of tiie foregoing 127 14564. (R. F. Davis, M.A.) — Given a conic and a circle having double contact, prove that the envelope of a variable circle, whose centre lies on the first and which intersects orthogonally the second, consists of two fixed circles 133 14563. (R. Knowles.)— From a point T tangents TP, TQ are drawn to the parabola y' = 4:ax. Prove that when the circle TPQ touches the parabola the locus of T is the parabola y^ =» 4« (2a— a?) 128 14651. (Professor G. B. Mathews, F.R.S.) — Let a, B be any two given complex quantities, and let t be such that (a + t$)l{l+t) is real. Prove that, if t '^ x + iy^ the locus of {x, y) is, in general, a circle. How is this to be reconciled with the fact that the line joining two imaginary points (a, ^), (7, S) contains only one real point? 121 14682. (Professor E. N. Barisien.) — Soit ABC un triangle. Calculer le rayon d'un cercle tangent k la fois au cercle inscrit et aux cdtes AB, AC 107 14683. (Professor P. Leverrier.) — Etant donnes un triangle ABC et un cercle O, on demande de couper le triangle par une transversale a^-y, telle que les cercles a/SC et a7B soient egaux et que leur axe radical soit tangent au cercle O 108 APPENDIX I. On the Geometry of Cubic Curves and Cubic Surfaces. (W. H. Blythe, M.A.) 137 APPENDIX II. Note on the Reduction of Formulas in Factorization, affording an easy means of Factorizing Composite Numbers, especially those whose Factors are of known form. (D. Biddle, M.R.C.S.) 147 CORRIGENDA. Vol. Lxxni. P. 27, 1. 12, for "A" read "A.'* P. 49, 1. 15, for " greater than" read *'less than.'* P. 66, 1. 10 (from foot), for *« OQ" read *' OS." P. 68,1. 4 {from ioot), for ''y„^i'' read'* y""*^.'* P. 69, 1. 8, for " + x"y'' read " + (a/'y" P. 106, 1. 17, /or **^a(mi+m2 + m3), faCmi' + mj^ + fWa') " read '* |a (mj + w, + m,), |fl (wi' + m^^ + m^) . " P. 112, 1. 3 (fromfoot), for **-y sin a^)" read**-yBme)V Vol. lxxiv. P. 73, 1. 13 (from foot), for '« DGB, EGC " read " EGC, DGB." P. 78, 1. 15 (from foot), for "and in general" read "is in general"; and for ^'Zahraduick " read ** Zahradmck." P. 102, 1. 9, for **flC08^(d-flr)" read "aco8i(d + ir)." P. 73, 1. 11, for " cos 2a sin Z<p " read '« cos 2a sin 3a." Digitized by Google MATHEMATICS FROM THE EDUCATIONAL TIMES, WITH ADDITIONAL PAPERS AND SOLUTIONS. 14424. (Professor Nevbbro.) — Trouver le lieu des centres des hyper- boles ^quilat^res qui ont une corde normeile commune MN. Solution ^Professor E. J. SanjAna, M. A. ; Professor A.Droz-Fahny ; and H. W. Ctjrjbl, M.A. The angle between the chord MN and the tangent at M is equal to the angle subtended by MN at the extremity R of the diameter through M. Thus MBN is a right an^e. Draw CO parallel to IIN to meet the chord. Then MCO is a right angle, and MO is half of Hence the locus of C is the circle on MO as diameter. 14501. (Rev. T. Wiggins, B.A.)— Given a triangle ABC, find a point D within it such that DA^ + DB^ + DO^ is a minimum. I. Solution hy J. H. Taylor, M.A. The general problem of which this is a particular case is given in Williamson's Biff, Cal. (1877), § 167. VOL. LXXIV. B Digitized by Google 26 Let A have coordmates h^k; B, 0, ; G, a, ; D, x, y, ^ (u) « 3a;2 + 3y8— lax - 2Aji; —S/ty + a« + A^ + A:?. rf«/<to -» 6x—2a-'2h =» 0, ««iQ>«Be. d«M/dic« « 6 (A), du/dy = 6y-2A:, d»«/rfy2 ^ 6 (c), tPul{dxdy) « (B). Therefore x =\{a + h), y ^ \k for the minimum, indicating clearly the eentroid. II. Solution by Professor Jan db Ykibs, Ph.D. E being the middle point of BC, we get BD« + CD^ = 2DB8 + iBC^. Again, G being the centre of gravity, we have, by the theorem of Stbwakt, 2DE« + AD« = 3DG« + |AE«. Combining with AB* + AC « 2 AE^ + JBO», we obtain AD« + BD' + CD^ = SDG' + i (a^ + 4> + e^). Hence AD' + BD' + CD* will be a minimum if D coincides with the centre of gravity. III. Solution by F. H. Pbachbll, B.A. If A, B, C, Scc.f be any number of points, O their eentroid, P any other point, then the sum of the squares of distances of P from A, B, C, &c., exceeds the sum of the squares of distances of O from these points by n . OP^, where n is the number of points. TlLen, for a minimum in the question given, D must evidently coincide with the eentroid of the triangle. [See Cabby's Sequel, 6th edition, p. 26.] 14410. (Rev. T. Wiggins, B.A.)— Inscribe in a given triangle the triangle of least perimeter. Solution by J. G. Smith ; W. J. Gkbbnstubbt, M.A. ; G. D. Wilson, B.A. ; and many others. The pedal triangle is that vequired. For, because KOPW is cyclic, therefore RPB » BOB ; because ABOQ is cyclic, there- ft)re ROB = A ; because ABPQ is cyclic, there- fore QPC = A. Therefore RPB = QPC. Therefore, if Q, It be fixed, QP + BP is minimum. Similarly for the others. Therefore any change in the position of any vertex of PQR increases the perimeter. Therefore PQR has minimum perimeter. Digitized by Google 27 14444. (P* HiLNEs.) — A conic cuts the sides of triangle ABO in D, D', E, E', F, F' respectively ; AD, AD' intersect the conic again in d, <f ; BE, BE' in «?, es' ; OF, OF' in/,/. Show that the intersections of <W', e^f ff with the polars of A, B, O respectively are collinear. (Solution by R. P. Paranjfye, B.A. ; and Professor Sanjana, M.A. With the triangle ABC as the triangle of reference, let the conic he = a3i^-\-hy^-^ez^-¥2fyz-¥2gzx-\-2hxy (1). The equation of the lines AD, AD' is by^-\-cz'-\-2fyz « (2). The locus ax^ + 2gzx-\-2hxy » passes through the intersection of (1) and (2). Therefore the equation of dd' is ax + 2gz + 2hy = 0, and we get similar equations for ee^ and ff. Now we easily see that the line xjf-^- yjg -k-zjk = Q passes through the intersection of dd' and BC, &c. Hence the three points in question are collinear. [Mr. CuRJBL solves the Question as follows : — Let the polars of A, B, C cut dd\ ee^, ff' in P, Q, R. Now the polar of A cuts AD, AD' in the harmonic conjugates of A with respect to J>dy J>'d\ Therefore P lies on BC. Similarly Q and R lie on AC, AB. Let O he the pole of DD*. Then OR. OQ, OA are the polars of 0, B, P. Therefore OQ, AR and OR, AQ cut DD' harmonically. Therefore OA, QR cuts DD' harmonically ; but OA, AP out DD' harmonically. Therefore QR cuts BC in P.] 14445. (Rev. J. Cullbn.)— Prove that g2''^-l = [mod(y.2' + l)], if ^. 2* + 1 be a prime. Solution by Lt.-Col. Allan Cunminouam, R.E. ; and'EL. W. Cubjel, M.A. Let g-. 2* + 1 = JO (a prime). Then, provided ^> 2, 2*^^"^^ = + 1. Also (i?-l)/22' = 2«"^ and 2«.^=-l (modi?). Therefore (2«)(''-i)/23.(^)(i>-i)/23 = (^i)(^-i)/2« = + 1. Therefore 2*^""^^ ^(i»-i)/2« ^ ^^^ whence j2«"^ = + 1 (modi?). The proof fails when q^1\ but, when q = \t i? ** 3, and it is obviously true ; and when q — 2, p = 9, which is not prime. Unfortu- nately, the form (g'.2' + l) gives no small primes, none in fact when 2'>20 (except 1? « 3). Digitized by Google 28 14493. (J. H. Taylor, M.A.)— If A', B', C are vertices of similar isosceles triangles described all externally, or all internally, on the sides of any plane triangle BOA, the straight lines AA', BB', CC are concurrent. Solution by W. L. Thomson and the Pboposer. Let A'BG be one of the isosceles triangles, the base angles being 0. Let AA' cut BC in D. Let AX, AY be perpendiculars from A on BA", CA'. Then BD ^ A ABA' ^ AX ^ gsin(B + e) DC*AACA' AY A sin (0 + 0)' Similarly, if BB', CC cut AC, AB in E, P, respectively, CE^ gsin(C + g) ^^AF^ ^8in(A+e) EA*<Jsin(A + d) FB^a sin(B + 0)* BD CE AF Therefore — -- . =— - DC BA FB 1. Therefore AA', BB', CC are con- current. 14278. (!• Arnold.) — Two non-concentric spheres intersect, forming a ^eU. Find the centre of gravity of the larger shell and its distance from the centre of the larger sphere, the distance between the centres of the spheres being d, and the radii of the spheres being B and r. Solution by the Proposer. Let E^ represent the weight of whole concentrated at centre of gravity C of larger sphere, and r^ the weight of part at A, the centre of g^vity of the smaller sphere; d being equal to AC, the distance of their centres. The centre of gravity of shell will be in AC produced. Let it be G. Then, by a well known theorem, CG:CA::r3:R3_,^^ or CQid ::?-3:R3_,5; therefore CG = rfr3/(R3 _ r^) , which determines the centre of gravity of the shell KBLE 14512. ^ (Professor Neubekg.) — Trouver dans le plan du triangle ABC un point M qui soit le centre de gravity de ses projections A', B', C sur BC, (5a, AB, pour les poids donnes a, /8, 7. Digitized by Google 29 Solution hy H. W. Curjel, M.A. ; and Constance I. Marks, B.A. Let A'M meet B'C in A", and let Z BAM = a and Z 0AM = <^. Then B'A" : A"C' = 7 : ^. Therefore sin : sin <^ = 3 sin C : 7 sin B. Hence AM may be drawn as the line joining A to the intersections of two parallels to AB, AG at distances /SsinC, 7sinB. Similarly, BM and CM may be drawn. 14201. (R. Tucker, M.A.) — P, Q, R are points on a parabola, such that PQ, QR are normals to the curve. If SP, SQ, SE are denoted by '•i* »*a. ♦'s, prove (»*2 - n? = (>*i + ^2? (Sra-n-rs) ; hence they cannot be in A. P. Show that the circle PQR is given by I»2(m2 + 2)2(a?2 + y8)-2p^aa:-4iw»fly + w2(w3 + 2)(m' + 4)^a2 « 0, where P is the point (am^, 2am) and p^ q stand respectively for in^ + 4m^ + 2, m^ + 6m^ + 4. Find also when the circle passes through the focus. Solution hy F. L. Ward, B.A. With the usual notation, let PQ meet axis in Gj, and QR in G2. Drop PMi, QM3, RM3 perpendicular to axis. Denote P, Q, R by (:r|, 1^1), &e. Now AMjAM, =» AGi^ or x^x^ = (xx-¥2a)\ since MjGi = 2a. Also X2Xi = {X2 + 2a)3, 49ince M2G2 » 2ei. From these, and since rj = a^i + a, &c., Xi (fj— rj) = 4a)*i and x^ O3— ''a) ™ ^arj ; therefore (r^ — r^ I{r2— »'i) = ''a^Pi / ?'i ^2 = ^'2 ^1" / **! (^1 + 2«r . But Xi{r2—ri) = iaf'i ; therefore (a;i + 2a)/a;i = (rj + r2)/2ri ; therefore ('•s-^'a) / (>*2- n) = ^rjrg/ (rj + r^)^, which is the same as the result given. Digitized by Google 30 This expresses the geometrical fact that MjMs : M1M3 » ratio of harmonic mean of SG3 and SGi to their arithmetic mean. The equation to the line joining {xi, t/i)y (x^y y^j is (y-yi)(%-^i)-(^-ari)(y2-yi) - o, which reduces to y (yi + ^2) — 4<'^ — yi ya * . The line through {x^^ y^) making supplementary angles with the axis to the ahove is therefore y(yi+y8) + 4«a;-y3(yi4y5 + ys) « 0. Therefore the required circle is {y(yi+y2) + 4«r-y8(yi+yj+ys)}{y(yi + y2)-*«a:-y,yj}-\(y«-4«r)«o, where the condition for a circle gives A. ■■ (yi+yj)"+16a* ; and the equation to the circle becomes 16<i«(a:8 + y2)-4ac{yi« + ys' + ys' + yiy2 + V2ys + ysyi + 16««} +y{(yi+y2)(y3'+yiy2+y2y8+y3yi)} -yiy2y3(yi+y2+y8) = o or (a;2 -^-y^—ax (m^ + m^ + m^ + ♦», Wj + wij Wj + WI3 Wj + 4) + ay i (*»! + wi,) {m^ + Wj mj + tn^m^ + m^ Wj) — a%i«»j»t8(«Wi + iW2 + *n3) » 0. But ma = — (2/wi + mj and iwj « — (S/wj + twj) ; therefore •«*« ^9a.«,2^ m.*n. - (m/ + 6mt« + 4) mi<.t-6wi»-t-4 miWg = - (2 + Wi^, msWj « -— , m^mx = ^.. * » Substituting these values, we get the required result, i.e,^ mi2(wi2+2)«(a;' + y')-2^«a:-4iwiiay + mi2(mi2 + 2)(mi2 + 4)^a2 ^ q. If (a; = a, y = 0) lies on this circle, therefore 1 — (mi* + m^ + m^ + mim^ + m^m^ + wsmi + 4) — mim^ms (wi + mj + ma) « or miS + mj^ + ms^ + mimj + Wjms + mami + S +mim2m8(t»i + m2 + m3) « 0, which reduces to .^^^f ^^ - (gg-till ^ i or 2i?^-m2 (m2 + 2) (m* + 4) ^ « m* (m* + 2)2 or a^ + lOo* + 29o3 + IBo*- 20o- 16 - 0, where a « m\ two solutions of which are o«-l orm* = — 1, one solution of which is o = — 4 or m2 « — 4. T he other tw o are m* « - 2±2^/2 ; the only real solutions being m = ±\/2(V2— 1). 14476. (Professor E. J. Nanson.)— If ?!z± + ^^+fz£*„0, then .^ + _l- + -^«0. rt A e a^-bc b^^ca e^^ah Digitized by Google 31 Solution by J. Blaikib, M.A. ; L. E. Reat, B.A. ; and many others^ If 2 {{a^'^he)la} - 0, 2 {aHe^V^i^) « 0. Multiplying by a + i + c, we get 2{a3ic + a^<j2-a3(^ + c2)} = or 2«(^-c«)(c2-a*) - 0. Therefore 2 {al{a^^be)} = 0. 14458. (J- A. Third, D.Sc.)— XYZ is a triangle inscribed in ABC and having its sides proportional to the medians of ABO. Show that the envelope of the circumcircle of XYZ is the Lbmoinb ellipse of ABO. Solution by Rev. J. Cullbn and G. N. Bates, B.A. The sides of the pedal triangle XYZ of the symmedian point K are proportional to the medians. Hence, if the lines KX, KY, EZ revolve round E through an angle e, the triangle formed by joining (EX«, BC)^ &c., is similar to XYZ, the modulus of similarity being sec 0. Therefore the envelope of the cirde X« Y«Z« is the in-conic whose foci are K and G, the oentroid, since in general, if the ranges forming the pedal triangles of A and its isogonal conjugate A' revolve through an angle B in contrary directions, the six points are concyclic, the envelope of the circle being the in-conic whose foci are A and A^ [Mr. Bates observes that this is a particular case of Mr. Cullsn^s Quest. 14182; see Vol. lxxii., p. 65.] 14443. (B. Enowlbs.) — F, S are the foci of a rectangular hyperbola ; from a point T on the circle whose diameter is FS, tangents TP, TQ are drawn to meet the curve in PQ ; the circle TPQ cuts the curve again in CD ; prove that (1) the diagonals of the quadrilateral PQCD intersect in Uie axis ; (2) two of its sides are parallel. Solution by Professor A. Duoz-Farnt. Soient a^^y^ ^ a^ et a;^+y' = 2a' les equations de Thyperbole 6qui- lat^re et du cercle FS. La polaire d'un point (a/, yQ par rapport h. rhyperbole 6tant xaf^yy' = a^^ Tequation d*un cerde TrQ sera de la forme (a;a/ — y/ — »') (^^ + yy ' + A.) + m (^' - y' - «') "• 0. Oette conique sera un cercle passant par le point (^, /) du cercle ic2 + y2^2a2 si \ = /u--a2. Les droites PQ et OD auront done respectivement pour Equations xx'—yxf « a'^ X3^ + yt/ — <»^' Oes deux droites 6tant sym^triquement dispos6es par rapport k I'axe transverse de 1' hyperbole, le quadrilatdre PQCD est un trapeze ayant cet axe comme mMiane orthogonale. H en resulte immMiatement les deux questions propos6es. Digitized by Google 32 14499. (Lt.-Col. Allan Cuxmikoh am, R.E. )— ProTe that the continiied product P = (aT=*)(a3T*»)(«*¥i*)...(a*-'T**'"*) 18 diyiaible by {(2ii)!-i- (2". «!)}"*' if (aTb) is divimble by (2ii)!-j-(2".ii!). SoUftum by Rev. J. Gullkn ; H. W. Curjel, M.A. ; and the P&opobsb. Let N - 1.3.5...2n~l = (2n)!-i-(2".fi!). Now, if r is odd, we haye a'TA' - (aT*)(a*"*±a''-'*+...±0. and, since a = ±6 (mod N), a-»±a--»*+...±ft-^ = rft*'-^ (modN). Therefore a'^F ft' is divisible by rN, and hence P by 1 . 3 . 6 ... 2fi— 1 .N", U., by N***. 14516. (Professor Jan db Vribs, Ph.D.) — For each conic of a given pencil the oithoptical circle (circle of Monob) is constructed. How many of these circles will pass by a given point P Solution by the Pboposbr. Denoting by Au the minor of the determinant A = «21» «22» «5S «S1» <*»» «» i belonging to au, the centre of the conic Oji j;2 + ^iay^xy + a^y* + 20,30; + 20,,^ + fljs = is determined by Xq = A,, : A33, yQ^ Aj^l Ajs. If the same conic is represented by the equation we have 633 — A : Ass > ^^^ ^e quantities ui=-b^:bn, u^^-b^lb^ will be determined by Aastt^ + («ii + «a) ^38 At* + a2 =. 0. Since the radius p of the orthoptic circle is determined by p2 = tti + «2 = -(0,1 + 0,2) A : A33, the equation of that circle will be (a:-A,3/A33)2+(y-.AWA3,)« --(o„ + Oa) A/Ai. It may be verified that Ai3 + A28+(aii + a22) A = (^-11 + A,,) A33. Hence the orthoptic circle is determined by A33(«^ + y^-2Ai3a;-2A23y + (An + A23) = 0. Digitized by Google 33 Now let the pencil of conies be determined by (flu + A*ii) a;2 + 2 (ai2 + A.ii2) ^ry + . . . = 0, then the corresponding orthoptic circles are represented by + 2 [ («i3 + \*i8) (a22 + ^^22) - (''is + ^^12) («23 + ^*23) ] ^ + 2 [(flii + \bu) (023 + ^^23) - («12 + ^^12) («13 + ^*13)] y + (^22+ ^^22) («33 + ^^33) — («23 + ^^23)^ + («11 + ^^11) («33 + ^^33) - («13 + ^^is)- -= 0. Hence by any point {x, y) will pass the two orthoptic circles belonging to the conies whose parameters A. are determined by the latter equation. The system of orthoptic circles has therefore the index 2. 14027. (J. J. Barniville, M.A.)— Prove that 24-1 ^ 34-1 6^-1 84-1 ••• ^^* Solution by Professor Sanjana, M.A. Let Un denote the nth term of 1, 1, 2, 3, 5, 8, ... . Then «i + 2-t4 + l ^ («n+2 + Wn + l) (w,» + 2-«n + l) («n+2 + «n + l) = «n + 3«n {«n+2(Mn + 2-2w„+i) + «„+i(«„+i + 2Wn+2)} = ««n + 3«n {«nt2( — ««-l) +«,» + ! («„ +4)} = «n«n + lWu + 3«« + 4-«n-lWrt«/i + 2Wn + 3. Also, wl + 1 — «i = «»»-lWuMrt+2W« + 3--«n-2M«-lWii + ltt«+2 ; and so on; wj— «i " ^» Hence «J+2— «i = «»«M+i«n+3«»i+«. Thus «»» _ 1 - 1 2«H+2 = i: «n+2— 1 «n + lW« + 3«»*+4 W»+lWn + 2«fi+3Mn + 4 « A «n^4-^^n^-l ^ , (" 1 1 7 «n + lWn+2Wf» + 3«'n + 4 CWn + lMn+2Mn + 3 «»+2«n + 3Wn+45 As the given series is 2* -r-^- — , its sum reduces to 4 , i.e,, 4 + 2-1 «2<*3W4 to A. 14474. (R* Knowles.) — Tangents from a point T meet a parabola in P, Q ; the circle TPQ cuts the parabola again in 0, D ; the sides PC, QD of the quadrilateral PQCD meet in E ; the diagonals in G ; M is the mid-point of EG; MN^, ENj, GN3 are drawn at right angles to the axis ; MNi meets the parabola in K. Prove that KNi^ = ENj . GN3. Digitized by Google 34 Solution by R. Tuckbb, M.A. Let P, Q, C, D be (mi, m^tfn^fn^; then 2 (m) » 0. Now PC is y{mi-¥f>h)'^^ "* 2o»iim,, QD is y{m2+m^—2z = 2am^m^.„{i,) ; therefore 2«e ■■ — a(*/i4in8 + mjf»4) and 2a?(j «■ — a(wim4 + m3*W8) ; therefore i»u — — a (wi + mj (i»j+ m^). Hence KNj* = 4<ia?M = «' (mj + wijj^. From (i.), ^e =■ » (mi*W8— »W2W4)/(mi + i»a) '^ a{mi + m^) {m^ + fn^)l{fni + iwj) and yo — » (»«3»»3 - m, m^l{m^ + i»j) « a (mj + wig) (mj + m^jim^ + wis) • Therefore y^yo = KNi*. 5703. (Rev. W. G. Wright, Ph.D.)— A chord is drawn through the point ^2, 0) in the ellipse whose semi-axes are 3 and 2. Find the locus of the intersection of normals from the ends of this chord. Solution by F. H. Peachbll, B.A. The equation of the ellipse is x^j^ + y'/^ " ^* ^^^ coordinates of the intersection of normals at the points whose eccentric angles are a, fi will befoundtobe Zx « 6coso.cos^.cosi(o + ^)/cosi(o— ^) (1) — 2y — dsina.sin^.smi(a + ^)/sin|(a— ^) (2) If the chord joining a, jS passes through (2, 0), we get cos|(a + i8)/co8i(o-^) =|. Squaring, we get l + cos(o + i8) « f {l + cos(o-i8)} (3) From (1) |a; « f cos o . cos j8, or fa; = cos o . cos ^, or |a; = cos (o + ^) + cos (a -i8). Substituting in (3), l + cos(o+^) = f {l+|a;— C08(a + i8)}, or ^cos(o + ^) «|a; + i, or cos (a + jS) = Jf^ + -ft. cos(o-^) = fr:-C08(o + i8) = H^-A» whence sin o . sin ^ « i {cos (a— jS) - cos (a + i8) } Substituting in (2), -fzV^ = (-H^-A)'sinH(a + i8)/8inH(a-^), = {^(2a;+5)}2{l-cos(a + «}/{l-cos(a-^)}, = {A(2aj + 5)}2(20-18a;)/(45-8a:). Therefore locus is 338y«(45 -8a;) « 26 (10-9a?) (2a;+ 6)2. 14434. (Edward V. Huntingtow, A.M.) — An astroid, two nephroids, and four cardioids are drawn on the same fixed circle of radius a, their cusps lying at the quadrantal points of the circle. Prove : a line of length Digitized by Google 35 2a sliding between either pair of opposite cardioids enyel<^s tliat nephroid which has the same cusps ; and a line of length Za sliding be- tween the two nephroids envelops the astroid. (Nephroid >- two-cusped epicycloid; astroid » four-cusped hypocydoid.) Solution by R. C. Archibald, M.A. This theorem follows at once from the following modes of generation of the nephroid and astroid : — I. The nephroid is the envelope of the diameter of a circle roUiag on an equal circle. n. The nephroid is an epicycloid generated by a point in the circum- ference of a circle rolling with internal contact on a circle of two-thirds its radius. A diameter of this generating circle envelops an sistroid. 14449. (Paul Gibson.)— driven that, in reducing 1/N (N prime) in scale 17 to a pure circulator, five consecutive remainders formed are 1«, 4, 9, 21, 5 (in scale 17), to find N. I. Solution by R. P. Paranjpyb, B.A. Let p be the remainder just before \e. Using, for convenience, the ordinary decimal notation, we have, obviously, 28 + \N -i?-17, 4+/*N = 28-17, 9 + vN«4-17, 35 + pN- 9-17 (2,1), 5 + <rN= 36-17, where A, /u, v, p, o- are the successive fig^es in the decimal. They are obviously integers. From (1) and (2) we obtain 9.9-35-4 + (9v-4p) N - ; therefore 69 « (4p-.9v) N ; therefore, since 59 contains no factor, N « 59 (i.^., 38 in the scale of 17). It is ^isily seen that, with this value of N and suitable values of X, fi, ..., all the above equations are satisfied. II. Arithmetical Solution by the Proposer. Lemma, — If r^, rj, rj, ... are the remainders formed when 1/P (any scale),, and P a prime in that scale, is turned into a pure circulator, then, if r„+i-r,» - r^, r„+2— >n+i = f^+i, where the sign of equality means that the same remainder is obtained on division by r. For, if r„+i-r„ « r„., then 10r„+i-10r„ = lOr^. But 10r«+i = r«+2, &c. ; therefore r«+2— f,i+i « r«+i. By inspection we see 9— 4 » 6. Therefore by the lemma 4 — 1^ must « 21, or Itf— 4 = the complement ; therefore 17 + 21 — 38 = N. [Mr. CuRJBL remarks that any two consecutive remainders out of the five given would have been sufficient to determine N.] Digitized by Google 36 14404. (Rev. T. Roach, M.A.) — Along the hedge of a circular field of radius r are placed 2n heavy posts at equal distances. A man brings the posts together, one at a time, to one post. Show that the product of the 2«— I walks multiplied together - 2^^'* . r^**'^ , n. Solution by R. Chaktres and Lionel E. Rbay, B.A. Evidently the product = (4r)^""^8inir/2«sin2ir/2n ... sin {(2w-l)ir}/2» 14439. (H. MacColl, B.A.) — There are five possible hypotheses, Hi, Hj, &c., of which one must be, and only one can be, true ; the chance of each being one-Jifih, Each of the three H^, Hj, Hg implies that the chance that a statement A is true is '52 ; whereas H4 and H5 lead each to the conclusion that this chauce is *06. From these data prove the paradoxical (but not absurd or impossible) conclusion that it is probable but not true that A is probable ; and show that the chances that A is probable and true^ probable but not true, true but not probable, neither probable nor true, are respectively '312, -288, -024, -376. Solution by the Proposer. Let the symbol A'' assert that A is probable ; and let A'" be the denial of A**, as A* is the denial of A. Let A"^ (as usual) mean (A*)^ whatever the predicates a and fi may be. Also, let A** (when « is a proper fraction) be synonymous with ( — = aj, and assert that the chance that A is true is a. \ € / Since Hi + H2 + H3 + H4 + Hs = € (a cei-tainty), we have A^ ^ (HiH-H2 + &c.)AP ^ Hi A^^H2 ^ A^^^^. 6 6 € Hi € H2 € € \ Ell H2 / = |(-52 + -62 + -62 + -06 + -06) = -336. Thus, the chance that A is probable is f ; whereas the chance that A is true is only '336. Hence, the paradoxical conclusion that it w probable hut not true that A is probable. This conclusion implies no inconsisteucy ; for do we not often find probable predictions falsified by the event ? The explanation in this case is as follows :— The a prion chance that A is true is either '62 or '06 ; and the former value is more likely to be true than the latter. In this sense, we may say that it is probable that A w probable. But, on the other hand, if we repeatedly take one of the five Digitized by Google 37 hypotheses Hi, Hj, &c., at random (see note), and each time (assuming- the H that turns up) try whether A turns out true or false ; though the odds will be more frequently in favour of A than against, they will nerer be much in its favour ; and when they are unfavourable they will be so heavily against it (94 to 6) that, in the long i^un, A will turn out false more frequently than true. In this sense we may say that it is not true that A w probahU, Next, to find ^^, —^ , &c., A^A A" 6 6 ■t.-\ -A^ = !(••«) = -312. A^A* A'' 6 C •^ = *(l-^)-*(l-52) = -288, AA"* A''* 6 € •^. = *-^- »(•««)- -o^*. A^^A* A"* 6 6 •f^--*(^-:^)=*^^-"^^=-"^- Note, — By a simple experimental method, I found as the result of 100» trials that A turned out probable 64 times, and true only 28 times ; which (so far as it goes) confirms the conclusions M"* and A***. The statements A^A, A^A*, AA"*, A^*A* turned out true 28, 26, 0, 46 times respectively,, instead of the theoretically most likely numbers, 31, 29, 2, 38. 14406. (O. H. Hakdy, B.A.)— Prove that 2 0-2 (w) (Ti^ {u) ff^ (v) (TgS (v) =- f {p (m + v) +p («~t;)} <r2 (w + v) a^ (w-v) ; the notation being that of Weiebstkass's theory of elliptic functions,^ and the summation applying to the six possible divisions into pairs of the functions o-, tri, 0-3, tr^. Solution by H. \V. Cukjel, M.A. Consider the function flu\ - 2^!M^i' M ^^ (^) ^»' (^) . NM raav^ •'^ ' (r2(i* + f;)(r2(w-v) D («) ^^^* Since o- (u), 0*1 (m), &c., are integral transcendental functions, all the poles- of /(w) are zeros of D (w), i.«., ±f; + 2m«i + 2«o>2 (each a double one). Also, if we change u into u + 2<aa (a = 1, 2, 3), we get therefore f{u) is an elliptic function of periods 2«„ 2«2- Again, limit of (»±«)« f (») = 6o'(.-W.'(»>W,Mt>).r,'(.) ^ ^ . u= ^t> (T* (2v) Digitized by Google 38 therefore the infioite parts of the developments of /(u) near the poles ivare 3 ^ B 3 ^ A . 2(w— v)' u—v 2 («* + !;)* tt + f;' therefore f{u) - Af (tt + f;) + Bf(w-i;)-f {f (w + t;) + f (w-t;)} - Af (1* + 1;) + Bf («-t;) ^^{p{u'¥v)^p («- 1^)}, where A + B =0. When m = 0, we get Zp{v) = Af (f?) + Bf (-v) +3p(i;) ; therefore (A-B) ({v) = ; therefore, if f (v) ^^ 0, A— B — ; therefore A — B « ; therefore if ({v) ^ 0, Bubstitate t; + 2». for t; ; this does not change f{u), and, as ^(t;+ 2«a), ^(f) cannot both be zero, the above proof can then be applied. 14452. (Professor Umes Cuand&a Ghosh.)— If n and n' are the Bbocabd points of a triangle ABC, Ai, A2, A3 and a'i, A'2» A's <ure the areas of the triangles OBC, nAC, XlAB and n'BC, n'AC, n'AB, show that .. V sin^w Ai _ Ag A3 __ A^i __ A^2 ^ A^s . ^^•^ 4A "" aV «2^ = ^2^2 a2*2 ^^^ aV * r- ?!5-^ ^ nA ^ nB ^ nC _ n^A _ n^B ^ a'G , where u is the Brocard angle and A the area of the triangle ABO. Solution by R. Tucker, M.A. ; A. F. van der Hbydbn, B.A. ; and others. With the usual notation /• X . r^-an 1 2A.a2c2 a, A 4a2 sin^a \2 ' a2*2 4^ /•• \ -. * M/> ^A 1 sin 00 (u.)nA = .*2/^; .-. _-_- = _, ^d 5:^ = J- = ?yi^, andsoon. ic* A 2a (iii.) The sinister side - abel\^ -/(»**- ^^) = abceJK ; where is the eccentricity of the Brocakd circle. Now £i£i' « 2fli^ (Milne, p. 109) = 2R^ sin « = 4RA(?/\ = aheJK ; . • . &c. The above results are easily got from my " triplicate-ratio " circle papers. Digitized by Google 39 13978. (J. J- Barniville, M.A.)— Having formed the series 1, 1, 3, 7, 17, 41, 99, ... (un = 2tt».i + «,.»2) and 1, 1, 3, 11, 41, 153, ... (tt„ - 4m„_i-«„.2), prove that 1 ^ + — —+ — ... « — . *^ 1.3^3.7 7.17 17.41 ^2* l+v'2 3+ a/2 17+ a/2 99+ a/2 a/2' and a/3 J^ + -JL_+_J_ + _i__ + 12 + 2 32 + 2 112 + 2 412+2 4 Solution by Professor SanjIna, M.A. Denoting the first sum by Si and transforming the series into a continued fraction, we get Si- 32 212 1192 _ 1-3 3 + 21-21-119 119 + 697- 1 9 441 172.72 I 2-18 98- 578- JL i_ J_ ^ 1 ^1 L+ 2+ 2+ 2+ '" "" 1 + (a/2-1) a/2' Again, for the first series, «n-«n-l«n + l = «n («'n+2-2f#,. + i)-«« + i («„+l- 2f«H) =- «»w„+2-t4+i - -(«'?i+i-ttHWii+2) ; but t4— «i*'8 ■■ ~2 and ul^u^u^ = 2. Hence wj- ««-if«M+i « — 2(— 1)**, Thus, the second sum fi 1-^2 . 3- a/2 ^ 17- ^^2 ^ 99- a/2 12-2 + ... 32-2 172-2 992-2 = -1+ — + J^ +...+ v/2-a/2J^— + -^+...) 1.7 7.41 (.1.7 7.41 J 2 2(3*1 7/ 17 \7 41/ S = _i. + V2-^(J L + _l L_+..1 2 211. 3 3. 77. 17 17.41 > = _± + V2-^fl-JL) = -L. 2 2 \ ■^2/ V2 Digitized by Google 40 For the second series, but ?^— W1W3 « —2. Hence wj— »<„_iMh+i = —2, always. Hence 1.3 1.11 3.41 11.153 therefore ^83 == ^ . | . jl|^ . 1 . 3^!^ .... _ 1 , 1 , 1 , 1 , 1 , 1 , 1-33 3.1l'^3.1t 11.41"*"* = J_ -L -?- ^Q8Q = i_ J_ JL JL " 1-4- 36- 484-"* 1-4-4-4- * " 1 ^ 1 _ V3 + 1 l-(2->v/3) ^3-1 2 * Therefore S3 = \^/Z, 14441. (Rev. T. Mitchbson, B.A.) — A regular polygon of an even number of sides is inscribed in a circle, and lines are drawn from one of the ang^ular points to each of the others. Show that the sum of these lines = (acotT/2«)/(8inir/n) (« being a side of the polygon), and if the lines be A^, A2> ^z* ^^'t then i (An-l + ^n-2 + ^n-8+--.)~{^i(»»-2) + ^i(»»-4) + ^4('»-6) + "') " R- Solution by the Rev. T. Roach, M.A. ; and the Proposer. If n be the number of points, we have S = 2R (sin T/n + sin 2ir/« + . . .) to n— 1 terms = 2R [ { sin i« . ir/« sin i (» — 1) v/n} / sin ir/2»] = 2Rcotir/2« = (acot jr/2n)/(8inir/«). Again i(^n-l + ^n-2+...) — (^4(n-2) + Aj(„_4) + Aj(„.6) + ...) = R (sinirM + sin 2ir/« + ...)J"^-2R (sin 2ir/n + sin 4ir/« + 8in 6>r/w)}^""*^ « R [{sin|M . ir/» . sin^ («— 1) ir/n} / sin ir/2n] -2R [{sin J« . 27r/» sin^ («-2) 2ir/«}/sinir/n] = Root ir/2«—2R cot V «■ Rtamr/2«. Digitized by Google 41 14481. (H. M. Tatlob, M.A., F.B.S. Suggested by Quest. 14382.) —On the sides of a triangle A'B'C, triangles B'G'A, G'A'B, A'B'G are constructed similar to three given triangles. Having given tiie triangle, ABC and the three triangles, reconstruct the triangle A'B'C. Solution by the Pbopobbk. ABC (Fig. 1) is the given triangle having been formed by construct- ing on the sides of the triangle A'B'C the triangles AB'C, BCA', CA'B' similar and homologous to the given triangles LXY, MXY, NXT Fig. 4. Fig. 3. (Fig. 2) respectively ; it is required, from these data, to reconstruct the triangle A'B'C. [The triangles ABC in Figs. 1 and 3 are identical, but they are drawn apart merely for the sake of clearness.] dmstruetion, — Take any point P (Fig. 3), and construct aPCQ similar and homologous to aXNY ; and aQAB similar and homologous to aXLY; and ABBS similar and homologous to aXMY. Next take a point o (Fig. 4), and describe lipoq similar and homologous to aPCQ, a ^09* similar and homologous to aQAB, a fM similar and homologous to ABBS. VOL. LXXIV. c Digitized by Google 42 Now (in Fig. 3) describe aPOS similar and homologous to Apos of Fig. 4. We shall prove O to be coincident with A'. Froof. — A rotation round C, through an angle equal to XNY, accom- panied by a dilatation (or contraction) in the ratio NX : NY, would shift P to Q and A' to B'. Next, a rotation round A through an angle XLY, accompanied by a dilatation in the ratio LX : LY, would shift Q to B and 3' to C ; and a rotation round B through an angle XMY, accompanied by a dilatation in the ratio MX : MY, would shift R to S and C to A'. Hence by the three operations A' is unmoved and P is shifted to S. Hence, by a rotation round A', through the sum of the angles XNY, XLY, XMY, i.e,f through the angle pot (Fig. 4), accompanied by a dilatation compounded of the three specified dilatations, «.«., in the ratio op : 08 (Fig. 4), P would be shifted to 8. But, from Fig. 3, a rotation round O, through the angle pos, accompanied by a dilatation in the ratio 4>p : cw, would ^ift P to 8. Therefore O is coincident with A'. The points B' and C can now be easily found. 14511. (John C. Malbt, M.A., F.R.S.)— If, in the sextic algebraic equation afi—PiX^+PiHH^—Pz^+PA^^—Ps^+Pe = 0, the sum of three roots is equal to the sum of the other three, (1) prove 4;^6Q4-Q4 08^-/^602'^ + JP6Q2Q3 -V - 0, where Q2=Pi-iPi^ Q3^Pz-'kPlP2 + vPl^ Q^=P4-'kPiPz + iPM-^Pi*'^ (2) solve the equation. I. Solution by the Proposer ; H. W. Cubjbl, M.A. ; and Prof. Sanjana. (1) Let x^y x^f x^f x^t Xft x^ be the roots of the equation ; then a?! + aJj + arg = 3:4 + iTj + :r8 = ^p^. Let now XiX^+x^x^ + x^ar^ = m,, x^x^ + x^r^ + x^x^ = i^, XiX^X^ = fi, x^x^x^ = Vj; and we find «i + «3 = Q, (i.). Vi + ^2 + ii?i («! + ttj) = jt?s ; therefore Vi + VjasQa (ii.). UiU^ + \Pi{t\-^Vct\ =i?4; therefore WiWj — Q4, t'i««2 + «'2«i = Pi> ^i^i - P^ — (i"-» i^'» ^O- From (i.) and (iii.), we have 2m, - Q3+ V( 08^-404), 'ifij - Q2- ^/(Q32-4Q4) ; and, from (ii.) and (v.), 2v, ^ Q«+ y(Q82-4jt?6), 2^2 - Qa- AQ^^-^P^) ; Substituting for t*,, M2, v,, v^ in (v.) and ratio aalizing, we get the required condition. (2) The roots of the sextic are the roots of the cubics 2^-i?i^'+{Qs±^(Q3^-4Q4)}^-{Q3±^(Q8'-4i?e)} = 0. Digitized by Google 43 II. Solution by G. H. Habdy, B.A. It is interesting to consider this equation from the point of view of the Galois theory. The function ii>i = Xi + x^+x^= [123] has, in general, 20 values, vi^. : [123], [124], [126], [126], [134], [136], [136], [146], [146], [166], [234], [236], [236], [245], [246], [266], [346], [346], [366], [466]; and satisfies an equation ^20 (0) = ^ (i.). The group of <f»i is of order (3 \)^ » 36 ; and the solution of (i.) involves the complete solution of the sextic, for, if Xi + X2 + x^, ... are known, so are ^1, ^Tj, x^, If we know ofi^ root of (i.), e.g., 4>i, and cu^'oin it, we can determine Xi, x^f x^ by the solution of a cubic, for x^x^ + x^Xi+XiX^ and XiX^x^ are rational functions of 4>i. As we also know X4 + Xi + Xq rationally, we can •determine x^, x^, x^ by the solution of a second cubic. Thus, when the coefficients of the sextic are so conditioned that <^i is a rational function of them, the equation can be solved by means of square and cube roots. This remains true in the present case, although two values of 0, viz., [123], [456], are numerically equal. Let afi^Pisfi+PiX^^p^sfi+p^x^-^p^x+pi = (a?'»-ipiira+ Aa;-B)(a:»-ii?ia;2 + A'a;-B') ; -then p^=^ ipi^ + A + A\ j?, == ipi (A + A') + B + B' \ .. Pa = ii?i(B+B') + AA', p^ « AB' + A'B, p^ = BB' J *** A + A'=i?2-ii?i2--Q3, B + B' ^Pi-kPiPi + iPi^ - Qs, AA' " P4-hPiPz + iPi^P2—^Pi* - Q4, (A-A')(B-B') = Q3Q8-2i?5, {Qi,Q^-2p,y - W-iQ^(Ck^'^P6); ».«., ^PeQ>4-Q4Q^'-P6Q2'+P5Q2Q^-Pi^ = (ii.). To solve the equation, we have to calculate A,A'-i{Q5±V(Q32-4Q4)}, which must, in virtue of (ii.), be rational functions of the coefficients ; and then determine B, B' by equations (a). The roots are then found by solving the two cubics. 14222. (Professor Elliott, F.R.S.) - If P + OoQ, in which P and Q are free from fl?o> is annihilated by Oq— +2«i — +303— + ... to 00, doi do, dos «how that ~ Q s« 0, and that, when m > 1, Otti Solution by G. D. Wilson, B.A. By hypothesis, (00^"+ 2 (« + !)«« t^)K<^ + P) = 0» \ d«i n-l dOn + l/ Digitized by Google 44 and P and Q are free from Oq, Hence - 0, 1 (» + l n-l (i., ii., iii.). Therefore « -— |^ 5 (« + 1) a« -^ [from (iii.)] Wrd«m-ri»-l Wn + 1 d«P + (l»— r+1): _, . dardOm-r+l Therefore n.l oaa^i \ rmi Oardam^rf r-l dtfr+ld«m-r — 2 2 r dordam^ r»2 dOrWin+l-r Bttt,from(u.), 5 (« + i)«.cr- It" — ^^T-'C-'+i)^ : hence 5 (n+l)«„ -^("i' -^ + 2^) = -2 5.^-^2 2(«. + l)-^ r-l OOrOam+l-r Odm + l ^r-1 tf«rd«m>l-r dOm + l/ Putting m » 1, 2, ..., &c., in this, we obtain the result stated. 14454. (Professor SamjJLna, M.A.)— Solve, in rational numbers, the equation M^— 2zliP » d;^~ 1, where x stands for any one of the natural numbers 2, 3, 4, 5, 6, ... . [The solution gives N^+ 1 as the difference of two squares. I have reason to believe that 5 is the only small value of X admissible. For the method see Chktstal, xxxiii., {{ 15-19.] Solution by H. W. Ourjbl, M.A. ; and Lt.-Gol. Allan Gvnnimoham, R.E. X cannot be even, for M^+l cannot be divisible by 4; and, since M* + 1 s 0, mod Xf all the prime factors of x must be of the form 4n + 1. A large number of values of x will be excluded by the condition that 2x must be a quadratic residue of all factors which occur in ^s— 1 to an odd power. This condition excludes the following values less than 100 : — 13, 25, 29, 41, 53, 61, 73, 85, 89, 97 ; a: = 5, 17, 37, 65, 101 give solu- tions ^ a 5, M a 8, N » 2, the remaining solutions may be deduced with the help of 19>-10 x 6- « 1 ; a; » 17, M - 72, N » 12, the re- Digitized by Google 45 maining solations with the help of 35^—34*63 » 1 ; a; » 37, all solutions are easily deduced from 432— S' . 74 « _i ^^^ 62—74 = -38, since 74 is a non-residue of 9; :r = 65, all solutions from S'^— 130«— 66 and 672-62. 130 «-l; a; « 101, all solutions from 102-202 =-102, and 31412- 202 X 2212 = — 1, since 202 is a non-residue of 25. 14534. (W. S. CooNBY.) — Let Oi, Oj, O3 be the centres of squares described externally, and oii, w^t 0)3 the centres of squares described internally, on the sides a, b^ Cy respectively, of triangle ABO. Join O^ to a>3 and wz, meeting side BO in P, F ; O3 to 003 and «i, meeting CA in Q, Q'; O3 to »i and «3, meeting AB in E, E'. Prove that A', B', C, the intersections of P'E, QT, E'Q are the centres of the insquares of ABO, and that, if AA', BB', CO' meet sides of A'B'O' in o, fi, y, then triangle afiy is similar to ABO. Solution by t?ie Pkoposbk. From Quest. 14473 and from figure, AAOjttfs =* /^^afii\ therefore the perpendic- ulars AD and AE are equal. Draw OjS per- pendicular to AO and OiTtoAB. ACQO2 is similar to aCP'Oi, and ADQ is similar to OSOi. Therefore OQ/OP' - OOa/OOi - AO/BO ; therefore QP' is parallel to AB, and, similarly, PE' and RQ' are parallel respectively to OA and BO. ADQ and CSOi are similar. Therefore AQ/AD « 00,/SOi ; also AE'AE = BOi/TOi; but AD - AE and OOi = BOi ; therefore AQ/AE' - TOi/SOi therefore diagonal of completed parallelogram P'QAET passes through Oi and also through A' (since Q'E is paraUel to PP'), and bisects QE' in a. Similarly, BO2 and OO3 pass through B' and u, bisecting P'E and Digitized by Google 46 PQ' in fi and y. Therefore sides of a^y are parallel to sides of ABC. Ai^jles of aBR'Oj are 46°, 90*^-3, and 46° + B. Therefore BR'-(ccobB)/(B), whore (B) - sinB + cosB. Also AR » {e cos A)/(A) and RR' » (^oosC)/(A) (B). Similarly, QQf = (*cosB)/(A)(C), PR' =» (3/c)BR' - (*co8B)/(B). sin BAB' R'B' AQ ^ PR^ TOj ^ (A)(C) sin(B + 45°) ^ (A)(C) (B) sinCAB'"" B'Q AR' QQ' SOi (B) sin (0 + 46°) (B) (0) . / A \ _ sin (A + 45°) _ sinBAOa " ^ ^ ~ sin (46°) sinCAOj' Therefore A.BB'CO, is a harmonic pencil, as is also C.BB'AOj. There- fore lines drawn through B' parallel to AO3 and CO3 will make angles of 46° with AO, and have the parts intercepted by the sides AC and AB, and AG and GB bisected at B'. Therefore B' is centre of insquare to AC. Therefore, &c. 14188. (Salutation.)— Bisect AB ( = unity) in C, and AC in D ; on AB describe a semicircle ; from A, D draw parallel lines intersecting the semicircle in P, Q respectively ; S, T being the projections of P on AB, and of S on DQ, prove that 4ST is the sine of an angle « 3PAB. Solution by G. Bihtwibtlb, B.A. ; Professor T. Savaob ; and many others. AP^cose; therefore AS » cos^ ; therefore DS » cos^ 0-| ; therefore ST - sin 6 (cos* 0-|) ; therefore 4ST » sin (3 - 4 sin* 9) — sin 30. 14532. (Rev. J. CuLLBN.) — Let A be any conic in the plane of a given triangle ABC. A point P is taken on A, and parallelfl through P to BO, CA, AB meet A again in A', B', C. Prove that AP, BP, CP intersect B'C, CA', A'B' in three collinear points L, M, N. (A parti- cular case is that the intersections of the symmedian lines with the corresponding sides of Brocard's triangle are collinear.^ Prove also that, if A be the circumcircle, then LMN is at right angles to the SiHsoN-line of P. Bemarks by Professor SanjJina. . The first part is proved readily by trilinears. I append a geometrical proof of the last part. Digitized by Google 47 The triangle A'B'C is inversely similar to ABC. Hence L MA'N - A - BAG = MPN ; therefore M, P, A', N are concyclic ; so also are N, P, L, B' and L, P, M, C Thus I PMN = 180**-PA'N = 180*»-PCL « 180°-PML, so that LM, MN are in a straight line. Let the arc PA subtend an angle a at the circumference ; let LMK meet AT in Q; and let POP' be the diameter. The angle made by LMN with BC = ZPQL = PLM-APA' = PB'A'-APA' = o « JPOA. But the angle which the SiMSON-line of P makes with BG = }F'OA ; hence LMN is at right angles to the SiMsoN-line. See the Lemma on p. 73, Vol. Lxvii. 14299. (^v- T. MiTCHBSON, B.A.)— Let PiQiRi be an equilateral triangle such that Pi is on one side of a square, Qi and Ri on the adjacent sides, QjRi parallel to the other side, and O the mid -point of Q^Rj ; and let PQR be any other equilateral triangle, whose angular points are the same sides, QR passing through 0, and let PiQi meet PQ in S, PiR^ meet PR in T. Then the circle passing through P, Pi, 8, T touches QR in O, and circles passing through O, T, R, Ri and O, S, Q, Qt, respectively, are each one Uiird of the first circle. (An echo of Quest. 14236.) Solution by I. Arnold and the Profosbb. Let ABGD be the square, and PiQiRi an equilateral triangle inscribed ; O the mid-point of QiRi, and QR passing through it meeting BG in R and AD in Q. From O draw OP perpendicular to QR meeting BA in P ; then is RPQ the other equilateral triangle. A circle described on OP as diameter passes through PSTPi, touching QR in O. Digitized by Google 48 It k idio ttTidMit that QO is the diameter of the oirole passing through Q8Q(0, and that OR is the diameter of the oirole passing through OBiRT, 'HJD^ '*S^^«^•« iVn,-*! > ^ .tM; ^ ><%jt>?* ^/u; r^tt J«ft 1 Digitized by Google 49 ^%3 + a^t/^ = a^t^ et celle de I'hyperbole d* Apollonius relative au point ojS, Cifcy-k-h'^fix^a^ay — 0, on trouve ais^meut ^^^2a^alc^; 2yi« -2*2i8/c2; X^i^ - 2fl2/^(a2o«-*2)32 + e?<) ; 2yi2 « 2b2le*(^pP-^a^a^ + c*). En poitant ces yaleun dans Tequation prdcedente eUe devient apr^ qnelques transformations a2(a?-26«) + )32(2a2-*2) ^ c2(K2_«2_j2). Pour a > ^^^2 le lien est une ellipse r^elle ou imaginaire. Pour a<b^/2 le lien est une hyperbole. Pour a = b^2 le lieu se compose de deux droites parallMes k Vaxe deBX. 14270. (H. MacGoll, B.A.)— If A; be a positive constant, and the variables x and y be each taken at random between and 1, show that the chance that the fraction k{l ^x—y) / {l—y ^ k$f) will also lie between Oand I is (A:2+1)/{2A;(A: + 1)} or (1 + 2A;-A;2)/{2(A:+ 1)} according as A? is greater or less than 1. Solution by the Pkopohbk. Let the symbol Q assert that the event whose chance is required will happen, while the symbol c (expressing certainty) asserts the assumptions or data of the question. The symbol Q/c (as in previous conventions) will then express the chance required. The symbols Xi, x^, &c., ^i, y^, &c., denote the limits of x and y^ obtained as in my solution of Quest. 14210 {Reprint, Vol. lxxiii., p. 69), of which we may use the result, simply putting y for b and x •'yr.o^i'.o*** yi-1 ya- \-x ya- 1/(^+1) y^^ l—k-^-kx x,^\ ^2 = A:/(A: + 1) x^={k-\)lk for a (see the accom- p!mying table). Using the symbol yx as an ab- breviation for the definite integral we get for the cases k^ and k^, respectively W« - (y2'.o +yi'.4)^i'.2 +yi'.2%.o +y4'.o^2'.s = (A:2+1)/(2A:2 + 2A:) ; Q/« - fe.o+yi'.4)^i'.2 + (yi'.2+y4'.o)^2'.o - (l + 2A:-A:2)/(2A: + 2). The symbol y^^ indicates that the integral is to be taken between the limits y^ and ; and similarly for the symbols y^, j^ and x^,^. In this solution the order of variation is y, Xy k. Another and fuller solution (with the order x, y, k) is as follows. Taking the order of variation x, y, k (see second table of limits), Digitized by Google x, = l y,-i *,-i ^-1-y y,-i/(*+i) :r,= (y^.i-l)/* n-i-i- 50 and using the rtatement NT)* (N - D)' + ISTD' (N- D)*, we get (see soln- tion of Quest. 14210) As the application of the formula m'.n -'w..(^--'.)* - ^ ' introduces no fresh limit or fiustor into either term, we multiply hy the certainty-factor ^i^j^Vyo* getting But ^,,.,.=^y; ^a.o-'aya + ^oyyi ^20 = ^1' Vi' = ^j'' y2'.i'=y2'' and yj ^j « y,. Therefore Q = ^^(^aya + ^oyaJyy.o+^a'.iyiM = ^a-.ayy.j.o + ^y.oyy.i'.o + ^y.ayi'.s- But y, <> = ya*i'+yo*r a*^^ yyi* * ^y Therefore S'l' yr.o =yy.o^i'' ^^*lo yy.3» y^.o* *^^ yiM introduce no fresh factors. Therefore Hence, for the case ki we have ''*2'.3(y2'.o+y2'.i) = (**+l)/(2Ar2 + 2A:); and for the case h^ we have ^* -- V3y2'.3+^2'.oy3'.0 +^2'.3yr.l =- ^:^.3(y2'.3 +yy.i)+^2'.oy3' = (1 +2*-**)/(2*+2). Note, — In working out integrations of multiple integrals with several limits, the following abhreviations (which I proposed some years ago in my paper on the "Limits of Multiple Integrals" in the Troc. Lond, Math. Soe.) will, I think, be found useful : — The symbol <f>{x)x^,^ means ["""dxiffix); whereas the symbol JXn ar^„4»(a;) means <p{x^) — <p{xj. Thus, for the case Atj (that is, the case k>l) we get W« - ^v.ziyv.o+yr.i) ^ K-^3)(y2'.o+y2'.i) - l/Ar{l-(A:+l)y} (y^.o + ^i-. ) -= (ya-.o + y^.i) V^ {y-H^+ l)^'} = 2/A:{y2-i(A:+l)y|}-l/A;{l-i(A:+l)l4 1 ^k-l_ k*+l k{k+l) 2k 2k(k+iy The working for the case ky (that is, A; < ) may be similarly simplified by abbreviated notation. Digitized by Google 51 14436. (Rev. T. Roach, M.A. Suggested by 14376.)— If I, Ij, Ij. I* be in- and ex-oentres of a triangle ABC, and Oi, o^, o^ circumcentres of Iljlj, Iljli, Ililj respectively, prove that OiliOjTjOiIj is an equilateral hexagon, and find the value of its angles. Solution by A. F. van dbr Hbydbn, B.A. Let the sides of A^i^a^s ^ cut in B, S, T by III, Ha, 11,. Then Hi, &c., are thereby bisected at right angles. Also, OjR = Ros fresuUs required for Quest. 14376) ; therefore Ojli » IiO,. But I1O3 » 03IS by definition. Thus O3I1O3I2O1I3 is an equilateral hexagon. Also o^T^^ is a parallelo- gram, &c. Thus opposite sides of the hexagon are parallel, and opposite angles equal. Now ZViIs-2ir-2ZljIl3, - 2ir-.2(A + JB + JC), = ir-A-B + C. Similarly, Zli^sls « ^"^ * A + B, and Z IiOjIs = ir-B = A + C. and H. W. Cuejbl, M.A. 6630. (Professor Nash, M.A.)— If three tangents OP, OQ, OR be drawn to a semi- cubical parabola from any point O, prove that (1) the circle through P, Q, R meets the curve in three other points P', Q^ R',. the tangents at which will meet in another point O' ; (2) the middle point of OO' always lies on a fixed straight line ; and (3) the lines joining 0, O' to the cusp make equal angles with the axis. I. Solution by Rev. J. Cullen and G. D. Wilson, B.A. (1) For the semi-cubical parabola we have x « a/*', y = Ufi^, whence the tangent is Ufi^^ Sfiy + 2a; = — ^ (^), say ; therefore aii + /its + jUs == 0, 2iui/i2 - -(3yi/«)» Mif*2Ms == -(2a;i/a), where (a;,, yi) is the point 0. Now the equation of the circle PQR is of the form therefore aii + /uj + /*3 + /*4 + /*6 + /xe = ; therefore fi^ + fXi + fiQ — O. Also the tangents at P', Q!, B/ are <f> (/i4), <f> M, <f> (/ig), but therefore these three tangents meet in a point 0' {x^, yz). (2) Since ^au =- 0, we have 2^11/42 + 2/44/45 ■■ 1 ; therefore 3(yi+y2) + «* 0. So that the mid-point of OO' lies on the line 6y + a » 0. (3) /41/42/43/44/46/46 - € - e - ^+-^^1 3yi (Mi/*2M3)' ^8' *i — Xi z Digitized by Google 62 hence the lines yxi—xyi » and yx^^xy^ = are equally inclined to the axis. (1), II. Solution by F. E. Gave. Let O be a, j8, and curve ay^ = afl. P, Q, R lie on 2a$y = 3<tr2-a^ and therefore their abscisssB are the roots of a;(a;-3o)2 = 4oi82 (2). The sextic which gives the abscisssB of the intersections of the curve (1) and a circle is satisfied by all the roots of (2) if the equation of the circle ^2 + 2/2+2yi3^-3^«?^,-^+4i83?«-*-'' 0. 3o " 2a2 ■ "" 3aa The same equation is obtained by starting with the point a', fi\ provided 3 (o + o') + a = and jS'/jS = —a /a. 14460. (R- F. Davis, M. A.)— Given the base of a triangle in magni- tude (« 2a) and position, and also the length {=> I) of the line bisecting the vertical angle (vertex to base^, prove that the locus of the vertex referred to the base as axis of x and a perpendicular to the base through its middle point as axis of y is (a;2 + y^ + a^^ = ix^a^ + l*xy{P - y^). Solution by F. H. Peachbll, B.A. ; R. Tucker, M.A. ; and others. Let BO be the given base, D the middle point ; and let PDQ be perpendicular to BC, AS perpendicular to BO. Then AO (the bisector) meets QD on the circle at P. Now PO.OA = BO.OC, and, from similar triangles, PO : OA : : DO : 08. Therefore 0A2. DO/OS = BO.OC, or ^^ZllillU^ = {a + x-.(P-y^)^} {a^x + (fi-y^)^}, or t^xjiP-y^^-P = a^-a^ + y^-P + 2x{l^-y^^, or {Z2aj/(r^-y2)*-2a;(/2-y2)l|2= {(^x2 + y2 + a^-2x^}\ or l^xyiP-y^-il^x^-^^x^il^-y^) = {x^ + y^ + aY+^x^-ix^-ixY-^a^x^ So locus of vertex is (a?2 + yS + «8)2 ^ ia^^n + /4^/(p _ yi). Digitized by Google 53 6587. (W. E. Wbight, B.A.) — From a point on a curve of the second degree tangents are drawn to another curve of the second degree. Find the envelope of their chord of contact. Solution by A. Hall, A.R.C.S. Let equation of curve on which the point lies be aa;2 + %2+2Aary+2^a; + 2/y + <?= (1), and of the other curve be Aa:' + By2= i (2). Let coordinates of point be {xi^ yy) ; therefore equation of chord of contact is Aara^j + Byyi — 1 (3). Eliminate y^ from (1 ) and ax-? -¥ by-f + 2hxiyi -¥ 2gxi + 2fyi + (; » 0, and we get a?i» (aBV + *A'a;8- 2hKxhy) + 2xi {gBh^^-fABxy-bAx + hBy) + (tfBV + VBy + *) = 0... (4). Now the condition for envelope of (3) when Xiyi is on (1) is the same as condition for equal roots of (4) or (aBV + *A2iB«-2AABa:y)(<jBV + 2/By + ft) - (ffBY-fABxy-bAx)^, which, on simplification, reduces to AV iP-bc) + BV iff^-ae) + 2AB:ry (ch-fg) + 2Ar (fh^bg) + 2By (^A-a/) + A2-aft - 0, which is the envelope required, and is a curve of second degree. 14402. (R* C. Archibald, M.A.)— Show that (1) the locus of the fourth harmonic point to P, S, P', where PSP' is any cuspidal chord of the cardioid r » 2a(l— cos9), is the Cissoid of Dioclks r = 20 sin 9 tan 0; (2) if r and r' are the radii vectores respectively of the cardioid and cissoid for a given 0, rir' ^ taD^O : tan0; (3) referred to (—a, 0) as origin, the equation of the cissoid becomes r/a «= ( 1 + tan' i^) / (1 — tan' \ 9) . Solution by the Pboposbb. (1) If (r', a*) be the coordinates of the fourth harmonic point, we have 2fl(l-cosO :2a(l+coea') « r'-2a(l-cos0') : r' + 2a(l+cos90 or r' a- 2a sin a' tan d'. (2) r : r' » 2a (1 —cos 0) : 2a sin tan = sin^ ^0 : sin \e cos ^0 tan^ ; therefore r xr^ » tan ^0 : tan 0. (3) In Cartesian coordinates the polar equation r = 2a sin tan becomes y*(2a— a;) = a:* ; or, on changing the origin to (—a, 0), y'(3a— a;) = (a?— a)', whi(^ in polar coordinates may be written f^ cos 0— 3ar* + 3a*r cos - a' = or (1 + cos 0) (r—af- (1 — cos 0) (r + of = 0, which becomes {r—af = tan^ ^0 (r + «)« or r/a - (1 + tan' \e) / (1 - tan' ^0) . Digitized by Google 54 14491. (B. TvcKEK, M. A.) — Squares are described externally on the sides of the triangle ABC, and tangents are drawn from their centres to the incircle of the triangle. Proye that 22 (tangents)' = 2 A (2 + 3 cot «) - 2 {be) . Solution by J. H. Taylob, M.A. ; the Proposbr; and many others. A', B'y Q' are the centres of squares described on the sides BC, CA, AB -of the triangle ABC ; a!^ Vy <f the middle points ; and Oyb^c the points of contact of the incircle with those sides respectively. CD is parallel to AB. Bc = «-ft, Bc' = ic; therefore tf'c«i(a— 5). Denote by CP a tangent to the incircle tc» t = C'02 -0P« = C'D2 + c'c2-OP» =i(<i? + 42+c2-2aA)+tfr. Therefore 2(jf« + 4 + ^?) * | (a' + ^ + c*) +2(a + * + c)r-(aft + ^+ <?a) = 6A cot » + 4A—(a6 -!-&<; + tf0), which is the result required, since cot« « {R(a2+ J« + <j«)}/a*<j and A - a*tf/4R. 14437. (R- !*• Pauanjpyb, B.A.) — Show that there are six conies passing through three given points and having contact of the second order with a given conic ; and, further, that these six conies all touch a -quartic having the three points as nodes. I. Solution by Professor K. J. SanjAna, M.A. Taking the triangle formed by the three g^ven points to be triangle of reference, and the given conic to be S = aa^ + b0^+cy^ + 2f»y + 2gya + 2ha» = 0, we have one of the required conies of the form S' = 2ffiy + 2y'ya + 2h'afi « (1). The invariants of (1 ) and the given conic are A = abe + 2fgh — aP^ bg^—eh^^ A' = 2//A', e = 2/ (^A - af) + 2/ {Jh - bg) + 2h' {fg-eh), And e' = - af^^b^-eh'^ + 2//A' + 2gh'f + 2hfg', Digitized by Google 55 For contact of the second order we must have e+e' » 3A+e » e'-i-3A' ; these give 4(F/' + G/ + HA')2 « 3A(-a/=-*y'2_cA'a+2//A' + 2^Ay' + 2A/y)...(2) and 2 (W + ^9' + HA')(-a/S- J^-S-cA's + 2//A' + 2^*'/' + UfY) - 18A//A' ... (3). From (2) and (3) the ratios f I ^ : hf are found by a sextic ; hence, in general there are six conies like (1). The quartic which has the triangle of reference as nodal triangle must be of the form /i8V+««y«^ + ««'i82 + 2i?a«)37+2y/8V + 2»yai8 « (4). The condition of tangency of (I) and (4) is found by the usual method to be /2L+^'2M + A'»N+2/AT + 2A7Q + 2/yR = (6) ; but (2) may be written in the form /2 (F« + 3BC) + /» {(P + 3CA) + A'« (H2 + 3AB) + 2/A' (GH + 3AF) + 2h'f (HF + 3BG) + 2f'g' (FG + 3CH) = 0. Hence, if we find /, m, », py q, r from the following six equations, m»-;?»= F« + 3BC, In-q^ ^ G2 + 3CA, Im-r^ = H«+3AB, qr-lp = GH + 3AF, rp-mq = HF + 3BG, pq-nr = FG + 3CH, we determine the quartic (4) uniquely, and the proposition in the second part follows from this. II. Solution by Professor A. Dhoz-Fabnt. Faisons une transformation homographique de mani^re k ce que deux des points donnas deviennent les ombilics du plan, il s'agira de prouver que par un point donn6 on pent mener six cercles osculateurs k une conique et que ces six cercles sont tangents k une quartique bicirculaire ayant un noeud au point donn6. Or dans son Treatise on the Analytical Geometry^ second edition, p. 316, 0a8bt a demontre que par un point quelconque du plan d'une conique on peut mener six cercles qui esculent cette conique et que leurs centres sont sur une autre conique. Mais on sait que tons les cercles qui passent par un point fixe et ont leurs centres sur une conique enveloppent une anallagmatique du quatri^me ordre, podaire de conique par rapport au point donne ; d'oti le theor^rae. [The Profosbr sends the following method of solution, which is inter- esting : — If we transform the general conic in homogeneous coordinates «, y, z by the substitution | = 1/a;, iy « l/y, ( » l/«, the transformed equation is that of a quartic with nodes at the angular points of the fundamental triangle. By the same substitution a line is transformed into a conic about the triangle of reference, and vice versa. Hence the given proposition can be reduced to the following : Show that a trinodal quartic has six points of inflexion and that the tangents at these points touch a conic. Now this is a well known property of these quartics. See Salmon {Higher Plane Curves), Hence the given proposition follows.] Digitized by Google 56 14455. (Professor Cochbz . ) — Ck)urbe p* — 3p tan » + 2 = . Solution hy Bev. T. Roach, M.A. By Cardan's solution, p= {-l+v'(l-tanS«)}i + {-l-^(l-tan3«)}*. 14265. (R F. Dayis, M.A.)— If O be the centre of inyersion (con- stant » ^, investigate the formula of transformation tangent from point P to the circle C — A (tangent from inverse point P' to inverse circle C), and show that A - OP (or ic«/OF)/ tangent from O to C. Apply this to Quest. 13801. (See Vol. lxx., p. 73.) pfp, TT^-e) Solution by E. W. Kbbs, B.A. OP-p, OP'-p', OC = a, OC « «', CA = r, CA' = r'. Take O for origin and line CC for axis of x, « p*— 2ap cos a + a'— r* -(icVp")-2a2(icVp')cos» pijya +a'--r*; « p'2- 2a' p' cos + a'«-r'2. But OA.OA'- -(« + r)(«'-rO = OB.OB'« -(a-r) (a' + r^ = -ic«; therefore a + r = icS/(a'-0» «-r = icV(a'+rO ; therefore, by addition, a « <cV/(a'2— j-'S) ; therefore PT^ - ic*/p'2 {l + (2aVcose)/(a'2-r'-) + p'2/(a'2-r'2)} = [icV{p''(«''-»^}]('^ +2a/eose+«'2-r'«) - [fV{p*(«''-»^')}] P'T^ - {ir4/(p'V2)} P'TS, where ? = tangent from O to C, i.e., VTI?T = icVpY - A, where A = OP (or k^/QP')/ tangent from O to C. [The Pbopobbb remarks as follows : — If XT' be the inverse of T (lying on C), then the line PT inverts into a circle OUT' touching C at U'. Taking a point-circle at XT', the tangents from V* to the circles C\ XT' are in the same ratio as the tangents from O. Hence P^y : P'U' = A' : OU'. But PT : FU' = OP : OU' ; therefore PT/FT = OP/A'.] Digitized by Google 57 12561. (C. E. HiLLYBK, M.A.)--(l) A, B are two fixed points on a circle whose centre is 0, and C is any third point on the circumference ; BC meets the tani^ent at A in D, Oa the tangent at B in E, and AB the tangent at C in F. Prove that DEP is a straight line which envelops a conic of eccentricity €, where €- = SOA^/AB*. (2) Generalize the above and Quest. 12462 (solved in Vol. lxii., p. 89) by reciprocation. Solution by the FiioPOiBR. Draw GB' equally inclined with GB to AB ; then ^FBG-GB'A, zFCB » CAB'; therefore the triangles FBC, CB'A are similar; there- fore FC/FB = GA/B'C « CA/BC, and similarly DA/DO = AB/OA, EB/EA » BC/AB ; therefore D A . EB . FC « DC . EA . t B. / But, since DB.DC = DA2, we have DB/DC - DA'VDC^ and similarly EC/EA = EBVEA2 and FA/FB = FC^/FB-; therefore DB.EC.FA = DC.EA.FB. Therefore DEF is a straight line. Let the tangents at A and B meet in G, and draw GG meeting DE in X and AB in H. Draw from X Xo, X6, Xy, and from G GP, CQ, CR perpendicular respectively to GA, GB, BA ; then G, X, C, H form a harmonic range ; therefore XG/CG =« \ (XH/CH) ; therefore Xa/CP Xi8/CQ = J (X-y/CR), and therefore Xo.X/B/CP.CQ = \{X.y^lGB?) ; but, pince C is on the circumference of the circle, CP . CQ « CR* ; therefore Xa.XjS =» ^Xt^; therefore the locus of X is a conic touching GA, GB at A and B ; also, since F, B, H, A form a harmonic range, GH is the polar of F with respect to the conic; therefore DEF touches the conic atX. To find the eccentricity, let GO meet the circle in K, L, and AB in N ; take y the harmonic conjugate of N with respect to G, K, and Y" with respect to G, L, M the mid-point of W, and draw MT perpendicular to VOL. LXXIV. D Digitized by Google 58 W meetinjf GB inT ; then V, V are the positions of X when coin- cides with E, L respectively, and therefore V V is the major axis of the conic. Thus, if a, 6 be the semi-axes of the conic, a-=MN.MG and ft2-NB.MT; therefore €2- l-CftV**) = 1-(NB.MT/MN.MG) (NB [N.l Now it can be shown* from the properties of a harmonic range that MO/MN = f (OG/GN) ; therefore €2 = f (OG/GN) = f (OG.ON/GN.ON) « f (0AVAN2) = 30A2/AB2. If GN = NL, the point V is at an infinite distance, and the conic is a parabola ; if GN < NL, V and V are on opposite sides of G, and ihe conic is a hyperbola • the least possible value of c is ^ ^3, and this will occur when AB is a diameter of the circle. This proposition is the reciprocal of Quest. 12462 with respect to the given circle. By reciprocating either with respect to any other circle we can extend the other to the case of any conic. (See Vol. lxii., p. 89.) * Lemma,-— li C, D be harmonic conjugates with respect to A, B, P the harmonic conjugate of B with respect to A, C, and Q the harmonic con- jugate of B with respect to A, D, and O the mid-point of C, D, and M the mid-point of PQ, then OM/MB = f (OA/AB). Take R the harmonic conjugate of B with respect to OA. Take any point K in the plane, and let KP, KC, KB, KO, KR, KD, KM, KQ meet a parallel to KA in ^, <?, 5, o, r, rf, w, q. Draw KL parallel to AQ, meeting ^ in L. Then pe = eb = bd '^ dq. Also bo = or. Now, since M is the mid- point of PQ, and KL parallel to PQ, p^ m, q^ L form a harmonic range ; therefore hn . bit = bq', similarly bo.bh ^ bcP ; therefore bm/bo = bq^/bd'^ = -i, therefore bm = 2^, and therefore A, B, R, M form a harmonic range. Therefore OM/MB - J (OR/RB + OA/AB), but, since A, B, O, R form a harmonic range, we have OR/RB « J (OA/AB) ; therefore OM/MB = 1(0 A/AB). Digitized by Google 59 14394. (Professor Thomas Sayaob.)— Discuss, n being integral and positive, (l + l/ar)'*<2, but (1 + l/a;r**>2. a^3 - 1/^3 ^4 = IM yi-2^^"-l ! Solution by H. MacColl, B.A. This question, as I understand it, will afford an instructive example of the application of symbolic logic to ordinary algebra. We are required to find the real limits of x. Let A denote the statement (1 + \lxf<2, and B the statement (1 + 1/a;)**** > 2. My result is »'(AB » ary i) + n*'(AB = iTy i + ar^^^), which asserts (see Definitions and Table of Limits) that either n is even and AB equivalent to the statement that x lies between the superior limit x^ and the inferior limit Xi, or else n is odd and AB equivalent to the state- ment that X lies either between x^ and Xi or between z^o and the negative inferior limit x^. Dejinitions. ^ The symbol «' Table of Limits. asserts that n is even ; n^ that n is 0^^; a** that a is positive; a" that a is negative ; x^^^, that ar^ is a superior limit to a; ; a?^ that x^^ is an iw/ijrior limit to ar; a:^,, ^^,^ ^ that x^^ and a;^ are superior, and a?^ and a;^ inferior, limits to a;. It will be convenient to put y for 1/a;, and first find the limits of g as follows : — n'A = «'{(l+y)''-2}' - n'{(l+y)-2^/"}'{(l +y) + 2''»}" = «'{y-(2^/"-l)}'{y+(2^/~+l)}" - n\,,; «*B - «•{(! +y)~^^-2}« « n'{(l+y)-2'/("^^^}" «^'{y-(2'/(-^)-i)}« = nV,; .-. «*AB = «Vi..3.3 =* «Vr.3' ^°' ^8 implies y,. ^^A « »^{(l+y)«-2}'' « «-{(l+y)-2^/"}* n'B « w*»{(l+y)''*i-2}'* = *»**{(l + y)-2^/tH*i)}« « «^{y-(2^/^-^>-l)}« + n^{y + (2^/^«*»> + l)}« = n^' (y. + g,,) ; .-. »**AB = n\(i/^ + y^;) = '»**(yi'.8 + yi'.4') = »*(yi'.3 + y4')» for y^, implies y^,. Thus the statement for the limits of y is «'(AB = y,,3)+n*'(AB-yi,,3 + y„). From this statement, the limits of x are readily found ; for, since y = 1/a;, Digitized by Google 60 jind yj and y, are poBitive and y^ negative, it is clear that y^^ = x^,^^ and that y^, = x^,^. The above contains every step— more than would be needed in actual practice— of the symbolic process for finding the limits. The whole reasoning presupposes but an elementary knowledge of common algebra. 14479. (Salutation.) — I is the incentre of the triangle ABC, of which A is the greatest angle. P is a point on the incircle, and through P lines are drawn parallel to the three sides of the triangle, and meeting the incircle again in Q, R, S, respectively. QR, RS being joined, prove that the quadrilateral PQRS is a maximum when AIP is a right angle, and find its mean area. Solution by J. H. Tayloh, M.A. ; Rev. T. G. W. PHESTON, B.A. Roach, M.A. ; and The greatest quadri- lateral must have the centre I within it. a, b, c are points of contact. Take a point P between el and BI, and draw PQ, PR, PS parallel to AB, BC, OA, respect- ively. Let therefore QIP - 2e ; SIP = 2 (A-e). . aQIP = J>'2 8in2a; ^ ^ ASIP « ir2sin2(A-e); aSIR - JrS8in2C; AQIR = if^sin2B. The convex quadrilateral PQRS is the sum of these four triangles and, since 6 is the only variable, is a maximum when sin 29 + sin 2 (A— 0) is a maximum, i.e., when sin A cos (29— A) is a maximum. These factors increase together until 20~A » 0, i.e., till » J A. But ZcIA = 90°-iA. Therefore, when the quadrilateral PQRS is a miximum, ZPIA- 90^ [The second part of the Question remains still unsolved.] 14504. (R. Knowles.) — The circle of curvature is drawn at a point P of a parabola ; PQ is the common chord ; an ordinate from P to the diameter through the focus meets the parabola in R, and a diameter Digitized by Google 61 through Q in O. If T he the pole of PQ with respect to the parabola, prove that TO, PQ, and the tangent at E are parallel. Solution by Professor A. Dkoz-Fahny ; afid J. H. Taylor, M.A. Soit PL la tangente en P a la parabole. S, le sommet de cette demi^re, est le point milieu de la soustan- gente. D'apr^s le theo- r^me suivant lequel len couples de c6t6s opposes d'un quadrilatere ayant pour sommet les points d'intersection d'un cercle ayec une conique sent egalement inclines sur les axes de la conique, on obtiendra la direction de la corde PQ en construisant la symetrique de PL par rapport k I'ordonnee. Le centre de gravite des quatre points d'intersection d'un cercle avec une parabole coincidant avec I'axe de cette derni^re, comme trois des points d'intersection du cercle de courhure coincidant avec P et un avec Q, on obtiendra Q en prolongeant PN d'une longueui NQ - 3PN. Soit M le point milieu de PQ; comme PN = NM, le diam^tre passant par M contiendra R et coupera PL au pole T de PQ. Comme TR = BM, TL = LP et PR = RO, il est Evident que les droites PQ, LR et TO sent parallMes. Remarques. — LR est la tangente en R car elle est symetrique de PL par rapport h. Taxe ; TQ est la tangente en Q ; on verrait que TQ divise RO dans le rapport de 1 : 2. 10358. (R- W. D. Christie.) — If cvj and 0^3 are irrational cube roots of unity, prove that, if n + 2 is a prime number, 2! «— 1 . n .« + 1 ■ 4! 3! («J-6+<w'3'-6) + &C. « 0. Solution by H. J. Woodall, A.R.C.S. The series /(»+2). But, o>2, a>3 being iiTational cube roots of unity, we Have (»,'+ir^ -(-«,)»*» »J*» and (»3»+l)"*'--.^''' Digitized by Google 62 {n being odd). The proposed series becomes = _ C_J?2 =l_ + _?5 izi- f/(« + 2) - 0, 1 H + 2/ « + 2 ,N M+2/ n+2 i\ C/^ ' C«2 («*2 —1) «3 («3 —1)-'* because the numerators are each zero, tjie denominators being always ^ 0. [The PaoposBR observes : — ^The general term is W + II / n-2m-»-2 , n-2iH<h2v w— «j + zl ml Now (ft>2 +«?) = 2 if « = 3A:, and =-1 if ft = 3A;±1 {v. Todhunter's Algebra, p. 213), And by a well known theorem (t;. Carr's Synopsis j 284, p. 94 Algebra), S » if ft be of the form 6m it 1 . The result necessarily follows.] 14425. (Professor U. C. Ghosh.)— Prove that fx^ (sin x)dx ^ ^V ^ (sin x) dx, Jo and hence evaluate I ^-^ — ^ 7" — — dx. Jo 1— sina? Solution by H. W. Cubjel, M.A. ; and Constance I. Mahks, B.A. 1 X <l> (sin x) « I ' [^ ^ (sin x) + (ir— a:) ^ {sin (*— ar)} ] dx Jo Jo = x\ (p (sin a;) = iir r ^ (sin x) dx, Jo Jo Hence f- xRmx(l-8in"a;) ^ Jq l—sina; _ ^ fi' Bin^d -RiD-^)^^ _ ^ l»'(sinx + Mn»*+... + 8m«*) Ar J, l-nna: Jj = »{» + i+l •♦+••• + iir(i+t-i+i-i.i+-)} where f* = 2m + ^ db i. 14372. (H. C. Archibald, M.A.) — Parabolas with a common focus pass through a fixed point. Show (1) that the locus of their vertices is a cardioid whose cusp is at the common focus and whose vertex is the fixed point ; (2) that the locus of the points of intersection with the parabolas Digitized by Google 63 of the lines through the focus making a constant angle with their axes is a cardioid. Solution hy the Propose k. (1) S is the common focus, A the fixed point, P the vertex of any parabola passing through A and with focus at S. If SA = 4rt, L PSA = e, SP = r, we have at once from the equation of the parabola, as the locus of P, r = 2a (1 +cose), which defines a cardioid. Geometrically, it is well known tiiat the circle on the radius vector of a parabola as diameter is always tangent to the tangent at the vertex of the parabola. Hence, the locus of P is the pedal of the circle on SA as diameter with respect to the point S of its circumference : a well known cardioid definition. (2) If F be a point of intersection and a the constant angle made by SP' with SP, we easily find the locus of P' to be the cardioid defined by 4a{l+co8(e + o)} the equation r — — ^- ~ . 1 + cos a 14315. (B. N. Oaha, M.A.) — If parabolas be described cutting an equiangular spiral orthogonally, and having their axes in the direction of the polar snbtangent, the loci of the focus and thf vertex are copolar spirals whose linear dimensions bear a constant ratio. Solution hy the Peoposbr. Let S be the focus and vertex of one such parabola. Then, clearly, OS = i(Oa-OT) « i (r tan o— r cot a) x r. Also the vectorial angles of OS, OP differ by a right angle. Therefore S describes a spiral copolar with the given one. Also OA = iOT - ir coto oc r. Therefore locus of A is also a copolar spiral, the ratio of the linear dimensions of the two loci depending upon the angle of the original spiral, and therefore constant. 14461. (Rev. W. Allen Whitworth, M.A.) — If a straight line be divided at random into three parts x, y, z, show that the expectation of the volume {y + z) {z + x) (x + y) is 14 times the expectation of the volume xyz. Digitized by Google 64 I. Solution by H. W. Ccrjbl, M.A. If the straight line is taken of unit length z » l-.^:— y, and if the in- tegrals are taken over all positive values for which x + y "jf* \, exp«?ctation of ir (y + z) expectation of xyz , SS {y + zyz + T)(x + y) dy dx _ & ^^ i^z dy dx ^ ,^ jjryzdydx jjxyzdydx J 1^2 2x^ + X* ^ .._ 2 '^'' ,^_i^ !!{yx{l-^x)^xy^dydx (^ x-dX' + S^-x* ^^ Jo 6 II. Solution by R. Chautkes. Let the length be one unit, and the parts x,yyZ; then 1' = (x-^y + z)^, which has ten terms each being of the same mean value, -^ ; therefore mean value of xyz = ^, and mean value of {x-i-y){y + z){z + x), or 5(arV) + 2:ryz = | + ,V = A; therefore mean value of the latter =14 times that of the other. [See the Proposbr's Expectation of Farts.'] 14464. (Edwahd V. Huntington, A.M.) — The angle between the principal axes of two given concentric ellipses is 90°, and a-hb = a'+V. Show that a line of length a—b' (or af^b) sliding between these curves envelops an astroid ; and that any line rigidly connected with this sliding line envelops an involute of an astroid. (Astroid = hypocycloid of four cusps.) Solution by R. C. Archibald, M.A. Inside a circle of radius a + b rolls a circle of half this radius. The ends of any g^ven diameter of the rolling circle trace out two perpen- dicular diameters of the fixed circle, and in these diameters lie the axes of the ellipses traced by the points in the given diameter, distant from the circumference b and o', or b' and a. The diameter of the rolling circle envelops an astroid with the perpendicular diameters as axes of symmetry. Whence the theorem. 14432. (R. Tucker, M. A.)— PSQ is a focal chord of a parabola, and PQR is the maximum tiiangle in the segment cut off by PQ. Prove that the equation to the circle PQR is S{x^ + y^-2{7p^+20)ax+p{Zp'^4)ay + ep^a^ = 0, where p = m^]/m (P is fl/w^, 2am). Digitized by Google 65 The locus of the centre is a cubic, and, if is the fourth point of section, the locus of the mid-point of OR is a parabola, and the envelope of the chord OR is another parabola. Solution by the Proposer ; Constance I. Mawks, B. A. ; and F. H. Peachell, B.A. The triangle is a maximum when R is the vertex of the dia- meter correspondiog to the chord PQ. The coordinates of P, Q, R are (awi^, 'lam), {a/m^j — 2ff/»»), (iap^y ap)y where p = w — 1/w. By substitution it is readily- verified that the circle through PQRis 8(«3+y2)_2(7p2 + 20)fla; +p (3jt?'- 4) ay + 6p^a^ = 0. The coordinates of the centre are A, k where 8A = {7pl^+20) a, \6k = -i?(3jt?2-4)a; therefore the locus is 343rtF - {2h-oa) {lla^Zh)^, The fourth point (O) of section is found from therefore y = — 3«jo, and hence x. The midpoint of OR is g^ven hj y ^ —ap, x ^ fop^ ; therefore its locus is the parabola t/^ ^ ^x. The equation to OR is 2py + 4a; « Zap^ ; hence it envelops the parabola y^-\- Vlax — 0. 2am — 2a/»i + «p + y ; 14419. (Lt.-Col. Allan Cunningham, R.E.) — Find three sums of successive cubes which shall be in arithmetical progression. Solution hy the Proposer. Let l?n= 1^ + 23+ 33+... +«3, T„-i»(«+l); then S = T? ... (1). Then S, + S* = 2Sy, if S,, Sy, S^ be in arithmetical progression. Hence Tl + T? = 2Tj, 2Tj-lt - Tj (2). Assume T, = |.Ts, Tj, = t^.T^, where ^, ij are both integers (3) (the possibility of this latter assumption is to be justified by the result). Any solution of the Diophantine (2), which also satisjies both of (3) — for the same value of a— will be a solution. Here «, -n) - (1, 1), (7, 5), (41, 29), (239, 169), &c., are solutions of (2). The first (| *» 1, 1? - 1) gives T, = Ty =. T,. Digitized by Google 66 The next case (^ — 7, »? « o) gives T,«7.Tx. Ty«5.T,. The lowMt solution is now easily found by ti-ial. For, taking z ^ 2 gives Ta =3, T, = 21, Ty = 16, whence a? = 6, y — 6, whence (1' + 25 + 3» + 4»+ 5»+ 63j + (15 + 2») « 213 + 38 « 460 - 2 . 16« = 2x(l» + 2» + 3» + 45 + 53). Other solutions probably exist, but seem difficult to find ; thus ft is easy to solve the equations T, = f . T,, T^ = ij . Tx separately (a general mode of doing this will be shown in the present writer's solution of Quest. 14413), but it is difficult to solve them for the Fame value of Tg. It is worth noting that, if any second solution of (2) be found, say T^ + T^* = 2Ty^, the two solutions together give a solution of the more difficult problem where the sums of cubes do not start from 1', for they give (T^-T;^) + (T|-T;') = 2 (T^-T;*) or (Sx-S,.) + (8,-SxO = 2 (S,-S^). 14284. (Professor Nbxjbbro.) — Soient 0, I, !«,, Ia, Je les centres des cercles circonscrit, inscrit et exinscrits au triangle ABC ; soient D, E, F les pieds des hauteurs et A^, B^ Cj les pdles de BO, OA, AB par rapport au cercle O. Les quatridmes tang-entes communes aux cercles (I, la), (I, Jb}, (I, le) forment un triangle a&y homoth^tique aux triangles AiBjO,, DEF. Le centre d'homothetie des triangles a/37, ^iBiOi partage la droite 01 dans le rapport R : r, et est le conjugu6 isogonal du point de Gbugonne de ABC ; ses coordonn6es normales par rapport au triangle afiy sont 1/0, l/bf ije. Le centre d^homothetie des triangles oBy, DEF a pour coordonnees normales, dans cos triangles, tan }A, tan }B, tan }0. Solution by Professor San j Ana. On AB take Axy = AC, on AC take Ayi = AB ; then x^yi is the fourth tangent common to (I) and (I«). Draw the two lines similar to Xiy^, and let the three form the tri- angle afiy. As lAxyyi =- C, lAyiXi — B, x^y^ is anti-parallel to BC, and therefore parallel to BiCj. Thus the triangles AjBiCi, a$y have corresponding sides parallel, and therefore are homothetic ; so also are DEF, a$y. As Xiyi, ..., are tangents to the incircle, the triangle afiy has I for incentre ; Digitized by Google 67 and the triangle AiBjOi has O for incentre. Hence 01 is the axis of perspectiye for these, and the centre of perspectiye divides 01 in the ratio of the inradii of AiBiCi and afiy, i.e., in the ratio R : r. The distance of this centre from BC = (Rco8A.r4r.R)/(R + r) = a(«-a)/2 (R + r) ; thus this point is a(s—a) : b(8-^b) : e{8—e), and is therefore conjugrate to !/«(«-«) : !/*(«-*) : l/e(8-c), the Geroonkb point of ABO. The distance from A to ^i^i is the same as that from A to BC ; thus, B^Oi and $y are at a distance pi apart. Hence the distance of the homothetic centre from j87 = {(j»i— R) r + r.R}/(R + r) «Pir/(R + r), and is therefore proportional to 1/a ; so also for its distances from 7a, aB. The distance of this centre from Bfii is similarly seen to be PiR/(R + r) ; so that the point is 1/a ; l/b : lie in regard to AiBjOi also. Similarly, if r' be the inradius of DEF, q^ the distance apart between EF and By, and H the incentre of DEF (the orthocentre of ABO), then the distance of the second centre of perspective from fiy «^ir/(r + r'). Now ^1 « 2R sin B sin 0(1 -cos A) = 2RsinBsin0 8inA(l— cosAj/sin A a tan ^ A ; hence this point is tan ^A ; tan ^B : tan ^0 with regard to afiy. So also with regard to DEF, the actual distances being now q^r'Hr-k r^), ... [The following analytical results are easily proved: — The equation to J87 is aa+(*— <?)(/3— 7) = ; to BiOj, ^ + <?j8«=0; to O17, <j(a— 3) +7(a— *) = 3 ; the centre of perspective of AjBiCj, 0)87 is a(»-fl)/2(R + r), * («-*)/2(R + r), c (»-(?)/ 2 (R + r), or cos2 JA : cos^ iB : cos^ JO. The equation to EF is — a cos A + jS cos B + 7 cos = 0; to F7, o cos A — jS cos B + 7 cos (a - b)le — ; F7, Do, EjS meet in the second centre of perspective cos^iA/cosA : cosHB/cosB : cofc'iO/cosO.] 6419. (The late J. J. Walker, M.A., F.R.S.)— Three lines in space are determined each by a pair of planes m, = Bjy + OiZ+l = 0, a; + «| « 0, (Wj = Diy + Ejz)... . Prove that the equation to the pair of planes through the axis ^ — 0, 2 =n 0, and one of the two lines meeting it and each of those three lines, « 0. 1 1 1 ♦»! m^ »»3 Wl «2 «3 Solution by Professor Jan db Vries, Ph.D. ; and W. H. Salmon, B.A. Let qy =pz, X '^ry-^s be one of the lines meeting the axis OX and «ach of the lines B^y + CjZ fl = 0, a; + Djy + E^s « 0. Digitized by Google 68 If it intersects the line X: = 1 in the point (a?i, f/i, aj), we have (Di + r)yi + Ei2i + » - 0, Biyi + Ci^i + l -^ 0, qi/i-pzi = 0. Hence, by elimination of i/i, Zi, i?r-(Bii? + Cig)» + (Dip + Ei^) = 0. Also we have Eliminating pr and «, we get I Bip + Ciq Di^ + E,^ 1 B2p + C2q 'B^p + Ezq 1 Bsp + C^q Dsp + E^q which agrees with the given result. = 0, 6514. (W. J. 0. MiLLBR, B.A.)— Find, to 4 decimals, the value of 1 rl-» fl-ar-y frf JoJo J{ 1 /•!-« rl-x-y dxdydz ZxyziX —x—y — z) [\-x){l-y){\^z){x^y + zy Solution by J. O. "Watts. dxdydz - ^xy {l—x—y^z) JoJo Jo (l-x){\-y){l-z){x + y + z) ^ f [''^,,dy-^ f.+ ^±^ log JL^i_V-^- JoJo ^ {l''X){l-y)\ :r + // + l *^a: + y + Jo JoJo ^ {\-x){\^y) Jo l-^rl ^ l-yl ^ Jo l-a:\ 2 * y * Jo (1-^) fi 3a;^loga:<fy ^ i^ 3 (l-a;piog (1 -a;) e/a; l-x ^z['^2^Sy^Z^^^-Q[\o^{\-x)dx^z['x\og{\~x)dx Jo ^ Jo Jo -ir« Therefore I = 5-iir2 = 5-4-9348 = -0652. 14219. (!• Arnold.) — If a and b be the two parallel sides of a trapezoid, and h the line which bisects those sides, the centre of gravity G of the trapezoid is in this line. It is required to find the distance of Or from a in the line h in terms of a, b, and h. Digitized by Google 69 Solution by the P&oposbb and Rev. T. Mitchbson, B.A. Let AB, CD be a, b, respectively, /^ j the parallel sides of the trapezoid ; df be h, the line joining their mid-points; and G the centre of gravity. From Abohimedbs' theorem, G- lies in df and divides that line in the ratio of 2a + ft : 2i + a ; there- fore /a : Grf:: 2rt + ft : 26±<», or /rf: Grf:: 3a + 3ft : 2ft + a; therefore Gd^\h(a-\- 2b)/ {a + b) . 13668. {^' BiDDLB.) — It is known that the opposite edges of a tetra- hedron are equal, and the trilincar coordinates of the projection of the apex on the base are given. Find the cubic contents. Solution by W. C. Stanham, B.A. Let OP be the perpendicular from the apex O on the base ABC, and let PD' (» a) be drawn perpendicular a to BC. Then aOCB is clearly similar - 10 A ABC and equal to it. Therefore CD' (which is clearly perpendicular to BC) is equal to the perpendicular AD (= pi) from A on BO. Therefore, if OP = A, and P is (o, )8, 7)* A' + a'=xjp,' (A); ^ similarly h^ + fi^^p^^ A- + 7--P8^ ^ ^ ? C where p^ and p^ are the perpendiculars from B and C on AC and AB. And since aa + bfi + ey ^ pia » p^b =» p^e « 2 A, A being the area, and a, b, e the sides of ABO, Combining this with (A), we can find A, pj, p^, p^ in terms of a, iS, and 7. Also, if 2» ■= a + ft + tf, * = A (l/pi + l//?2+ 1/P8)» «-« - A (l/i?3+ l/i?8- \lpi)y with similar expressions for «— ft, 8—e\ ^^ ^ 8(8-a)(8-b)(8''e)\ therefore A can be found in terms of a, /3, and 7, and therefore the cubic contents (= Aa/3) can be found in terms of a, )8, 7, the only difficulty being the finding of h from the equation a/(A3 + a2)* + i8/(A2 + ^2)* + ^/(^a + ^2)* « 1 . Digitized by Google 70 14203. (V. R. Thyaoaraoaiyar, M.A.) — Show that the roots of the equation 32a^ + 16a:*- 32a^- 12a?« + 6a? + 1 = are C08y\ir, C08-j*jir, cos-ftir, cos-^yir, and cos^ir. Solution by F. L. Ward, B.A. ; Professor Sanjana, M.A. ; and others. Expanding, cos 5^ = 16 cos* 6—20 cos* $ + 5 cos 6, cos 40 = S cos* e- 8 cos^ e + 1, cos 30 = 4 cos^ e— 3 cos 6y cos 26 = 2co83e- 1, cos = cos ^. Therefore 2(cos56 + cos40 + ...+cose) + l = 32 cos5 a + 16 cos* e- 32 cos' 6- 12 cos= 6 + 6 cos 6 + 1. But 2(cos6g4.....fcose)^l ^ sinV^e-smie ^^ ^ ginj^ Bin^e sin id This is zero for the values = r^ir, -^ir, -^jt, ^Sj-ir, \^ir. Therefore the cosines of these five angles are the solutions to the equation. 14456. (Professor N. Bhattacharyya.) — There are n smooth rings fixed to a horizontal plane, and a string, the ends of which are fastened to two of the rings, passes in order through them. In the loops formed by the successive portions of the string are placed a number of pulleys whose masses are m, im, ^m, -^m, \my &c. If, in the subsequent motion, all the portions of the string not in contact with the pulleys are vertical, show that the acceleration of the rth pulley is {(»— 2r)/«}y. Discuss the case when n is even. Solution hy H. W. Gurjel, M.A. Let T be the tension of the string, and Xr the downward acceleration of the rth pulley. Then (m/r)^;, = Kr)y-2T, and 2l!{*Xr — length of string = constant ; .-. («-l)^ = (T/m)«(«-l); .-. Xr '^ 9 {n-2r)ln. When n is even ( = 2»»), then the mth pulley remains at rest until one of the pulleys cannot move any further up. Then the equations of motion are as if that pulley were left out of the system, and the new accelerations can be found the same way as before. If the pulley which first stops is the (n— l)th pulley, then the new equations are formed by substituting n— 1 for n in the old equations, thus : '^r^9{n-l-2r)l{n-\). Digitized by Google 71 14262. (Professor Sanjana.)— Prove that 1 11 11 1-2,1 1.2.3 1*4 2 '4.6 3 • 4.6.6 4 * 4.5.6.7 6ir2-72 36 and show how to find the value of 11,1 1,1 1.2 I ' n ' 2 '«(«+!) ' 3 ' n{n+\){n + 2) ' •• > where n is any positive integer. Solution by H. W. Cubjbl, M.A. The rth term of the former series ^ 6(r-l)! ^ 6 ^ r{r-^^)\ ^r«(r+l)(r+2)(r+3) r2 r r+1 (.r(r+l) (r+l)(r + 2)) ''ir(r+l)(r+2) (r+1) (r + 2) (r + S))' therefore sum to infinity «Z!-J__J-_-L« 6ir2-72 6 V 22 32 " 36 * Again, the rth term of the second series ^ (fi-l)!(r-l)! ^ (w-1)! * r(r + ft-l)! »^(r+l)(r+2) ...(»• + «-!) = ± _ (r + l)(r-«-2). .(r + »~l)-(w-l)! . rs r2(r+l)(r + 2)...(r + «-l) ' therefore sum to infinity = i*"'~*'w where «„-«, , = y , ,<"-f>' j. = ,-!- * r(r+l)...(r-^»-l) («-l)2' But .,.^^^.._^; ... ,; = :s'-l; sum = :^-i--^ 1 1 6 12 22 32 • («-l)2 which can be found ; in fact, the series = — - + ^ Vvo + ••• *o infinity. «2 (« + l)2 ^ 13391. (H. W. D. Christie.) — Prove that the solutions of X2-5Y2--4 are 'S. = a^-b^^Ub, Y = fl2 + j2^ where a and b are any two successive terms of a continttant. Digitized by Google 72 » Solution by the Pkoposbr. Solving 6Y2-X2 = 4 by X = a2-ft2 + 4a* and Y-a«+iS we get Now, if we assume afb — j9n/?„ the convergents of the continued fraction, 1+1+1+ * (i+V6)»»-.(i«v'd)'* it will be found that these convergents possess the property a^^i^^ab «±1 <v. Solutions to Quest. 13480 and 13633). 13339. (R. W. D. CHRi8TiB.)--Prove that in general any integral square may be resolved into three integral squares in six ways at least. Solution by the Proposer. Assuming the truth of Fbrmat's theorem that every integer may be resolved into four squares (proved by Cauchy), we have N2 = (aS + ft2 + e2 + cP)2 = (^,3 + ^_c2-£^;2 + (2atf + 2W)» + {2ad^ 2beY = {cfi + ^-c2-rf2)2 + ^2flk?-23rf)2+ (2«rf+ 2bcY = (-.^3 + J2_c3 + ^^2+ (2fli + 2^rf)2+ (2arf-26<?)« = (-a2+^-<.2 + <fS)9+(2ai-2crf)2+(2(jr<; + 2ftc)2 = (-a2 + i2 + c2-(fO' + (2fl* + 2«f)2 + (2a^--2M)« = (-a2 + i2 + ^_^^8^.(2aJ-2c(i)2 + (2fl<j + 2W)'. E.g., 3432- 2372 + 226*+ 1022== 2372+1382 + 2062 - 93-+3142+ 102- = 932 + 2682+206- = 32 + 3142+1382 . 32+2682 + 2262 (accidental) = 2942+ 147'' + 982 « 2792+1862+ 622 „ &c. There are six other forms, but they give the same results as above. 12055. (Professor Clayton.) — ABCD is a quadrilateral figure formed by four lines of curvature on an ellipsoid. If p^ q, r, s be the central perpendiculars on the tangent planes at A, B, C, D, respectively, then Solution by the Proposer. Draw tangents to each line of curvature at the vertices A, &c. Let «i, ^1 be the semi-axes of the central section parallel to tangent plane at A ; 03, ^2 corresponding lines for B, &c. Then, by a property of the surface Oibip >= a^b^q ^ a^b^r ^ a^b^a, and by a property of lines of curvature, aiP = a^q, b^q « ^r, a^r = a^a, b^s = bip ; hence, at once, pr = qa. Digitized by Google 73 14520 (Profeasor N. Bhattachaktta) and 14670 (E. W. Adair).— Required a direct proof of the old problem : — If the bisectors of the base angles of a triangle, being terminated at the opposite sides, be equal, show that the triangle is an isosceles one. (See Todhunt£u*s JEuelid,) I. Solution by Rev. T. Roach, M.A. Let A- 26 + 24), B^20-2<1>; Bm{2e + 2<p) ^ BE ^ AD ^ au{'20^2<f>) . sin (30 + <^) BA AB ^ sin (30 - </>) ' .-. cos 26 sin 2<f) {sin (30— </)) + sin (30 + </))} = sin20cos2^{8in(30 + </>)— sin(30— 4>)} ; .'. sin ^ » or cos 20 sin 3^ . 2 cos^ <p = sin 20 cos 30 cos 2<p ; 2 cos^ ^ sin s — 2 sin cos 9 cos 30 ; .*. sin = 0, which is impossible, or cos 2(^+ 1 — —(cos 40 + cos 20) ; .'. cos2<^ + cos20 = — (1 +COS40) ; .*. 2cos(0+4>) co8(0--<^) «— 2cos'20, which is impossible, as each factor is positive ; therefore sin^ = 0, and A ^ B. II. Solution hy R. Chaktkbs. (i.) Indirect. — By Etuf. vi. B, ac—ab^cKa + e)^ « ab — abc^Ka+b)^, or a{e'-b) ^abc{b/{a + cy'-c/{a + b)^}y of which <J— i is a factor; therefore 6 = tf, or the triangle is isosceles. Indirect proofs are given in Todhuntek's Euclid, p. 317, and in Nixon's Euclid Revised, p. 383. The following is submitted as a direct proof : (ii.) Direct, — Since BD = CE and they subtend the same anp:le A, there- fore the circumcircles of ABD and AEC are equal ; and F is clearly the incentre of the triangle ABO ; therefore AF bisects A and passes through both P and Q, the middle points of the equal arcs DGB, EGO. .-. rect. PF.FA = CF.FE (Euc. in. 35) and rect. QF . FA = BF . FD, .-. rect.FA(PF-QF) - (CK«-FK2)-(BH2-HF2) (ii. 6) =. HF*-FK«, and, since PK = QH, . = FQ2-FP2, 8 C which evidently requires PF = QF, or P and Q coincide at G, that is, ABO is isosceles. [Dr. J. S. Mackay observes : — " A direct proof of this Question will be found in the London, Edinburgh, and Dublin Philosophical Magazine (Fourth Series), Vol. xlvii., pp. 364-7 (1874)." Mr. R. TucKEK further observes : — ** This Question was proposed as VOL. LXXIV. E Digitized by Google Quest. 1907 in The Lady^s and Gentleman's Diary for 1856, and is Bolred on p. 68 (1867) by Messrs. T. T. Wilkinson, J. W. Elliott (the Pro- poser), and (analytically) by others. Mr. Wilkinson returns to the problem in his 'Note GeometricsB' in the Diary for 1859 (p. 87). A historical note is added on p. 88 which traces the Question back to the Nouvelles Annates for 1842. Professor Stlvbstbb drew attention to the subject in the Fhilosopkical Magazine for November, 1862. Dr. Adamson further discusses the matter in the Fhilosophical Magazine for April, May, and June, 1863. The best article I know on Quest. 1907 (Diary) appears in } 11 of Wilkinson's 'Horse GeometricsB,' in the Diary for 1860, pp. 84-86, with a neat proof by the Rev. W. Mason. I find that the above references are given in Dr. Mackat*s Euelidy p. 108. In the Key to this work Dr. Mackat prints a proof by M. Dbscubb (ef. p. 92)." And Mr. W. J. G-bbbnstrbbt adds the following interesting informa- tion: — ''For this and the similar theorem for two symmedians, v. IntermSdiaire des Mathematiciens^ Vol. ii. (1896), pp. 151, 326. If the external bisectors of B and C are equal, it does not always follow that the triangle is isosceles. The data lead to 4Rrj « a^ + 3^ in the triangle sides a, ^ e {v, Mathesis, p. 261, 1895).^] Editorial Note on Quests, 14620 and 14670. As an alternative geometrical solution of the above, the following, not- withstanding its simplicity, may be deemed sufficient : — Let the bisectors of B and C meet in P ; then AF bisects A. About AF as axis, let AC be revolved until it coincides (at least in part) with AB ; AG being con- sidered shorter than AB, it is clear that FO must take up some such position as FC ; that is, FC < FB ; also, since Z CFD'^ CFD = BFE, it is clear that FD must take up some such position as FD'; that is, FD>FE. Thus, we have FB + FD > FO + FE, contrary to the hypo- thesis. Therefore AC is not less than AB. If AB be supposed shorter than AC, revolve AB to lie on AC, when B' and E' will be above C andD respectively, and thQ same reasoning will apply. When A > 60°, the longer line from F will lie below-the shorter at E and D as well as at B and 0, so that the same method of proof cannot be utilized. But we may aidopt the same principle, by taking two axes of revolution, namely, BF ^ and CF, bearing in mind that BC is now g" longer than either AB or AC Revolving CB about CF, it will lie on CA, and stretch some distance beyond A ; and the same when BC is revolved about BF. But, if AC < AB, then BC - AB < BC - AC, and C lies nearer Digitized by Google 75 A than B' does. Consequently CF(- CF)<BF(« B'F). Again, if AG < AB, it is clear (CE being equal to BD) that E is nearer to A than D is, that is, in this case, nearer the foot of the perpendicular from F ; therefore DF>EF. Thus, as before, we have BF + DF>OF + EF, or one bisector of a base angle greater than the other, which is contvar]^ to the hypothesis. Hence, under the conditions mentioned in the Question, the triangle is isosceles. [It is the language alone, in the foregoing, that is not strictly Euclidean.] 14251. CR» KxowLBs, B.A.) — Prove that the sum of the first r coefficients in the expansion of (1— a;)"** is {r(r+ 1) ... (r + n— l)}/tt!. Solution by Rev. T. Mitchbson, B.A. ; and others, TooHUNTBR shows that the sum of the first r + 1 coefficients in this expansion is (n + 1) (n + 2) — (« + r)lr\. For r put r — 1 ; then we have («+l)(«+2)-(« + r-l)/(r-l)! « (« + r-l) !/«! (r-1) ! « {(n + r-l)(« + r-2)...r}/«! 14250. (RoBBRT W. D. Christie.) — Prove the following very general theorem:— ar.lO'^*** - Vm-k- x (modP), {Tiiy(XP+l)}*modP' where a?, «, h are any integers, P any odd prime, p the period of 1/P, m any integer required to make the remainder an integer (always possible). Ex, gr.—il) a? » 3, A: « 6, P = 7, X = 1, 3, 7, 9, when P ends in 9, 3, 7, 1» respectively. Therefore S.IO'"**- -p^(modLl) = ?^?^ « 1 mod7. 6'mod7^ ' 3 Thus 3.10*"-* « 1 mod 7. (2) fi « 7, A: = 1, P - 19. T.IO*^""* - (19m + 7)/2* - 13 mod 19. Thus Solution by Lt.-Col. Allan Cunningham, R.E. ; and the Proposbr. Let a be any base prime to the prime P, and let jt? be the least exponent giving a"* =\ (modP). Let X be a number such that (XP + 1) -4-a « integer. Then a;. a"'**. {(XP+ l)/a}* - j:.a.rf„(XP + 1)* = x (modP) [because a^'sl, and (XP + 1)* = 1 (modP)]. Now substitute the residue of (XP + l)-i-a to mod P in the sinister, as is clearly admis- sible; therefore a?.***"** [residue of {(XP+ l)/a}*modP] =a;(modP) - (wP + a;). Now make a « 10, and divide by the expression in the brackets [...]; this gives the required result. Digitized by Google 76 14430. (J. A. Thihd, D.Sc.)— a conic, whose centre is 0, touches the sides BO, OA, AB of a triangle at X, Y, Z, and O' is the point of ooncurrence of AX, BY, CZ. Show that O hears to ABC the same rela- tion that the isotomic conjugate of O' hears to the anticomplementary triangle of ABO (the triangle formed hy parallels through A, B, to the opposite sides). Solution by Professor A. Dhoz-Farnt ; and Professor K. J. SanjAna, M.A. II suffit de demontrer que O est le point compl^mentaire du conjugu6 isotomique de O', point de Gekoonnb de la conique inscrite. Soit P un point quel- conque du plan du triangle ABC et g une transversal e quelconque passant par P. Cette droite coupe les cdtcs en A', B', C. Soit A" sur BC I'isotomique de A' ; les trois points A", B", C" sont sur une ligne droite / la transversale reciproque de g (nomenclature de M. db LoNOCHAMFs). Commc ou le demontre ai&ement, lorsque g toume autour de P, g* enveloppe une conique inscrite au triangle et touchaut les cotes aux points X, Y, Z isotomiques des points d' intersection de PA, PB, PC respectivement avec BO, AC, AB. Les points O' et P sont done conjugues isotomiques. Soient a le point milieu de BO et a' celui de AX. D*apr^8 une proposition connue, cas particulier du theor^me de Nbwton sur le lieu des centres des coniques inscrites dans un quadrilat^re, O est le point de croisement des droites oo', jS/S', y/. Or, G etant le centre de gravite du triangle, soit u le point d' intersection de PG avec tui*. Les triangles AGP et tmG 6tant semhlahles, Qtu : GP = Go : GA « 1 : 2 ; « est done le complementaire de P et par consequent un point fixe sur PG par lequel passeront de m^me /SjS' et yy' ; u coincide done avec O ; d'oii la proposition. 14522. (J. H. Taylor, M.A.)— If A, B, C are vertices of equilateral triangles described all externally, or all internally, on the sides of a triangle A'B'C. and Aa, B6, Cc are diameters of circles circumscribing tnose equilateral triangles, then AA', BB', CC are equal and concurrent, and a, by e form an equilateral triangle and are middle points, each of a pair of arcs, on sides of the triangles ABC, A'B'C. Solution by the Proposer and Professor SanjXna. AA', BB', CC are concurrent, since the equilateral triangles are a particular case of similar isosceles triangles. (Quest. 14493.) Digitized by Google 77 B'A', A'B « CA', A'C, each to each, and ZB'A'B «CA'C'; therefore B'B — CO' = AA' in like manner. a, by e are middle points of arcs containing angles of 120°, and therefore are centren of equilateral triangles described internally on the sides of the triangle B'C'A'; therefore abc is an equilateral triangle. (Quest. 1.4412.) It has been shown (Quest. 14382) that AA', BB', CC intersect at 60^ Z aOB' « aCB' « 30® = aOO ; therefore a is the mid-point of arc con- taining angle of 120® on GB, and it is also mid-point of a similar arc onC'B. [Regarding this Question as well as Quest. 14412, Mr. Grbbnbtrbbt observes : — For complete discussion of the numerous properties connected with these triangles, with copious bibliographical references, v. Proe. Ed, Math. Soe.f Vol. xv., p. 100 (Dr. J. S. Mackayou ** Isogenic Centres**).] 14412. (H. A. Webb.) — Three equilateral triangles are described outwards on the sides of any triangle as bases. Prove geometrically that the centres of these three equilateral triangles form the vertices of a fourth equilateral triangle. Digitized by Google 78 Solution by 3. G. Smith ; W. J. Gkeenstrbet, M.A. ; and many others. The three circles round the equi- lateral triangles meet in a point. For let two meet in O ; then BOG » GOA * 120*. Therefore BOA - 120^ Therefore O is on the circle ABR. Join OA, OB, 00, and the centres of the circles X, Y. Z. Then OA is common chord of circles ARB and GAQ. Therefore YZ is perpendicu- lar to O A. Similarly, XY is perpen- dicular to OC, ZX to OB ; hut OA, OB, 00 are equally inclined to one another. Therefore YZ, ZX, XY are equally inclined. Therefore XYZ is equilateral. Extensions, — (1) Same holds if the triangles are inscrihed inwards. (2) This theorem may he extended thus : — ^If any similar triangles he described inwards or outwards on the sides of any triangle so that each angle may be in turn vertical angle, then the centres of the circles round those triangles form the vertices of a new triangle similar to the described triangles. The circles will meet in a point, and so on. 14463. (R- G. Archibald, M.A.) — Express the coordinates of any point on the cardioid as rational functions of a variable parameter, and show that the locus of a point which moves such that the triangle formed by joining the points of contact of the tangents drawn therefrom to the cardioid is of constant area and in general a curve of the eighth degree. [This theorem is due to Professor Zau&adxtik.] 11427. (^« Lachlan, M.A.) — If the points of contact of the three tangents which can be drawn from the point P to the cardioid r s a (1 4* cos 9) be collinear, prove that (1) the locus of P is a circle r+aco9$ — ; and (2), if the feet of the three normals which can be drawn from P be collinear, the locus of P is the circle 3t* » a cos 6. Solution by Professor SanjAna, M.A. We have a? « r cos0 =« 2acos2iacosa - 2a (l-^*)/(l +<«)', and y '^ rsind = 2a cos^ iS sin — 4a^/(l + f^)^, where t » tan ^6. The normal makes with the radius vector the angle ^6 ; hence, its in- clination to the axis of a; is fO, that of the tangent ^ + ^6* Thus the equation of the normal at t is 4at _ Zt^t^ I 2aa-^ \ Digitized by Google 79 or, on reduction, y ( 1 — Se') - « (3< — ^) + 2«< « 0. Similarly, the equation of the tangent is From any point P {hk), let three tangents be drawn to the curve ; then the three points of contact are given by the cubic kfi+^ht^'-Ut + 2a'-h - 0. If these points are ^i, t^, h. we get 2^j = -3A/A;, 2^2^5 -1-3, and ^j^s =■ -(2a-A)/A;. The area of the triangle formed by any three points on the curve is \ (i + v)2 (i + ^jV (1 + V)' (i + ^r)»/ " (i^^.¥(ir..y(i...^)^ ^^^--^-^(^^^-^-»^-^-'^' (1 + ^I'J' (1 + tj^r (1 + ^a)^' ^ ^^ '^ ^ ^^ ^' ^'^'^^ If the three points are those of contact, (l+^l^). (1 + ^^2)2 (1+^3)2 = { 1 + (2^i)2-22^2^s + (SVj)* - Wih'^h + ^Wi*}* 1 6 (a«-fa»-4aA-»- /!'»)» ^^ ' and ('i-<j)(<.-'iH'a-',) - y'(-§(G» + 4H»)] (see BuBNBiDB and Panton, { 42) ^yjl^^{h^ + 2A»A;« + k^-a^k'^ 2aM'- 2flA»; ] . If this triangle has the constant area %a\ we g^t for the locus of P When the area is zero, the three points t are collinear, and the locus of P is the circle A'+ A'+ «A = 0, or r + a cos0 « 0. If the three points are feet of normals, and {t.^m^-mt-t,) "Jlji (27A* + 64A«A« + 27A<- 64aAA;«- 8a A + 12fl2A»- 64aA3 + 9a«A«) j . Digitized by Google 80 If this triangle has the constant area 8a'd, we get for the locus of P l6(a» + 4A2-4aA + a2)2 ^ ' When the area is zero, the three points t are collinear, and the locus of F is the circle 3A'+ 3Ar*- aA = or 3r— «co8 6 « 0. In the general case the locus of P, for hoth tangents and normals, is a curve of the eighth degree, as is evident on squaring. [For another solution of Quest. 11427, see Vol. lviii., p. 42.] 11069. (J. J. Barnivillb.) — Prove that l* + (l« + 25)2-* + (15 + 25 + 3»)2"*+... -2744, 13+(l»+3S)2-V(l8 + 3»+63)2-%(l» + 3» + 6»+108)2-'+... - 6416, 12 + (l2 + 42)2'^ + (l2 + 42+102)2-^ + (12 + 42+102+202)2-'+... =2016; «.l8 + (w-l)23+... + 2(«-l)3+l.«» «^^«(M + l)(«+2)(3«2 + 6« - 1); in the figurate series 1, 7, 28, ..., 66M,» + 26(M»+i + tt„.i) + «„+2 + tti»..2) = a sum of consecutive fifth powers ; the ultimate term of the series 1» i, |, i» |-> -i^» ... is 2 sin -^v. Solution by H. W. Curjbl, M.A. Let B(x) = 15+ (1' + 26) a; + (18+ 2» + 3») «»+..., and S„ = 1" + 2»a;+ 3«a;2+ ..., So = r~, Si=— L- , 82 = -!-(--So + 28i)- ^ + * S3«^(3S,-3S. + 8.) = i±i^f, 8, = ^ (4S3-6S» + 4S,>So) = ' ^ ";iy ^"" ^ 8. « _L (5S,-1083+10S,-6S, + 8o) » 1 ^- 26^ -^ 66^^ ;^ 26^ -^ ^ . 1— a? (1 — a?)* therefore 8, - 1 + 26^ + 66^^ + 26^3^^ therefore firet series « 8 (i) = 4328. Again^ if se » i, ■ecpud seriea 1 f ^ . 3S,\ ^ l + 20g + 48x'4 20g' + g« ^ g... (l-Ji)»U 4 ) (l-a;)8 • Digitized by Google 81 Third series [l—X)^ \ 12 J ^ , (2Sfi + 5S, + 4S3+S,) = l + 9£±9^;±^ = 2016. 12(l-:i:3):!V o -» a x, (l-a:)8 Fourth series - coefficient of a:»-» in (1 + 2x+Zx^ + ,..)(l^+2^x + Z^x'^+ ...), «.^., in SiSa or 1 6x _ n (n -f l)(n +2) (n~ 1) w (;t + l)(»i + 2){n -f 3) (l-a;)4 (l-x)^^ 6 20 ^ n(n-H)(n^2) 3^,^^^^ 60 ' Again, expanding 1/(1— a;)7 in (1), and equating coefficients of a:"*^, we get 66Wn + 26(«„ + i + «n-l)+Wn + 2 + Wn>2 = ^S^*^ (*»*). Also 1| i, I, I, &c., are the convergents of the continued fraction J_ J_ J . 1+ 1+ 1 + ...* therefore ultimate term of the series is x, where x « -i- ; therefore x = =» 2 sin — . l+x 2 10 14478. (Bey. T. Mitcheson, B.A.)~P, Q are the ends of conjugate semi-diameters of an ellipse, and a straight line drawn from the intersec- tion of the normals at P and Q, through the centre C, meets PQ in S, whilst the tangents meet at the point {h, k); show that CS - «'^^ Solution by R. Tuckbk, M.A. ; F. H. Peachbll, B. A. ; and other's. Let the normals meet in O ; then the equation to OC is <m:/(cos^— sin<^) = iy/(cos ^ + sin ^) (i.), and to PQ is hx/a^+ky/b^ ^ 1 (ii.), whence (A/a)(cos4> + sin^) = (A:/*) (cos^— sin4>) (iii*)* From (i.) and (iii.) a^kx^b^hy = 0, but (ii.) gives b^hx + a^ky — a^b^ ; hence OC is perpendicular to PQ ; whence CS-V(a;2 + y2)-&c. Digitized by Google 82 AUematwe Pro©/ ©/Ptolbmt' 8 Theorem. By R. F. Davis, M.A. Let P be a point on the circumcircle of ABC, between A and C (say). Produce PC to Q 80 that the angle PBQ = ABC. Then, obviouslyi the triangles PBQ, ABC are similar; and so also the triangles CBQ, ABP. Thus PQ-(*/.)PB; QO-(a/ir)PA. Since PQ-QC + PO, *PB-aPA + cPC. 14407. (Bev- T. MiTCHESON, B.A.)— If a, jS, y be the distances of the incentre from the ang^ular points of a triangle, the diameter of the indrcle - ^(a-'ooB|A-^3-lco>l^B■f ^-^ cosjC) a cos i A + iS cos ^B + 7 cos ^C Solution by Rev. T. Wiggins, B.A. ; J. G. Smith ; W. J. Gbbbnstreet, M.A. ; and many others. Expression jSy sin i (B + 0) + oy sin i (A + C) 4 a/8 Bin i (A + B) AF+BD+CE 2(BTC.|.AIC.t.AIB) ^ 2S 2r. 14338. (Professor SanjAna, M.A.)— In Quest. 14110 denote (2tf— I)a2 + ft2 + fl2 |,y <fjj take *i, Cy similarly; call e the ratio of the Tucker circle, and let A = (1-^) tan «. Then prove that (1) the equation of the Tucker circle (whose ratio is e) XX'YY'ZZ' is $ylbe + yalea-¥ a$lab-{ala + $lb-¥yle) (l-e) taai»-¥ {l-eytaji^ w « 0, or (o—flX) 08— *X) (7— <?X) — aBy ; (2) the envelope of this circle, as its ratio varies, is c?/a^ + 0^lb!^-¥'^le^-2afilab'-2$ylbc-2yalea « 0, or ^/{a|a) + ^{B/b) + ^/iy/c) = 0, the Bkocard ellipse ; (3) the radical axis of two Tucker circles of radius/ and g is a/a + fi/b + yje =* i^i—f-^g) tan 00, so that the radical axis of the Tucker circle e with itself is a/a-i-fi/b + yje « 2a., which is also its chord of double contact with the envelope ; (4) the radical axis of the Tucker circle e and the circumcircle is ala+fi/b + yje = A, and the chord of contact of the circumcircle with the Brocard ellipse is the Lbmoinb line ; (5) if f+g = constant = l+e, the varying Tucker Digitized by Google 83 circles / and g hare a fixed radical azis, which is the radical axis of the circumcircle and the fixed Tucker circle whose ratio is ; (6) if f^g a 2 sin^w, the varying Tucker circles / and g are of equal area ; (7) the polar of the symmedian point with regard to the circle e is a/a + ^jh + yje ^ (4^— 1 )/ {2e) X, and, if >^ — i, the varying Tucker circles /and g have the same polar for this point ; (8) the radical centre of the circles ronnd AY'Z, ABC, and the Tucker circle, lies on BG, the radical axis of the first two being ^bi + ybci » ; so also for BZ'X, GXT ; and these three radical centres, on BC, GA, AB respectively, are situated on the line {aa^ja + {Bbi)/b + {yei)le = ; and (9) the radical centre of the circles round AY'Z, BZ'X, CY is the point o/(*iC, cos A) = iB/(<?iai cos jB) — 7/(«i*i cos G), which lies on the curve fiy sin 2A sin (B— G) + ya sin 2B sin (G— A) + a$ sin 2G sin (A— B) <- 0, that circum -hyperbola of ABG which is the isogonal transformation of Evlbr's line. [The last result has been obtained by Rev. J. Gullbn in Quest. 13921.] Solution by G. N. Bates, B.A. For figure and notation see Quest. 14110. Wehave BX.hX' = ep{a^(lfi/c) p] - (2AX-^X«)/sin2B ; therefore equation of Tucker circle is abe^afiy » 2aa.5 {(2AX-a''A.2)/sin» A} oa or ^fiy/be-^KXa/a + X^ =. 0, or {a'-a\)($-'b\)(y-c\) ^afiy (1). From these equations (2), (3), (4); and (5) follow immediately. Now Rj - R2 {/2 + (1 -/)8 tana „) and Rarin--/- R' {(2 Bin2»-/)2+ (1 - 2 sin^ cp +/») tan^w} ■■ R/ on reduction (6). The polar of K (}a tan w, &c.) is i tan i»2ala - ix^a/a^i (Z\ tan ») + x' » 0, or 2o/«- (4<?-l)A/2« (7). Now BA.BZ-.{M*W,».{.-2(l^.|.g, and GA.GY'-i'<ri/2a2; therefore AY'Z is a2(^^.^$y — ^aa {bifi$ + eiby). Similarly for BZ'X, GX'Y. Therefore radical axis of ABG and AY'Z is bicfi + e^by » 0. Radical axis of AY'Z and T* is Y'Z, which intersects radical axis of T« and circumcircle on BG (see Quest. 14159, Vol. lxxi.). Therefore radical centre of AY'Z, ABC, and T« is on BC, and is 0/0 =« Bjbci — 7/— <?*!• This and the two corresponding points lie on the line aia/a+bifilb-^ciy/c = (8). Digitized by Google 84 The radical centre of AY'Z, BZ'X. and CXT is given by (cbifi-¥beiy)la =» (ca^a-¥ac^y)lb ^ {baia + ab^fi)/c, and is therefore the point byCi cos A ; a^Ci cos B ; aibi cos 0. To find the locus of this point we have to eliminate between icco8A/o--2a2-2(e-l)a* = and two similar equations, giving 2 {b^^c^ cos A/a = or 5 sin 2A sin (B-C)/a « 0, the isogonal inverse of Eulek'b line. 14525. (J- Maoleod, M.A.)— KL is a diameter of the circle KML. From L any two chords LM, LN on the same side of KL are drawn and produced to meet the tangent at K in Q' and O. Through O a line is drawn parallel to MN, and LQ' is produced to meet it in Q. QQ' is bisected in Y, and the straight line OY in P; through P a tangent is drawn to the parabola which is touched by OQ, OQ! in the points Q, Q', and meeting OQ, OQ' in R, R'. Prove that the angle KOL is equal to the angle of the focal distances of P and R. Solution by Cokstancb I. Mabks, B.A. ; and the P&oposbk. Draw QT parallel to YO. At Q and P make angles 0Q8, RTS, respectively equal to angles TQO, OPR'. Therefore S is the focis of the parabola. Join RS and NK. Now Z KOL is complement of Z KLO. Therefore ZK0L = ZNKL = ZNML «Z0QQ' = Z0RR' — difference between Z R'PS and ZRQS « difference between Z R'PS and ZPRS ( A8 PRS and QRS being similar) - Z RSP. 13480. (It. W. D. Christie.) — Find a series which gives all the odd primes in order as factors. Digitized by Google 85 SoltUion by the Fropobbr. The series is 1, 4, 11, 29, 76, 199, 521, &c. The law of f onnation «»» + 1 = 3a» — «„ . i , (ft,2 + «8)^'*"^ + («4 + «6f*'^ = 1 mod 2«-.l = On, if 2n— 1 be prime and »« ftn unreal root of rr^ + 1 ss 0, i.^., {J(l+ V6)}-"-V{i(l->v/6)f'*-^« lmod2n-l ^ I l^^^^-^f^ol^l + jlziMllN^} = 1 niod;,, where j9 is any odd prime, .2l^M or i±^-lmodi,. 2P 2""^ Now, by Fbrmat's theorem, 2**"^— 1 ^p,qt when and when only j* is prime. If jE? = 2, we have i(l + 2A/5 + 5)+|(l-2V'6 + 5) = 3- 1 mod 2. Thus the theorem is universally true for all primes. 13496. (Professor SanjIna, M.A.)— If pr denote the cofficient of x" in (1 — a;) "*, prove that log 4 = j?i + ip, + to + .. . , log(v/2 + l) -j»i + to + iP5 + ..- = ^'-hPl + \Pi-]Pi"" Solution by Kev. J. Cullen ; H. W. Cukjbl, M.A. ; and others. Since pr = 2/irl sin2»-erf6, first series = - 2/ir [ 'log (1 - sinS 0}de^- 4/ir [ "log coBdde ^ log 4, Jo Jo second series = 2/ir (*'log ( j '^"!"^^ ) dd - log (a/2 + 1), jQ \ 1 — Sin / third series = 2/ir r'tan-i (sin 6) dd/ain 6, To integrate this, let I = f 'tan-i („ gin ^) rfe/sin d; Jo therefore dljda = p'fl?e/(l +a2 sin^ e) = iir (1 +a2)* ; therefore I = Jw log (a + a/i + a^j, putting a = 1, we have socond series = third series = log ('/2+ 1). Digitized by Google 86 11864. (Professor Lucas.) — Mettre le nombre a?'"— 5*y*°, ot ;? et y sont des nombres entiers positifs, sous la forme d'un produit de trois £Eu:teiirs rationnels. SolutioH by H. W. Curjbl, M.A. ; H. J. Woodall, A.R.G.S. ; and othert. «io_5»yio « («S-5y«)(«8 + 5-p8yj + 25a:<y<+125a;V+625y8) - («2-6y«) {a!* + 6^y + 16a:y + 26iry» + 25y<) X (a:*-6;r>y+15a;V-25ary» + 26/). 11075. (R. P- Kathbrn, M.A.) — Solve the equation tan'a + asecO + dtane + sec'a = 0. Solution by H. J. Woodall, A.R.G.S. tan'tf + asec0 + &tan0 + Bec'0 = 0; therefore (2 tan*0 + * taii«+ !)« - a'sec^tf == a«(taii««+ 1) ; therefore 4tau4a + 4*tan>0+tan«6(4 + 6«-aa) + 2*tan«+ (l-a«) - 0, which requires solution. 11852. (Professor Zbrr.) — ^If S represent the length of a quadrant of the curve r*^ » a, cos m9, p the radius of curvature of the path of the pole when the curve rolls along a straight line, Si the length of a quadrant of the first pedal of the curve, prove that (1) SSi « fywa^l2r ; also (2) p/r » (m + l)/m » a constant. Solution by the Proposer. The equation to the first pedal is r"*/(»*+») » o^/C^+i) cos"»/(«'»-») 6 ; hence (Wili.iam80n's in^. Cal., Art. 166, Ex. 3) S = 2m 'iDi'm^ «^ss,.f!4S-r(i)/r(.,i).SM^. The {Pf r) equation to the curve is r"»+^ — a^p ; by the theory of roulettes (djdp) (p/r) = l/p ; dpKrJ ^ ^^l+mKm + lJ r\m + ir .-. p-«r(m + l)/w; .-. ^ = (m + l)/w = a const. ; .-. SSi - ^. Digitized by Google 87 10364. (L* BiN^zBOH.) — Si b est inf^rienr k ^/a, a et b 6tant des nombres entien positifs, Pexpressioii a{l + a+2b) ne sera jamais un carr6 parfait. Solution by H. J. Woodall, A.K.G.S. Since b < \/«, we may pat « « ft* + A;*. Therefore a (l+a + 2ft) = (« + *)» + A;" — (« + * + <?)', say; therefore *« — 2<? (« + ft) + c* « «— ft» ; therefore ft* + 2ft<j + c» = a (1 - 2<;), i.e,, (ft + <?)«««(!- 2c) (a negative quantity). This is impossible. Therefore a(l+a-f2ft) cannot be a perfect square. 11762. (Professor Lucas.) — Demontrer I'identit^ x-xP xP-xt^ x^-x^*^ (\^x)(\-'XP) (l-«P)(l-a?iO •" (l-a;0(l-^0 Solution by H. W. Gxtrjbl, M.A. (l«a;P)(l-«p"*V The given sen 1 l-x 1 1-a; Les r 1 1 -XP 1- 1 1 1- -ay*+» 1 -x^^' (1- -x){l-x^-S 14413. (RoBBRT W. D. Christie.) — Find integral values of n to satisfy the equation T« « nTy, and give general values for x and y. (T a triangular.) Solution by Lt.-Col. Allan Gunninoham, R.E. ; the Froposbr; and others, T.-Ja;(a;+l)-J(X'-l), T, = Jy(y+l)-J(Y2-l) ...(1), where X«2a;+1, Y=.2y+1 (2). Then T, « «.Ty gives wY'— X» « w— 1, (« supposed integral) ... (3). Now suppose Xq, Y© to be awy solution of this Diophantine (Xo = 1, Yq » 1 is the obvious solution), and let |r» i7r be any solution of ^—nr^ s 1, all the solutions of which (infinite in number) are known by the ordinary rules. Then all the solutions (corresponding to Xq, Yq) are given by Xr - |rXo + «i?rYo, Y^ = |r Y^ + i;,.Xo. From these x, y are given by (2), and T„ T^ by (1). Digitized by Google 88 The following are examples for the simple cases of n = 2, 3, 5, 6 given hy Xo = 1, Yo = 1 :— «-2. »» 3. n • = 6. « = 6. 1 = 3, 17, 99, &c. 2, 7, 26, 97, &c. 9, 161 5, 49 V « 2, 12, 70, &c. 1, 4, 15, 66, &c. 4, 72 2, 20 X = 7, 41, 239, &c. 5, 19, 7, 265, &c. 29, 521 17, 169 Y = 5, 29, 169, &c. 3, 11, 41, 155, &c. 13, 233 7, 69 X = 3, 20, 119, &c. 2, 9, 35, 112, &c. H» 260 8, 84 y = 2, 14, 84, &c. 1, 5, 20, 77, &c. 6> 116 3, 34 T.= 6, 210, 7140, &c. 3, 45, 630, 8778, &c. 105, 33930 36, 3570 T,= 3, 105, 3570, &c. 1, 15, 210, 2926, &c. 21, 6786 6, 595 A similar procedure is applicable when n is a. fraction, say n = v-i-fif so that the problem is to solve /tTx — pTj, , which gives /tX*— vY* = /*— v (3a). If now Xo, Yo be awy solution of (3a), (Xq = 1, Yq = 1 is the obvious one) ; and if fr, Vr be any solution of f^^^^.^a _ i^ all the solutions of which are known by the ordinary rules ; then all the solutions of (3a) corre- sponding to Xo, Yo are given by Xr = frXo + i^rYp, Yr = /lATjrXo + ^ Yq. The values of x, y, T^r, Ty are thus given by (2) and (1). The two fol- lowing examples are for the simple case of « = f , taking X© = 1, Y© = 1 ; here /* «= 3, v = 2, /av = 6. i V X Y X y T^ Ty 6, 2 ; 9, 11 ; 4, 6 ; 10, 15 ; 49, 20 ; 89, 109 ; 44, 64 ; 990, 1485. Cor. — From the above may be solved the following more general problems (4) Tx = Ta.Ty; (5) T;.Tx=T«.T,; (6) T;.T;.T;...T, = Ta.T,.Tc = Ty. It suffices in (4) to take T« = w ; in (5) to take Ta = /ia, T„ = v ; in (6) to take T^.Ti .Tc... « ju, T^.T^.Tc ... = v ; and the preceding solu- tions now apply. 11888. (Professor Zbrii,) — A paraboloid floats in a liquid which fills a fixed paraboloidal shell ; both the paraboloid and the shell have their axes vertical and their vertices downwards : the latus rectum of the para- boloid and shell are equal, and the axis of the shell is m times that of the paraboloid. If the paraboloid be pressed down until its vertex reaches the vertex of the shell, so that some of the liquid overflows, and then released, it is found that the paraboloid rises until it is just wholly out of the I i fluid, and then begins to fall. Prove that (1) the densities of the pfiTabokud and the liquid are in the ratio 2 [m- + ;» + !- {m + 1) V)t7^-\} : 3 \/(;m+ l}/(m-l), r Digitized by Google 89 the free surface of the liquid being supposed to remain horizontal tit rough- out the motion ; and (2), if cone and conical b e read for pa raboloi d and paraboloidal, the ratio is 3 {m*— 1 — (m^-l) v^m^— 1| : iVfn^—l, sup- posing the vertical angles of both equal. Solution by the Pkoposbb. In the two positions in which the velocity of the paraboloid is zero, the heights of ^e centre of gravity of the paraboloid and liquid are equal. Let p be the density of the liquid, o* of the paraboloid, h the height of the paraboloid, x the height of the surface of the liquid in the second posi- tion. Then, if ^^ » 4iax is the equation to the paraboloid, we get 2airx^ « 2airfn«A«- 2awh^ = 2«ir (mS- 1) A* ; therefore x « \/m*— 1 A ; hence |mA . 2airm' A'p — | A . 2airh^ (p - tr) « |v/»i-- lh.2aw (w2- 1) AV + (%/w*- 1 + J) A . 2airAV ; 2 {mS- l-'\/(m2- 1)3} p = 3 \/w«2- 1 ff ; therefore 2 {m2 + m+ I -(m+ 1) v/m^-l} p == 3 a/(w + l)/(m- 1) <r ; ^ira:3 tan2 i8 = ^irm^AS tan* jS - ^ir A3 tan* jS ; 2^3 = vertical angle ; therefore x = V^«^— 1 ^ ; hence |wA . ^inrt3A3 tan* jS . p - |A . ^irA3 tan* jS (p- o-) = A ^^^3-1 A. I {ir (m3-l) A3} tan* jSp + ( Vw3 - 1 + f ) A . ^irA' tan* jB<r ; therefore 3 {m*-l- (»»3- 1) Vw»3- 1} p = 4 J/m^-l o-. 13663. (The late Professor Wolstbnholme, M.A., DSc.) — Normals to the parabola y* = iax at the points P, Q, Q' meet ia a point, and QQ' passes through the fixed point (— «, 0) : the envelope of the circle PQQ' will be the pedal, with respect to the origin, of the parabola 4ay* * tf* (^— 4a + o). The origin (A) is a double focus of the envelope (which is a lima9on),and there are two single foci S^ Sj at the points (— «, 0), («— <r*/4a, 0) ; the vector equation being, for the infinite branch, (c - 4a). SiP + 4^.82? «= c.AP. When e » 4a, the envelope is the pedal of the parabola itself with respect to the vertex (a cissoid), which is tiien a triple focus, and the vector equa- tion becomes nugatory. When e > 4a, there is no oval branch. When c = 8a, the envelope becomes the straight line a; + 4a « 0, Si, 8^ coincide, and the vector equation is SiP — AP. VOL. LXXIV. F Digitized by Google 90 Solution by the Proposer. The normal at {xy) passes through (XY) if Y--y=-y/2«(X-yV4a); so that, if yii y^t y^ he the roots of the equation, yi + ya + ys^ 0, yjys + yayi + yiya =• 4a(2a-X), y^y^^ « Sa^Y; and the equation of QQ' is 4flar-y {y^ + y^) + y^i = 0, whence y^^ = iae. Hence, if we put y^ + y^^ 2«\, X — 2a"€ + a\^, Y =—e\, and the equation of the circle PQQ^ ^2 + y2_a;(2fl + X)-iYy«0, hecomes x^-\-y^—x[4a—C'¥ (t\^ — ^\cy = ; so that the envelope is <;V + I6ax (a;' + y^—iax + ex) = 0. The equation of any tangent to the parahola 4ay^ = (^{x—ia + e) is y => m (ar--4tf + c) + c*/16«w, and the perpendicular from the origin is a; + my = 0, whence the equation of the pedal is 1 6ax (y^ + a;" - 4flj? + ci;) + t^y^ = ; or coincides with the envelope. The straight line x + iy ^p will he a tangent if the equation I6ax {{x—iy) p—iia- c) x} + c-y^ = has equal roots giving the equation 16<wj2 {p-ia + e) + 64fl2jy2 ^ o, whose roots are — c, c—cyia. The origin, heing a node, is itself a douhle focus. Also, at any point P (a?, y) on the curve, ^ c^+16ax* ^ ^ ^ ^^ e'-¥l6ax <?+Uax Hence -AP _ S,P _ -S^P x{%a^c) (c+8a)a; + <;2 (3c-8a)a;— tf3 + c5/4a' the proper signs for the infinite hranch heing determined hy putting X « — c2/16rt, which is the value at infinity on the curve. Thus we have the vector equation \ . SiP + ii . SjP = v . AP, where \ (^ + 8a)— /i* (3<;-8«) + v (8a— c) = and X = ju (e/^a— 1), giving A/(4a - c) = ju/( - 4a) = — v/c, or (4a-c) S)P-4a . SoP + c . AP = 0. This will he the vector equation for all values of <; if we suppose that when c < 4a, and P is on the loop, AP shall he counted negative. When c — 8a, the origin is the focus of the parahola 4ay2 =x (;2 (a;— 4a + c) or y^ « 16a (x + 4a), Digitized by Google 91 and the pedal is the tangent at the vertex x»—ia. The vector equation becomes SiP + SjP » 2AP, but, Si, S3 being coincident, this is equivalent to Si? = AP, the straight line bisecting the distance between S|, A at right angles. 14184. (J. J. Barniville, B.A.)— Prove that -l-+^L. + _2__ _^ _^_ ^17 28 33-1 63 + 1 83-2 13» + 3 *" 90' 2.3 6.8 13.21 34.66 *** ' 7 31 183 835 ^ _3^ 3.6 8.13 21.34 66.89 "* " 2 * Solution by Professor San j Ana, M.A. The first series \8 126 2200 **" 39312 "J \26 610 9266 "'/ The first part 1.8 8.66 66.377 3.21 21.144 144.987 = |-^ y-_ y-^ . . . + ^_YZ fr i^^yf^^ ^8ual f ormul89) 1 ^ 1 _ 3--v/5^7-3V6 8-^(7-3^5) 63.i!,7^3 ,5. 6 18 9 " 3 * and the second part 63 -.-|- (7-3^/6) 2.13 13.89 89.610 5.34 34.233 V5-2 6a/6-11 26-2(7-3^/6) 170^26 ^7^3^5j 6 30 ^5-2 5^/5-11 ^ V5 7_ 6 30 * 3 ^10' Thus the whole series "9 10 ' 90* Digitized by Google 92 The second Beriee 3-2 12-5 70-13 319-34^ " 2.3 5.8 13.21 34.55 _J_ J_.J 1 ,_ig_ ^ . 29 _1^ ""2""3 2.6 8 3.13 21 5.34 66 = l-.l_l + l-l-i+±-l-l+i-i-i + ...-l. 2 3 2 5 8 3 13 21 6 34 56 The third series 10-3 39-8 204-21 ^ 890-56 ^ "" 3.6 ^8.13 21.34 65.89 ■"3^58 13 21 34 55 89 '" 3 5 2 8 13 3 21 34 6 66 89 2 2 [Mr.- G. D. Wilson, B.A., sends the following interesting method of summing the first of the above series: — Denoting the series 1, 1, 2, 3, 6, ... by ui, uj, ... vn, ..., the law of formation being wn+i = wn + wn-i> we have the relations wn-iv»+i — v» = (— 1)** (i.)f and v,»un+6— w»+iu«+4.= (— 1)*** 3 (ii.). From (i.) we deduce that uii+a— ( — ir^n-i « w«u»+iw»+5.« (iii.)' From (ii.) and (iii.)^ 5 -il n-i)"'^n (-ir^n.4 i *-l*^+l— ("irWn-1 w-li "« + l «'»» + « i n-l Vn-t-l 14S01. {J- J. Barnivillb, B.A.)— Sum the series 1 , 1 1+32 2 + 42 -i-+ ' 2 + 4» 6 + 7» 1 1 1 as+3 32+7 ■^3 + 52 , 1 8 + 102 , 1 42+11 fi2 + Z ' 92+15 ' 132 + 23 "** *'" ■ K Digitized by GoOglC 93 SohUion by G. D. Wilson, B.A. ; Professor Sanjana ; and othert. 3 1 Vn+1 « + 4/ 36 9 1 v « «+iy 9 ' 6 1 \ » « + 6/ 120' --Lir± -_L^-l. 6' + 7 9»+15 13» + 23 161 \n « + l/ 16 1 , 1 1 + 3* 2 + 4' 1 + 1 2 + 4' 6 + 7' 1 + 1 2» + 3 3' + 7 J_+..^. *3 + 8' , 1 8 + 10' , 1 4'+ 11 + 1 11830. (Professor Lbbon.) — Deux plans parallMes k un cercle commun 4 plusieurs spheres et 6quidistanta de ce cercle d^terminent dans les spheres qu*ils coupent des segments Univalents. Solution by H. W. Curjbl, M.A. Let h be the distance of each of the planes from the common section (radius a). Consider the sphere whose radius is B and whose centre is at distances »xt x^ from the two planes. Then Therefore volume between the planes = \irh{Za^-¥^{Zx^-k-^x^x^-\-'6x^'^Xi^-^x^X2-ix,^] = |irA(3a'-A^) « constant. 14158. (Lt.-Col. Allan Cunningham, R.E.) — Express (3 . 2» + 1) in one or more of the forms (c*;i7 2rf2), (A^rt^SB^) ; or show that it does not admit of this expression. Solution by the Phoposbb. This number has been recently resolved by the Bev. J. Cullbn into two prime factors : N » 1893029 . 13613. These, being both of form p ^ 24tr + 5, cannot be expressed in any of the proposed forms ; so that their product (N) is also not so expressible. 14400. (R. JF. Davis, M.A.) — Find positive integral values for N, X, y which will render N2-3a:«, N2-3y^ N2-3(a; + y)^ N2-3(a:-y)* perfect squares. [A special Christmas puzzle.] Digitized by Google 94 Solutum by the Proposer. The following values satisfy the conditions : — 2662-3.88' =» 2182, 266«-3. ISS* = 232, 2662-3.«62- 2412, 266«-.3. 232 = 2632. 14367. (Professor N. Bh attach aryya.) — Show that the product of three immhers representing the sides of a right-angled triangle is divisible by 60. Solution by H. W. Cubjbl, M.A. ; and many others. The numbers are of the form m*— «', nfi-¥n^j 2mn. Their product is divisible by 4, since at least one of m, «, m*- w' is divisible by 2. And tnn (m2— f»2j| jg divisible by 3, since every square number = 1 or mod 3, and the produce of the three numbers is divisible by 5, since every square number = it 1 or mod 6. Therefore product is divisible by 60. [Mr. Paranjpyb and Mr. Muoobridob direct us to a solution of this Question in Smith's Algebra, § 384, p. 489. The Question also appears in Todhuntbr'b Algebra (1879) among the '* Miscellaneous Examples," No. 268.] 14515. (J. A. Third, M.A., D.Sc.)— X, Y, Z are three points in the plane of a triangle ABC, such that the pairs AY and AZ, BZ and BX, CX and CY are equally inclined to the bisectors of the angles A, B, G respectively. Y moves on the straight line »&, and Z on the straight line Ue. Prove the following statements : — (1) the locus of X is a straight line Ua ; (2) if Ub pass through B, and tf« through C, Ua passes through A ; (3) if ui, be perpendicular to CA, and Ue to AB, Ua is per- pendicular to BG ; (4) if L, M, N be the points where Ma, «6, t*c meet BC, CA, AB respectively, AL, BM, CN meet in a point P ; (6) AX, BY, GZ are concurrent in a point whose locus is, in general, a conic circumscribed to the triangle and passing through P ; (6) if m^, Ue meet on the cubic circumscribed to the triangle, and passing through every pair of isogonal points whose join passes through P, viz., / {j/i^z^lyz + m {z^-x^lzx + n {x^-y^)lxy - 0, where /, m, n are the coordinates of P, Ua, u^, u^ are concurrent. Solution by G. N. Bates, B.A. Let the equations of AY, BZ, and CX be y— ica =« 0, a-Aa; = 0, and «— fiy = ; «* = ^a: + i»iy + »,2 = 0, «c = ^c + my + »2S. Then X is the point (A/a, A, /a), Y is (ic, /uk, /h), and Z (ic, A, kA). Digitized by Google 95 The locus of X is found by eliminating between tiK + miKfi + nifi =0") ,. . t^ + m^K +«2^ic = 0) and x/y » /u, x/z = \; ♦.*•, (miiwj— «in,)a; + ^«i^ — ^j«i« — (1). If Ml s and fi2 « 0, this reduces to a straight line through A ... (2). If fHi =» /i cos C + wi cos A and Wj = ^ cos B + m^ cos A, (1) reduces to (/im^ cos C — tiil^ cos B) a? + litn^ - l^iz « 0, or (/imj cos C — wj/^ cos B)(«a? + Ay + cz) + (c/imj— iwi^ (y cos B— « cosC) - 0, a straight line perpendicular to BC (3). AL is ^i»«sy - l^iz = 0, i.e,, li {l^ + m^) — Ij. [l-^x + n-^z) « 0, a straight line through the intersections of BM and CN, viz., the point (»V>i» -^«i, -^i^a) (4). AX, BY, CZ meet in the point (ic, A., /t). The locus of this point is obtained by eliminating between the equations (i.) and xJk — yj\ » 2;//a, giving «2y ( h^ + Wi«) =" *«i« (^' + ^y)i a circumconic passing through the intersection of /ia? + «i« as and /j-r + mgy = 0, ».«., P (5). «a, wj, «c will be concurrent if lix + ftt,y + n-iz = 0, ?2iP + Wjy + m = 0, (WiWj — «i»2)i; + ^i/rt2y — /2''i^ ■■ ^> t.0., the point of intersection will be on the curve obtained by eliminating between these equations and Ifm^ni = ml'^l^ni = nj—litn^, i.e. , { — «»2 {lix + «i«)/y + «i (^^ + m^yz] x + ^imjy - l^niZ = 0, or l(y^-z^)/i/z + m(z^-'X^/zx + n{x^—i/^)lxy='0 (6). 14518. (Professor A. Goldenbero.) — Eesoudre le syst^me {x + 2i/)(X'i-2z) « a^, (y + 2«)(y+2a:) = b^, (« + 2:c) (« + 2y) = <j2. SoltUion by James Blaikie, M.A. ; Rev. T. Eoach, M.A. ; and many others. (a; + 2y)(a? + 2«)« {(iP + y + a) + (y-a)} {(a? + y + a)-(y-a)} = (a; + y + 2}2_(y_2j)2^ Thus, if we write « for a; + y + 2, the system becomes whence 5 { ± a/(«'— a^) } = ; solving, we find «* » i (a2 + d2 + c2)±| ^(^4 + j4 + c4_^^2_^V-a2^2), Digitized by Google 96 Also 3a? = t^iz-x) + {x-y) = »=F >/(«'— 62)± ^/(«2—c«), with similar yalnes for 3y and 3^. J?.y. a=l, 6«2, tf = 3; a:-f^/21±i^/3, y - f A/2l±f a/3, r«fA/2lTA/3. [Professor Frank Mohlby, of Haverford College, U.S.A., refers to Vol. XXXII., p. 90, of the Reprint^ where this Question (as 6028), having been set by himself (wrongly called T. Mokley), is solved by the present Proposer.] 14482. (Professor Neubbrg.) — Soient rt, h, <?, d les c6tes AB, BC, CD, DA d*un quadrilatire spherique ABCD circonscrit a un petit cercle. Demontrer la relation sin a sin b sin^ J6 = sin sin d sin^ ^D. Solution by Professor SanjAna and H. W. Curjbl, M.A. Since the quadrilateral is circumscribed to a small circle a — b = d—c^ therefore cos a cos & + sin «( sin 6 s cos <; cos <2+ sin c^ind; but COS a cos d + sin a sin i cos B =» cos AC » cos<; cos ^+ sine; sin ^ cos D. Hence sin a sin & siu^ JB ^ sin <; sin <2 sin^ }D. 14528. (R. P. Pahanjpyb, B.A.) — Show that any triangle can be projected into an equilateral triangle whose centre of gravity is the projection of a given point. Solution by G. D. Wilson, B.A. Let ABC be the given triangle, P the given point. Let AP meet BC in A'. Take X so that (BA'CX) ^^ — 1, and similar constructions for Y and Z. Then X, Y, Z lie on a straight line. Project this line to infinity, and two of the angles of the triangle into angles of 60° ; the projection of ABC is then an equilateral triangle, and has the projection of P for its centre of gravity. 14508. (W. H. Salmon, B.A.) — The frustum of a pyramid with quadrilateral base is such that the intersections of the opposite faces are coplanar (A) ; prove that (1) the diagonals of the frustum are concurrent (O) ; (2) each diagonal of the frustum is divided harmonically by O and its point of intersection with A ; (3) the diagonals of each face are divided harmonically by their point of intersection and the plane A. I. Solution by the Proposer. The principle of continuity will allow us to project figures in three dimensions as in two dimensions. Hence in this problem the plane A Digitized by Google 97 may be projected to infinity, and the problem is reduced to : The diagonals of a parallelepiped are concurrent, and bisect each other ; and the diagonids of the faces bisect each other. II. Solution by H. W. Curjbl, M.A. Let O be the vertex of the pyramid, and BCDE the base, and E'C'IVE' the opposite face ; GB'B, GC'G, GD'D, GE'E being straight lines. Since the opposite faces cut in a plane A, any two pairs of opposite faces cut in a point on A; t.^., B'C, ED, E'lK, BG meet in a point K on A, and B'E', CD', BE, CD in a point H on A. Therefore the intersection of B'D, G'E is on the line joining G to the intersection of BD, GE, and B'D, EG' cut on the same straight line ; therefore CE, B'D, EG' meet in a point 0. Similarly, BD' passes through O. B'E, G'D cut in M on the line of intersection of BB'E and GG'D ; therefore M is in A, and therefore MK is in A. Therefore G'E, B'D are divided harmonically by O and A. Also B'E, B'E are divided harmonically by their point of intersection andGH. 6648. (H. FoKTEY, M.A.) — In a plane triangle ABG, bisect AB in D, and take DBA' opposite to DA, or DAB' opposite to DB, each equal to half the sum of AG and BG ; and prove that the semi-perimeter AA' or BB' will subtend an acute angle at G if the base AB does not exceed half the sum of AG and BG. Solution by D. Biddlb. It will be convenient to take A A', the semi-perimeter of ABG, as unity, and on it describe a semicircle. Then, if we divide A A' into three equal parts by Bi, Di, and on ABi construct the equilateral triangle AB|Gi, we have J (AGi + BiGi) - Di A' - AB, which (as base) is thus at the limit stated. Now, DjA'-l and DA = 1/a/3. Therefore AA'.DiA' « 2l>fi^\ and the circle on A A' as diameter, namely, AIjA', is the circle of curv- ature at A' to the ellipse of which DjA', DiGi are the semiaxes. Gonsequently, of ellipses having one focus at A and the other in AA', and whose major axes terminate at A', A'GiGi is the ellipse of greatest eccentricity, capable of touching the circle AI^ A' at A' and wholly enclosing it. But this ellipse is the locus of Gi when J (AGi + BjGi) « Dj A', and, since in every position Gj is outside the circle, therefore ZACiA' is acute. Taking ABj<ABi, and making AG] » B2G2 » DjA', we can describe the ellipse of which Digitized by Google 98 ' Ds A% DsG] are the semiaxes, as the locus of G^, and so on, nntil the fool coincide in A, and the ellipse becomes a circle. All these positions of are clearly external to the circle AIi A' ; therefore, within the limits stated, L ACA' is always acute. [Mr. G. D. Wilson, B.A., sends the following trigonometrical solution of this Question: — Let the straight line bisecting AG at right angles meet AB in P. Then ^AGA' is acute if AP>JAA', i.tf., if JsecA>«. This condition becomes, after some reduction, sin A > tan |G, or 2»(«— c)>Jtf >ac — («— A)c or («— i) ff>*(3ff— 2«) or («— A)<j>2«{c-J(a + *)}. The angle ACA' is therefore certainly acute if c<i(a + *), *.«., if AB<DA'.] 14447. (H. W. CuRJEL, M.A.)— Tf /(a:) is finite and continuous for all positive finite values of x except a finite number of values, then r sin {/(a;)} rfj? and cos {/(«)} da; are convergent or divergent according as limit ^^' is infinite or finite ; Xm» ax I except in the case where limit f{x) = or wir, when I sin {/(a;)} dx may «»« Jo be convergent, and the case where limit f{x) = Jir or (2«+ 1) Jir, when «■« I cos {/(a;)} dx may be convergent. Solution by the Pkoposek. First consider the case where limit -^ • is infinite. Here it is obvious x>aD dx that to any assigned small quantity € there corresponds a finite quantity m such that f(x + c)—f{x) > ir if ar > w, and that the smaller € is the greater the least value of m is. Hence, if x>in^ as x increases sin ( cos 5 c(S j-^(^) C *^^*^8>®^ signs at intervals which continually decrease and the greatest of which is less than e. Hence r^vM < a series of terms whose signs are alternately positive and negative, of which Digitized by Google idx\ < € ; 99 the first and second are less than c, and the third, fourth, fifth, &c., are each less than the preceding one ; therefore therefore the integral is convergent. Again, if limit *^^' = a, a finite quantity not zero, then the interval COD dx hetween the changes of sign of J^ j/ W [ 's always finite, and the inte- gral can be reduced to a series of terms which are alternatively positive and negative, but never become indefinitely small ; hence the integral is clearly divergent. This reasoning can be applied a fortiori to the case where limit -~~^ = 0, except in the cases where limit *^^ 5 / W f = 0, «BQo dx «.<B cos ^ y ».«., when limit f{x) — iT" ^^J*\i , when the convergency of the in- tegral clearly depends on the form of the function /(x). Note on Quest. 14447. JBy 0. H. Habdy, B.A. Mr. Gubjbl's solution does not seem to me to be entirely satisfactory. *' The smaller € is, the greater the least value of m is '* ; but it does not follow, so far as I can see, that, when m is fiixed, the intervals between the points a^ at which sin/(:r) changes sign in (m, oo) will diminish continually as i increases, though they will certainly all be less than c ; and, even if they did, it would not follow that 1 ** sin {/(*)} d^ diminished continually as i increased. And, in fact, the theorem is un- true. df(^) f/(H) For, if -*—-' be continuous and > for all finite values of x. dx Now suppose /* (a;) = y"- a sin y (0 < /* < J) ; Joif^aBlTLI/ Then I sin {/(«)} dx converges or diverges with p^in^^ (0<M<J). Jo y^—aamy as Xf y become infinite together. But it is not difficult to show that this integral diverges. It would take too long to enter into the proof at present ; but it is easy to see the reason. The oscillations of the de- nominator coincide in phase with those of the numerator ; so that, to speak Digitized by Google 100 roughly, all the poaitiye elements are increased by the presence of the oscUlating term in the denominator, and all the negative elements de- creased. The second part of Mr. Gubjel's theorem is trae. Bat the integral may be determinate if /' (x) does not tend to a limit at all, for x ^ on . But it w a sufficient condition that f {x) tend to oo steadily ; i,e., with- out oscillations. [Mr. GuRJBL having seen the above, and having been requested to send any further remarks he might deem requisite, says : *' I have no- thing to add, as Mr. Ha&dy'& criticism is quite correct ; my proof only applies when /' {x) increases tteadily to oo , and consequently the theorem to a comparatively limited class of functions ; i.«., the condition given in the Question is not by itself sufficient for convergence, but that for divergence is sufficient."] 14509. (I* Arnold.) — Given two circles, one within the other, a point can be found such that the extreme portions of any right line cutting both circles shall subtend equal angles at the point. Solution by J. H. Taylor. Proof by Coaxial Circles. — PAF, QBQ', the given circles, are cut by chord PQQ'P'. Let O and C be the limiting points of the system of coaxial circles (Fig. 1) to which the given cir- cles belong, and letanother circle of the system touch PP' in M. Then, since the last circle, the limiting point C, and the circle PAP' are circles of the same coaxial system, PC : PM - P'C : P^M. (Pitt Press Muelid, p. 481.) Therefore CM bisects Z PCP'. Similarly QC : QM = Q'C : Q'M. Therefore ZPOQ = FCQ'. Similarly it may be proved that PQ, P'Q' subtend equal angles at the other limiting point 0. If a point be taken within the circle QBQ', and a chord PQQ'P' cutting both circles be made to revolve round this point, the segments PQ, P'Q' always subtend equal angles at both the limiting points. Proof by Inversion.— Let PAP', QBQ' (Fig. 2) be the given circles cut by the chord PQQ'F, and the circle OCF, with its centre on their common diameter, cut them both orthogonally. Therefore CM bisects Z QCQ'. Digitized by Google 101 [The points O, G, D, and the radii vectores in Figs. 2 and 3, are identical : but, for the sake of clearness, the iu verses are drawn apart in Fig. 3.] Fig. 2. Fig. 3. Invert with as pole and OT as radius of inversion ; then circle QBQ inverts into itself ; circle PAP inverts into a concentric circle.* The line PP' inverts into circle pqp\ touching at the pole a line parallel to PF. Since this circle cuts concentric circles, the arcs pq, j/q' are equal. Hence the angles pOq^ pfOq' are equal, and these are identical with the angles POQ, FOQ'. Similarly, it may be shown that the segments PQ, P'Q' subtend equal angles at G. [Mr. H. W. GuBJEL, M.A., solves the Question as follows : — Let AB be a straight line meeting the two circles in AB, CD ; the locus of points at which AG, BD subtend equal angles is the circle on E, F as diametpr, where E, F are the common harmonic conjugates of AB, CD. This circle cuts the two given circles orthogonally, and therefore passes through the limiting points of the two given circles, either of which is the point P required. The Proposer remarks as follows : — If PQQT' be produced to cut the radical axis of the two circles in B, and a tangent be drawn to either of the circles from R, then with R as centre and the tangent as radius a circle be described, it will cut the line joining the centres or its produc- tion in two points either of which fulfils the conditions of the Question. The points thus formed are the foci of involution. Of course they will correspond with G and 0.] 14526. (R. G. Archibald, M.A.) — With reference to the centre of the fixed circle, the corresponding tangent and normal pedal curves (positive or negative) of the cardioid are similar. * Because the circle OGF, which cuts the given circles orthogonally, inverts into a straight line cutting OD at right angles ; and a straight line cannot cut two circles, whose centres lie in a line at right angles to -it, both orthogonally, unless those centres coincide. (Fitt Frees Euclid^ p. 461.) Digitized by Google 102 Solution by the Proposbb ; J. H. Taylor, M.A. ; and othert. This theorem is evident, from the fact that the centre of the fixed circle of a cardioid is also the centre of the fixed circles of all its evolutes. Jfote, — The /irst positive tangent and normal pedal curves of a cardioid, with respect to the centre of its hase, are similar to the cardioid' s radial curve. {Cf. R. Tucker, Ed. Times, Feh., 1863 ; M, Timet Beprint, 1864, 1., v., 16 ; 1865, II., 28.) If a he the radius of the base, the polar equation of the tangential pedal may he written r » 3a cos ^0 ; of the normal pedal r = a cos } (0 — ir) ; and of the radial curve, r » \Sa cos \e. 14344. ('^' J* Barnivillb, B.A.) — Having ttn-i + t*n^i » 4Mn, prove that + + + + ... — — t 1 + 1 3 + 1 11 + 1 41 + 1 2 ' -J- +-^ +-!- + -^ + «i^ 1 + 3 3 + 3 11 + 3 41+3 •" 10' 1 + 11 3 + 11 11 + 11 41 + 11 2+v'6 4+V6 14+^/6 V3-r J: 1_ . J L_ + =i 1 + 2 2 + 2 7 + 2 26 + 2 *** 6* J L. +_J L^ +...=-i-. 1 + 1 5-1 19 + 1 91-1 2a/3 Solution by Professor SanjJLna, M.A. The first series 2 1.4 3.4 3.14 11.14 11.52 = 1+ 1 48 32. 4' 3M4a 11».14« 2 4-4.4-3.3.6- 14.14-11.11.6-"* = i + _LJ__LJ_J__ =-L+ 1 _ 1 , 1 ^ a/8 2 4-1-6-1-6-*" 2 4-(3-V3)*2 ^/B + l 2* The second ^1.1.1. 1 .i+JL+JL +i_+i + 4 44 2134 110774" 6 166 7956"* 14 574 "* 1.4 4.11 11.194 194.671 2.3 3.52 52.163 ... + -L+-i- + — 1- + ... 1.14 14.41 41.724 Digitized by Google 103 42 42.112 112.1942 ^ 1 22.32 32.522 4_ 44.3_ 112.18- 1942.3- . 6- 3=^.18- 622.3- 1 142 142. 412 ...+ 14- 142.3- 412.18- _1_J__1_JL +_L_±_ JL_1_JL +_i 1 1 1 4_ 3_ 18- 8- **' 6- 18- 3- 18- 3- " 14- 3- 18- 3-' 4-(9-6y3) 6- 18-(9-6a/3) l4-(9-5^/3) 1 3 1 ^3^/3 "5(V3-1) 10a/3 5(-v/3+1) 10* The third series 12 14 22 52 164 582 2142 7964 29692 110782 413414 may be written as the sum of five series 12 582 413414 14 2142 22 7964 "*" 52 ■*■ 29692"^"* "^164 "^110782 ■'■**• 1 32.42 32.1942 _ ^l 142 142.1532 L4-3.3.66- 194.194.11- 14- 14.14.11-153.153.66- 1 22.112 1 522 '" 2.11- 11.11.66- *" 1.52- 52.52.11- * 1 42.412 ...+ 4.41- 41.41.66- _1_J6__1 1_ _J 1 1 1_ _1 4 1_ 12-66- 11- 66- *** 14- 11- 66- 11- '** 22-- 66- 11- _1 1 1_ _1 16 1 52- 11- 66- "* 164- 66- 11- 1 .1.1 12-16(33-19V3)/6 14-(33-19V'3) 22-4 (33-19 >v/3)/6 1 1 52-(33-19a/3) 164-16 (33- 19 a/3)/6 3 .1.3.1. 3 76(2-%/3-3) 19(v'3-l) 38V'3 19(-%/3 + l) 76(2^/3 + 3) 38 * Digitized by Google 104 The fourth series a/6--2 ^ 4~a/6 ^ 14- a/6 52- a/6 194-^/6 2 10 190 2698 37630 2 2 15 95 1349 / I 1.10 10.19 19.142 142.265 / 2 2 I 5 5 19 19 71 / - ^e i JL IQ' 10". 19^ 19g.l42« \ I 10- 102.2- 192.8- 142^.2- *' » = :^ - i. - ^6 f 1 ) ye + a/2-2 2 2 110-(4-2a/3)/ "° 4 The result giveu in the Question seems inaccurate. The fifth series «l+i.+JL+_l- +...-(i+_L+ J- + -!_+...) 3 9 99 1353 V 4 28 364 5044 / «_LJ«J-J_ _^JL_L>i-JL \ 3_ 4- 4- 4- *** \4~ 2- 8- 2- ***/ 1 I a/3+1 a/3 1 3-3 (2- a/3) 4-(4-2a/3) 6 6 6' The sixth series may be similarly summed. [Mr. H. J. WooDALL, A.R.C.8., makes the following remarks regard- ing the last series : — ttn-i + Mn+i = 4wn gives 1, 5, 19, Vl, 265, 989, 3691, 13775 ; whence series is ±_J.+i_l + J L+_i L_. 2 4 20 70 266 988 3692 13774 Comparison with series — ± — + — ± — + ... «i *2 ^z ^4 (C. Smith, AlgebrUy p. 466) and its transformation _ 1 a^ a.? gives the equivalent continued fraction 1 2^ 42 202 702 2662 9882 35922 2+ 2+ 16+ 60+ 196+ 722+ 2704+ 10082 + ^12 111 2+ 1+ 2+ 1+ 2 + " N^^ rT2-VrT = ^ = IT^ gives y=. A/3-1; therefore series = ?— ^^""^ = ^^^^.^ 2+ a/3-1 2a/3~2 + 2 2a/3 -" Digitized by Google 105 14541. (John C. Malbt, F.R.S.)— H the roots a^i, x^, x^, x^, a?,, x^, Xjy Xi of the equation sfi-PiX^+p^-'PsX^+P4X*''P63fi+PePP^''P'jX+p^ «= are connected by the relations Xi + x^ + x^ + x^^ x^ + Xq + Xj + Xs and XiX^x^ ^ x^ftfspsjX^y (a) prove Pj = ^/pB (Pn-kPiP^ + iPi^)* ((^Pr-^Ps^Ps+^PiPs? = (Q«'-4Q4) {pj'-ipsQ^), where Qj = P2-JM Q4=P4-^iP7K^^PB)-^^Psf Qe =J»6-Qj -/Ps; (^) solve the equation. Solution by the Proposer and H. W. Curjel, M.A. We have Zti - Sa^s = Ji?i ; 3:12:2^3^4 =- x^x^XjXg = Vpg. Let now Xfirrj = Wi, Xcs^e = "2* S^i^a^s = ^'i, S^s^e^y = «^2> andwefind Wi + W2=Q2 (1)» therefore «i + «2 = l^-ii?ii?2 + ii'i' (2), ttiU^ + iPl(Vl'^V2) + 2Vps'-P4j U^V2 + UiVl=Pi-Pl^/pt... (3, 4), (ui +«3)V>8 + «i«2 "i'e; therefore ri«?2 = 0^ (5); K + V2)v'i98=i?7 (6)- From (6) and (2) we find Pj = Vpfi{p3-^PiP2 + iPi^}y one of the required conditions. We have from (3) and (6) uitit ^Q4 (7). Hence, from (1) and (7), 2«i - Q2+ v'(Q2»-4a»), 2«2 = Q2- ^{Q^^-^Qa) ; and from (5) and (6) 2vi = PjI Vps + ^^iPjVPs- -tOe), 2^2 = i?;/ v^g- ^{Pt^lP8-^Q>6)* Substituting in (4) for t^, u^, v„ i^j, and rationalizing, we find the second required condition and the roots of the given equation are the roots of the quartics 2jc4_j,j^+ {Q^i V(Q22-4Q4)} x^^'-{p7IVps±AP7VPs-^Q6)} x+2^/ps » 0. Fui'ther Note on Quests, 14520 and 14670 {triangle isosceles if bisectors of base angles equal) . Dr. J. S. Mackay observes : '< A direct proof of this Question, by F. G. Hessb, will be found in the London, Edinburgh, and Dublin Philo- sophical Magazine (Fourth Series), Vol. xlvii., pp. 364-7 (1874), and another by Mossbruggkr in Grunert's ^rcAiv, Vol. rv., pp. 330-1, 1844. "The Question seems to have been proposed for the firat time by Prof. Lehmus, of Berlin, to Jacob Steiner in the year 1840. A proof, with (extensions of the Questiou, was given by Stbineb in Obblle's Journal, Vol. xxvin., p. 375-379, and many other proofs will be found scattered through the volumes of Gbunbrt*s Arehiv. See, for example, Vols, xi., VOL. LXXIV. o Digitized by Google 106 XIII., XV., xYi., XVIII., XX., xLi., xLii., &c. The proof given in Tod- hunteh's Euclid is Stbinek's." Mr. B. Ghartbbs sends the following foither direct proof , which seems quite satisfactory : — Place the triangle AEG in the position A,BI> ; then a circle will circumscribe BDAA,, and the bisectors of BAD, BA^D will meet at G, the middle of the arc BGD, by a well known rider. Since a; = y = iA4 4>, and therefore = GAA|, therefore a circle circumscribes F'FAAi; and, since the chords of it FA, F'Aj are identical, therefore {Eitc, HI. 14) GA = GAi, or ^ + iA = a + ^A ; therefore 6 = <^, or ABC is isosceles. Cor. — If the base angles be divided in a given ratio, then, if the dividing lines terminated by the opposite sides be equal and intersect on the bisector of the vertical angle, the triangle will be isosceles. Professor K. J. San j Ana remarks: *<In the algebraical solution by Mr. R. Chart&es (p. 73), when the factor c^b is removed, there is left 1 ^ hf^{a^■^■l^ + c^■^.be^¥2ea^^^1ab) (a + A)2(a + t'}=' ' or a* + 2cfib + 2a^c + aH- + aV + baHc + Ul^e + 4aJc3 + ^c + 2ft2c2 + jc3 = 0. Removing the factor a + b + e, we get a^ + a^ + ah + Zabc + b'^c + *c2 « q. This result is more readily obtained by Euclid vi. a. and trigonometry. We may write it thus a2 (2») + 2abc + be (2») = 0, or a^ + bc + abcjs = 0, i.e. , a^+bc :=— 4Rr. Thus there is no solution but c = b. If the external bisectors terminated by the sides produced be equal, we shall get by a similar process a^^a^-ah + 3abc-b^e-bc^ = 0: Digitized by Google 107 this may be written -a=. 2(»-a) + 2a*<?-*<j.2(«-a) « 0, or cfi + be — abel(8^a)y i.e., a' + Atf«4Rri. Thus in this case the triangle is not always isosceles.' ' The accompanying diagram, supplied by Mr. Gubbnstrbbt, affords an instance in support of this view. Mr. TxjCKBR sends the following :— The above Question reduces to this : Construct a triangle on a given base, with a given vertical angle and given bisector of that angle. Let BG be the base ; on it describe a segment cod- taining the given angle, and let F be the mid-point of the arc remote from A. PEA is drawn so that AK = given bisector of angle. Now, by Euclid iii., from P there can be drawn only one other line = PLA' ( — PEA) ; hence there can be only two con- gruent triangles, fulfilling the data of the problem ; heuce, &c. Editorial Note on Quests. 2810, 3250 (the late Professor Sylvbstbu). Dr. Thomas Muib, of the Educational Department, Capetown, draws attention to the fact that a paper of his, ** On a Class of Alternating Functions," intheTrans. Eoy.Soc. ^rfiwi.,Vol.xMiii., pp. 309-312 (1887), contains a wide generalization of the late Professor Sylvbstb&^s theorem as enunciated in these Questions, which are identical. A solution by the late W. J. CuR&Ax Sharp, M.A«, is to be found in the Reprint, Vol. XLVU., p. 21. 14384. (W. H. Salmon, B.A.)— If a chord of a circle S subtend ii right angle at a fixed point O, show that its envelope is a conic S' ; and that of the common tangents S and S' two pairs intersect on the polar of O, one pair at the centre of S, and the other on a fixed line. Show also that O has the same polar for S and S'. Solution by the Proposbr. Reciprocate with respect to O, and we have the well-known theorems : — The chords of intersection of a conic and its director circle which are not diameters are directrices. A conic and its director circle are concentric. 14682, (Professor E. N. Barisxbn.)— Soit ABC un triangle. Calculer le rayon d'un cercle tangent h. la fois au cercle inscrit et aux cdtes AB, AC. Digitized by Google 108 Solution by Professor A. Dkoz-Farnt ; Jambs Blaikib, M.A. ; and many others^ Solent r ]e rayon du cercle inscrit, pa, p\ les rayons des deux cercles, le premier int^rieur au triangle, tangents k la fois au cercle inscrit et aux cot^ AB, AC ; I, O, (y les trois centres ; T et T les deux points de con- tact. Menons les tangentes communes qui coupent le cdt^ AB en D et jy ; dans les triangles semblables ODI et ID'O' on a OT : TI - IT' : TO', pair^rip'^; d'oil pap„ = r3. Dans les triangles rectangles ODT et TDI on a OT - DT tan J (180- A), DT = TI tan J (180 - A) ; d'oii p« = r{tenJ(x-A)}-, p;=r/{tan2J(ir-A)}. Rbma&qub. — De la relation 2 {tan J (ir— A) tan i (ir— B)} = 1 on d6duit ais6ment les deux relations pour les groupes de circonferences tangentes dans les trois angles du triangle 14683. (Professor P. LEVEaRiBR.) — Etant donnes un triangle ABO et un cercle O, on demande de couper le triangle par une transversale a$y, telle que les ceroles oiSC et ayB soient ^gaux et que leur axe radical soit tangent au cercle O. Solution by Professor A. Droz-Farny and R. F. Davis, M.A. On salt que les quatre circonferences ojSC, oyB, ^A et ABC se coupent en un m^me point F, foyer de la parabole inscrite au quadri- latere. Les deux cercles aiSC et ayB 6tant egaux, on a FB = 2r sin FoB, FC «- 2r sin FoC ; done FB - FC ; F est done le point milieu de Tare BC. On en d^duit la construction suivante : — On m^ne de F ime des tangentes k O ; cette droite coupe BC en o. La circonference FaB d^rmine y sur AB. 14540. (Professor G. B. Mathews, F.R.S.)— Prove that, if Q « 2 j*(3»*i)'4. 5 g^% then Qi = AA_^, -» -00- 16kk where X, X' are the moduli into which k, k' are transformed by the change of q into ^. Solution by H. W. Otjrjbl, M.A. If ^0 = n (1 -^2«), q,^n(i + q^% ^3 = n (1 + ^«'-»), 1 1 1 y3 = n(i-y*»->), 1 Digitized by Google 109 and Qo, Qi, Qs, Qs are what q^ qi, q^ q^ become when q ib changed into ^. Then Q = 1 1 « 1 « 2!2j . 1 + 2^^3,.' ^Us^ Qa»* 1 therefore Q" - ?*?2"/Qj»«, whence the result in the Question. therefore X»A' (1 + aOV^^Bick'Q" ■ince qiq^q^ « 1 and QiQsQs « 1, which is clearly not equal to unity. [The PaoposBR observes : ** The proper result ^ 16ic«' easily comes from Mr. Ctjrjbl's result Q}^ — ^^lr> Q5M because, with Webbu's notation {El, Funct.j pp. 63, 149), ^2" - ?*/'» («) « ^*, Q»^ - y*/^ (3«) = ^, whence Q» = ^^^."1 13746. (LioNBL E. Eeay, B.A.)— Find the area of the triangle with sides equal to the medians of a given triangle. Show whether such a triangle is always possible. Solution by W. J. Gbbbnstkbbt, M.A. Since m^ » \^/{2l^^'\■2<^—a'^y the area may be found in the usual way. The condition -%/(2*2+2tf»-a2)+ -/(2c3 + 2a2-.i3; > V(2a» + 2*»-<J») reduces to b-¥c>a. Therefore, &o. Digitized by Google 110 14543. (Professor Mohlrt.) — The greatest number of regions into which n spheres can divide space is 2n + ^n (n— l)(n— 2). Solution *y H. W. Curjbl, M.A. The corresponding theorem for space of m dimensions is easily proved. If in space of tn dimensions we call the hypersphere of highest possible dimensions a sphere, and a space of (m— 1) dimensions a plane, the theorem may be stated : The greatest number of regions into which n spheres can divide a space of m dimensions is {2(«-l)(n-2) ... («-m)}/m! + 2 [l + (n-l) + {(»-l)(«-2)}/2! + ... torn terms] = 2Vfn,n + Um,n (using the notation of Quest. 13395, Vol. lxviii., p. 39). The greatest number of regions into which m planes can divide space of m dimensions is shown in Quest. 13395 to be Vm,n + t^m, n ; if we invert with respect to a point in none of the planes, we see that the same is true of n spheres passing through a point. But the number of regions that are made to vaniidi by making the n spheres pass through a point is easily seen to be Vm,n ; therefore the gpreatest number of regions If we put m = 8, we get the result stated in the Question. 14546. (Professor Nbubero.) — Si les angles des triangles ABC, A'B'Cy veriiient les egalit6s A + A' » t, B » B', les cdt^s sont'Ues par la relation aa' ^^ bl/ + ccf , Solution by R. P. Paranjptb. B.A. ; Professor Ignacio Beybns, Lt.-Ool. du Genie a Oadix ; and many others. With centre C describe a circle, radius CA, cutting AB in D. Then, obviously, DBC is similar to the triangle A'B'C' of the enunciation. Since the relation to be proved is homogeneous vn a^ b, c as well as «', *', cff we may prove BC2 = A02 + BA.BD, which is a well-known proposition (see Cabby's Sequel). Hence the required relation. 14547. Professor Lanohornb Obchabd, M.A., B.Sc.)— Showjthat, if n be any positive integer greater than unity, l8 + 2» + 3a + 48+...->-w8-(l6 + 2S + 35 + 4S+...+n5) ^ ^ (l + 2+3 + 4 + ... + n)a-(l6 + 2* + 3* + 4S+. ..+«*; " Digitized by Google Ill Solution by H. W. Curjel, M.A.; Lt.-Col. Allan Cunningham, R.B. and many otheis. If 8r =- "2 «^ «3 = »i', and »5 = i (2«2 + 2rt-l) 53, therefore ""i^ = ^(2.2^2.^1)-! ^ ^ 14536. (!• Arnold.) — In any triangle the radius of the circumscribed circle is to the radius of the circle which is the locus of the vertex, when the base and the ratio of the sides are given, as the difference of the squares of those sides is to four times the area. Solution by F. H. Peachell, B. A. ; Baounath Rati, B.A. ; and W. J. Greknstkbet, M.A. The locus of vertex, when the ratio of the sides is given, is the semicircle on DF, where D, F divide AC internally and externally in ratio of the sides. Therefore CD : DA - a : c, or CD : AC = a : c + a ; therefore CD = fl^*/(c + a), and similarly CF = «i/(c— a) ; therefore radius of locus =» abcl{(^—a^). Radius of locus : circum-radius = dbcjif—a^) : abe/iA = 4A : (^—a^. N'ote on Quest. 6144 {Reprint, Vol. lxxiii., p. 113). By Rev. Charles Taylor, D.D., Master of St. John's College, Cambridge. For "circumscribed" read **circum-inscribed.*' This term is used in the Aneient and Modem Geometry of Conies (pp. 139, 140) in the sense circumscribed to one curve and inscribed to another. 14467. (G. H. Hardy, B.A.)— Prove that r {<^(aJ-a)-4>(ar-3)}<«a;=«(*-rt){<^(oo)-4>(- »)}, provided each side of the equation represents a determinate quantity. Deduce the values of r dx j,r dx J.« coBh(a;--a)cosn(a;— ft)* ]_„ 8inh(«— a) sinh(a;-ft)* Digitized by Google 112 Solution by the Pbopobbr. f {^ («— a)— <^ (^— *)} «?«? = Urn I QH-a /"H-i -I ^ (u) du-- <l> (u) du -H-a J-H-6 J = lim If A (u) du- \ d> lu) du\ H-w LJh-6 J-H-6 J «(J-a){4>(oo)-<^(-a>)}, if both sides of the equation be determinate. If ^ M '^ tanh u, <t>{x-a)-<l>{x--b) 8inh(^~a) ^^ ' ^^ ^ cosh («-a) cosh (ar-*)' and L dx h-a cosh {x — a) cosh {x—d) sinh (*—«)' If ^ (u) =: coth u, we find •r. sinh {x—a) sinh (a? —A) sinh (b--a)' It is easy to see that the proof remains valid, although in the latter case only the principal value of the integral is determinate. Euclidean Froofof Pascal's Theorem. By R. F. Davis, M.A. Let ABCDEF be a cyclic hexagon. Produce AB, DE to meet in G, and AF, CD in K. Let BO and the circumcircle of DFK intersect GK in H, P respectively. Then (1)P,D,B,G are eoncyclic, for 180°-DPG = DPK = DFK - ISO^'-DFA = DBG. Also(2)P,F,B,Hare eoncyclic, for FPH - 180°-FPK « FDK « FBC =» 180°-FBH. From (2) BFH « BPH = BPG -BDE from(l) = 180°-BFE; therefore EF passes through H. . . j . j [In most geometrical conies Pascal's theorem for the come is derived Digitized by Google 113 from the theorem for the circle by conical projection. The above proof is strictly EucUdean: it neither involves anharmonic ratios (Gasbt's Sequelf Nixon's Cfeometry Revised, &c.) nor Mbnelaub' Tranwenal Theormn, which is employed by Catalan.] 14312. (Professor N. L. Bhattachabyya.)— A parabola slides be- tween the two foci of an ellipse, such that the focus of the parabola always lies on the ellipse. Find the envelope of (1) the directrix, (2) the axis, of the parabola. Solution by A. F. van dbr Hkyden, B.A. ; H. W. Curjbl, M.A. ; and Rev. J. Cullbn. Let SL, S'L', CH' be perpendiculars to the directrix of the parabola, in any position, from the foci and centre of the ellipse. Then SL + S'L'«2CH'. But SL+ SOi' = SH + S'H [if H is the focus of the parabola] - 20 A ; therefore OH' = OA ; therefore the envelope of the directrix is the auxiliary circle of the ellipse (1). Let H, H' be corresponding points ; HT, H'T tangents. Then SH : S'H « ST : ST = SL : S'L'. Hence, if H is the focus, TH' is the directrix of the parabola. Let the Digitized by Google 114 axis ZH cut the axes of the ellipse in P and Q. Then, if CD be con- jugate to CP, we have CP « (a-*)/*N'D and CQ = (a-*)/* HN ; therefore PQ ^ a—b. Therefore the envelope of the lais ZH of the parabola is the four-cusped hypocycloid generated by the rolling of a circle, radius i (a— &), upon a circle, centre 0, whose radius is a—b (2). [Mr. CuKJBL remarks that there are two parabolas with focus H passing through S, S', the directrix of the second being the other tangent from T to the auxiliary circle ; consequently the part of the axis intercepted between the axes of the ellipse ^ a±b ; and its envelope is the two four- cusped hypocycloids a;* + y* = (a±i)*.] 14453. (Professor A. Droz-Farny.) — Constraire un triangle, dont on connait la base, la hauteur correspondante et sachant que sa droite d'EuLBB est paranoic au c6te donne. Solution by Rev. J. Cullbn ; Professor Jan db Yribs ; and many others. Let BC be the given base ; through its mid-point a, draw aH perpendicular and equal to the given height ; trisect aH in O, and through O and H draw parallels to BO ; with O as centre and radius OB, describe a circle cutting the line through H in A and A'. Then ABC (or A'BC) is the required triangle, since oA is tri- sected by the line through O in G, Eulbr's line being OG. 14503. (Robert W. D. Christie.)— Show that the primitive roots of 331 are connected with the associated roots by the modular equations r"*= » mod 331, rf =-»2mod331, where r is a primitive, and r^ an associated, root ; also w signifies one of the roots ol tx^-kl^ 0, namely, i {l + a/(-3)} or ^ {l- ^(-3)} ; and generalize the result. Solution by the Proposer and Lt.-Col. Allan Cunningham, R.E. We have r^"* + l = 0mod6m + l. + 1 ^rr-rT + l -0mod6m+l. ir^-r?) = 0mod6m+l. -*1 Digitized by Google 115 .-. r'" + rr-l -0mod6m+l and r**"-r"*+ 1 -: mod6m+l. r^ + r? = mod 6m+l or (r»»)" + rf « mod 6m + 1 ; i.e,, w2_jp2 -_ mod6m + l. Thus r** = «mod6m+l, rj* --w^ modem +1. 14238. (Rev. W. Allen Whitworth, M.A..) — If a straight line be divided at random into any number of parts, the expectation of the square on any part taken at random is double of the expectation of the rectangle contained by any two of the parts taken at random. [This can be proved by algebra without the integral calculus.] Solution by Rev. T. Roach, M.A. The expectation of the square of any part = 2«'/{»> (**+!)} (Whit- WOKTH, Choice and Chance, Supplement). The expectation of the product of any two parts = 8^l\^n{n + 1)} . Therefore ex. squares — 2 ex. product. 14529. (Lt.-Col. Allan Cunningham, R.E.)— Showthat ^= 1 (mod p) where a; = ^ . 210<Q, Q = ^^ i? = Q • 210*Q+ 1 = prime. Solution by the Proposer. Here {p- 1)/8Q = \ . 210^^ = x, and \x is even. Also i? = 1 + (3' • SIO^^)-* ; therefore 2\0^^ .q^ = -l (modi?) ; therefore (210^ . q^)l' = 2100*. ^ = ( - l)i' = + 1 (mod p) ; therefore (2.3.5. lyp-^)!^, ^ = + 1 (modi?). Now p = 12+(2102Q.^2)2= (^2,2102<i-l)2+2(y.210<i)3. These two partitions of i? suffice to show that 2, 3, 5, 7 are each 8-ic resi- dues of p when p is prime, [See two papers by Mr. C. £. Bickmorb "On the Numerical Factors of («'»—!)»»" in Messenger of Mathematics , Vol. XXV., p. 18, and Vol. xxvi., pp. 16, 18, 21.] Therefore (2.3.6.7)^''"^^^® = +l, leaving q' = +l (mod p) . 14524. (R. F. Davis, M.A.)— If A, B, C, D be the angles of any convex quadrilateral, sinA{sinC + sinB— sin(A + D)} : sinC {sinA + sinB— sin(C + D)} — sinA + 8inD-sin(A + D) : sinO + sinD— sin(C + D). Digitized by Google 116 Solution by F. H. Pbachbll, B.A. From the question A + B + G + D « 2ir; therefore Bin ( A + B) = — em (0 + D) , &c. Also 8ini(A + B) = sm|(0 + D), &c. Now 8mA{BiiiG-i-8iiiB — 8iii(A-i-D) » 8mA28m|(B + C)cos|(B-C)-i-sm(B + G)} = 2smA8mi(B + G){co8i(B-C) + co8i(B + 0)} = 48inA8mi(B + G)coBi6co8iG. Similarly, sin G {sin A + sin B — sin (G + D) } — 4 sin G sin i (A + B) cos ^ A cos ^ B. Therefore ratio of the two expressions is sin |A sin i (B + G)/sin iG sin i (A + B). Now sinA + sinD— 8in(A + D) « 2sini (A+D) {cos 4 (A-D)-co8i (A + D)} :=4sin^(A+D)8iniA8iniD » 4 8in^(B + G) sin^Asin^D. Similarly, sinG + sinD— sin(G-»-D) » 4sin|(A + B)sin|Gsin|D. Therefore the ratio of these two expressions is the same as that of the first two. 14132. (!• Arnold.) — Describe a square in a given sector, haying two angular points on the arc and the otiier two on the radii. Solution by the P&oposbr. (1) Let OEG be the given sector ; make Oe = O*, and join be. Take Ime O'A' « OE, and on it describe the segment O'B'A' containing an aline angle O'B'A' equal to a right angle + O^;^. Also on it describe another segment O'G'A' containing an angle O'G'A' equal to \ right angle + Oeb^ Fig. 2. Fig. 1. and complete the circles. Draw O'K, cutting off a segment KA'B'O' containing the angle KB'O' equal to Oeb or Obe. Also draw the line Digitized by Google 117 O'L, cutting off from tlie other circle a segment containing the angle CC'L equal to Obe. Join KL, and produce O'C, O'B' to meet the circle described with O'A' as radius in the points E', G'. Then is B'C the side of a square described in the sector E'O'G', which is equal to the sector EOG. Next draw GB parallel to eb and equal to CB'. Lastly, draw BA, CD each perpendicular to CB, and join AD. ABCD is a square, and it is inscribed in the sector EOG. The analysis of the problem is derived from Fig. 1. (2) When the sector AOB is greater than a semicircle. Make 0/= Oe, and join/<?. Onfe describe the square edef. Join 0^;, and produce to meet the circumference in C. Draw OF parallel to ef, and FE parallel to fe. Also draw CD .parallel to FE, and join DE. CDEF is a square, and it is inscribed in the sector AOB, which is greater than a semi- circle. The quadrilateral figures CFEO and cfeO are symmetrical, and, since fe = fcy then FE — FO, and, the angle cfe being a right angle, EFG is also a right angle. Hence CDEF is a square, and it is inscribed in the sector AOB greater than a semicircle. The foregoing are proposed as geometrical solutions of the problem. [For another solution of this Question, see Vol. lxxi , p. 103.] Fig. 3. 6498. (J- W. Russell, M.A.)— Show that is divisible by (a + d + cf + abe. Solution by H. W. Curjbl, M.A. ; Professor Sanjana, M.A. ; and J. O. Watts. If, in the given expression, we put a + * + c=— « {abe)^ = x (say), where ft>' =» 1, it becomes ' \ {x-ay ) - n(« + J)» (6;r»-2«J?!ir|S\ _:rn(<. + J)« [6:r+2iI^±^J I {x—a^ ) (. x-a ) = xU{a + b)^ ^bx + 2(-a+ ~\ I -=2a;2n(a + i)2(2 + 2-^) = 2x^U{a + b) {2 {x-a) (x^b) (x^c) + -Za {X'-b){x^c)} = 2x^n{a + b) {2x^-2x^ + 22bcx + 2x^ + a^-:^a{b + c)x^Zix^} = 0. Therefore the expression is divisible by (a + b + e)^ + abe. Digitized by Google 118 14538. (Salutation.) — Arrange in one plane two triangles of given dimensions in such manner that two specified vertices may coincide, and the other four he concyclic. Solution by the Pboposbu and H. W. Ctjbjbl, IA., A. ^ jointly, Analy8i8,^ljet ABC, ADE (Fig. 1) he two triangles fulfilling the conditions, A heing the common vertex, and B, C, D, E heing on the Fig. 1. Fig. 2. circumference of a single circle, of centre O. Let BC = 20}, DE = 2a« ; FH = Aj, GI - Aa ; HA = f/j ; lA = rfj; OF = arj, OG - x^. Then we clearly have and {Xy ■(«). 08). The construction is therefore easy. Thus (Fig. 2), let IG — Aj, GK = W-a^^^y found as in Fig. 1, where DK = BF = Oj ; also let IL = [d^—d^^y found in the same way, GK and IL heing perpendicular to IG. Moreover, let the arc QSP he drawn from centre K with radius = FH = Ai, and let IL he produced both ways, IM -= IL. Then, with centre on IG, draw any circle passing through LM and cutting the arc QSP, say in Q, P ; join PQ, and produce to meet IL in N ; on KN de- scribe a semicircle cutting QSP in S, and produce KS to meet IG in T. Then KT = x^, the required height of O above BC (Fig. 1). For T is the centre of a circle passing through L, M and touching the arc QSP ; therefore TL = TS = arj— Aj, and TI = h^—x^; and we have the con- ditions stated in (o) and (0) fully carried out. {Cf. Euc. in. 36.) Of course the example given is a particular case out of many, but the principle of construction is the same in all. Digitized by Google 119 14109. (Professor Cochez.)— On donne les deux paraboles y^ = Ipx. x^ = 2qy. Lieu des points M tels que les tangentes issues de ces points a chaque parabola soient rectang^aires. Solution by W. J. Grebnst&ebt. y—mx—pftm =: is a tangent to y' = 2px ; my-^x-^-h = is a tangent to a;' = 2qn \( k = ql2m ; therefore for intersection we have ^- ^ y ^ —1 ^ x- ^n—y . pm + q qm^p 2m (1 + m*) q{l+ »»^) * therefore m » (^^^ and I x + ^-^±M A ^(P^^ ^ ^ ^ q ; {qx—py) \ qx—py I qx—py therefore 2 {px + qy) (a;- + y^ + {qx—pyY = is the locus required. 6701. fProfessor Cbofton, F.R.S.) — Find a function X such that, whatever F be, F (<j«D) X = F (a) X. Solution by Rev. J. Cullbn. Let u=^-e"; then e'Dy^dy/du, Therefore F (D') U « F (a) U. Now consider a term prX'^ in F (») ; hence D'*"!! = a*'V, which is satisfied by U — e"» or X = «-««"*. [Mr. CuBJEL remarks : — '* It is necessary and sufficient that ^DX = «X, !>., X = C^-««"'."] 6632. (Professor Gbnbsb, M. A. ) — Find the envelope of the asymptotes of conies inscribed in or circumscribed about a given quadrilateral. Solution by W, J. Gkebnstbebt, M.A. Take the asymptotes of the equilateral hyperbola xy + K = (1), which passes through the four points, as axes of coordinates. Then ax^ + by^ + 2ffx + 2fy + e^0 (2) is any conic through the four points, with axes parallel to the coordinate axes; and ax^+by^ + 2\xy + '2gx + 2fy + e + 2\K ^ (3) is the general equation to conies through the four points. If (1) were one of the hyperbolas through the four points, the axes of (3) would be oblique, but the form of (3) would be unchanged. The locus of the centres of (3) is found to be ax" -by^ + dx—ey =z (the nine-point conic) (4), Digitized by Google 120 and it is worth while noticing that, if (3) is a parabola, X « ± ^/ad, and the directions of the axes are given by x^/a^ ±yv^d. Hence the asymptotes of (4) are parallel to the axes of these two parabolas. The tangent at the origin is ^a;+/y = (6). y ^ tnx + n meets (3) at infinity only if 2\m + bm^ 4 e; «= « An + nbm +fm +^. Hence, eliminating X, we get «*m' + 2/m' + 2^w— «» — 0; and, writing fn a —u/vj » » 1/v, we get for envelope in tangential coordinates 2uv(fu~ffv) + bu^-av^ = (6). This is a curve of the third class and fourth order, 'and has three cusps. It touches the line at infinity where bu'^ = av^y i.e., the points of contact are on the axes of the parabolas mentioned above. It also touches the axes of coordinates. The third tangent from the origin has for ang^ular coefficient —u/v = — ^//, «.<?., this tangent is the conjugate harmonic of (6) with reference to the axes of coordinates. If (3) is a series of equilateral hyperbolas, (6) touches the line at infinity at the circules, and is a three-cusped hypocycloid. When the quadrilateral is inscriptible, we have, if 0^, ^2» ^3» ^4 ^® *^® angles made by the normals with the axis of x, $1 + 6^ = 03 + 64. Therefore, in V^a^m*- 2a$b^^ + *»» (aV + b^$^-e^ r- 2a«a/3m + a^fi^ = we have (mi4m3)/(I— fWiWj) = (»»3 + w4)/(l— W3W4) {a). Writing this pH^-Q) = »V(1 -«), we have p + q ^ 2j8/o, pr + q + s = (a^a^ + b^g^- e^yb'a^ JW + jT « 2a^fi/b^ay qs = a^ff^/b^a^. Therefore p/il-q) = r/(l-*) = {2&la)l{2-{q + s)}. Therefore d-.Ul-^) 26 ^ aV.^g^^^ ' a{2-{q + ,)} W <(l~(y-)2g ^ q{l-s)2$ ^2a^B ]- (A). «{2-te + *)} a{2-{q + s)}^ l^a qs = a2)3V(J2^») Eliminating q and s from (A), we get a' \ b^a^ l\ b'^a I \ b^a^ J This gives )3 = 0, a^fi^- b^a^==0; ▼is., y = 0, aV-J%2«o (1), (2), and {x^ + t/)^aH^—c^{a^ + b^) {ll^x^-ah/) = (3). We see that, if a^0^ = i^a-, then mym^m^fn^ = 1, which may be incon- feistent with (a). Hence the locus is composed of the axis major of the ellipse and (3), which is the pedal of a^x^—h^y^ =- (a< — ^) with respect to the origin. [For another solution of this Question, see Vol. iii., p. 80 (I860).] Digitized by Google 121 6642. (J* B- Ha&ris, M.A.) — When two or more spherical soap- bubbles, blown from the same mixture, are allowed to coalesce into a single bubble, prove (1) that we obtain for the radius of the bubble an equation of the form ri, r^, ... being the radii of the bubbles, and a some positive quantity; and (2) verify (what one would infer also from physical consiaerations) that this equation implies a reduction of the total surface. Solution by H. W. Cukjbl, M.A. ^t i^i, p^^ p^t &c., be the pressures within the bubbles, and p the pressure within the single bubble of radius x. Then P^ = PiT^ -^-p^r^ + i?3» 38 + . . . , and p = 2tlx + », Pi^ Itjri + », &c., where x is the atmospheric pressure and t is the surface tension ; therefore a^-r^^-r^-r^^...^ 2tl'K{r^'\-r^ + r^^-k- ...-x'^, where 2^yx ^ a is obviously positive. If the total surface is not reduced, ^r^—x^ IS zero or negative. Therefore x is greater than any of the r's ; therefore 3^ ^ xl.r'^ > 2r3 ; * therefore a^'-'S,*^ is positive, which is impossible if 2r'— ^' is zero or negative. Therefore the total surface is reduced. 14651. (Professor G. B. Mathbws, r.R,S.)— Let o, B be any two given complex quantities, and let t be such that {a-¥tff)l{}.'¥t) is real. Prove that, if t — x + iyy the locus of (Xy y) is, m general, a circle. How is this to be reconciled with the fact that the line joining two imaginary points (a, )3), (7, S) contains only one real point? I. Solution by Professor E. B. Elliott, F.R.S. A *<real" line contains 00 > points, of which 00 are real; a **real^ plane 00 ^ points, of which 00 ^ are real ; and a '* real " space 00 * points, of which 00 8 are real. The connector of two imaginary points on a *' real " line is that line, and contains all its real points ; that of two imaginary points in a ** real '* plane, but not on a '< real " line in that plane, contams one real point ; that of two imaginary points in *<real*^ space, but not in a ^'real'^' plane, contains no real point. Finft on a '* real " line, if ^ -i- »y : 1 be the ratio in which a real point divides the intercept between two points whose distances from a real origin are a •¥ ia\ h + iV, we have, from the reality of a-¥iaf-¥{x-¥iy){h-\' ib') a + bx—b'y _ a' + b'x+ht/ l+«+«.v ' 1+j; y i.e.y the equation of circular form V (a;2 + y^ + «) +«' (1 +a;) + (*-a) y - 0, VOL. LXXIV. H Digitized by Google 122 as the only relation limiting x and y. One of these may be taken at will, subject to the requirement that the resulting quadratic for the other have real roots, and the infiniteness of the number of ratios of division for real points is apparent. The solution a; + l— 0, j^aOis excluded. Next in a "real** plane, if («i + i«'i, aj + ti's) and (^i + ift'i, b^-k-iV^ be the coordinates referred to *'real" axes of the points, the intercept between which is divided in the ratio x-^-iy l 1, we have, for the reab'ty of the dividing point, b'l {x^ + y2 + a;) + a\ (1 + a;) + (*i— ai) y = and i'a (x^ + y' + x) + a\ (1 + a-) + (*s — «s) y = 0, which give, besides the irrelevant a? = — 1, y « 0, which refers to tho point at infinity on the connector, a single pair of real finite values of x and y, and so a single ratio of division. And generally in " real " space, referring to three " real " planes, we have three such equations of circles with a common point x ^^l^y— 0. They have as a rule no common second intersection ; and so as a rule there is no real dividing point of our intercept. II. Solution by the Proposer. Let o =a a + ^i, fi ^ c + dif t ^ x + yi. Then a + tB ^ {a + cx-^dy) 4- (b -h dx -k- cy) % l + t (l+a;)+yi and this is real when (I + x) {b + dx + cy) ^y (a + cx^dy) «= 0, or when d {3^ + y-) + {b + d) x + (c—a) y + b = (i.). Suppose now that 7 = 0' + b'i, B >= c' + d'i. Then {y + tB)/{l + t) is real when d'(x^ + y^) + {b' + d')x+{i/--<^)y + b' = (ii.). The circles (i.) and (ii.) intersect at the Jixed point (—1, 0), and at another point whose coordinates are rational functions of a, a', &c. The &rst point gives ^=—1, and this makes {a + t0)l{t+l) and (7 + ^^)/(^ + ^) hoth infinite ; the other leads to the one real point on the line joining (a, 7) to 03, 5). 14028. (G. H. Hardy.)— Reduce the evaluation of f S^pIiL<PJ^ }ol + 2tC08(l>-\ t' where Pf q are integers, p<q and ^< 1, to the integration of a rational fraction. Prove, in particular, that Jo 1 + 2^ cos <^ + ^2 ~ i^i ^7^ ' and deduce (and also prove independently) that ftan-i (2^)-,^ = 8 tan-i^^ tanh-i^/. Jo \ l-<2 / 8m4<^ Digitized by Google 123 Solution by the Pkoposbb. ^^ rT2rir*+«=-'-^2l(-o"oo8«^, t<i, 1 — ^ (a 1 \« + o «— a/ ) if a<l. That is to say, _ Bin ai r C-,p ^*~'<f^ ,, p T'^fQ If a » 1?/^, this is q Bin (p/q) IT C ipl9) {^'^ r^dr .pi, ft^^' r^"'"'^ ) 1-^ I Jo 1-T« Jo l-r' )' If 1? » 1, ^ =* 2, we get ^ 2 tanh-^ ^^t l+t a/^ * (This integral and the corresponding integral with a sine in the numer- ator may also be easily obtained by contour integration applied to f«-^ \ That is to say, r cos i^rfi^ ^ 2 tanh-i Vt ^ (2 ten-i ^^t). Jo l + 2^C08<^ + /2 dt^ ' Similarly, T ^^^^^^ , - 2 tan-i Vt 4- (2 tanh-i ^^t). '' Jo l-2<oos4> + e2 ^v / Adding and integrating from to t, we find f'tan-i (2^?^^^) JiU = Stan-i^^tenb-i-v/*. Jo \ i-f^ I 8ini4> This may be verified independently, as follows : — Tten-i (2l8in^\ J<^ ^ 2 2-^ p sin (2n 4 1) <ft ^ Jo V 1-^2 /sini4> i2« + lJo 8in^<^ ^ = 8f(l-*-.%V^l)2^1 = 8[^+(l-i + i)i^ + (l-i + i-| + ^)it^..] = 8tan-»y*tanh-i'v/^ Digitized by Google 124 14473. (W. S. CooNBY.) — Construct the triangle, being given any three of the following six points : — the centres of the squares described externally and internally on the sides. Solution by the Proposer. Let Oj, Oj, Og, »i, ws, ws be the centres of squares described externally and internally on sides of ABC. By Quest. 13716, or easily from figure, O1O2 is perpendicular and equal to CO3, for CO2 and Cwi are proportional to AC and BC, and t OjCwi =- Z C ; there- fore AO2CW1 is similar to ABC ; therefore O^^ = AO3. Similarly OjWi « AOj ; therefore A02a>i03 is a ^rallelogram ; as are also BOi^sOs, COgo^Oi, AwjOiVj, "Rfofi^i, and CwiOsttfj ; therefore evidently OOa = Oi02> and C»8 is iJso equal and perpendicular to <ai<o^ ; therefore, if Oi, O2, Og or x^i) x's* ^ ^6 §»i'^6i^} the per- pendiculars of the triangles being drawn, the construction is obvious in each case. If »i, O^, O3 or 0|, <a^^ w^ be given, the completion of the parallelogram in each case gives A. If Oi, »|, w^ be gfiven, B and C are known, which disposes of the twenty cases ; therefore, &c. This construction shows that the triangles O1O2O3 and teiw^u^ are so related that the perpendiculars of each bisect the sides of the other, and pass through A, B, C, for CO3 bisects »xa>3> ^^^ ^<^z bisects 0|0). 14329. (J. A. Third, M.A., D.Sc.)— L, L', M, M', N, N' are points on a conic. LL', MM', NN' form the triangle ABC ; MN', NL', LM' the triangle A'B'O' ; and M'N, N'L, L'M the triangle A"B"C". The straight line AA'A" meets BC, B'C, B"C" in X, X', X" respectively ; the straight line BB'B" meets CA, C'A', C'A" in Y, Y', Y" respectively ; and the straight line CC'C" meets AB, A'B', A"B" in Z, Z', Z" respec- tively. Show that the following are triads of concurrent lines : — YZ, Z'X', X" Y" ; ZX, X'Y', Y 'Z' ; XY, Y'Z', Z'X" ; YZ, Z'X", X'Y' ; ZX, X"Y', YZ' ; XY, Y'Z", Z'X' ; and that the points of concurrence lie on a conic. Solution by the Proposer. The pairs AB and A'B', BC and B'C, CA and C'A' meet on the same Digitized by Google 125 Pascal line u. Let P be the point of intersection of AB and A'B\ Join TX'\ PY''. ThoD, since m is a diagonal of the quadrilaterals ANA'M' and BLB'N', PX" and PY" are harmonic conjugates of u with respect to PA and PA', and therefore coincide. Thus, AB, A'B', and X"Y" are concurrent. Similarly, B'C, B"C'', YZ are concurrent, say in Q. Hence the triangles X'Z'A' and QZN are copolar with respect to C Therefore Z'X' and YZ intersect on the same line as the pairs A'B', AB and X'A', ON. Thus the first triad consists of concurrent lines. A similar proof holds for each of the other triads. Again, the triangles XYZ and X'Y'Z' are obviously in perspective. Therefore the six points of intersection of the sides of the one with the non-corresponding sides of the other lie on a conic. 14173. (D. BiDDLB.)— The sides of a triangle being given. a> b> Cy draw a line parallel to one of them, such that the quadrilateral formed shall have the maximum area possible in proportion to its perimeter, and find both area and perimeter. Digitized by Google 126 Solution by W, 0. Stanham, B.A. Let PQR be a triangle whose sides are a, fit 7- Parallel to PQ ( « «) draw FQ', so that P'Q' = \a. Then, if A and 8 denote area PQR and ^(a + fi + y) respectively, and if a/s = /*, area PFQ'Q - A(l-\2) (i), perimeter PP'Q'Q = 2« (1-x) + 2\a...(2). The ratio which is to be a maximum is therefore (l-\2)/(l-x + /iX), which for any value of A. is clearly a maximum when a = c, the least side. Diflferenti- ating, the value of A. which gfives a maxi- mum is found to be [1/(1-^)-{1/(1-m)'-1}»]. Substituting this value of A. in (1) and (2), and putting a ^ e, fi-. s ^ ^(a-{-b + c)y the required values are obtained. R c/8, 14519. (Professor U. C. Ghosh.) — Find the sum of the products of the terms of the geometric series a, a^y «', a^, ..., «**, taken r at a time, r being less than n. Solution by Lt.-Col. Allan Ounnikoham, R.E. Let »i=a + a2 + «8 +...+««, Sr = a*' + a"'' + a^+ ... + a*^*-^ Let Sr = required sum of products of the terms of «i, taken r together. Let Xr == sum of terms in Sr containing a particular term a". Let Sr = sum of terms in Sr free from a particular term a^. Let 2 denote summation with respect to x. Here *i « 5«* = a . («** - 1) /{a - 1) , and Xr = a* . Sr-i, Sr = 2 {a* . S^. i) . Hence Sj = «i, S[ = Sj— aj*, Sa = 5flMSi-a*) = Si.2a*-2«2* = S,«i-*2 = «i-»2, Si = Sa-a* . Si', Sg = 2a' (Sg-a* . S() = 5 («' . Sj-a^* . Sj + a^*) ; therefore S3 =* siSj— ^a^i + «3 = »J— 2«i»2 + *3» Sa = Ss-a*. si, S4 = Sa'Sa = 2 (a'Sg-a^^Ss + aS'Si-a*') ; therefore S4 = s^S^ — ^a^a + ^gSi — *4 = «} — 3«J«2 + «2 + 2*153 — s^. The law of formation of each sum (Sr) from the preceding (Sr.i) is now clear, all the terms being of equal weight (r) Sr = »iSr-i-»2Sr-2 + *38'-3-&C. ... +(-l)»-l*,.. Digitized by Google 127 14549. (J- A.. Third, M.A., D.Sc.) — K is a conic circumscribed to a triangle ABO ; P is a point on it ; Q is the isogonal conjugate of P with respect to the triangle ; R is the point where PQ meets K again ; L, M, N are the points where AR, BR, CR meet BO, OA, AB respec- tively ; X, Y, Z are variable points, Y lying on QM and Z on QN, such that the pairs AY and AZ, BZ and BX, OX and OY are equally inclined to the bisectors of the angles A, B, respectively. Prove that the locus of X is QL, and that the locus of the point of concurrence of AX, BY, OZ is E. The construction usually given for Kiepert's hyperbola (see Oasby's Analytical Oeometry^ p. 442) is a particular case of the foregoing. Solution by Rev. J. Oullen. Taking ABO for the triangle of reference and P to be the point (:r, y, i), we have K = 2//^ » 0. It is easy to see that R is the point axHy^-7^)ll=^...^ (1). Now, if X, Y, Z be the points (a^i, y^, «i), (a:,, y*, «,), and (arj, y,, a,), then AY = ift53-7y2 - 0, AZ = )823-7ys = 0, .... These lines are equally inclined to the bisectors of A, ..., if ViVz - HH> HH = ^i^3> ^1^2 = y\y2' L, M, and N are given by putting, successively, a — 0, i8 — 0, and 7 = in (1). Qis cu; =... = .... Hence ^2» y2» ^2 1 1 I X y z I QM = ;r2(y2-z2)' 0, = 0si?2a:2 + ^2y2 + *-2«2» z^x^-y^) and QN =^a:3 + ^3ya + r3Z3 =« 0, where p^y ..., are the coefficients of x^y ..., when a similar determinant is expanded. Eliminating x^ and y^y we have p^z^y^-^- q^x^z^-^r^x^y^ « 0. Using x^x^ - y\y^ and 2^2 ^2 = 0, we get X\ (»V3-^2?3)-yi?aP2 + «i'*2/'3 =- (2) as the locus of X ; or, expressed in terms of x, y, z and /, m, n, (2) may be written 1 1 1 X y z « 0, ^ ^ which is the equation of QL. From the equations AX = $zi—yyi - 0, BY = yx^-az^ =» 0. OZ = ayj-i3«, = 0, we find the point of concurrence to be ayiz^ = jtoja^ = yx*Zi, Therefore l/a+m/fi + nly >= lyiz^ + tnzix^ + nx^yi (3)- Digitized by Google 128 EliminatiDg x^ and z^ by means of the preceding equations, we find that tihe dexter of (3) is the sinister of (2) multiplied by the factor y(«'— a;^/af(y'— «^). Therefore the point of concurrence lies on X//« = 0, t.*., K. j]The Fboposer adds : '' The theorem is true not merely when the pairs P and Q, AT and AZ» &c., are isogonal conjugates, but also when they are isotomic conjugates, and generally when the members of each pair are derived the one from the other by any reciprocal transformation (a, ^, y becoming l/pa, l/qfi, l/f'y). Quest. 14516 could be similarly generalized."] 14563. (H. Enowlbs.)— From a point T tangents TP, TQ are drawn to the parabola y- «= iax. Prove that when the circle TPQ touches the parabola the locus of T is the parabola y^ « 4a (2a— a;). Solution by R. P. Paranjpye, B.A. ; G. D. Wilson, B.A. ; and many others. Let T be the point It;. The equation of PQ is yri == 2a {x + 1). The tangent at the point where the circle touches the parabola is parallel to the line yq — — 2ax (since PQ and the tangent are equally inclined to the axis on opposite sides), and is therefore y = -(2a/,,)ar-ii,. Therefore the equation of the circle is of the form {yi,-2a(a: + ^)} \yn + 2ax+ \ri^] +\{y'-4ax) - 0. Sinee this is a circle, ri^ + \ = ^4a*; therefore x = — (ly* + 4<i') ; therefore the equation of the circle is {yri-2a{x + i)}{yr, + 2x-i-W} = {v^ + 4a^){y^-4ax). But this passes through T (|, ri) ; therefore, omitting the factor vi^—ia^y we get fn' + 2^1 = »?2 + 4^a . therefore ij* « ia {2a - fl, whence the proposition. 4953. (W. J. G. MiLLEK, B.A.) — A king is placed at random on a clear chess board, and then, similarly, (1) a bishop, or (2) a rook. Find, in each case, the charce that the king is in check so as to be unable to take the attacking piece ; and find also (3) the chance of check, with or without the power of taking, for any combination of two or three of the pieces. [If we estimate the powers of the pieces (a) by the chances of simple check, as investigated in the solution of Quest. 3314, Seprint, Vol. XV., pp. 50, 51, in January 1871 ; (/8) by the chances of safe dieck, as shown in an interesting paper by H. M. Taylor in the Fhilosophieal Magazine for March, 1876 ; (7) by the results given in the JBerliner Schaeh- zeitung, we have the relative values of the knight, bishop, rook, queen as (a) 3 : 5 : 8 : 13; (jS) 3 : 3} : 6 : 9J; (7) 3 : 3J : 4J : 9J.] Digitized by Google 129 Solution by Professor SanjAna. d c e c c d d e b b b * e d c h a a a a a a 1 b b e c c h a a c b a a a a b a a b ; b e e b a a b e e d e b b d d e e € e d Simple check, — 1. Elnight. — When the king occupies one of the 16 squares marked a, this piece can check from 8 squares ; on the 16 marked by from 6 squares ; on the 20 marked Cy from 4 squares ; on the 8 marked d^ from 3 squares; and on each of the 4 comer squares, from 2 squares : alto- gether 336 squares. Thus the chance of checking the king is 386 -^ 64 X 63 «X. 2. Bishop. — When the king occupies one of the four squares marked a, this piece can check from 13 squares; on the 12 marked by from 11 squares ; on the 20 marked Cy from 9 i, Enioht. squares ; and on each of the 28 border squares, from 7 squares: altogether 560 squares. Thus the chance of checking the king is 560 -^ 64 x 63 » -f^. 3. Rook. — This piece com- mands 14 squares when placed on any square. Hence the chance of checking the king is 14 -^ 63 = f 4. Queen. — This piece com- mands, in any position, the squares commanded by both bishop and rook. These being mutually exclusive, we have 560 + 14x64 squares. Thus the chance of checking the king is (560 + 14 X 64) -h 64 X 63 = ^. Multiplying each of these chances by 36, we find the relative values to be 3:5:8:13 (a). Safe check. — 1. Knight.— As this piece id never close to the king when checking, the chance of a safe check is still -j^j> 2. Bishop. — For each of the squares marked a, by Cy we have now to diminish the checking positions by 4 ; for the 24 border squares by 2 ; for the 4 comer squares by 1. Hence the chance of a safe check is (4x9 + 12x7 + 20x54 24x5 + 4x6)-^64x63 = 1^^. 3. Eook. — For each of the internal 36 squares, we have now to diminish the checking positions by 4 ; for the 24 border squares by 3 ; 1 1 c c c ^ c 1 <r 1 e b b b b C c b a a b C 1 c b a a b ^ i c b b b b e I ' e c c c € 1 c , 2. Bishop. Digitized by Google 130 for the 4 comer squares by 2. Hence the chance of a Bafe check is (36 X 10 + 24 X 11 +4 X 12) h- 64 x 63 = i. 4. Queen. — ^The chance of a tafe check is iWr + i ^ -^h- Multiplying each of these chances by 36, we find the relative values to be 3 : 3i : 6 : 9J (i8). I have not seen any of the papers referred to, and nothing is given as to the principle on which the method of the Berliner Sehaehzeitung is based. SFor further remarks on the Relative Values of the Chessmen, see . XL., pp. 85-97.] 14213. (Robert W. D. Christie.)— If A» s= m»— «.m»»-2+ — ^ m"-* ^ ;«"-«+ ... 2 ! o ! for all integral values of m and n, then X*"-A,»X" + l«(X3-mX+l)(«iX^-Va2X^*"^ + a3X^^-*+...+«3X+l), where <in = a series allied to A„. E.g. — If w = 5, « «» 3, then «6-110a;»+l = («2_6a;+l)(a:4 + 5a:3 + 24a:2 + 6a;+l). There are two other allied theorems for positive values of A,* and m ; it is required to establish them. Solution by the Proposer . Let o * m+ V'(w2-4)/2, jB = »»— >/(/»'— 4)/2 ; then we have A,»«a" + i3» 2! 3! r!«-2r! (which is the generalized form of the ** continuant " series 1, 1, 2, 3, 5, 8, 13, &c.). Also -t —A" = m«-»-n-2.m'»-3+ * ^^'^''^ m"-'^ ... n dm 2 ! «— r— 1 .»_o^_i r! «-2r-l! (i.) We have to show that s^'^—AhX'' + 1 = {x^—fnx+l) (M), where (M) =ooara*-2 + aia;2»»-3 4^a2a;2»-4+...«„_irc»»-i + ... + aaa;2 + ajar» + ooa;». On multiplying this series (M) by x^^mx-^lf we find the law of the coefficients, excepting the middle one, is ctn—fnan + aH-2i fti^d for the middle one man - 2a,i-i. Now the values of these are respectively and o" + /8*». Hence the theorem. (ii. , iii.) The theorems x^** + A,»a:» +1 ^ (x^ + mx + l) (M) when n is odd, and x^—A*Xn + 1 =• {x^+mx + 1) (M) when n is even, are solved ina similar way. The above is a generalization of several known theorems, e.g., m - 1 and n « 6A;±1, we have a?*»— 4f« + 1 = {x^—x+l) (M) ... (1), « = 6A;±2 „ x^ + x** + l = (r«-a;+l)(M) ...(2), w= -land««3A;±l „ x^+x''+l = {x^ + x+l){U) ... (3). Digitized by Google 131 For odd powers w& can establish cognate theorems, «.^., (iv.) ai2»+i-A«(a^+i + a^) + l = (a;«-«Mr + l) (M), where An is the »th term of a cognate series. (y.) In Fbbmat's theorem we easily get q in terms of p (a prime), vi£., 2^-'-l ^pq^p{'^-^ P-^.P'^.p-i.p-d, P-^ oP'6, p-i.p- "2!"^ + ---3j- Other theorems are : — (vi.) ^^' + 1 = (a;2-a:+l)(M) if «= 3A: + 1, (vii.) (^*^ + l){x**+l) ={x^-^x+l){M.) if «- 6A; + 2or6^ + 3, (viii.) s^*^-{x''*Ux*') + l = (»2_a:+l)(M) if « = 6A:or6Ar-l, (ix.) «^*^ + 2a:"*U23:" + l = {x^ + x + l) (M) if n = 3A;+1, (x.) (ic***^-l)(a^-l) - {x^ + x+l)(M) if « = Zli or 3A;-1, &o.f &c. 14495. (R. C. Ahohibald, M.A., Ph.D.)— The points Pi, p^^ p^, where any three parallel tangents to a cardioid cut the double tangent, are joined to the centre O of the fixed circle. Prove that the angles P\Op^f pjdp^ are each equal to 60°. Solution by the Proposer; H. M. Taylor, F.R.S. ; J. H. Taylor, M.A. ; and Professor SSanjAna. The cardioid is generated by a point P of the circumference of a circle, with centre rolling on a fixed equal circle with centre O ; their point of contact E formerly coincided with P at the point S on the fixed circle or base; S is the cusp of the cardioid. The line 8P makes an angle B with OS produced, and is parallel to the line OEC, which, when produced, meets the generating circle in B. Hence Z EOS = ECP = a, and zOBP«Ja. But the line BP is tangent to the cardioid at P ; it therefore makes with OS produced an angle |0. Hence the line joining the points of ; contact of parallel tangents to a cardi- ' oid subtends an angle of 120° at the cusp. The line OS is produced to A, so that 2SA = OS ; as is well known, the line through A perpendicular to OA is the double tangent of the cardioid traced by P. The perpendicular to OB at its middle point D meets the double tangent in jt?i, the same point as that where the tangent Digitized by Google 132 to the cardioid at P meets it. For, joining Opi, we hare the triangles PiOD, PiOAy equal in all respects, since OD » OA. Hence iPiOA = jt?iOD - DBpi » ie. Since, then, Z SO^i » ^0 = } Z PSA, we deduce the required result from the theorem just given alM>ve. It may be noted, finaJly, that the points O, S, Pi, P lie on a circle. 14484. (Professor A. Dkoz-Farny.) — On joint un point A de la directrice d'une parabole au sommet S de cette demidre. AS coupe la courbe en un second point B. La tangente en B rencontre en P le diam^tre de la parabole men6 par A. Tirons la deuxidme tangente PC. La droite CB est normale en B ^ la parabole. I. Solution by Lionel E. Heat, B.A. ; H. W. Oukjbl, M.A. ; and Professor SAiuiNA. It is a known theorem that, if THB be a focal chord and BSA be drawn, AT is parallel to 8H. Hence, if ASB be drawn and AT be diameter through T, TB passes through H, the focus. Therefore tangent at T meets PB at K, on the directrix, and TK is perpendicular to BE. But TE is parallel toCB. Therefore GB is perpendicular to PB. II. Solution by J. H. Taylor, M.A. ; and F. H. Pbachbll, B.A. Equation of parabola y^ =^ ^ax (1). Take point A (— «, b)\ equation of AS : y » — (bja) x. Coordinates of Bare U^jb^, -ia^/b. Equation of tangent BP : {-2a/b)y = a; + 4a8/*» (2). Therefore m = — i/2a. Hence, coordinates of P are Therefore equation of chord of contact BC is by ~2a{x-2a-ia^lb^) (3); here w' = 2a/*, and mm' = - 1 ; therefore Z CBP = 90°, or CB is the normal at B. Digitized by Google 133 14554. (I^* F* Dayib, M.A.) — Given a conic and a circle having double contact, prove that the envelope of a variable circle, whose centre lies on the first and which intersects orthogonally the second, consists of two fixed circles. Solution by C. E. McYickbk, M.A. Let OQ be radius of fixed circle, P the centre of the variable circle, PR, PG the tangent and normal at P to the conic. Draw OR perpendicular to tangent, meeting the focal radii SP, PH in T', T respectively. OT : OH - GP : GH. OT : OS = GP : GS ; therefore OT.OT' : OS. OH = GP» : GS.GH =:: constant =. 0Q« : OS . OH ; therefore OT. OT' « 0Q«, proving that T, T' are inverse with respect to the fixed circle. It follows immediately that PT (or PT') must be the radius of the variable circle. Again, ST'/SO « SP/SG = Ije = SQ/SO ; therefore ST' «» SQ. Hence the variable circle, centre P and radius PT', touches the fixed circle, centre S and radius SQ. Similarly, it touches another, centre H and radius HQ. The complete envelope is therefore a pair of circles concentric with the foci and co-intersecting at Q, Q'. 13754. (Professor TJmbs Chandra Ghosh.) — A<i, B^, and Qc are the perpendiculars from the vertices of the triangle ABC on the opposite sides. O is the orthocentre of the triangle ABC. Oj, O], and Os are the points of intersection of the opposite connectors AO, bo ; BO, ca ; CO, ab of l^e tetrastigms AJbOe, BaO^, and GbOa. AOji, AOj cut BC at a^ and oj; BOj, BOs cut AC at ij and b^', and COj, CO2 cut AB at c^BXi^Oi. Prove that (i.) Qc (l/eCi + l/ee^) + Bb {l/bb^ + l/bbi) + Aa (l/a«i + Ijaa^) ^ 60c/ Ac Bb/Cb Aa/Ba; (ii.) C(?(l/ctfi-l/<JC2) + B*(l/**i~l/W2) + Aa(l/<iai-l/rtfl',) * 0. Solution by W. J. Gheenstkbet, M.A. In the triangle AaC we have aoi.Cb.AO =» Oa.Ab.afi « Oa.Ab.{aO-^aa{). Substituting obvious values for CA, AO, Oa, AA, aC, we get, after Digitized by Google 134 arranging, Aa/{aai) ^ ban C/{aai) » tan B + 2 tan C. Similarly, ^aj^aa^ » tan C -I- 2 tan B. Therefore 2Aa (1/001 + l/aoa) = 23 (tanB + tan C) » 62 tanA » 6n tanA » SnAa/Ba, and XAail/aai-llaa^) » 2(tanB-tanG) » 0. 6731. (R- A. RoBBRTS, M.A.)— If from (1) ^rV^' + yV*'--* - 0, (2) a«j;« + 4»y'-(«' + **)'-0, tangents be drawn to aj'/a^ + yS/iS— i = S = 0, prove that they form, with their chord of contact, a triangle whose (1) centre of gravity, (2) intersection of perpendiculars, lies on S » 0. Solution by Professor SanjAna. Let the tangents he drawn from T, and let a and fi be the eccentric angles of the points of contact. Firstly, let x, y be the coordinates of <^e centroid. Then T is given by xla = cos i (a + o)/cos J - a), yjb = sin J (jB + o)/cos J (i8 - a) ; hence, from equation (1), we get 1 = 4cos*J(/3-o), or C08j(i8-a) = J. Now 3iP — acosa + acos/3 + acos J(8+a)/co8j (i8-a) = 3acosJ(/3 + o), on reduction ; so also 3y = 3isinJ()3 + o); therefore ^/a^+ySp = i, or the centroid lies on S. Secondly, let ;r, y be the coordinates of the ortbocentre. As T now lies on the locus (2), we get fl4cos2J(i8 + a) + i4Bin3j(^ + «) ^ (a2+*2)2cos2 J (i8-o) (3). The perpendiculars from a and /3 on the tangents at )3 and a are respectively (y— b sin a)l{x—a cos a) = a sin fijb cos jB, {y-b sin 3)/(a;— a cos /3) =* a sin a/i cos a ; 80 that ;r = J- ^^^i^4 {«' + «' sin a sin i8 + 6^008 a cos )3}, C0Bi(i8-a) *- ■' and y«4- ?^5i_0^±iO/i8+^cosacos/3 + a«sinasin/3}. b cos 4 (/3— a) *- ^ It will be fouud that Digitized by Google 135 Bimilarly, y = -^^^ cos i (3 -a)' Therefore x^ld' + iPIl^ - a^coB^j^i^ ^^a)-^M8in^^(3-^«) ^ , Iheretore x'/t^ + i^ /o^ (a^ + b^y^coBm^^a) or the orthocentre lies on S. 6679. (Rev. T. R. Tekry, M.A., F.R.A.S.)— Show that the value of the continued fraction ^ ,— * ^ ^ —, ^^^ere N = fw and 1+ 1+ 1+ 1 + *■ (2r-l)(2r+lj * (1 +d;)» + (l-j;)»' Solutiofi by Professor Saj^jana, M.A. Take Eulbr's expression for tanwa;, viz., wtana; (w^-l^) tan"a; («2--22)_tanrr I- 3- 0- ■"' and put tan x ^ isf \ thus we get — ttan»(tan-*«d?') = ; — - -^ — ^ . ' — ... ^ 1+ a+ + 1+ 1+ 1 + Hence the given continued fraction = — « tan n (tan-^ ix). Let a? = tanhu; then tan"^ ix — tu^ and the fraction = —ttannm » tanhfiM _ Oj tanh H + C3 t anh^ u >- Ca t.anh^ u-\- ... ~ I + C3ta!ih'M + C4tanh^M + ... ^ i {(1 +taTihw2"-(l--tanhfi)"} ^ (i + ^)'»- (!-,;,;)>> i{(H-tai.hw)"+(l-tanhM)*'} (1 +3?)" + (1-ar)" 14235. (R. Tucker, M.A.)— ABCD is a square. P, Q, R are points on AB, AD, BC, respectively, such that PQR is an equilateral triangle. Find maximum value of triangle. 8how also that locus of intersection of AR, BQ, as P moves along AB, is a parabola. Solution by Rev. T. Mitcueson, B.A. ; and the Proposer. (1) Let Pi be mid-point of AB. With Pj as centre, radius AB (= 2«), describe circle meeting AD in Qj, BC in Ri. Complete triangle P|QiR|, which is obviously equilateral. Digitized by Google 136 Join Pi and Oj (the mid-point of QiRi)» and through draw any intercept QsBj and OPj perpendicular to it, and complete triangle PsQ^Bs. Since L QiOQg = z PjOPj (by Euc. vi. 4), OQ2 : OQi = OPj : OPi or OQj : OPj = OQi : OP, ; therefore (by Euc. vi. 6) OPiQ, and OPjOq are similar, and OPjOg is equilateral. Let Q3 be at D, and complete (as before) the equilateral triangle PsOgHs. This latter is the maximwn^ while PiQiR, is the minimum. As L PaOa A = L RsQsC, each = ^k and QaRs = 2rt sec -^ir, hence area of PsOsRa = 2a' sec' -^-k sin ^ir. » P3 Pz P, B (2) Let ^1 «> Q1Q3 « RiRst and take O as origin. The equation to AR is y-yi = {(aA/3-yi)/-2a} («-«), thatof BQis y + yi = {(aVZ+y{)/2a}{x + a). Eliminating y,, we obtain a:'V3 « fl(2y-aV3), which is the locus of a parabola whose axis is coUinear with P,0, whose vertex cuts PjO at its mid-point. Digitized by Google APPENDIX. ON THE GEOMETRY OF CUBIC CURVES AKD CUBIC SURFACES. By W. H. Blythb, M.A. When investigating the properties of a cubic surface on which lie twenty-seven real straight lines I found that it was quite possible to define all such surfaces as the locus of the common vertex of six tetrahedra on six fixed bases among the volumes of which existed certain relations as to ratio, and that the plane sections of such surfaces could all be described by Grassman's properties of a cubic curve. It cannot be asserted that all plane cubic curves can actually be described by this method, nor is it fair to say that every cubic curve — for example, y- = a^ — has a real point of infiexion, nor yet that every cubic curve may be projected from one curve or one series of curves, but theoretically, and by use of imaginary points, certain projective properties true for one cubic are true for all. To put bhortly, by means of analysis, what has been assumed, the equation of every cubic surface having twenty-seven real straight lines may be put into the form afiy — /xScvj, where o = 0, i3 = 0, 7 = 0, 5 = 0, € = 0, ry = represent the equations to planes and ^ is a constant, and that any one of its plane sections has an equation of the same form taking the equations of straight lines in a plane, instead of planes in space. The geometry of cubic curves and surfaces has been thoroughly investigated by Kb ye, in what is known as **' Geometry of Position,*' but not, to the best of my knowledge, by any one by Euclidean geometry. I suggest the following scheme : — I. Plane cubic curves treated geometrically. (1) Circular cubics and their projections. (2) Central cubics. 11. Cubic surfaces having real straight lines. (1) Without nodes. (2) "With nodes. III. Cubic surfaces on which lie no real straight lines and their sections. [Note. — In the following paper the first part of the first section only is given. I have also many interesting propositions on circular cubics, and a few on the second section, but little or nothing on the third, to which must be added a full investigation into the five different kinds of plane cubic curves from one of which any other can be projected by means of shadows, and the forms they take when projected. J VOL. Lxxrv. I Digitized by Google 138 1. Definition (Grassiun). — A point P moves bo that the straight lines PA, P£, PC, joining it to three fixed points A, £, G meet the sides of a triangle BDF in points lying in a straight line ; then the locus of P is a cubic curve. It will be found more convenient to assume that AB, CD, £F meet in a point E, and that A, B, G, B, £, F lie on a conic. This is only equivalent to assuming that the cubic curve has one real point of inflexion. 2. To show that PTABEF] « PrCDEF].— Let PA meet BF in L, PG meet DF in M, and PE meet BD in N ; then it is well known that the ratio compounded of the ratios of the alternate segments of the sides of the triangle BDF made by LMN is unity. Observing that LB : LF : : area APB : area APF, MF : MD :: area GPF : area GPD, KD : NB :: area EPD : area EPB ; alto area PLB : area PAB :: PL : PA, area PLF : area PAF :: PL : PA, so that LB : LF :: area PLB : area PLF; and similar results for the points M and N, therefore the ratio compounded of the ratios area APB : area APF, area GPF : area GPD, area EPD : area EPB is unity ; from which, by Euclid vi. 1, we arrive at the result P [ABEF] - P [GDEF]. 3. It is possible from the result of the last article to determine when G&assmam's definition leads to the degeneration of a cubic curve into a conic and straight line. For example : suppose the straight lines AB and EF coincide so that the four points A, B, E, F lie in one straight line ; then the ratio P[ABEF] is constant, so that the ratio P[GDEF] is also constant. Therefore P moves in a fixed conic through G, D, E, F, or in the straight line ABEF. 4. A series of conies is described through four fixed points A, B, E, F, and the straight line EF is projected to infinity, one of the conies becoming a circle ; then the series of conies becomes a series of circles described through two fixed points A, B. Then, if P [ABEF], P' [ABE Fl be two anharmonic ratios determining any two conies of the series, and Gi, Gj be the centres of the circles corresponding to the conies, P[A.BEF] : F[ABBF] :: OGi : OGj where O is a fixed point for the series. Let M be the intersection of the diagonals of ABEF and let mtif be one of the diagonals of the quadrilateral circumscribing one of the conies and touching it at A, B, E, F ; then the direction of mW is the same for all conies of the series. Let mtif meet two conies of the series in P and P' and EF in p, and let t and ^ be the poles of EF with respect to the two conies ; then it will be found that P [ABEF] : F [ABEF] : : [«>m] : 1 . This problem being projective, it is more convenient to suppose ABEF a rectangle in the first case, when the above statement becomes evident. Digitized by Google 139 Now project EF to infinity, and the conies into circles; p goes to infinity, t and fy the poles of the line at infinity, become the centres of the circles, and m a certain fixed point for the series. Therefore P [ABEF] : P',[ABEF] :: OCi : OCj, for the values of anharmonic ratios are unaltered by projection. 5. It is clear from the above that, if two series of conies be described through two sets of four points A, B, E, F, G, D, E, F, and each conic of one series intersects one of the other, so that P [ABEFJ : P [CDEF] is a constant ratio, then P traces out a curve which projects into one described by the intersection of two series of circles each passing through two fixed points A, B and C, D, where, if L is the centre of any circle of one series and M the corresponding circle of the second series, then EL : HM in a fixed ratio, where E and H are fixed points in the straight lines along which L and M move. 6. One very simple well known case is that in which the ratio P[ABEF] = P[ODEF]; for now E and H may be taken as the middle points of AB and CD after projection, and EL : HM :: AB : CD. Therefore, if EF is projected to infinity and the conies into circles, P becomes the locus of tho intersection of similar segments of circles described upon two fixed straight lines AB and CD. Digitized by Google 140 7. If we take Gkassman's definition of a cubic curve and the limitation that A, B, C, D, E, F lie on a conic and that AB, CD, EF meet at a point, and further project the conic into a circle and EF to infinity, we obtain the following construction for the projection of the cubic curve : — (Art. 6) A point P movea so that z APB = CPD, where AB, CD are two parallel chords of a circle, and, therefore, bisected at right angles by a fixed straight line OH. The curve is symmetrical with respect to OH, which may therefore be called its ** axis." AC and BD meet at S in OH ; S is a point on the curve. A circle with centre S and radius SH, where SH=^ = SA.SC, may be called the "con- struction circle" of the curve. AB and CD meet OH in L and L'. OY is drawn at right angles to OH so that O and S are equidistant from the middle point of LL'. J and J' are two points on LL' such that CJ' is parallel to AJ, so that, if circles are described with centres J and J\ and distances JA and J'C, then we have similar segments on AB and CD, and the point T at which they intersect is on the curve. Let SA meet the "construction circle" in Q and B, and complete the figure as indicated, by perpendiculars and parallels to OH. TS is supposed to meet the curve again in T' and the construction circle in q and r, and perpendiculars are drawn from T, T', q,r to meet OH in V, V, m, and mf. V, V, m, and m' are related to the chord TT' as L, L'y M, and M' are to AC. 8. To prove that T^.Tr : SH» :: OS : OV.—We find by £uc.y Bk. ii. that SC2 - ST3 = 2 J'S . VL' ; 8 A2 - ST^ = 2 JS . VL. .-. SC2-ST2 : SA2-ST2 :: J'S.VL' : JS.VL :: SL'. VL' : SL. VL. .-. SC2-SA2 : SA2-ST2 :: SL'. VL'-SL.VL : SL.VL ::LL'.OV:SL.VL. Now, if we remember SC.SA = SH^, and that SA : SC : : SL : SL', we find that "SH^-SA^ : SC^-SAS = LL' : SL. Compounding this ratio with that just found above, SH2-SA2 : SA2-ST2 : : OV : VL ; .-. SH2-ST2 : SH2-SA2 :: 0V + VL : OV : : OL : OV. We know that SH2-ST2 = Tq,Tr and that SH^-SA^ : SH2 : ; OS : OL ; .-. Tq,Tr: 8W : : OS : OV, and OV - TU. Cor. — This result may be written : The edges of a rectangular solid block being equal to TU, Tp, Tr, the solid is of constant volume. It is clear that OY is an asymptote to the curve, for as T becomes more distant from S the rectangle Tp,Tr can be made to increase indefinitely, and, therefore, TU diminishes without limit, but can never actually vanish. 9. A straight line meets the curve in three points, two of which may be imaginary. Digitized by Google 141 Let a straight line meet the construction circle in e and / and the asymptote in d. Then, if a point T be on this straight line and also on the curve, and if TU be perpendicular to OY, we must have rect. Te.T/: SH^ : : OS : TU and TU : Trf a fixed ratio for this straight line. ez = SB, and dy of such a length that TU : T^ : have rect. T^.T/ : ez^ \\ dy, Td. (Art. 8) If we now measure : OS : df/, then we Now, if we suppose T to be near e, Te,Tf ia nearly zero, while Td is not infinitely greater than di/y but as T moves in the direction of z the ratio Te . '£/ rapidly increases. "We may assume, therefore, that there is at least one position of T for which this ratio is true, which we may call Tj. [A more obvious way of stating this is as follows : — The volume Te.Tf.Td increases as T moves from e in the direction of z from zero to any possible value, however great, and therefore, once at least must have the value ez^.dt/.'] If T be any other intersection of the line and curve, we find by com- pounding ratios rect. Te,Tf: Ti« . TJ : : T^d : Td ; .-. rect. Ti* .Tj/ : Te.T/-Ti^.T,/ : : Trf : TTj ; rect. Ti«.Ti/= Trf.Ta = ce^-Tc^, where ef is bisected in d, ab = Tib, and ea is bisected in c. If Ti^ . Ti/> ce'^y there can be no real position of T. If Tj^. Ti/ = ce^t there are two coincident positions of T at e. If Tie,Tif<ee^y there are two real positions of T measured at equal distances from c so that Tc^ = Te.Tf—ce-. If X be any other fixed point in the line, it is clear that xTi + 3fr^ + xT3 = xd+xe + xf (1), afri.xTi + xTi.xT^ + afT^.afri = xd,xe-\-xe,xf+xf,xd (2), icTj.ajT, : xe.xf-xt^ iixdiafTi (3), where dy : xd :: xl^ : ez^, so that xl is a length independent of Ti, Tj, T3, and known when d, e, x, y, z are given. Digitized by Google 142 10. To show that, if P and T he two points on the curve and TV, PVj he perpendiculars to OH, and T^, Tr, Fq\ P/, segments of the chords of the construction circle through T and P, and x a point on OH the centre of a circle through T and P, then Pr'.P}'* 2Sa;.OV2 and Tr.T^ = 2&F.0V. It has heen shown hy Art. 8 that T^.Tr rSH*:: OS : OV and and, siuce T^. P^.Pr'zSH'rrOSrOVj (0; .Tr : P^.Pr' : : 2S:c.OV9 : .S&c.OV T^.Tr =SH«-ST», P^.Pr'^SH^-SP*, Tj.Tr-P^.Pr' = 8P*-ST^ = 2Sa:.OV2-28ar.OV ^by £uc, II. 12 and 13). Hence, from (1), T^.TrlT^.Tr-P^'.Pr':: 2Sa;. OVj : 2Sj: .OVs-28a:.OV ; .-. T^.Tr «2&f.OV5, and P^.Pr'-2S:c.0V. 11. If TS meets the curve again in T' and T'V is at right angles to OH, to show that OV = 8V, OV = SV, and OV'-OV = OS, also ST.ST'-SH». We know (Art. 8) that T^.Tr :SH«::OS:OV and T'^.T'r : SH^ : : OS : OV ; and, as Vm, Vm', Sm are proportional lengths to T^, Tr, S^ made hy perpendicuhurs to OH, ; OV :: Vm.Vw' :8m^ (1), : OV-OS : : Vm . Vm' : Vim . Vm'-Sm^ ; : SV:: Vm.Vm':SV»; Vm . Vm' = OS.SV (2). Vm.Vm' = OS. SV; Vm': Vm.V'm': :SV:SV'; Ym,Ym' : Vm . Vm' : : OV : OV ; SV:SV::OV':OV; : SV + SV : : OV : OV + OV ; ■■ O V and similarly OV = SV and OV-OV = SV'-OV = OS. Since SV = OV, .-. hyc2) Vm.Vm' = OS. OV; .-. hy(l) OS:OV:: 0S.0V':Sm2; .-. OV.OV = Sm«; .-. SV.SV = Sm«; ST.ST'-:Sj2 = SHs hy proportionals. .-. OS for SH « S^ ; .-. OS .•. OS • • Similarly ,*, Vm. hut, hy Art. 8, Vm. ,.^ SV: ,', sv = Digitized by Google 143 12. From the results of Art. 11 it follows that the curve consists of an oval and an infinite branch, each being the reciprocal of the other, with regard to the construction circle for ST . STY = SH'. That one part of the curve is an oval is dear from the fact that the curve cannot cut the construction circle ; therefore, if there be any points on the curve within the circle, they must form one or more closea curves within it, and there can be but one closed curve, or a straight line could be drawn to cut the circle in more than three points. The curve cannot cut the construction circle, because T^ or Tr cannot vanish for T^.Tr:SH2::0S:0V. That the part outside the construction circle consists of an infinite branch is not only evident from the fact that OY has been proved an asymptote, but also because it is the reciprocal of the oval which passes through 8, the centre of the circle with regard to which reciprocation is performed. 13. Any circle is described on the chord T3T3 joining two points on the curve, and cuts the curve again in P3 and P3. Show that the chord PaPb passes through a fixed point P^. Let the chord P^Pg meet TjTg in x and construct as in Art. 9. z is not actually a fixed point, but may be considered so for the chords PjPs and TjTg. Let xY be perpendicular to OY. By supposition TV :Td :: 08 : dy; xY :xd::08:dy: xY :0S:: xdidy: again di/ : xd :: xt^ : ez'^ ; OSixY ::xP:ez^. But ez is a fixed length — SH ; so is OS ; and xY is the same for both chords ; therefore the line known aa xl is equal for both chords arPj and a;T,. Therefore afr^ , afTi : xe . xf -xl^ ::xd : xTi and a^Pj . xF^ : xci.xfi^xl^ : : xd^ : a;P, (Art. 9). Now the construction circle passes through ef, e^f^ ; therefore xe,xf ^ xe^ .xfi. Again a circle passes through T^, T3, V^^ P3 ; therefore a^Tj . afTs = xV^ . x\\, ; therefore we obtain xd : xd^ :: r'Vi : xVi\ therefore Tj Pj is parallel to the asymptote, and, T^ being a fixed point on the fixed chord TiTjTj, the position of Pj is also fixed. 14. Two series of circles are dcFcribed on two fixed chords, one circle of one series intersecting one of the other so that their common chord passes through a fixed point. It is evident from the last proposition that their intersection traces out a cubic curve, for, if we (uaw any second fixed chord through T^ and describe circles to meet the curve, these circles will also intersect the curve in chords passing through Pj ; that is, in Digitized by Google 144 the same chords at which the first series intersect it, and, therefore, they may he looked, upon as tracing out the curve. It is required to show that these circles may he described by the rule stated in the latter part of Art. 5. Let ED be a fixed chord upon which a series of circles is described and chords be drawn from a fixed point P to cut these circles. Let A and B be the centres of two such circles, and PHL, PFG chords through P. Now PH . PL = PH2- AHa = PA'-AD^ and PF . PG = PB^ - BF^ = PB^ - BD's ; therefore PH . PL-PF . PG - PA^-PB- + BD2_ AD2 = AN2-BN2 + BC2-AC2; where C is the middle point of ED, and PN perpendicular to CAB ; therefore PH . PL - PF . PG = 2NC . A B. If there is a second series of circles described on another fixed chord having common chords through the fixed point P, each with one of the first series, it is evident that the rectangles under segments of chords drawn through P are equal for two intersecting circles ; therefore, if a similar construction be made for the second fixed chord, writing A', B', C, &c., for A, B, C, then 2NC . AB = 2N'C' . A'B' ; therefore ' AB : A'B' : : N'C : NC ; therefore AB and A'B' bear a constant ratio to one another. This can only be the case if the centres move along the lines CAB, C'A'B' so that their distances measured along CAB, CA'B' from some fixed points in these lines bear a constant ratio to one another. Hence we arrive at the conclusion that, if a curve be described by Guassman's definition, other points can be taken on the curve as described Digitized by Google 145 in Art. 6, such that P [AiB^EF] : P [GiDiEF] is a constant ratio. Conversely, a curve so described is a cnbio curve. [NoTB. — This proposition is used for constructing cubic surfaces.] CoR.~The cubic may be described by a series of similar segments of circles described on the double ordinates through T and ly, the extremities of any chord through S. 16. The Auxiliary Parabola, — A parabola described with focus S and directrix OY is closely related to the circular cubic (Art. 7), and points on it may be used for drawing tangents and normals to the cubic. Since OV = SV (Art. 11), the middle points of all chords of the cubic through S lie on the tangent at the vertex of the auxiliary parabola. Let ajb,e be three such middle points of the chords AC, PP, XT', and let adghf bgek^ chief be tangents to the parabola touching it at dy «,/. Now it is evident, since ag^ bg bisect AC, PP' at right angles, that a circle centre g passes through A, C, P, and P^ Hence the cubic curve may be very simply described. Take any fixed chord AC and the fixed tangent to the parabola, namely, a^; by drawing any other focal chord as PP' to meet dbemb, and bg at right angles to it to meet agin, g^ we find g, A circle centre ^, distance ^A or yC, cuts ihe chord PF in P and P'. This construction assumes its simplest form if we take the tangent at tUe vertex as the "fixed tangent," for suppose the curve cuts the axis in u ; then circles described with centres 0, 3, e and distances am, bu, cuy respectively, each cut the focal chords aS, ^S, <^ in points on the cubic. 16. To draw tangents to the cubic. Digitized by Google 146 Construct as in Art. 15. Suppose b to move up to a and ultimately to coincide with it ; then P ana P' move up to and coincide with A and G 80 that a circle described with centre d touches the cubic curve at A and 0, for as b moves up to and coincides with a the point y, the centre^of the circle, moves np to d. Since a circle centre d touches the cubic at A and G, the normals at A and meet at dy and the tangents may be drawn at right angles to them. Cor. — It is known as a property of tangents to a parabola that hg.hf»hd,kf. If we then regard the points hyd,f aa fixed, but ffk a variable tangent to the parabola, we obtain the following : — ff and k are the centres of two circles which move along two fixed straight lines, and each passes through two fixed points A, C, and T, T. hg :fk is a fixed ratio - hd : hf. Then the intersection of these circles is a cubic curve, namely, the variable points P, P'. This is in effect the same result as Art. 14. Digitized by Google APPENDIX II. NOTE ON THE EEDUCTION OF FOEMUL^ IN FACTOEIZATION, AFEOSDUra AN BAST MBANB OF FAOIOBIZING COMPOSITB ITUMBBSS, BSPBGIALLT THOSE WHOSB FAOIOBS ABB OF KNOWN FOBM. By D. Biddlb, M.E.C.S. It is possible by careful inspection, without any elaborate process, to leain a good deal about any number. Let N-S«+A-(S + «)(S-v) « (2Ap + l){2Aq+l), where A, 8, A are known. Then we easily arrive at the following feusts: — (i.) A+fw-S(«-t>). (ii.) Since N is always odd, S, A are of opposite character, one odd and the other even. (Ui.) Since S-¥u, S-^v are both odd, their sum is even, and it follows that M— V is invariably even. (iv.) In consequence, u, v are both even or both odd. (v.) From (iii.) it follows also that A + ut; is necessarily even. (vi.) When A is odd, uv is odd also. (vii.) When A is even, uv = (mod 4). (viii.) Since S + w = 2A/>+ 1, we have « = — (S— 1) (mod 2a). (iz.) Since S-v ^ 2Aq + 1, we have v= (S— 1) (mod 2A). (x.) When A is odd, it follows from (ii.) and (iii.) that A + uv=0 (mod 4) . (zi.) Since all odd numbers are of form 4»± 1, we can at once find out (if A be odd) which of these A belongs to, and we then know that uv is the opposite. . (xii.) If uv be of form 4w + 1, then «, v are both of one form, either 4»+ 1 or 4» — 1 ; and in these circumstances u—v = (mod 4). (ziii.) If uv be of form 4n— 1, then u is of one form and v of the other; and in this case u—v = 2 (mod 4). (xiv.) Since A + ut; is a multiple of S (see i.), we have iw = —A (mod S), which we can combine with any congruence already discovered. Thus, having given uv = fi (modB), =7 (mod G), we have the equation Btt + /3 » Gk + 7, and, the lowest value of k' that will make fi integral being found and substituted in the equation, B/4' + /3( — Cue' + 7) will be the residue of uv (mod L.G.M. of B, 0). Notation. — Let A«, A^ denote that A is even, that A is odd, respectively. Then So, S* will be corresponding characters of S. Let 0, « equal ^A«, 4s, respectively. Let «i, Vi represent i«, ^v respectively ; Mj, »j represent i»i, ^i ; and so on ; and let the same suffixes supply to any quantities, even when bracketed. Digitized by Google 148 Let H, h represent respectively the half-sum and the half-diilerence of the factors of N. (XV.) H -8 + i(f«-f)- A(» + g) + l. (xvi.) A = 4(f« + f>) « A{p—q), (xvii.) First values of H can conveniently be found from the following formula :—pq «• { J(N + 1)— H}/(2a'), where the right side is evidently integral, and in which H advances by steps of 2a'. (xviii.) M-t; » 2(H— B). Therefore the trial values of m— v advance by steps of 4a^. (xix.) "When x + ^ iB divisible by 2, we know that i;— ^ is even also, and vice versa. But this reciprocity does not extend to higher factors, unless Xy y be both multiples tiiereof or of the half. (xx.) During the process of reduction, uv becomes a function of a value which we may call /(, and u-^v becomes a function of a value which we may call k. But all is done by known steps, so that the trial values of uv and u—v corresponding with those of /a and k can readily be found, and it is possible at once to compare any composite trial value of uv which is arrived at with Uie corresponding trial value of u— f, in order to see whether they suit each other. (xxi.) It can be shown that /i and k are functions of a common value p, so that uv and u—v can be represented as functions of one unknown, namely, p. Consequently, u can be eliminated, and a quadratic in t; be formed, having under the radical sign, in the root, the one unknown, p. (xxii.) Since the <][uantity under the radical sign referred to is in reality A', we know, by (xvi.), that it is A'(^— ?)*. This fact enables us to find the residue of p in respect of the -modulus A. (xxiii.j Taking p = A( + d, we can greatly simplify the quantity under the radical sign, and other means of shortening the process occur in particular cases. The process of reduction may be set forth as follows : — Take first the case of Ae, So ; then we have A, + «t; = So(«— V), A« + 4miVi = So2(wi— t>i), a+2«it;i = So(«i-Vi). If a be even, Mj, v^ are both even or both odd ; but, if a be odd, they are one odd and the other even. Each case requires its 0¥ai treatment, but we need describe only that in which a is odd. Take So from both sides ; then 2«iVi— (So— fl) = So(«i— Vj— 1), and «iVi— (S©— a)i = S(,(tti— t>i — l)i. We know that UiV^ is even, and we are guided in our subsequent treat- ment by (So— a)i. But an example will best show how the process is conducted. Let N = 669 = 232 + 30 ; then 30 + wv * 23 (w-t;) ; 30 + 4Mif;i = 23 . 2 («*i — Vi) ; 15 + 2mjVi « 23 («i — Vj) ; «j, v, are one even, one odd. Take 23 from each side; iu^Vi — ^ = 23 («!- Vj— 1) ; «iVi — 4 = 23 («i— Vi— l)i ; (Mit;,)i— 2 =s 23 (wi— Vi — 1)3. We therefore know that wv -= 8/a, and that tt-t; = 8k + 2. Consequently 30 + 8/u «= 23 (8k + 2), whence /u = 23ic + 2. Taking k = 1, we have fi « 26, uv — 200, «— 1> = 10, giving u = 20, t;= 10, N = 43.13. Digitized by Google 149 We might have had to take more trial values of k. But the foregoing shows how to set about investigating the nature of K, when we have A*, S«. Take next the case of Ao, S«. Here we have Ao + uv — Sc(u— v), and we know that the right side is a multiple of 4. Consequently, we know that Ao + Mt; = (mod 4), whilst u, v are both odd. By reference to (xi.), (xii.), (xiii.), we can now render the formula as follows: — Ao ± 1 + 4/i = 4« (2/c), or 4« {2k ± 1), and we can tell which is the correct quantity to put on the right, by examination of A and thence of uv. Let «i' = i (Ao ± 1). Then Oj' + /* =* » (2k), or » {2k ± 1), and we know which. We also know whether a is odd or even, likewise a/, so that we can often learn much regarding ft beyond what yet appears. To take another example, let N = 1843 « 42*«+79; then 79 + ttf; « 42(u— v). Clearly 79 = 4«— 1 ; therefore «v =» 4»+ 1, and u—v = (mod 4). Hence wr— 1 = (mod 8), since S is even. SO + tw—l = 21. 8k; 10 + ^ = 21k; ^=21k-.10; wt; = 8/a+ 1 « 168k-79. The value on the right must be factorizable like that on the left. But, for most early values of k, it is prime. However, for k = 3, and next for k == 8, we obtain composite values. For k » 3, we obtain 425 » 17.25; but 42 + 25, 42-17 yield only 1676. For k - 8, we obtain 1265 = 23.66; and 42 + 56, 42-23 are the proper factors of N » 1843 == 97.19. But we might improve upon the process in this second example. Thus, 79 + uv = 42 (m - v) ; therefore, taking « =* 4^ ± 1, t> = 4Ar ± 1, we have (79 + I) + 16yAr ± 4 (y + A;) = 42 . 4 (^— Ar). Since every term except 4(^ + Ar) is divisible by 8 without question, that term itself must be, so that (^ + Ar) is clearly divisible by 2. Therefore {g—h) is divisible by 2 likewise (see xix.). Consequently 10 + 2yAr±(^ + Ar)i = 21 {g-k), 6 +yAr ± (y + A:)2 = 21 (y- Ar)i, showing that (^ + Ar) = (mod 4), whence A; = — y(mod4). We now know that kv— 1 = (mod 16), and have wv = 16/u+l == 336k — 79, which gives us the correct value on the fourth trial of k. If into this example we introduce the consideration of A, which here is 3, we find that u-v = 2Ar + 2 = 4(^— A:), or Ar+ 1 = 2 (^— Ar). Now we know that y— A; is even, and 2{g^k) is therefore a multiple of 4. In order to make 3r+ 1 similarly a multiple of 4, we must take r = 4n+ 1. Takins: r — 1, we have g-^k = 2, and u—v =» 8; but this makes uv — 257, a prime, which is impossible. Taking r = b^ g—k = 8, «— t; — 32, and uv — 1266, which is the correct value. Thus we require two trials only. Of course, wo is found by (i.) when u—v is given. Let us next investigate a large number and see what can be known regarding it. Let N = 329664457. Here So = 18153, and A« = 23048. So = 32. 2017 ; A, = 23. 43 . 67. Moreover, the factors of N are of known form,2.11m+l, Abeingll. A can be written 28(2.11.2-1)(2.11 .3 + 1); and, since S — 1 = 2(mod 2. 11), we have, from (viii.) and (ix.), «= -2 (mod 2. 11), r = 2 (mod 2.11), whence wv = — 4(mod 4.11). Further, by (xiv.), we have uv = 13268 (mod 18163). Thus we have 23(2. 11.2-1) (2.11.3 + l) + 4.11m-4 = 32.2017{(2.11.y-2)-(2.11Ar + 2)}, 22(2.11.2-1) (2.11.3+ l) + 2.llm-2 = 32.2017{(lly-l)-(lU+ 1)}, 2 (2.11.2-1) (2.11.3 + 1)+ Um-l -3V2017{11(^-A;)i-1} (o) Digitized by Google 150 From (a) we find that m = 1 (mod 3) ; therefore uv - 132/i + 40 « 18163IC + 13268, and /i « (18153k + 13218)/182 ; therefore 18153k = - 18 (mod 132), that is 69jc = - 18 (mod 132), which is satisfied by k' « 38, whenoe f/ *« 5326 ; therefore uv = 132 . 5326 + 40 (mod 2\ 3«. 11 . 2017) = 703072 (mod 798732) (seexiv.) — . (3). From the above equation (a) we obtain by reduction m- 3«.2017(^-Ar)i-2174, whence /4 = 3.2017 (^-Ar)i- 725, in which (^-A:)i = («-f; + 4)/44. Therefore we have wt; = 32.2017 (t«-» + 4)-95660 « 2«.3Ml. 2017 («-r-40)/44 + 703072, which agrees with (j3), and shows that (u— v— 40)/44 is the multiplier of 798732 involved in (/3). Let p be this multiplier ; then u—v ^ 44p + 40, and u = Up + 40 + v. Consequently we have the following quadratic : — uv = (44p + 40)v + v2 = 798732p + 703072, whence i; + 22p + 20=(4.1lV + 799612p + 703472)*=2.11(p»+1652-jirp + 1453TV)» (7). But 22p + 20 — i(u—v), and v added to this makes it ^{u + v) = h — A{p'~q). Therefore 11 is a proper factor of both sides of (7), and the quantity under the radical sign must needs be integral ; therefore by considering the two fractions under the radical sign we obtoin p = —5 (mod 11). Let p = 11^+6. Then we have v + 22p + 20 = 2.11(11^^+18305^+11402)* « 2.112(42+151 3^^^94^^J» (5). It can be proved that in the particular instance (see below) h = (mod 112). Therefore the quantity under the radical sign in ($) is integral. Therefore 34^ = -28 (mod 112), or { = 49 (mod 112). Let 4 = 121t + 49. Then we have v + 22/>+20 = 2.112(11*.t2+30163t + 9908)* («). The quantity under the radical sign in (t) is found to be a perfect square when t = 4. Therefore t>+ 129138 » 2.112.604, whence V = 17030, and N = 1123.293459. In order to prove that in the present case A = (mod 112), Iq^ ^^^ y^q rendered thus, 8.6.1l2-4.11-2 4.llm-l =S.11(^-A;)i-S 8.6.1l2-4.11 + llm = S.ll(y-A;)i-(S-3) = S.11(^-A;)i-2.112(M). Since S = 3 (mod 11'), we have 8.6.112-4.11 + llm-3.11(^-A;)i = 0(modll2), that is, t»— 4-3(^-A;)i = O(modll). Digitized by Google 151 It is required to prove that (^-f^)], a factor of A, is similarly divisible by 11. If so, m— 4 + 6*1 = (mod 11). Now m ^ llffk+g—k, and 2*1 » k. Therefore, in order to fulfil the conditions, ll^jfe+^-jt-.4 + 3* = 0(modll); and finally A: = 4(modll). This would render t? = -31 (modi I'). Let us then refer to (xvii.), bearing in mind that H~A » S— r. We find that H = 34(mod 11') ; S = 3(modlP), as we have seen; and 3-h31 > 34. Therefore it is right to take h == O(mod 11'), in the particular instance. Knowing that his given by r + 2Ap + (;, as on the left side of (S), we have 2Ap-^e » H~S, and since we are able, by aid of (xvii.), to find the residue (mod 2a') of H— S, we are able to find the residue (mod A) of p without difficulty, and thus to check the result obtained as in the last example. It would be useful to be able to find the residue (mod a') of h when it is other than 0. Since h = A (/?-;), we can take h = Ad;(mod a'). Consequently, the quantity under the radical sign at the particular stage of the process, being presumably a perfect square, will, when diminished by a;' (at 0, 1, 4, 9, Ac), be divisible by A. Owing to the combination of fractions (left by this division) amounting to an integral value, we are able to find the residue (mod A) of |, the unknown quantity specially concerned. We know that x =p-q (mod a). Let f/ = p + q (mod A), which can readily be found, because N » 2 Am + 1, and m » 2^pq+p + q. In fact, y = m (mod A). Therefore the sum of the residues of p and q, taken positively, cannot exceed A + ^ ; nor can their difference, namely, X, be outside the range ±i(A — 1). Consequently there will be only i (A — 1) possible values of a?' when a? is other than 0. But p—q ^s ff + kf and it can be shown that p + q = ff—k + D (mod a) ; for « » 2a^— (T, and f»2AA; + o', where, taken positively, ff = S— l(mod 2A); moreover H « A(iP + 8^)+ 1 » S + A(^— *)— (t. In order to obtain D, we first divide S — 1 by 2A — this gives remainder a ; we then double the quotient and divide again by A — this gives remainder D. Let p^ q^t P'f ^ he the residues (mod A) of the separate quantities. Then ^-q' ^ g' + k*, and p^ + q' ^ /-A:'4 D. It follows that p^-g" ^ ^^k' ^ \D ^ ±df say. The accompanying diagram brings this out, and will exjdain more readily the relations between the various quantities. Digitized by Google 152 Now we have the four following equations : — j/ + /-m'; /-Ar'.m'-D; j/-/-iD; / + A^-iD. Let a ^ f/ ■¥q* •¥g'-¥lc'y the sum of the unknown quantities. Then, although they are (us shown by Kronbckbr, Molk, and Q-. B. Mathews) severally and collectively indeterminate by algebra, we obtain n « 2p' and n— m'ssy— 5^ — / + A:'. Frequently, eitiier / or Je' is identical with p\ and the two remaining quantities sum to ; but, whether this be so or not, we have ft » 2^ in all cases. When m' « 0, we have ^ - — y, and p'—V = q'+ff* « — JD. Again, when m' = D, we have / = f/y and p'—q' = IV. When wl « ^D, we have q' -^g' ^ 0, and hence p' = V. When m' ^ |D, we have / + A:' = /— Ar' « ^D, and g'+/ = D ; also /-/ = /-Ar' = ^D, and ^'-Ar' = D. When D = 0, we have j/ = ^', and / = — A:'. When D =» i»»', we have p'—V = ^+/ = |m', and /-/ = 2/-J»t', whilst n = 2/ + Jw'. When p^ — q'j each = im'. When / -• Ar', each = JD. Let p' = |w' + i7, ^ = iw'— ij; then / = ly + e. A;' « ij— tf, where = ^(m'-D). Thus we have n = m' + 2ij ; Z-^' = / + A:' « a-m' -2„. The following particulars regarding certain specified numbers will afford a useful exercise : — N 8 A A m <r D m' / q' / k' a y-^ 61811 248 307 7 4415 9 6 -2 -3 1 1 2 1 3 65869 256 333 11 2994 13 2 4 -2 4 2 -3 -6 71107 266 351 7 5079 13 1 4 3 1 -1 3 -1 2 138659 372 276 13 5333 7 2 3 4 -1 3 2 -5 5 217801 466 645 U 9900 3 9 -5 5 -4 6 1 1 447639 668 1315 13 17213 17 11 1-6-6-6 6 10 Since v= v (mod 2A), we may take v = 2AA + o-. Then by substitu- tion in the equation just above (8) on page 150, and division of both sides by 2 A, we obtain an equation in A and |, two unknowns. But by this means p is eliminated, and we are able to define the value of | in terms of A, arranged in fractional (though integral) form, such as I = (a'+/a— *)/(<?— 2aa), where b, e,f are known. This enables us to find Umits, upper and lower, to A, and greatly shortens the labour in cases which are not of the privileged order of the example referred to. [Rev. J. CuLLEN, referring to equation (c) in the foregoing, says : " It seems to me, if this form could be obtained in every case, we should have a very rapid method even if t were large ; for we could apply the prime moduli to the equation ar^ + br + c^ n» giving on an averstge JQ?- 1) cases to mod j9. For instance, in {§) we have t' + 3t + 3 = 0, 1, 4 (mod 5), and a very large number ought to be manageable when r is so small foB N = 329554457.] Printed by C. F. Hodgson & Son, 2 Newton Street, Holborn, W.G. Digitized by Google MATHEMATICAL WORKS PUBLISHED BY FRANCIS HODGSON, 89 FABBINGDON STREET, EC. In 8vo, cloth, lettered. PROCEEDINGS of the LONDON MATHEMATICAL SOCIETY. Vol. I., from January 1866 to November 1866, price lOs. Vol. II., frt)m November 1866 to November 1869, price 168. 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