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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/I M ^^^^^H MATHEMATICAL QUESTIONS, ' SOLUTIONS. FBOM THE " EDUCATIONAL TIMES." VOL- XLI. J miinp 6000303151 MATHEMATICA.L QUESTIONS, WITH THRIK SOLUTIONS, FROM THE "EDUCATIONAL TIMES," WITH MAMT t.^^^ t» "^Ufttfi mi Siohxtiam not fnVlh}itii in tJ^je ** (thxcvUaml 9%mt% BDITED BY W. J. C. MILLER, B.A., REGISTRAR OF THB QENfiRAL MEDICAL COUNCIL. VOL. XLL LONDON: '" FRANCIS HODGSON, 89 EARRING 1884. *^* 01' this series forty -one volumes have now been published, each volume containing, in addition to the papers and solutions that have appeared in the £ducational Times, about the same quantity of new articles, and comprising contributions, in all branches of Mathematics, from most of the leading Mathematicians in this and other countries. New Subscribers may have any of these Volumes at Subscription prices. LIST OF CONTEIBUTOllS. Auiia, J. S.. M.A. : UM. Inspector of So)ioult AlJ.EN,Eev. i. J.C.M.A. iSt-pBter'aColL, Cam! Allhin. ProfEBaor Geo. J., LLD. ; ObIwby. FrofB ■A.i Queeii'aColl..Gawuj- «.A.jT;li(!Elins. HBrafoni >rj PegaiD, ,. LL.D.. P.l Bau., Edbt. Siawbli. ^^..,., . .^..j. , . .„.™.u of Astronomy in the DniTersity of Dublin. Bahu, SAtlsB CsASDBA;FreBid. (!aI].,CgJcaIla, BATTAGUiti, QicBgpPBj ProftjKBore di MatE- niaticho iiall' TTiiiTorailA di Bouuu BAYijaB,t«eh., l^vis. Hareb, BASA.H ; Cuabiidge Street, H;de Pu-k. TJie Panwon, Bicbmond-on-ThHme>. allI>CH!H,aTU.,H.A.: Prof, in Cooper's Hill Coll. UiiCHEeoii,T.,B.A..L.C.F.: City of London Sch. MOBCi, Hbjtet Stahibi, M.A. i Prof, or Moral Philotwpb; in the Uniiereity ot Dublin. UoHCoDBT, Proftwor: Paria. HooiI,KuBBBT,H.A.iBi.Fel].Qii.ColL,Csrab. MOOBB, H. K.. B.Aj Tnn. Coll., DubUn. UoBBL, Profeasar : PunB. MosGAN. C, B.A.1 Salisbury Si'iiool. MoBLEr, T., L.C.l'-i Bromley. Kent UOBLEY. F., B.A. 1 Woodbridie. BitfTolIi. MOBEICB, R. G.. B.A. ; The Hill, Suiiabury. MonLTOJi,J.F.,M.A.iFBll.ofOh.Coll.,Ciiuib. Mdib. TnoHAa, M.A., F.E.8.B. ; BiahopUin. UDEHOPADii;jkr,ABiii(>ss,U.A.:Bhowaniuor?. HaSq, A.M.,M.A.: lTor.lnP»9.ColI.,CBloutt]i. NsLaoH. B. J., M.A. I Naval School, Loudon. .. B. Prof. SraoM. M.A. 1 Wa»hin§' „...„^i O'Ebqas, ifoi _ Obchabd,H.L.,M.A..L.C.P.{ Bumham. OwBS, J. A., B.Sc, 1 Tennyson St., LiysrpooL PiKTON, A. W., M.A. 1 FaU.of Trin. Ooll-.tlublill. Pehulkbdbt, Rev. C, M.A.: Lnndoi Pbkeihah, w.; r-"' ' " " PniiLipe, F.B^W. ToLiasik, Prince Cahillb d_ . PoLLBirBH.H., B.A.i Windermwe Collide. POTIBB, J., B.A.i lUehinond-on-Thamea. Pbfddbh, FeaucbbE.i Lockport, New Tork. PuBBBB,Pror.F.,M.A.;que«n^sCoilexe,BelAut. Pdisam, K, S., M.A. 1 Borne, New York. Kau, B. Handuahta, B.A. ; Head Mast«r of the Normal School, U&dnia. Bawbos, Bobbbt ; Havant, Hants. Baihofd, K. Labcelot, B.A., Surbitoa. Rbau, H. T„ M.A. ; Brasenose Coll., Oiford. Bbevxs, 6. U., U.A. 1 Lee, Kent. BSBBHAW, S. A-! NottinBham. Bbyholdb, B., M.A.; Netting Hill, London. BiCHABDSOH, uev. G., M.A^ Wiucheater. KriPPiH, Chabms B., M Jl.i Woolwich Common. BoBEBT8,B.A3M.A.;Schol.orTHn.Ooll.j>ublin. BOBBRie, 8JtlLA.,F.R.S. 1 TofoeU Park,LaadOD. T. fl._ 4.7 » . „_.„ pgjiow of BoBBBis, Bev. Wl U'ji. Tiiniiy CoUeae. uubUn. iBrW.Bi;H.A.|Ei ' -Sch. of Trin. Coll. Jhit EcoflBBO, SiHOHBLiii CniTBTsttA di Bom*. BUBBBLL. i. W., M.A. ; Merton Coll., Oirord. BtrsBEi,!., B.. B.A.i Trinity College, Dublin. BVTTBB, BuWABnj Sunderland. .ai^KOjr, £«r. O., DJ>., F.KB.; BcfciuB Professor SAiinoT. C. H.,H.A.; Billiol Collegs, Oirord. Babdbbs, S, B. i Bloomlngton, Indiana. SiirnEt«OB,BeT.T. J.,HX; Boyston, CBmbi. Sabeab, Nileaetha, M.A.; Calautta. SAVAOE,THU]iAa,M.A.i FelLPeiiib.CoU,,0>mb. SCHBBBEtFrofessorj Mercersbury Coll., Pi. SciXTT, A. W., M.A. 1 St. Daiid'sColl., I^mpeter. Scorr, CUABLOTTB A. ; Girton College, Cnnh, Scott, B. F..U.A.: Fell. Bt. John's Coll., Camb. SBBBB7, Professor ; Paris. Skabp.W.J.C.M.A.: Hill Street, London. 8HABPB, J. W., M.A. ; The Charterhouse. SHASfB, Sin H. T., M.A.1 Chiirry Marham. Shephebd, Bev. A. J. P., B.A. ; Fellow of Queen's College, Oitord. BiHMOxa.Rev. T.C.,M.A.;Christ'8Coll.,Breoan. BiTKBLT, Waltebi Oil CiIt. Penusylvanik. BUITB, C, M.A. 1 Sidney Sussex Coll., Camb. Btabbbow, U., M.A. : New York. Stbooail, J. E. A., B.A. ; Clifton. Btbin, A.i Venice. aTBPHBH,8r,JoaH.B.A.iCaiusOoll.,Cambrid«8. Btbwaet, H., M.A.: Fra lUnKham, Suffolk. BwiFi, C. A.. B.A.iGrajomar ilch., WeybridgB. STi,rEaTBB, I.J., b.C.L., F.BJ.; Pn^issar of MathomaCics in Che Uniienity ol Oiloid, Memberofthe Institute of Francw, Ac. Stmobb.E. W^M.A.; Fell. St. John's ColL,Pion. Tait, P. G., M.A.; Professor of Natural Philo- sophy in the University of Edinburgh. Tankbb, Prol. H. W. L., M.A. i Bristol. Tabletoh, F. a., M.A. ; Fell. Trin. Coll., Dub. Tayiob, Bev. C, D.D. ; Master of Bt. John's Colleao, Camtridge. TirLO^B^.M., M.aTFcH. Trin. Cull., Camb. Tayloe, W. W., M.A. i Bipon Grammar SdiooL Tebai, SEPiiuas, B.A.: Famworiih. Bolton. TBBBY,BeY. T. B^ M.A^ Fell. Mn«e harmonically cut by the curve. (This is an extension of Question 6787, which may be extended to surfaces as follows) : — Through an ordinary point on a quartic surface lines may be drawn so as to be cut harmonically by the surface ; the points of section will trace out a quintic curve on the surface 31 6063. (The Bev. A. J. C. Allen, B.A.)— A prism filled with fluid is placed with its edge vertical, and a beam of light is passed through an infinitely thin vertical slit, and is incident normally on the prism infinitely near its edge. The emergent beam is received on a vertical screen. If the refractive index of the fiuid varies as the depth below a horizontal plane, find the nature and position of the bright curve formed in the screen 73 6878. (B. H. Ban, B.A.) — Given a concave spherical mirror, a luminous point, and the position of an eye perceiving one of the reflected rays ; find the point of incidence and reflection on the mirror 99 6S81. (For EoDncUtion, see Queetion 1904) Ill 6907. (S. Tebay, B.A.)— If A, B, C can do Bimilar pieces of work in a, b, e hour* reBpectively, (« < i < o) ; and they begin aimultaQeoualy, and regulate their ikbotir by mutual intercltanges at certain interralE, Bo that the three pieces of work are finished at the same time : finii the num- her of solutions 47 7040. (Rev. T. R. Terry, P.E.A.S.}— If p and j be two positiva in- tegers such that p>;, and if c be any poaitiTe integer, or any negatire integer numerically greater than p, show that , L _!_t ?( ? -' ) . '-('--I) &c. p-q + \ p + r-l ij,~si-l)(p~g + S) [p + r-l){p-tr~2] _y- p-q + 7165. (T. Woodcock, B.A.)— If P, Q be the points in which the plane through the optic and nty axes interaecta the circle of contact PQ of a tangent plane perpendicular to an optic axis of the wave aur&ce of a biazal crystal, and if a, c, the greatest and least axes of elasticity, be given * prove that, being the centre of the wave surface, 1 1) the triangle FOQ, (2) the circle of contact FQ, {3) the angle FOQ will have their greatest vunes respectively, when the square of the mean axis 6 is (i.) the arithmetic, (ii.) the geometric, (iii.) the harraonio mean of a' and i' ; and the cone whoae vertex is O and base the circle PQ will have its maximum volume when i' ^ J[o^ + e' + {o< + 14BV + e<)'] 46 7169. (E. Knowles, B.A., L.O.P.)— In a parabola whose latus rectum is 4u, if B be the angle subtended at the focus 8 by a normal chord Pft, prove that the area of the triangle SPQ = n= cot iff (tan ifl + 4 cot 1 fl)=. 64 7194. (Professor Wolatenholme, M.A,, Se.D.)— In the e for the Mathematical Tripos. Januury 2, 1868, Qusstion (G) ia as follows: — "It there be » straight linea lying in ooe plane so that no three meet in one point, the number of groups of n of their points of in- tersection, in each of which no three points lie in one of the m straight lines, is i (n — 1)." Prove that this is not true ; but that, if " n-sided polygons" be written for " groups of" points, ic," the result will be true; aod calculate the correct answer to the qQestioQ enunciated. ... 57 7230. (The Editor.)— On a square (A) of a chess-board, a knight is placed at random : find the probability that it can march (1) from that square (A) to a given square (B), as, for example, to one of the comer- squares, within a moves ; and (2) over i squares in less than c moves, for instance, ovor the four corneF'Squarea of the hoard 70 7236. (The Hev. T. W. Openshaw, M.A.)— On AB, a chord of an ellipse, as diameter, a circle is drawn intersecting the ellipse again in C, D ; if AB, CD are parallel to a pair of conjugate diameters ; sbowthat the locus of their intersection is Px + a^i/ — 44 7247. (Dr. Curtis.) — Two magnets, whose intenaitiea are Ii, Ij, and lengths a„ a,, are rigidly connected so as to be capable of moving only in a horizontlil plane round a vertical lino, which passes through the middle point of the line connecting the two poles of each magnet ; if 2a denote the angle between the lines of poles of the two magnets in the VIU CONTENTS. direction of opposite poles, while $ denotes the inclination to tfie magnetic meridian of the line bisecting this angle, prove that (1) the positions of stable and unstable equilibrium (discriminating between them) are given by tan $ ■■ (Iia^ + 1^) tan a/ (Ii^^— l^is) ; and hence (2), if the intensities of the two magnets be inversely proportional to their lengths, the posi- tions of equilibrium will be such that the lines of poles of the magnets will be equally inclined to the magnetic meridian 67 7273. (A. McMurch^, B.A.) — Prove that, if radii be drawn to a sphere parallel to the principal normals at every point of a closed curve of ocmtinuons curvature, the locus of their extremities divides the sphere into two equal parts 35 7287. (Professor Wolstenholme, M.A., D.Sc.) — Two circles have radii a, b, the distance between their centres is e, and a>b'i-e; prove that, (1) if any straight line be drawn cutting bol^ circles, the ratio of the squares of the segments made by the circles has the minimum value «{[(a + ^2-* + cos 01 cos r r • • 9 1 / A T»\ cos IP = — - -;— — - [sma sm2 f (AP) cos R sm i (a + d + 61 7418. (The Bev. T. P. Bjrkman, M.A., F.B.S.)— Prove that no polyedron can have a seven-walled frame of p«atagons 40 «^m— 7*21. (R. KnowleB, B.A.}— Two eqaal tangunts OP, OQ are drawn to a pBTBbola ; prove that (1) the angle POQ ie bisecteil by tho axis, and (2) the distance of the centre of thB oircle OFU from the vertex is constant and equal to one-half the lataa rectum 2B 7422. (For Enunciation, see Question 0878) 99 7-127. (Profesaor Townsend, F.H.8.) — A lamina, setting out from any arbitrary position and moving in any arbitrary manner, being Bup- pofled to return to ita original position aiter any number of complete revolutions in ita plane ; show that — (a) All syatems ot points of the lamina which deseribe cun-eB of equal area in the plane lie on circles ftsed in the kniina ; (ft) All ay^ems of lines of the lamina vhieh envelope curves of equal perimeter in the plane are tangents to circles fised in the lamina ; (e) The two aystems of circles, for different valuoB of the area in the former case and of the perimeter in the latter cttse, are concentric, and have a common centre in the lamina 112 74S1. (Piotesaor ■Wolatouholrae, M.A., Sc.D.)— If 2i=« + e + 7 + J, prove that sin i [B-y] sin 1 {=-8) [ain (.-3) + ain (,-7)~ain (.-«)-sin (s-?)p + sini (7-") sin l(a-») [sin {»-T) + 9in(»-a)-9in(,-^)-ain {,-«]» + Bin i (.-fl) sin i (7-i) [aiD (--») + Bin (i-B)-sin(.-,)-Bin (,-a)]' = -16siui(fl-7)Bini(a-B)8inl[7-«)8ini(fl-8)Bini(B-fl)flini(.),-S). 7+39. (R. Rawaon.)— Two inclined planes of the same height and intlination a, j9, are placed bacV to back, with an interval between them (2n). Two weights P, Q are placed one on each inclined plane, and kept at rest hy the connection of an IncxtenBible string, indefinitely long, passing ovpr two small tacks, one at the lop of each inclined plane. A wiight a, having a vorti<«.l velocity (c], is then placed on the string by a Bmooth ring St a point midway between the inclined planes. Show that the system thereby put in motion will come to rest at H point determined by a root ot the quadratic (*P=Bin'i.-«')^- S-(i«Pain = + K«r'.-(2Pve that the locus of P' is the quartic UV = TJ'^*, where V is the polar re

■ + 1 be drawn, prove that its remaining intersection with the given cubic is a fixed point.... 6S 7637. (Professor Townaend, F.H.S.) — An ellipsoidal shell being supposed, by a small movement of rotation round an arbitrary axis passing through the centre of its inner surface, to put into irrotational strain a contained mass of incompressible Quid completely filling its interior ; in- vestigate, in finite lorma, the equations of Qie displacement line-system of the strain 37 7638. (Professor Haughton, F.H.S.)— Show that the law of pro- pagation of heat in a solid sphere is ^ = n (^ + 1^\ 38 at \ ax* X ax f Xn CONTENTS. 7641. (Professor Wolstenholme, li.A., Sc.D.) — ^The coordinates of a j^oint being x ^ a {m^ + m-^), y ^ a (m— m*^), where m is the parameter, according to the usual rule the locus should be a ouartic, since we get four values of m for determining the points in which the locus meets any proposed straight line. Neyerth^ess, the locus is the parabola y* ■• (« — 2a) . Account for the discrepancy. Also, with the same values of (a?, y), the equation of the tangent is mH — 2i» (m' — 1) y + a (m*— 4w' + 1) * 0, which would make the class number 4 80 7542. (Professor Martin, M.A., Ph.D.) — Prove that for n » oo, ^{ r+...+ ^ r) »log.2. 66 2if [l+ v'2sin (i» + 5^) 1+ '•2sin f ^ + ^ j j 7543. (Professor Wolstenholme, M.A., Sc.D.) — ^In a rectangular h}*perbola, PQ is a chord normal at P, and T is its pole : prove that GT wiU be at right angles to CP ; tiiat is, T is the extremity of the polar subtangent drawn from the centre 0. [Otherwise : if be the mid-point of PQ, prove that the angle OOP will be a right angle.] 74 7544. (The Editor.) — Construct a triangle, having given the base, the vertical angle, and the ratio of the segments of a given chord of the circumscribed circle drawn parallel to the base, cut off between the circle and the sides of the triangle 68 7646. (J. J. Walker, M.A., F.R.S.) — ^Prove that the points on a right line have a (1, 1) correspondence with the rays of a pencil in the same plane ; show that the lines drawn from the points so as to make a given angle with their corresponding rays all touch a parabola, which is also touched by the given right line. [A generalisation of a theorem of Steinrr's.] 74 7647. (R. Tucker, M.A.)— PFR, QFS, are two orthogonal focal chords of a parabola, and circles about PFQ, QFR, RFS, SFP cut the axis in points the ordinates to which meet the curve in P, Q', R', S' : prove (1) locus of centres of mean position of P, Q, R, S is a parabola, (latus rectum ^L) ; (2) 2 (FF) + 2L « 22 (FP) ; and (3) if also normals at uiree of the points P, Q, R, S cointersect, then y^^ 2' (y" ) — — 24L-*. 114 7660. (J. Griffiths, M.A.)— If ^ » jf + 1 sn m . sn (K- ti) and modulus - ^^^3, K - f*'_— -^^__. ; show that r, — -- — ^, — ,,,, ^,.., -rfu. Jo(l-Jsin2«)» * [(<-2)(<-3)(/-4)(^-6)]* 90 7652. (The Editor.) — ^In a road parallel to a range, find, by ele- ioientary geometry, a point at which the sounds of tiie firing and of the hit of the bullet would be heard simultaneously 49 7666. (W. Nicholls, B.A.) — Two cubics XJ and V have the same points of inflexion. Show that the intersection of the tangent at any point on U and the polar of that point with respect to Y lies on U. ... 84 7668. (W. J. C. Sharp, M.A.)— If A', B', C, D', be the feet of the perpendiculars from any point on the four faces of a tetrahedron ABCD, show that AC«-BC^ - AD'^-BD'*, &c., and oonversdy ... 69 7664. (D. Edwardes.) — ^If the sides, taken in order, of a quadri- CONTENTS. XTU lateral insoribM in one circle, and oirctmuoribed about another, are a, b, e, d ; prove that the angle between its diagonals is cos "^ ^ ^ ■ j . ... 117 ae+ bd 7667. (Professor Sylvester, F.R.S.) — Let nine quantities be supposed to be placed at the nine inflexions of a cubic curve, then they will group themselves in twelve sets of triads, which mav be called colUnear, and the product of each such triad may be callea a collinear product. From the sum of the cubes of the nine quantities subtract three times the stun of their twelve collinear triadic products, and let the func- tion so formed be called F. With another set of nine quantities form a similar function, say F'. Prove that FF' will be also a similar function of nine quantities which will be lineo-linear functions of the other two sets, and find their values. [The inflexion-points are only introduced in order to make clear the scheme of the triadic combinations, so that the imaginariness of six of them will not matter to the truth of the theorem.] 63 7669. (Professor Townsend, F.R.S.) — In a tetranodal cubic surface in a space, show that — (a) The four nodal tangent cones envelope a common qiiadric. (b) Their four conies of intersection with the opposite &ces of the nodal te^ahedron He in a common quadric. (e) Tbe two aforesaid quadrics envelope each other along a plane having triple contact with the siurface 65 7671. (Professor Haughton, F.R.S.) — A solid body is bounded by two infinite parallel planes kept constantly at the temperature of melting ice, and by a third plane, perpendicular to the first two planes, kept con- stuitly at the temperature of boiling water. After the lapse of a very long time, show that the law of distribution of temperatures will be repre- sented by the equations (between the limits y ■■ ± iir) r =« a«-« cosy +*«"** cos 3y + &c., 1 « acosy + ^cos 3y + &c. ... 64 7673. (Professor Hudson, M.A.) — Parallel forces act at the angular points of a triangle proportional to the cotangents of the angles. Can they be in equilibrium? 60 7674. (Professor Wolstenholme, M.A., Sc.D.) — If we denote by F (Xf n), the determinant of the nth order a?, 1, 0, 0, 0, 1, a?, 1, 0, 0, .. 0, I, ar, 1, 0, .. 0, 0, 1, ar, 1 a, 0, 0, 1, X prove that F (ar, 2r + 1) s xF (ar«- 2, r), F(x, 2r) = F(a?3-2,r) + F(«a-2, r-1), F(x, n)s fa;-2cos-^Wa;-2cos— ^ ^ ' \ n+ll\ n+ll X fa?-.acos-^V..f«-2cos-^^V V . n+lj \ n+ll 112 7676. (Professor Wolstenholme, M.A., Sc.D.) — Two normals at right angles to each other are drawn respectively to the two (confocal) parabolas y' — 4a (» + a), y' — ib {x + b) ] prove that the locus of their common point is the quartic 2y =. (a* + **) [x-2 (aft)*]* + (a* - ftl) [or + 2 (aft)*]*, which may be constructed as follows^ — draw the two parabolas yi • (a + ft) it- 4aft ± 2 (oft)* («^a-ft). XVlll CONTENTS. and let a oommon ordinate perpendicular to the axis meet these parabolas in P, p, Q, q, reepectivelyy then the quaxtic bisects PQ, P^, j^, pq. Also the area included between the quartic and its one real bitangent is ^tAvfi (m + 1) (m — 1)', where a — bm*, and a > b. These resolte wSl only be real when ab is positive, or when the two confocals have their con- cavities in the same sense, but in all cases the rational equation of the quartic is (y»- 02? 4- 2ab) {y^-bx + 2ab) + oi (a - A)' « 0. \Th.e quartic is unicursal, but has only one node at a finite distance (2; . a •»- 6, y -i 0) ; there is singularity at 00 , equivalent to two cusps. The class number is 4, and the deficiency 0, so that 2S + 3k » 8, S + ic -* 3, or « - 1, jc « 2.] 117 7676. ^The Editor.) — Two houses (A, B) stand 750 yards apart on the side of a hiU of uniform slope, and at the respective distances of AC*- 600 j9x6a and BD = 160 yards from a brook that runs in a straight line CD along the foot of the hill. A man starts from the house A to go to the brook for water, which he is to carry to the house B. Supposing he can only walk 2 miles an hour in going up hill with the water, but 4 miles an hour in going down hill to the brook ; show that (1), in order to perform his work in the shortest possible time, he must strike the brook at a point P such that GP » 646*124 yards, the distance he will travel is AP + PB = 811-494 + 169-298 = 970-79 yards, and the time the walking part of his journey will take is 6*916 -i- 2*715 - 9-631 minutes ; also (2), if he start from B to return likewise to A, he will have to take the water at the mid-point (Q) of CD, the length of his return journey will be 460v^5 = 1006-23 yards, the time will be VW \/^ = 14*293 minutes, and the two parts BQ, QA of his path will be perpendicular to each other. 82 7678. (The Rev. T. 0. Simmons, M.A.) — If a number have the sum of its digits equal to 10, find under what circumstances twice the number will have the sum of its digits equal to 11 64 7579. (B. A. Roberts, M.A.) — Two uniform spherical shells attract according to the law of the inverse fifth power of the distance ; i^ow that, if they cut orthogonally, they will be in equilibrium under the influence of their mutual attraction 65 7581. (C. Leudesdorf, M.A.)— If A + B ^-0 = 180^ (y — a) cot iA + («- a:) cot ^B + (a;— y) cot \0 « 0, (y2-««j cot A + (a2-a;2) cot B + (a^-y*) cot =» ; that y'^ + g' — 2ygC08A z^ + a;'— 2ga;cOBB ar^ + y^— 2a;y cos C prveini ^^^^ = ^^^ - ^^..^ 62 7687. (Syama Charan Basu, B.A.)— If (T-f)(7^T) — 0' where a, ^ are the roots of ax^ + bx^ + e -* 0, show that a » iS » 2 82 7592. (S. Tebay, B.A.) — Find an integral value of a such that (m' + ffi)'^ + a and (m^ + m^)'— a shall be rational squares ; m and n being positive integers 66 7693. (R. Enowles, B.A., L.C.F.) — A circle passes through the ends of > chord I'Q of the parabola yi = 4(i* and its polo (Ai) i prove that (1) its eqnalion is r^ + ,j'-^t:^x- ^ (^a-h) y + h {ia-h) =0; (2)ifPQ IB purpeadicular to the axis, the focus is the oentro ; (a) if the dicle cuts the penthola again in CD, the middle point of tiie line joining the polps uf PQandCD, with respect to the parabola, is the focus 78 7591. (W. J. C. Sharp, M.A.) — It the circle inscribed in the triangle ABC touch the sides at the points D, E, F respectively, and P he the point of concurrence of the lines AU, BE, CF ; and again, if D', E', F", P bo the corresponding points for the escribed circlB opposite A, PD^PE^pp^, _?tJ'^?:e'^pt;_ , .J 2j Bhow that AD BE CF [In the second result, the lines are eonaiderod if regard ' '' ■' - -- '-^- -' 'J >— BE' cr e had to the signs, the — should be omitted.] nagnitudcB : 7687- (Professor Townaend, F.E.S.)— A syfltera of plane wavoB, pro- pagated by small parallel and equal ie<:tilinear Tibrations, being supposed to traverse in any direction an isotropic elaBtic solid, under the action of its internal elasticity only ; show that the dicci^tion of vibration is noces- sarily eiOier parallel or perpendicular to that of propagation, and deter- micD the velocities of the latter corresponding to the two casea 86 7598. (Professor Wolstenholme, M.A., So.D.)— 1. Circles are drawn with their centres on a given ellipse, and touching (o) the mujoi axis, IB) the minor axis ; prove that, if 2n be the major a nis, and i the eccentricity, the whole length of the arc of the curve envelope of these circles ia r:). "(' (l-fll< ) (».»)• 2. Circles are drawn with their centres on the arc of a given cycloid, and touching (u) the base. (B) the tangent at the vertex ; prove that tho curve envelope of these circles is (a) an involute of the cycloid which is the envelope of that diameter of the generating circle of the given cycloid which passes throngh the generating point ; (fl) a cycloid generated by a circle of radius jfii rolling on the straight line which Is the tuuus of the centre of the generating circle (rudius a) of the giveu cycloid. 3. Circles are drawn with centres on a given curve and touching the axis of z; prove that the arc of their curve envelope is z— 2|y dS, where #, ff are the coordinates of the centre of the circle, and -J^ = tan 8. ... IDS 7601. (Professor Hudson, M.A.)— The lenses of a common astro- nomical telescope, whose mngnilyijig power is 16, and length fromohject- glHBS to eye-glass 8^ inches, are arranged as a microscope tu view an object placed J of an inch from the object-glass; Gnd the magnifying power, the least distiince of distinct vision being taken to be 8 inches... 76 7602. (Frofeasor Hudson, U.A.) — A ray proceeding from a point P, and incident on a pkno surface at 0, is partly reBocted to Q and partly infracted to K : if the angles POQ, POR, QOR be in arithmetical CONTENTS. 7603. (The Editor.) — If on a rectangle AOBZ two random points (P, Q) be taken, P on the base OB, and Q on the surface OZ, show, by a Seneral solution, that, OA remaining constant, (I) as OB increases in- efinitely from zero to infinity, the probability that the triangle OPQ is acute-angled decreases from i to ; and (2) in the cases when OB — OA, OB = |OA, OB -20 A, 0B-40A, the probability will faU short of i by the approximate values •^, -^^'^, -^|^, -^j^, respectively 103 7605. (J. J. Walker, M.A., F.R.S.)— Referring to Question 1686, show that (1) the circles drawn on the common chords of three mutually orthotomic circles as diameters have not a common radical axis (as erroneously stated in that Question) but have the same radical centre as those circles ; and (2) their common chords are equal to one another, and (3) respectively parallel to the radii of the circle through the centres of the orthotomic.triad, drawn to those centres 77 7611. (B. Reynolds, M.A.) — A man, having to pass round the comer of a rectang^ular ploughed field, strikes across the field diagonally, at 46°, upon nearing the comer, to save time. If his velocity on the beaten path is u, and that on the field is u^x, where x is the perpendicular distance of the path chosen from the comer, find (1) where he should leave ^e beaten path, and (2) what value of x will make either route occupy the same time 75 7619. (M. Jenkins, M.A.) — Prove that the coefficient of x^ in where E f — j is the integral quotient, and R f ~ ] the remainder, when n is divided by ^ 107 7622. (Syama Gharan Basu, B.A.) — PSQ is a focal chord of a parabola ; tangents PR, QR intersect in R. Show that the third tangent parallel to PSQ bisects RS at right angles 73 7623. (The Editor.) — If a knight is placed in a given square on a chess-board, show (1) how to move it 63 times, so that it may not occupy any square twice ; and (2) how to solve the same problem when the num- ber of squares is 49 or 81 93 7628. (R. Enowles, B.A., L.G.P.) — If a, b, e represent the sides of a triangle, and 9j ^^s—a, &c., prove that ^— *i2=* flk?— «2' = a3 — tfj' « r (rj + rj + rj) 93 7631. (The late Professor Cliflford, F.R.S.) — ^A point moves uniformly round a circle while the centre of the circle moves uniformly with less velocity alon^ a straight line in its plane ; find the nodes of the curve which the pomt describes. 86 7633. (Professor Genese, M.A.) — ^A circle is inscribed in a segment of a circle containing an angle $ : the point of contact with the base divides it into segments h, k. Prove that (1) the radius of the inscribed circle is •; — - cot ^$ ; and hence (2) that the inscribed circle of a triangle h + k ' ouches the nine-point circle 123 CONTENTS. XXI 7636. (Professor Angelot.)--I)einontrer que t&n'^ ^ + isin-^ ^ + ta.Ji-^ j\ + ...+ tan- ^ —:^+ ,.. ad. inf. = iir 93 7638. (The Editor.) —If from a given point 0, in the prolongation through C of the base BC of a given triangle ABC, a straight line OPQ be drawn, cutting the sides AC, AB in P, Q ; show that, R being any point in the base, the triangle PQR will be a maximum when a parallel QS to AC through Q cuts BC in a point S, such that OS is a mean pro- portional between OB and OC 87 7644. (W. S. McCay, M.A.) — Prove that the three lines that join the mid-point of each side of a triangle to the mid-point of the corresponding perpendicular meet in a point 88 7648.* (D. Biddle.)— A series of isosceles triangles, beginning with the equilateral, is such that each in succession has two-thirds the vertical angle and two-thirds the base of its predecessor. Show that, when the base and vertical angle reach zero, the height of the last in the series is to the height of the first as 2^3 : ir 92 7653. (For Enunciation, see Question 6878) 99 7657. (J. Crocker.)— If an ellipse be described under a force / to focus S and /i to focus H, and SP = r, HP = ry ; prove that ^A_f^2(I-M 114 dr^ dr \r r^ ) 7658. (S. Constable.) — The vertex of a triangle is fixed, the vertical angle given, and the base angles move on two parallel straight lines ; construct the triaugle when the base passes through a fixed point. 115 7660. (R. Knowles, B.A., L.C.P.) — From the angular points of a triangle ABC, lines are drawn through the centre of the circum-circle to meet the opposite sides in D, E, F, respectively ; prove that AD ■" BE ■*■ CF " R" ^^^ 7666. (Professor Haughton, F.R.S.) — Prove the following formula for finding the Moon's parallax in altitude in terms of her true zenith distance, viz., sin j? = sin P sin 2 + i sin^ P sin 2« + 1 sin** P sin Zz + &o. ... 117 . 7669. (Professor Townsend, F.R.S.) — A thin uniform spherical shell being supposed to attract, according to the law of the inverse fifth power of the distance, a material particle moving freely in either region of its space external or internal to its mass ; if, in either case, the current velocity of the particle be that from infinity under the action of the force, show that its trajectory will be an arc of a circle orthogonal to the surface of the shell 101 7676. (J. J. Walker, M.A., F.R.S.)— If F (ajyz) = is the equation to any surface referred to rectangular axes, show that the equation to the curve in which it is cut by the plane x cos a + y cos )8 + 2 cos 7 — i?, referred to the foot of j3 as-origin, and the line in which the plane is cut by that containing the line p and the axis of «, and a line at right angles thereto, as axes, is obtained by substituting for a?, y, 2 in F {xyz) = 0, j3 cos o + (y cos )8 — « cos 7 cos a) cosec 7, l?cosi3— (ycoso + 2cosj8cos7) cosec7, j?oos7 + 2sin7 106 XXll CONTENTS. 7683. (R. Tucker, M.A.)— LSP and LHL' are a focal chord and a latus rectum respectively of an ellipse, and the circle LL'P'cuts the curve again in Q (<^) ; prove that tanHC^^) =» (1 +<^)V 0-^)"- 121 7696. (Alpha.) — Two guns are fired at a railway station at an interval of 21 minutes, but a person in a train approaching the station observes that 20 minutes 14 seconds elapse between the reports ; suppos- ing that soimd travels 1125 feet per second, show that the velocity of the train is 29*064 miles per hour 122 7699. (R. Enowles, B.A., L.C.P.) —Prove that in any triangle cos A cosB cosC _1^ ^,v (;sinB asinC isinA R 110 7734. (J. Crocker.) — If A and B are fixed points; find, on a fixed circle, a point P such that AP + PB is a minimum 121 MATHEMATICS FBOM THE EDUCATIONAL TIMES, WITH ADDITIONAL PAPESS AND SOLUTIONS. 7382. (By Professor Sylvrster, F.R.S.'j — li p and q are relatiye primes, prove that the number of integers interior to pq which cannot be resolved into parts (zeros admissible), multiples respectively of i? and q, ia *(i'-l)(?-l). [Ifl? = 4, fi' « 7, we have i (;?-l)(gr-l) - 9; and 1, 2, 3, 5, 6, 9, 10, 13, 17 are the only integers inferior to 28, which are neither multi- ples of 4 or 7, nor can be made up by adding together multiples of 4 and 7.] Solution by W. J. Curran Shasp, MA. If the product (1 +a:'' + a?2p + ... + »«) (1 +ic« + j?2«+... + /pfHr) be con- sidered, each term between 1 and x^^ corresponds to a number less than pq, and of the form mp + nq ; also 2xf*^ is the middle term, and the co- efficients from each end are the same. Hence twice the number of in- tegers of the form mp + nq^ and less than pq, is the value of the above product when a:=»l with four deducted, since the terms involving a;^, a^, x^Pi are not included ; and therefore the number of these integers is i(jP+l)(2 + l)-2, and the number of those which cannot be put into this form -i'^-l-H(i?+l)(?+l)-2] = ib^-i?-J+l] = i(i?-l)(^-l). 1585. (By the late Professor Clifford, F.R.S.)— If three circles are mutually orthotomic, prove that the circles on their common chords as diameters have a common radical axis. VOL. ZLI. Salulim by AsCtobh Let A, B. C be the centreB of the thres mutually ortliotoinic circles^ P„ P,. P», P„ Ps, Pj their pointa of intenection. Then, from the ordi- Djiry theory of orthotomic circles, we know that the centre of each circle lies on [lie common chord of the other two ; it can also be shown, from elementary geometry, that the line of ceotreB of any two intereect- ing circlee bisects their common chord at right angles; hence, we inferthat.in the triangle ABC, AAj, BB[> CC, are the perpendicnlars, and its orthocentro O is the radical centre of the Bystain. Moreover, calling the sides of the triangle ABC, a, b, n, aa usual, we hAte BA, -ocosBI CB, - ac ' C,B -oooaB Now, let ua call Si, S„ 8, He cirulea described on PiP„ P,P„ P,Pj rta diameters ; let their radii be p^, p,, pj. Then wo ahall have p^ — AjPi' = Bflj . AjC (because, the circles being orthotomic, ~ be cos B cc ZBPiC is a right angle) p,' = AC,.C,] Taking A, as origin, AjC, AiA a isC.cosA DsA.coeB the itangular axes of s and y, the , wiU be (0, 0). (ccoai B), reapectively. But, A| of radii ■ ' coordinatee of the points A,, acoa C sin C), t"* cofl A CM B,;fl c( are the centres of the circles 8,, Sj, 8, of radii p,, p,, p^, therefore the equations of these circles are x* + y' = faoosB oosC (1), {I - e cos A 008 C)' + (» - a COS C sin O' (iii-icosAcosB)'-i-(if-acOBBsinB}' = iiicosAco8B (3). Subtracting (2) from (1), attending to the relation sin A ; a — &c., and cancelling like ttoma, we get, for the equation of the radical azia, *coflAtj^sinA = i(i-cosC + AcosB-»coBA) (4). Subtracting (3) from (1), we get the same equation as (4), Therefore, the three circles have a common radical axis. [This Question In fact, " BolvituT P-P,, .. been discovered by Mr. Walker leando": if we draw (even mentally) the cirolea J we see they cannot have a common radical aj of the above solution lies in the sign of the abscissa of C] (at ' above) which should be ntgative, the abscissa being - The equations of the tliree radical axes are in fikct d;'co«A + y sinA - }( — acoBA-i-JcoBB + 23 and (& cos B + <; cos C) cosA2;+(— dcosB + ^cosC) sin Ay = ( — * cos B + (? cos C) ( — a cos A + * cos B + -''\<^\\ and that *, = fc 2j^cot9_ ^^^ I'COtO + i'iCotSj ccotf 4-i']Oot0i 24 giving in all oases the ratios of V and ^i to % in terms of and Oj and of V and vp and showing that ^a when vj cot 0) » y cot 0, supply in con- sequence the solutions of both parts of the question. The angle of incidence for which A;'» 0, with the corresponding angle ©f refraction 9^ for which fti - 1 , may be readily determined, in terms of the coefficients v and vx^ and of the densities p and pi of the two solids, as follows. Since, in that case, as appears from the above, v cot = y^ cot Oj, and since in all cases, from the known laws of wave refiraction, sine: Bin (C-i-ii^-Jiiyifiiy, where v and v^ are the velocities of propagation for transversal vibrations in the solids, therefore here (pv)* . cos B (pi yj)* . cos (K and (pvi)* . sin « (pi y)* . sin 0} ; from which, by elimination successively of 0^ and 0, it appears at once that tan'e- IL.^JLU^^ and that tan^ e^ = i . ^:^i:JlHl, . vi ypi-yip V vpi-vip which give manifestly the values of the two angles in terms of the four quantities in question. Multiplying together the two pairs of equivalents in the two general equations of condition at the separating surface of the media, viz., (A; + k') =s h^, and ycot d (Ar - ^) = vj cot dik^, we get at once the additional equation p cot 6 {k^—k'^) « y^ cot $i k^ ; or, remembering that y : vj « pi?« : p^ v^ « p sin^ e : pi sin* 6i, the equivalent equation /> sin cos d (A:'— Jl/*) « pj sin 0i cos di fej' ; from which it appears, by the usual mode of inference, that the via viva of the entire motion is the same after as before the division of the original wave system into two at the separating surface of the media : a result techni- ciilly termed the preservation of the vis viva in the division at the surface. (7481.) For perpendicular incidence, either for transversal or for normal vibrations, the two equations of condition at the separating surface of the media assume alike the same simplified forms, viz., (k + k') « kj and pv (fc— ft') = piVik^ ; where v and Vi are the velocities of propagation corresponding to the spocies of vibration, whichever it be, in the media, and equal consequently to (^iiy and (J^y for transversal, and to OiUl^y and nh±^y for normal vibrations. Therefore, for that incidence, in both cases alike, Hs appears at once by multiplication together of the two pairs of equiva- lents, pr (A?'—*'^ ■■ pi Vi k^ ; and therefore, &c., as regards the property in question. Solving for kf and k^ from the two equations of condition, we see also that, in both cases alike, jfc'- ;t'?^^iafl and ;&!«* -2££_ from which it follows at once that, in both cases alike, A/= and k^x^ k when the products pr and pi v^ are equal in the media. 25 7357. (By Professor A. Morel.) — H6soudre un triangle, connaissant une hauteur, le rayon du cercle inscrit, et le rayon du cercle circonscrit. Solution by Arthur Hill Curtis, LL.D., D.Sc. Let ABCbe the triangle, then, denoting by a, b^c, «, R, r, p the three sides of the triangle, its semi-peri- meter, the radii of its circumscribed and inscribed circles, and the altitude passing through the vertex C, we have pe = area = r*, therefore * : = weight in tons of unit length, i = weight of ballast, a; = required length of tail ; then \a ,wa = \x .wx + x, b, whence b « ^^ — \u}Xy and we 2x require the minimum of y = «t^ (a + a?) + \wx» Putting -^ « 0, we 2a? dx jret nw— iu> = 0, whence x'^ 2x'^ 2«-l 5350. (By S. A. Eenshaw.) — An ellipse and hyperbola have the same centre and directrices, and they have a common taugent which touches the ellipse in D and the hyperbola in E, and meets one of the directrices in H. Also from the common centre of the curves S'R is drawn parallel to the common tangent and meeting the same directrix in R. Tangents RW, RV are drawn to the auxiliary circles of the ellipse and hyperbola. Show that, if FH, /H be joined, F and/ being the foci of the curves be- longing to the directrix RH, DH . HF : EH ./H = WR'. : VR. VOL. XLI. D Draw any two parallels to BE, one cutting the ellipse id E, L and iii«etiiig the directnz in Q, tha other cutting Uke hyperbola in M, K and meeting the directrix in P. Join PF, Of, and draw BTT, R«' paniUel to them, and meeting the auxiliary circles in T, T", t, t' Then, since the aniiliary circles are generating circles of the curve, ve have by similar triangles KQ.CiL:/Q'-ffi.Ef :SR' and PF" : PM . FN = SH'iTR.BT'; therefore KQ . QL .PF= :/Q=.PM.rN = iE.Hi' ; TR.Rr- BW* : EV! when therefore the parallels both coincide with the tangent, thia becomes DH>.PF':/Q".HE'-EW>:EV= or DH. PF :/(J. HE - EW : ET. 7497< CBf ^- Hbfpbl, M.A.) — If four concurrent normals meet an ellipse in points whose eccentno angles are o, ^, 7, S ; show that a + ^-t-7-i-t — Sror S*, according as the ordinate of the point of con- Salatitm by the Bev. J. L. Kttchin, H.A. ; Prof. Matz, U.A. ; The equation to the normal at the point whose eccentric ang (a'-i») " tane y__tane.»= a(i + tan'e)i ' therefore „>- 2^tanfl+ ^'un'fl - ^Z±)!tf^; 81 whence a'a?' tan* 6—2 abxp tan' 6 + [aV + *V-(«'-*')Ttan«a-2a%tana + *»y8 = o. Whence, since coefficients of tan' and tan 6 are equal and of same sign, tan a + tan jB + tan y + tan 8 — tan a tan jB tan y + tan a tan y tan S + ... ; whence tana + tan^ ^ tany + tan^ ^ ^^ 1— tanatanjB 1 -tan 7 tan S or tan (o + 0) + tan (7 + 8) - ; whence generally a+fi + y + 9^nir; nan integer. Now, since the last term is positive, only two, or four, can he negative together; this consideration with the geometry of the figure will show that n can he only 3 or 6. 7488. (By Professor Hudson, M.A.) — If O he the circumcentre of ABO, and forces act along OA, OB, OC proportional to BO, OA, AB ; prove that their resultant passes through the in-centre. Solution by 0. Morgan, B.A., H.N. ; G. B. Mathews, B.A. ; and others. The force along OA is equivalent to two ^ forces proportional to cos B, cos along BA, cosByrkcosC OA respectively ; hence, resolving the other forces similarly as in the figure, tibe trilinear equation of the resultant is (cos B - cos 0) o + (cos 0— cos A) j3 + (cos A— cos B) 7 - 0, which goes through a ^ jS «. 7 ; therefore &o. ^ cwC coaB eosA 6787 ft 5945. (By W. J. 0. Sharp, M.A.) — From an ordinary point on a quartic five straight lines can he drawn so as to he cut harmonically hy two curves. - How far is this modified when the point is a node ? (5945.) From a double point on a quintic, a triple point on a sextic, or a p^c point on a (/> + 3)^°, prove that a limited number of lines can be drawn so as to be harmonically cut by the curve. (This is an extension of Question 5787, which may be extended to surfaces as follows) : — Through an ordinary point on a quartic surface lines may be drawn so as to be cut harmonically by the surface ; the points of section will trace out a quintic curve on the surface. Solution by the Proposer. The (»— j»)*** polar of O with respect to an n-ic curve or surface is the locus of a point B in OHn, cutting the figure in Bj, Bj, Bg,.. B^, such that 5COB1-OB) ... (OBp-OB) OEp+i ... 0B„ « 0. 32 Now, if Rj coincide with 0, and 0H| — 0, this becomes OR2(OR,-OR) ... (ORp-OR) OR^^i ... 0R„ « 0, and one value of OR is zero, as it should be. If Rj also coincide with O, the equation reduces to OR22 (OR3-OR) ... (ORp-OR) ORp^i ... 0R« = 0, and so on. Hence the 2nd polar of an «-ic, if Rj, R2 ... Rn-3 all coincide with O, has forits equation OR"-3[(ORn-2-OR) OR„-i . ORn + (OR„_,-OR)OR„.OR„.2+(ORn-OR)ORn-iOR„.2] - 0, and n^Z values of OR are zero, and the other satisfies 3 ^ 1 1 1 . OR * 0R,._2 "^ OR,._i 0R„' and hence, if R coincide with Rn-2, R»»-i> or R,,, the line is divided harmon- ically; i.e., if the line pass through an (« — 3)^*' point in the locus, and through an intersection of the locus and the second polar of the puint, it is harmonically divided by the M-ic. Now (1) a quartic curve is cut by the polar conic of a point on itself in five points, distinct from the point, the lines joining which to the original point are harmonically divided by the curve. (2.) A quintic is cut by the polar cubic of a double point on it in 3x6 — 4x2 = 7 points ; and so many lines can be drawn through it so as to be harmonically divided by the curve, and generally a (j? r 3)*'' is cut by the polar ( jo + 1)*« of a j^^^ point on it in (p + l)(p + 3) - (jp + 2) jp = 2jp + 3 points, distinct from the multiple point, and hence as many lines can be drawn through so as to be harmonically divided by the curve. (3.) Since an «-ic surface meets any plane in an w-ic curve, the above argument shows that the curve of intersection of the second polar of an (« — 3;^c point on it pierces the plane in 2w-6 + 3 = 2w — 3 points, and therefore this partial intersection is a (2» — 3)^<^ curve ; and therefore a quintic when the point is an ordinary point on a quartic surface. 5754. (By J. Hammond, M.A ) — Sum the series 1 1 o 1 1 2w(2w-l) 1 1 « u '2?n + n w ••■ 1 2m 1- w — 1 1.2 « r 2 2w + w — 2 where m is a positive integer, and the (r + 1)*** term is . .^ 2w(2m-l)... (2m--r + l) 1 1 1 . 2 . 3 ... r ' n + r 2m-t^ n — r /Solution hy W. J. C. Sharp, M.A. ; Rev. J. L. Kitchin, M.A. ; and others, -3. 11 1/1 I \ fSince . ; = — • . I + I , n^r 2m + n — r 2{m + n) Kn + r 2m + n-rJ the proposed series may be written -— i .rj__2;n.-i- +&C. + —i 2m,- -+&C.1; 2 (//* + n) [_ n n + l 2m +n 2m t-n—l J 33 Sum ^ -i- (±-2w. J- +&C. J = -J— (l-D)2«* ("i-V w + « ( « « + 1 » w -t w \ « / using the notation of Boole's Finite Differences, p. 17, (-n-" ...u 1 \, 1 (-1) 2m! m + » Vw/ w + « \ n ) m + n n (n + 1) ... n + 2m [The two bracketed series in line 4 are identical, and if we take twice the second, instead of twice the first, line 6 becomes (D being Boole's K) -L, r_i 2m ^ + &C.1 m + n L2m + n 2m + w — 1 J == _1_(D-1)2»*( L] = _L_A2m (l^ = &c.] m + u \ n J m + n \ n / 7489. (By Professor Wolstenholme, M.A., Sc D.) — Prove that, if 2* = o + j8 + 7i-8, the equation of the directrix of the parabola that touches the four tangents to the ellipse — — + " =» 1 at the points whose excentric angles are o, 6, 7, 5, is — [cos («— a) 4 cos(« — jS) + cos(«-7) + cos(*— 8)] a + jL [sin (« - o) + sin '\& — ^) + sin (s -7) + sin (« — 8)] b « {a^—lfi) cos a + (a2 + i-) [cos (*-o— 8) + cos (*-j8— 8) +cos (*— 7— 8)]. Solution by R. Lachlan, B.A. ; G. Eastwood, M.A, ; and others. Let the tangential equation of the parabola be A\-+ Wfi^ +2F fjLu+ 2Gv\ + 2B\fi = ; then the equation of the directrix in Cartesian coordinates is (Salmon's Conies, Art. 294) 2G:r + 2Fy = A + B. But the tangential equation of the parabola must be of the form [a\ cos ^ {a + fi) + bfjL sin i (o + jS) + v cos i (o — /8)] X [a\ cos i (7 + 8) + */i sin i (7 + 8) + v cos i (7 - 8)] + k [a^\^ + b^fi^ - 1^-] = 0. Comparing the two equations, we have k «= cosi(o — i8) 008^(7-8). A + B = fl'cos J (a + jS) cos i (7 4- 8) + b- sin i (a + jS) sin i (7 -r 8) + (a- + b^j cos i (a - i8) cos i (7—8), 2G = a [cos H^ + 3) cos i (7- 8) + cos ^{a — fi) cos i (7 + 8)], 2F =b [sin i (o + /8) cos i (7- 8) + sin i (7 + 8) cos i (a- 3)]. Hence the equation of the directrix is ax [cos (« — a) + cos (»- jS) + cos (»— 7) + cos (»— 8)] + by [sin (» — o) + sin (»— j8) + sin («- 7) + sin («— 8)] + {a^- b^) cos s + («' + b'j [r.os(«-o-i8)4cos (v- 0—7) + cos (* — 0-8)] = 0. 34 7519. (By R. Tucker, M.A.)— AP, A'F are two confocal and coaxial parabolas, the parameter of the former beinfif twice that of the latter ; prove that, if any chord QQ' be tangential to the inner curve, then the dis- tance of the mid-point of QQ' from the directrix of AP is tnsected where it meets AP, and the tangents at Q, Q' pass through the other point of trisection. Solution by R. Lachlan, B.A. ; R. Knowles, B.A., L.C.P. ; and others. Let QQ' be the chord touching A'P' in P', and V the mid-point of QQ' ; and let the diameter through V meet the curve in P, the tangents at Q, Q' in T, and the directrix of AP in Z. Then SZ must be perpendicular to QQ', and the tangent at P ; hence, if SZ meet QQ' in Y', and the tangent at P in Y, Y, Y' must lie on the tangents at A, A' to the parabolas ; and evidently since SA « 2SA', YY' = iSZ, therefore PV = \ZY ; and since PT = PV it follows that ZV is trisected in P and T. 7366. (By C. Leudesdorf, M.A.)— Two particles A and 0, each of mass m', are connected with each other by an elastic string whose modu- lus of elasticity is A and whose unstretched length is / ; and they are connected with another particle B of mass m by two massless rods, each of length a. The system lies on a smooth horizontal table, and is held 80 as to form a straight line ABC. The constraints at A and C are re- moved, and at the same instant each of the particles A and C is projected with velocity v in a direction at right angles to ABC. Find the stress along either rod at the moment when the particles form an equilateral triangle. Solution by D. Edwardes ; Professor Nash, M.A. ; and others. Since evidently m moves in a straight line, if x be its distance from a fixed point at time <, a/, y' the coordinates of w', and 6 the angle between the rods, then (w + 2w') 2aw' sin 6 — = 2w'f> (no external forces). dt dt Again, since xf^x-\-a cos B,y'^ a sin 0, the kinetic energy is 2m'2t;2-2«2;n'28in2 h , -^a ; + f» a*^. Subtracting m'v^ from this, and equating the result to the work done by the tension, we get at once wmV + 2a (w + 2m') ( -^ cos' 6— « + asine] ©2. w! (wj + 2m') a=^— 2a-m'2 sin^ Q 35 Differentiating, and putting » 30°, X (m + 2m') ^a{m + ^m") - -^ m | ^3 + mm'^v^y/Z e a^m'{m + ^m'y^ Now, when 6 - 30°, r^Lt^' x^ ^36^ + 0; am' whence, substituting, we have, when Q » 30°, s 2mw'i;2 + A. I -~ (m + |m') — 1am' \ ~7z (m + \n^f Let H be the stress along either rod, then m — - «■ B^/S, 2fMf»V + A I y (m + f m') - 2am' \ therefore R = m ^ a (m + f m')2 ^ 2 ^ww'^y^ + gX { 2gm + (9-l _i dx^-^ dx^^^ 2«+l^ Solution by D. Edwardes ; Belle Easton ; an^ others. Let y " (1 — a?^)**, and D s -—. Then, integrating by parts, dx u mm rD»»+»»-iyD'»-*»y— D'»+»»-2yD'»-"»+iy + ... ...(-)*»-» D»yD'»-v"[* +(-)»» r*D»»yD»y2» a>^i be the component angular velocities about the axis of the ellipsoid, and V the velocity -potential referred to axes which at any in- stant coincide with the axes of the ellipsoid ; then we have dx a' dy b^ dz (^ rf^Y rf^Y ^p^ along the surface, and — - + -— -r + -—r =» throughout the liquid ; dx^ dy^ dz^ whence we easily deduce b^ + cr c^ + a^ a^ + b'^ which, by giving different values to V, represents a series of similar and concentric hyperboloids, having their common centre at the centre of the shell. Transform this system to its axes, and let its equation be — + -^ + — = K, where K is the parameter ; then the directions of a fi y VOL. XLI. E 38 the velocity of any particle of fluid are given by the equations dx dv dz . . i. X* v^ zi — = -i « — , or, integrating, — « -^ = — ; ^ y_ ±^ A >t K a i8 7 where A, /*, v are arbitrary, and a, i8, y known constants. 2. Denoting by a, i, t? the three semi -axes of the inner surface of the shell, by jp, g, r the three components of the rotation with respect to their directions, by ar, y, z the coorinates of any point of the fluid with respect to the same, and by ^ the potential of the strain ; then since, as is well- known, ^S—c* e^—a^ a2_j2 we have, therefore, for the differential equations of the line-system in question, — ^ « ; — ^—- = — - — ; the complete integrals of which in gz-^hy hx+fi fy+ffx finite terms are, as is also well-known, where A,, Aj, A3 are the roots of theequation A'— C/^+^' + A^ A— 2/S7A=»0 ; h*'*i^u ^2»«2«2> ^*«3'»3 the corresponding values of Ij m, n as given by the equations Am+^fi — /A = 0, /» + A^— wA=» 0, gl+fm—n\ «= ; and fj, c^t c^ any arbitrary constants the ratios of which are given for each particular line of the system with any single point o^yV of its course. 7538. (By Professor Hauohton, F.R.S.J — Show that the law of pro- pagation of heat in a solid sphere is -^ = « ( — + — "T" 1 • at \ ctx X (tx I Solution by C. Graham, M.A. The general law of propagation of heat in an isotropic solid is dv [d^ , dh ^ d^v\ dt \dx^ dy^ dz^l' where a depends on the conductivity and specific heat of the solid. Trans- forming this to polar coordinates by the well-known transformation, it , dv fdh,2dv,ldh, 1 d^\ beoomeB _ = « ^_ + _ _ + - _ + ^^-j- _ j. but V is independent of 6 and 4> since the solid is a sphere equally hot at points equidistant from the centre, therefore the equation reduces to the form given in the question. 7513. (By Professor Minchin, M.A.) — Give a simple geometrical proafof the existence and fundamental property of the Instantaneous- Acceleration Centre in the uniplanar motion of a rigid body. Solulion bg C. Qbaham, M.A. Suppose A and A' to be two consecutive poaitiona of tlie centre of in- BtantttneouB rotation, and P any point. II the acceleration of P is lo bo lero, we must have the Telocitf of P the same in direction and nuignitude when the body is lotating about the two points A and A'. Therefore, if ■ and u' he the anguhir velocities in these two positions, we must hnva AP . ■ •■ A'P . ■' to mate the velocities equal. Therefore F must lie on a known circle, aincethe ratio of AP to A'P is known ; and to make the velo. cities parallel we must have AP parallel to AT' when P' is the aecond position of P, and therefore AA' am (PA'Aj — AP . ^ when fi is the aniall angle through which the body baa turned in going from the first to its aocond position. This determines the angle PA'A, and therefore the position of P on the circle already found. Bo we see there ia one position, and only one position, of P. Again, since the acceleration of P is zero, the acceleration of any point relative to P is its absolute acceleration ; "but, if Q is any point, its accele- ration relative to P along Pft- PQ . wV and perpendicular to PQ - PQ . j^' and therefore the angle which the resultant acceleration makes with PQis = tan-' (^/"' )> which is independent of the position of Q. 7483. (By Professor WoLsraNHOLiiE, M.A., 8c.D.)— In Waltob'h Mechanieal Problem! ISrd ed., p. 19, " Centres of Gravity of Solids of Eovolution," Ex. 10) it is slated that the centroid of the solid formed by scooping out a cone from a paraboloid of revolution, the bases and vertices of the two solida being coincident, bisecta the axis ; prove that (Ij this is true for the volume formed by the revolution of any segment, cut off by a chord PQ, from any conic, about an aiis of the conic, provided PQ does not cut the axis ; also, more generally, (2) if FSf, QN be drawn perpen- dicular to the axis, and a aphere be described on MN as diameter, the cen- troid of any part of the vnlume generated by the segment, intercepted between two planes perpendicular to the axis of revolution, ia coincident with tbe centroid of the volume o( the sphere intercepted between ths same two planea. Solution Ay T. Woodcock, B.A. ; Professor Mati, M.A. I. Taking the nxisMN for axis of X, and the middle point O of MN for origin, the equation to the conic may be written y' = Ax'*hx + C. Let PM - A, 'ON - b', MN = 2b. Consider two equally narrow strips RL, R'L', drawn paraltol to Oj/, on opposite aideii of it, and equi- distant from it, meeting the curve in R, E', the chord PQ in r, r", and the axis in I^ L'. Let OL = >. = OL'. Wb have A'i- Ao't-BatO, A'*-Ab»-B« + 0, EL»-AA* + Ba Al«, ,L = ^ X + i^' ; thcn^fore If L' - rh' - ^"^^f^" 40 Similarly R'L''— r'L'' xb equal to the same quantity. Therefore the Tolnmes generated by the elements Br and RV, when the fig^nre reTulvc^ round the axis of rr, are equal. Therefore the centre of gravity of the volume generated from the segment PQ is at O. 2. If the circle on MN as diameter meet RL in S, SL* « a'— A*, there- fore RL* — rlr varies as SL'. Therefore the volumes generated by the elementu Rr and SL are always in the same proportion. Hence the second part of the theorem follows. [Otherwise: * = f J^C^s'— yi')4), which, when a^ = a^ — 0, becomes a^ = 12 + S?«a6+»f On all the edges of a (6 + r)-gon A draw 6-gons, making 6 + r summits 665, and 12 + 2r summits 665 with a circle of 12 + 2r pentagons, within which draw 6 + r more 6-gons coUatoral with a central (6 + r)-gon A', and making with the same 5-gon8 the like summits. We thus get a (26 + 4r)-edron P, which has no fiame, but a circle C(i2+2r) = !• If a (6 + r)-waUed frame F is possible, F can be imposed on the (6 + r) -gon A', and equally well on any (6 + r)-gonal face of any polyedron. Let F be imposed on A', a face of P. The frame F will have a contour 6 besides these walls will contribute to that a^ r pentagons, and the compL ted solid can have no more than a^^l2 *-r-*-'R pentagons. In imposing F we have made no change in the 12 + 2r penta- gons of P, and we have added to them not fewer than R more. It follows that 12 + 2r + R > 12 + / + R ; Q.E.A., if r > 0. 41 7535. (By R. Lachlan, B.A.) — ProTe that, if a< ir, and n be posi- tive and < 1, f" ^"^^ " «^" ''^ Jo 1+^ and Jo 1 + 2(0 cos o + «' sin nv sin a ' X** - ^dx IT sin (l-n) a 2x cos a + a;^ sin nv sin a Solution iy "Professor WoLSTENHOLME, Sc.D. ; R. Knowles,B.A. ; and others. By a well-known formula, ^ = -r^ — «»-!, if «>0< 1 ; for Jq x + a sin«ir all values of a, real or impossible. Putting a = cos o + 1 sin o, and there- fore = — , and denoting the integrals proposed by x + a a^ + 2:rcosa +^ 1 o ^ xr j U, V, respectively, we have U + (cos a—i sin o) V = -7^ — [cos (w — 1) a + 1 sin (n— 1) ol ; sm WIT \ / J7 whence U + V cos a =* -r^ — cos (1 — w) a, V sin a = . ^ sin (1 — m) a ; Binnir Binnx whence the results stated. In these results, a is an angle determined as cos-^ (coso), and the limits are accordingly and ir ; but, since writing — o for a does not alter erther member, the results will be true from o = — ir to o = ir. Both results are included in I * x*"-^ , ^ v sin(l-»?)a x^ + 2x cos o -•- 1 sin wir sin o ' '0 if we take w>0<2, a>-ir < 2p, and joa > — » < ir ; and in this form, which is not really more general than the one proposed, the result will be found in Wolstenholme's Math. Problems [1919 (60)]. If we put a = — IT or IT (in U or V), both members become 00 , but pro- bably the limiting ratio of the two members as a tends to ir is not one of equality. [SdO De Mokgan's Cakulus, p. 666.] 7439. (By R. Rawson.)— Two inclined planes of the same height and inclination a, jB, are placed back to back, with an interval between them {2a) . Two weights P, Q are placed one on each inclined plane, and kept at rest by the connection of an inextonsible string, indefinitely long, passing over two small tacks, one at the top of each inclined plane. A weight w^ 42 having a vertical velocity (c), is then placed on the string by a smooth ring at a point midway between the inclined planes. Show that the system thereby put in motion will come to rest at a point determined by a root of the quadratic (4P'sin«a-«^«»- — (4^flPsina + w;c2)«-f 2P«sina+ — ^ — -0. Solution by D. Edwardes ; Professor Matz, M.A. ; and others. Let a fixed horizontal plane below the system be taken as plane of reference, referred to which, let V be the original potential energy of the system P, Q, and h the common altitude of the wedges. Then the whole energy at first is V + m^A + i — c^. Now, since the ring is smooth, the ten- 9 sion is the same throughout, and w descends vertically. Let j: be the dis- tance it describes before reaching its position of instantaneous rest. Then the sum of the distances described by P and Q along the wedges is 2 [(a:* + a')* - a]. There is evidently no impulse on the weights, and there- fore no energy lost. Also, since the weights P and Q are in equilibrium at first, P sin a = Q sin /8. Hence the increase of potential energy of P and Q is 2P sin o [(jj^ + a^)k - «]. Therefore 4/1 * V-J-wA+i — c^-. Y^hole potential energy at last iv (A-a^) + V + 2P sina [(a:2 + fl2)| _«], 10 . or i -^ c= + m;jp= 2P sin a [{x^ + a-*)* -«], 9 which, when rationalized, gives the required result. 7396. (By D. Edwardes.)— Prove that ( ' I 'f (1 -sin a cos 4>) sin e « iir f F (m) du. Jo Jo Jo Solution by Arthur Hill Curtis, LL.D., D.Sc. The limits of integration show that the integral is extended to the sur- face of a quadrantal trianglo traced on a sphere ot radius unity, or, taking as axes of ar, y, z the radii of the sphere drawn to the angular points of the triangle, and supposing the radius vector to any variable point on the surface to make with these axesangles a, jS, 7, we have a = 0,cosi3 = sia0cos^y cos 7 = ein 6 sin 4> ; therefore, if dO. denote the element of surface, I ' 1 'F(l-sin6cos<^) smededip = I ' I ^ ¥{l — coB^)da Jo Jo Jo Jo « (*'[*'F(l-cosj8)sin/8iAg) and (riiA^— o^i), we get (2), and by squaring we get (1). 7475. (By J. O'Reoan.) — The figures 142867 are arranged at random as the period of a circulating decimal, which is then reduced to a vulgar fraction in lowest terms; show that the odds are 119 : 1 against the denominator being 7. Solution by A. Martin, B.A. ; Rev. T. C. Simmons, M.A. ; andothera. There are 6 ways in which the denominator can be 7, viz., when the decimal is -142867, -428671, -286714, -867142, -671428, -714285. In all other cases the denominator is not 7. But there are 720—6, or 714, of these cases ; hence the required odds are 714 : 6, or 119 : 1. 7236. (By the Rev. T. W. Openshaw, M.A.)-— On AB, a chord of an ellipse, as diameter, a circle is drawn intersecting the ellipse again in C, D ; if AB, CD are parallel to a pair of conjugate diameters : show that the locus of their intersection is b^x + c?y = 0. Solution by the Fbofoseb. AB and CD must be parallel to equi-conjugate diameters, therefore their equations are of form bx + ay + k =: 0, bx^ay + !, cos i (a + /8) = «/, sini(7-8) = r, cosj (7 + 8) = r' ; then, if U = left-band member, we have U 5s ipu {qf/ + q'v) {ru?*—r^w) + iqv (no' + r'w) (pu'—p'u) + Arw {pt/ +p'u) (^v'— ^v), 4U ^ ruf [pq {uf/ + vfv) + «p (jo?'— p'?)] - r^w [jju {pv^ +p'v) —pv (?»'— «''«)] + rw( pi»* + y w) [qv' — q'v) ^ Now the following identities hold good : — «v' + m'v = — m; cos 7, P^—p'q « «^ eos 8, pt/ +yv =» r cos a, qu'—q*u = r cos 3 ; therefore — = — pqu/ cos 7 + tww* cos 8 — qur cos a + pvr' cos /8 ^*^*' +(pf/+/«)(^f;'-j't;) ■a pq \yv' — «?' cos 7] + uv [w' cos 8 -p^q''\ + ^« [ jjV - r' cos o] -pv [^V - r' cos j8], and every one of these four terms « —pquvy therefore U « — IQpqruvw, 7343. (By Belle Easton.) —If a debating society has to choose one out of five subjects proposed, and 30 members vote each for one subject, show that (1) the votes can fall in 6^ ways, and (2) the chance that up- wards of twenty votes fall to some one subject will be 6-20. Solution by W. W. Taylor, M.A. ; Sarah Marks ; and others, \, Let V, «;, aj, y, z represent the subjects, then the possible combinations of votes and the relative probability of each combination will be repre- sented by the coefficients in the expansion (t; + M> + a; + y + z)^ ; and the sum of these coefficients is (1 + 1 + 1 + 1 + 1)*' = 5**. 2. If upwards of 20 fall to one subject, 21 is the least number of votes that can be recorded for that subject ; the remainder of the votes is 9. These can be given in 6' diff'erent ways, and the one subject can be chosen in five different ways ; so, the conditions of this part of the problem can be satisfied in b^^ different ways, and therefore the chance required is 5~^. 6907. {By S. Tbbay, B.A.) — If A, B, can do similar pieces of work in a, ^, c hours respectively, {a 8(b—a){e-a) 8(b~a) tl'-Ztabe , ^ 8— She , ,^v *- —, 77 *» y- -7 ]T«* (6). 8{a-0) 8 {a—b) * Now «' - Vahc is always posit ivo ; and, since a <*<a{b + c}, 2ab *(a + *)...(2', 6', 3', i^. Here (2') and 1 6') are both possible, but only one is applicable. (3') and (4') are inconsistent, one only beiug applicable. There are therefore only two possible solutions. I£ b< , (2) and (4) are applicable ; if i > , (3) and (6) are applicable ; if * = 2ae a * c (2) and (5), and also (3) and (4), are identical. the work being completed in b hours. Examples. — The harmonic mean between 4 and 12 is 6 ; so that, if a = 4, ^ = 6, c=V2y the whole work is finished in 6 hours ; A and C changing works at the end of 3 hours. Take a = 4, ^ = 6, c =* 12 ; then X = 2yY«g hours, y — lyy hours ; x = 2y'y*^ hours, y = 2^ hours... (2, 4), the whole work being completed in 5f hours. Take « = 4, * ^ 7, c = 12 ; then X = 2|^ hours, y = 2\^ hours ; x = 2J^ hours, y ^ \% hour... (3, 6), the whole work being completed in 6^'^ hours. 7552. (By the Editor.) — In a road parallel to a range, find, by ele- mentary geometry, a point at which the sounds of the firing and of the hit of the bullet would be heard simultaneously. Solution by (1) D. Biddle ; (2) A. H. Curtis, LL.D., D.Sc. 1 . Let A be the firing-point, B the target, CD the road, and AF the distance the sound will travel before the bullet reaches B. Draw BC at 50 right angles to AB ; on AB the semi-circle AGTTQB ; CI with radius BC ; IL perpendicular to AB ; FG with radius AF ; and GK perpen- dicular to AB. Draw LMO with radius BL, and KH with radius BK. Join HB cutting liMO in M, and draw MN parallel to HA, that is, at right angles to HB. Make NP =-- 4AB, and draw PQ at right angles to AB, cutting the semicircle in Q. Join BQ, and make QB = \A¥. Finally, with radius BR, draw RS, cutting CD in S. Then AS « BS + AF, and S is the required point. For it will readily he seen that by oonstruotion, and the properties- of circles and right-angled triangles, if AB«1, then AK=AF«, BL = BC2, and BP = BQ«. Now, BIl = BS, and BQ - BS 4- 4AF; therefore BP = (BS + iAF)«. But BP = BN + iAB* ; hence we have BN - (BS + iAF)2-4AB«. But BM.BL-BC and BN : BM«AB : BH=AB2 : AB^-AF', therefore (BS + iAF)«-iAB2 : BC^ = AB« : AB2- AF^, whence BS - (iAB»+ ^^J*- ^AF, which is the identical yalue arrived at from the equation (BS«^BC-)* + f(BS + AF)« BC2]* = AB. Consequently, since it is plain that (BS«-BC'>* + (AS2-BC5)* = AB, it is equally evident that AS = BS + AF, and S ia the required position. 2. The question is immediately reducible to the following:— Given, of a plane triangle, altitude, base, and difference of the other two sides, to construct it. Let ACB be the triangle, FE the diameter of its circumscribed circle, which is per- pendicular to the base AB, and CG parallel to AB, then EG.DF -f GD.DF, = ED.DF = AD^ = iAB«, is known, and EG.DF, = ^ (AC-CB)-, is known ; therefore GD.DF is known, and GD, = altitude, is known ; therefore DF is known, and therefore GE. If therefore at D, the middle point of AB, we draw a perpendicular and measure off DF, DG, GE, on FE as diameter describe a circle, and through G draw GO pirallel to AB ; GC will intersect the circle in the required point C, and in another point C which is exchided, as, if B be the target, BC is the smaller side. [Question TiSo u related to this problem.] 7416. (By R. Rawson.)— In the Royal Society's Tramaetions (Part III., 1881, pp. 766, 767\ Mr. J. W. L. Glaisheu has shown, by the assumption of 2 Ar;«^+'' for all positive integral values of r, that (AU + B V) 51 is the general integral of ^ - a'w =b £iA±Jjft, whete fir* 4r2 I p + l 22 (i? + f)tp + f)2^2! • / Show that the restriction imposed upon r is unnecessary, and that, if w = « — 2p, the general integral of the above differential equation is ^ I (» + 2)(m+l) (» + 4)(» + 2)(». + 3)Cm + l) ) ^ n.m-l.A, ^...,^ ^ («-2)(m-3) ^ («-4)(«-2Km-S)(»».-3) , ^^ > Solution by the ProfoseIi. Let » = 1^ (a:) = 2 Ara?**"^^ «a;'»[Ao + Aia;'+A2a^ + &c.] + a?'»-2[A_i+A_2a;-* + A.8«"* + &o.]...(l). Differentiate (1)> then ^ = 2 (2r + /8) A^ a;2r+/.-i^ ^ = 2 (2r + 3) (2r + jB- 1) A^ a>a»'*^-«, ax ax* hence ^-a^o, = -^i^iHo, (2), dx^ a?« ^ ^' if (2r + i3+i?)(2r + /8-i?-l)Ar = a2A^-i (3), the summation extending to all positive and negative integral values of r. The general integral of (2) is, therefore,

sin(«— c) . . n Bin io sin i3 Bin i7 = : — ^ — ;; — . ^ . — r—r-^ ^> since tan r = -: — . sm r cos- r sm a sin sin c sin a ' heiice cos PD sin « cos {« — a) + cos PE sin b cos («—*) + cos PF sin c cos (« — c) _ 2 cos IP sin 8 cosr But, from the triangle PBC, we have cos PD sin « =s cos PB sin (a — c) -f cos PC sin (« —b) &c. Hence cos PA sin « + cos PB sin * + cos PC sin c « 2 cos IP sin # cosr Al^o (Todhuntbe's Spherical Trigonometry, Art. 144), ... . 2 cos 01 sina sm a + sm + sin c = — , cos R cos r .-. ^25^-^?^[sinasin3iAP + BinJsinHBPtsin(?Bin«iCP]=co8lP. cos R Bin* 53 7567. . (B}' Professor Stlybster, F.R.S.) — Let nine quantities be supposed to be placed at the nine inflexions of a cubic curve, then they will group themselves in twelve sets of triads, which may be callied coUinear, and the product of each such triads may be called a collinear product. From the sum of the cubes of the nine quantities subtract three times the sum of their twelve collinear triadic products, and let the func- tion 80 formed be called F. With another set of nine quantities form a similar function, say F'. Prove that FF' will be also a similar function of nine quantities which will be lineo-linear functions of the other two sets, and find their values. [The inflexion-points are only introduced in order to make clear the scheme of the triadic combinations, so that the imaginariness of six of them will not matter to the truth of the theorem.] Solution by K. Bussell, B.A.. The expression denoted by Fmay symmetrically be written down thus : — F = a3 + ^ + ^^. /3 + ,;i3^„3^^3^.yS^^_3^^^_ Zhnn — Zxyz — Zalx—Zbmy — 3c«2 — 3a(m2 + «y) — 3^ {nx-irlz) — Zo{ly + mj;) (I). A little consideration suggests the transformation (where «;* = 1) + ^ + {;s=a, /4-m + wssA, ar + y + « = {^ fl + *«? + tfu;2 = /8, I + mw ^ nw^ ss fA, x + yw + zto^ '^ ri i ...(2), a + bw^ + cto ^ y, 1 + mw^ * nto = y, x + yto^ + zw ss ( ) which reduces F to the very simple form, ctfiy + \fiy + lri(—cih^—fifiri^yv{ or F = a, f, ij /*» ^ 7 and proves the remarkable property that any determinant of the third order can be reduced linearly, as in (2), to the form of expression given in the question. F' can be expressed similarly ; and FF', being also a determinant of the third order whose constituents are lineo-linear functions of the two original sets of nine letters, can by the above be reduced at once to (1). [The theorem is the analogue to Euler's theorem that the product of one sum of 4 squares by another is also a sum of 4 squares. In a precisely similar way, any determmamt of the second order is reducible to a sum of 4 squares, with the aidof «' = — 1.] 4925. (By the late Professor Clifford, F.R.S.)— Let U, V, W « be the point equations, and u, v,to = the plane-equations of three quadrics inscribed in the same developable, and let u + v + w he identically zero. Then, if a tangent plane to U, a tangent plane to V, and a tan- gent plane to W, are mutually conjugate in respect of au-i-bv + ew ^ 0, TT V W they will intersect on - — — + + -„ « 0, ^ (b^cy {c-ay {a-b)^ ' which passes through the curves of contact of the developable with au+ bv + cw and one other quadric. VOL. XLI. o 54 Solution by W. J. 0. Shabp, M.A. Let aiX + fiif/ + 7i« + 8i«^ « 0, o-j^ + fi^y + 73^ + f^w = 0, and a^x + fi^y + y^z + ^w » 0, be the three tangent planes, and let (U), (V), (WJ denote the values of U, V, and W, when the cooxtlinates of the intersection of the planes axe substituted for (x, y, z, w). Also let Un, ti^,, t 2Vi3r3,f;31-Vl,r232-r33t;,2« = (V), 2tri2W23«^3i-«?iitr33«-«?22«<^'3l2 = (W), • ^^) +JZL+ (^ ^ O f t )2 / ' since u^^ I v^^ I to^ - «2s • ^'as • *<'23 "= &c. ^h — ei e—a : a—b. And if T and T be the ordinary oovariants of U and V (A and A' still being ■■ 1), the point equation to m + Ao == is U + AT + Arr + A^V-O (B), and, since u+v^to- 0, —W « U + T + T' + V; also, at the points where « + Ar « touches the developable, the equation (B) gives equal values for A, and therefore T-¥2\T + Z\^ = and 3U + 2AT + A^T' « 0, and therefore along the curve of contact U + V + W, 3a«V, 3TJ 1, 1, 2a 1, 2a, a' 0, or -3a2(U + V + W; + 3a2V(2a-a2) + 3U(2A-1) = 0, or U(A-1)' + Va2(A-1)« + Wa2«0, TJ V W and this is identical with — + ^ ,„ + - « 0, {b-ey {e-ay (a-by ' if A« : 1 : (A- 1)2 - {b-eY : {e^ay : (a-by, or A and u + W B» the same as au + bv + cto — 0. b^e 7571. (By Professor Hauohton, F.R.S.) — A solid body is bounded by two infinite parallel planes kept constantly at the temperature of melting ice, and by a third plane, perpendicular to the first two planes, ^ept con- 65 stantly at the temperature of boiling water. After the lapse of a very long time, show that the law of distribution of temperatures will be repre- sented by the equations (between the limits y » ± iir) V a a* -« cosy +**-** cos 3y + &c., I = acosy + 6cos 3y + &c. Solution by T. Woodcock, B.A. ; Prof. Nash, M.A. ; and others. We have to find v in terms of x and y, knowing — ; + -— - » 0, also dx' dy t;»0 when y » ± iir, and v=^\ when d7»0, as well as when x ^ co , Try i; — 2 M cos ry, where u is independent of y. We must have r = an odd integer « 2m + 1 say, and also -3-5 — r"f# = ; .\u — Ae^ + 2a cos (2m + 1) y, the limits of y being ± ^t. 7679. (By R. A. Boberts, M.A.) — Two uniform spherical shells attiact according to the law of the inverse fifth power of the distance ; show that, if they cut orthogonally, they will be in equilibrium under the influence of their mutual at^ction. Solution by Prof. Townsend, F.R.S. ; J. A. Owen, B.Sc. ; and others. The attraction, for the law of the inverse fifth power of the distance, of a thin uniform spherical shell, upon a particle in its space, either external or internal to its mass, being directed towards the point of its surface nearest to the particle, and varying directly as the radial distance from its centre and inversely as the sixth power of the tangential distance from its surface ; it follows at once that, for two such shells intersecting at right angles in a common space, if an elementary cone be supposed to diverge from the centre of either, it will intercept on the surlace of the other two elements of mass, whose attractions by the former pass in opposite directions through its centre, and are to each other directly as the cubes of the radial distances from its centre, and inversely as the sixth powers of the tangential distances from its surface — that is, directly as the cubes of the radial distances from its centre, and inversely as the cubes of the perpendicular distances from its plane of intorsection with the lattc^r ; and the two attractions being consequently equal in magnitude and opposite in direction, therefore, &c., as regards the property in question. 7569. (By Professor Townsend, F.R.S.) — In a tetranodal cubic sur- face in a space, show that — (a) The four nodal tangent cones envelope a common quadric. (b) Their four conies of intersection with the opposite faces of the nodal tetrahedron lie in a common quadric. 56 (e) The two aforesaid quadrics envelope each other along a plane having triple contact with the surface. Solution by Professor Malet, F.R.S. ; Prof. Nash, M.A. ; and others. The vertices of the tetrahedron of reference being the four nodes, the equation of the cubic is of the form — + — +— + — =0; and the X y z w equations of the four nodal tangent cones are ab ae be * ab ad bd ae ad cd * be bd cd which envelope the quadric \ a b e d I \ao ae be ad bd cd J and each of which meets the opposite face of the tetrahedron of reference in a conic situated on the quadric £?: + ££ + f + ^ + V^^'J^L = 0. ab ae bo ad bd do Now the plane ~ + -^+ — +-^ — is a triple tangent plane to the abed cubic, the points of contact being (a, 6, - c^ — d}, (a, — i, — c, d), (a, ^b,c, — rf) ; therefore, &c. Note on Inverse- Coordinate Curves, with Solution of Quest. 6969. By R. Tucker, M.A. The general discussion of the properties of these curves is given in Salmon's Higher Curves (Ist ed., p. 238, &c. ; 2nd ed., p. 244). We have xxf = yy* = <52, whence tan = cot b', i.e, + 0' = iw and rr' sin 20 — 2(?'^ ( 1 » 2) . If If = inxy then i -c. c. [i.e. inverse-coordinate curve] is a; = my, i.e. a line through the origin has for i.-c. c. a line through the origin equally inclined to other axis (rectangular axes). Therefore, if any number of points on one curve are collinear with the origin, there will be the same number on the i.-c. c. collinear with the origin. If y=mx + b, then i.-c. c. is bxy = c^ {x^my). Hence, if parallel chords be drawn to the primitive curve, the corresponding points on ** i.-c.*' lie on hyperbolas, the locus of whose centres is a straight line through the origin orthogonal to the above system of chords. If ip has the usual meaning, then cot i//' *%» cot if^ = cot 2Q '^ cot 26' = 2 cot 2d. This enables us at once to solve the following Question (6969) : — **If, in a parabola vertical chords AP, AP', complemontally inclined to the axis, make angles \(^, y with the tangents at P, P', then cot f ^ cot v^ = 2AAPF / L«." 57 The parabola y^ « \ax will be its own ** inverse '* if c = 4a = L, then 2a APF = AP . AP' sin PAF = n-' cos 2» = 2c2 cot 19 -= 32«2 cot 2« = L2 (2 cot 2a) = L2 (cot 4^' cot ^) . Similar properties are readUy obtained for other curves which are their own i.-o. c. as 2xy = o^ {a^ = 2c^), 3^ =■ ay^ (c = a), &c. [Another solution of Quest. 6969 is given on p. 114 of Vol. 37 of the Beprint.'] 7194. (By Professor Wolstenholme, M.A., ScD.) — In the examina- tion for the Mathematical Tripos, January 2, 1868, Question (6) is as follows: — **If there be n straight lines lying in one plane so that no three meet in one point, the number of grouj)8 of n of their points of in- tersection, in each of which no three points lie in one of the n straight lines, is Hw- 1)." Prove that this is not true; but that, if "fi-.'-ided polygons " be written for " groups of n points, &c.," the result will be tiue : and calculate the correct answer to the question enunciated. Solution byW. J. Q-keenstheet, B.A. ; A. MacMuuchy, B.A. ; and others. Denote the n straight lines by 1 . 2 . 3 ... n. Make a group of n intersec- tions in this way : — I and 2, 2 and 3, ..., «— 1 and «, n and 1. Then there are two, and only two, points on each straight line. Hence we must take two points on each straight line, for if not there would be more than two on some line or lines. So that we merely require now the number of ways the n intersections may be arranged in a ring, that is -^ (n - 1) ! 7247. (By Dr. Curtis."^ — Two magnets, whose intensities are I,, lo, and lengths ^i, a^j are rigidly connected so as to be capable of moving only in a horizontal plane round a vertical line, which passes through the middle point of the line connecting the two poles of each magnet ; if 2a denote the angle between the lines of poles of the two magnets in the direction of opposite poles, while $ denotes the inclination to the magnetic njeridian of the line bisecting this angle, prove that (1) the positions of stable and umtable equilibrium (discriminating between them) are given by tan B = (T,«i +I2"2) ^" «/(Ji^i~l2''2) I ^^^ hence (2), if the intensities of the two magnets be invirsely proportional to their lengths, the posi- tions of equilibrium will be such that the lines of poles of the magnets will be equally inclined to the magnetic meridian. Solution by W. M. Coates, B.A. ; Belle Easton ; and others. The moment of the fiisl magnet is Iia^, and the angle its axis makes with the magnetic meridian = — o; therefore the moment of the couple tending to turn it is proportional to Ija, sin {d- a). Similarly tlie moment of the couple tending to turn the second magnet in the opposite direction is proportional to Igf/jsin (& + o). Hence, when the system is in equilibrium, Ii«i sin (0 — a) = L^^.^sin ((9 + a), whence tan =s (Ijffi I- 12''2 tan a) / {li(fi - la'^g). [Whether the sign of the right-hand 58 side of this equation be positive or negative, as the tingle 6 is sought from its tangent, there will be two solutions 0iy 0], whose difference is 180°. The ttadle position is that in which the magnetic axis of each needle, in the direction of its north-seeking pole, makes an acute angle with the meri- dional line in its northerly direction, as in such position, if the system be turned through a small angle, the moment tending to return it to its original position is increased, and the opposite one diminished. The unstable position is derivable from the stable by turning the system through 180°.] If the condition Ijai ^I^O] be fulfilled, 0=90°, from which it follows at once that the axes of the magnets are equally inclined to the meridian. 7508. (By Professor Sylvester, F.R.S.) — If m, n be any two square matrices of the same order M » (mn— nm)>, N a {mht - nm-) {nhn — mn^ — {n^m — mn^ (m*n — nm^ , m', mn + nnif n^ in\ mn + nm, «' iw', mn •♦- nm, n^ ; and D the determinant to the matrix prove that D is an invariant to m, ft ; that is, remains unaltered when (supposing p^~p'q ^ I) pm-k-qn and p'm + q'n are substituted for m and n. Solution by the Proposer. Let m become m + en where c is infinitesimal ; then the increment of P divided by e is mn + nm, mn ^r nm, mn + nm. mn + nm, mn + nm, mn + nm. n^ ns ns m\ 2n«, n- + m\ '2n\ n2 m\ 2«2, n« i.e., 0. Again, for N, call (m'n — nm^ =« A, (nhn — mn'') = B ; then N - AB — BA , 8A aB and — = (mn + nm) n — n (mn + nm) « B, — — n'. n - n . n^ — hence 5N = 5A . B-BaA = (B--B2) 8e = 0. Finally, 5 (mn — nm) = € (n - n) = ; hence oM + jSN + 7P is unaltered by the change of m into m + en and in like manner of n into n + §n; whence it follows, by the same reasoning as in the theory of ordinary invariance, that w and n may be changed int^ p»n >- qn and p'm + q'n, provided pq*^ pfq = 1 without oM + jSN f 7P undergoing a change, and consequently without its determinant changing, so that this latter is a binary invariant of the matrices m, n, as was to be pr6ved. [Professor Sylvester caUs attention to the immense new horizon in the theory of Invariants opened out by this question, which forms part of a general theory of Matrices, including the algebraical theory of Quaternions as an insignificant single case ; and in which he connects the subject with the ordinary concomitants of Ternary as well as of Binary forms, and in such a manner as to react upon the ordinary theory of Invariants. The theorem in the question is paH of the solution of the prodigiously difficult subject of Involution of Matrices, now happily accomplished, or brought at least within a stone's throw of accomplishment.] 59 7558. (By W. J. C. Sharp, M. A.)— If A', B', C, D', be the feet of the perpendiculars from any point on the four faces of a tetrahedron ABCD, show that AC"^-BO'=» = AD'-'-BD'^, &c., and conversely. Solution hy W. G. Lax, B. A. ; Margaret T. Meter ; and others. Join AP, BP, where P is the point from which the perpendiculars are drawn ; then, since PCB and PCA are right angles, AC- - BO'S = AP2 - PC- - BPs + PC* = AP«-BP«. Similarly, AD'* - BD'^^ = AP* - BF?, therefore AC*-BC* = AD'*-BD'«, and so on, and conversely. 7364. (By W. S. M'Cay, M.A.)— If the line joining two points on two circles subtend a right angle at a limiting point, prove that the lo<;a8 of the intersection of tangents at the points is a coaical circle. Solution by T. A. Finch ; Sarah Marks ; and others. Let A and B be two circles, PN the fL chord subtending the right angle at F. Let PN meet the circles again in M and Q, and draw PL, ML, NL and QL perpendicular to L the radical axis of the circles and com- plete the Figure. Then RP _ sin RNP _ cos BNM KN *" sin RPN cos APQ ^ MN 2AP " 2BM • PQ FM.FNsinMFN AP FP.FQsinPFQ ' BM' and, from right angle, sinMFN « cosPFM ; sin PFQ = cos NFQ, also Z PFM - / NFQ (since angles MFN and PFQ have same bisectors) therefore RP» ^ FM8.FN« AP* ^ BF.ML.BF.NL AP» M^iAFi« const RN* FQ».FP»'BM* AF.QL.AF . PL* BMs" AP*.BM« Therefore, &c. [The Proposer remarks that this proof is much more elegant than his own. The theorem was originally derived by reciproca- tion from Question 5395, solved in Iteprint, Vol. xxix., p. 23.] 60 1945. (By the late C. W. Mbrripibld, F.R.S.) — Find a rectangular parallelepiped such that its edges, the diagonals of its faces, and the diagonals of the solid, shall all be integral. Solution by As^jtosh MukhopAdhyAy. Lot X, If, 2 be the edges of the solid ; then the diagonals of its faces are (^''' + y'-)*» (y' + ^^*» (i' + ar')* ; also, the diagonals of the solid are equal to one another, and represented by (x^ + y^ + ^ijj, "Wq i^ave, accordingly, to investigate whether it is possible to find positive integral values of X, y, z, which make j:, y, z, (x'-k-y^)^, {i/'^ + z^)^y (-' + ^")*, {x'^ + y^ + z^^ all integral. Let • ar'-ey^ = (fe2 + 1)2, y^ + gS = (/j^. ija (1^2), «'+ar3= (w2+l)2, a;2 + y2 + «-- (n^+lf (3, 4). Then it is well-known that the solutions of (1, 2, 3) are X ^2k \ y = 21 \ 2 =» 2m \ y = A-2-1 f z =/«-l j a;« m'-l j * Now, substituting in (4) from (2), we get x^ + (/-^ + 1)2 « («2 + 1)2. There- fore X = 2w, l^t i = w2— 1, therefore n'' — l'- = 2 ; hence the solution of the original problem depends on an equation of the form (^+,y)(^— y) = 2. Now, a moment's consideration shows that this equation has no positive integral solution ; for, assuming Xy y to be positive integers, and since {x + y) (x—y) a. 2, (x ♦.y), (x-y) must be each a positive integral; and since the composition of 2 is unique (2x1), we must have x+y « 2, x—y=sly which give j* «= J, y = i, — iractional values. Hence, it is demonstrated that the original system of equations has no positive integral solutions, and it is impossible to find the rectangular parallelepiped in question. 7573. (By Professor Hudson, M .A.) — Parallel forces act at the angular points of a triangle proportional to the cotangents of the angles. Caa they be in equilibrium ? Solution by (1) B. Reynolds, M.A. ; (2) W. J. Barton, B.A. 1. Yes : if a^ + b^= 3^2, and if BD'«^PA= *cosA, CD' being the direction of the forces. For cot C = cot A + cot B, cote sin C ^^ „^„p, ^ Bin«0 + &c. | . VOL. XLI. H 62 Let this be written p = H ^ (1 +N2 8iii2e + N4 8in< f N68in«0 + ...) ; then, since f*'sin» 0de « ?i=J . ^ . ^ ... i It, (n even), the mean value of p is a « £. (1 -^2 cos2 X + 1^ sin' A C082 A + ^V* sin' ^ co8*A + ...), nearly. b Thus the mean value of p at a point on the equator is b. Take a » 3962-824, b « 3949*585 miles ; then the mean length of the knot at a point on the equator is T * 1-14889 mile = 1 m. 1 fur. 42 yds. 108U0 Approximately, the length of the knot is -3^1 = 1-1566 (l-^cos'X + l^sin'Acos-A+A^sinSxcos* a) « 1-1666(1 --00667 0082 A). At 30**, 46**, 60° we have, respectively, p' = 3966-460, 3962-881, 3969-6, and length of knot 1-1508, 1-1527, 1*1647 mile. 7681. (By C. Leudesdobp, M.A.)— If A + B fC = 180*», (y— 2!)cotiA+(«-ar) cotiB+(a?-y) cotiC = 0, (y2_5;2j cot A + {z'^-x^) cot B + (x^—y^) cot C = ; i.1. i. y^ + z^ — 2vzco8A g' + a:"— 2arcosB a;2 + t/2— 2a:vco8C prove that •- r-^^^ = t-t^r — . .. i: . Solution by W. J. Barton, B.A. ; Margaret T. Meter ; and others. Let a, by c be the sides of a triangle having opposite angles equal to A, B, C ; then we have 2 (A - c) cot iA oc 2 (sin B — sin C) cot ^A oc 2 sin ^ (B — C) cos iA a 2sini(B-C)8ini(B + C)oc5[8ini(2B)-8inH2C)] « 0...(1), 2(*2-(j2j cot Aa Ssin (B + C) sin (B-0) cot A oc ^sin (B-C) cos(B+ C) oc2(8in2B-8in20) = (2); therefore, from (1) and (2) combined with the given equations, b-c ^ e_ja ^ a-b . b^-c^ ^ c^-a^ ^ a^-b^ ,, p-z z—x x—y y^—z^ ^—x^ x^—y^ therefore — = — = — ; therefore a triangle can be constructed with X y z X, y, z for sides and A, B, C for angles; hence we obtain -^ = const. = y^^-^^-2.yzcosA ^ z' + x^--2ztcobB ^ ^ sin^A sin^A sinSfi 63 3835. (By the Editor.)— The sides of a triangle ABC are BC = 6, CA « 6, AB - 4 ; and Q, R are points in AC, AB, such that CQ = 2; Bit = 3. Show (1) by a general solution, that the distance from B to a point P in BC, such that / CQP = BRP, is BP = i (601* -13) = 3-83843 ; and (2) give a construction for finding the point P. Solution hy D. Biddle. ; Prof. Ma.tz, M.A., Ph.D. ; and others, 1 . Draw (Fig. I) PM, QS parallel to AB,andPN, RT parallel to AC ; then the triangles QMP, RNP will be simUar, and PN : PM = NR : QM ; AC.BP T>., AB.PC but PN - NR- BC AB.PT PM QM = BC AC.SP BC ' BC ' Fig. 1. therefore AC . BP : AB . PC = AB . PT : AC . SP, BP : PC « AB<. PT : AC2. SP ; but PC = BC-BP, PT = 1^1^- BP, SP = BP+ 5^i^-BC; AB .-.BP : BC-BP=AB.BC.BR-AB BP : AC^P + AC.BC.CQ-AC^BO, , gp^ AC . BC . CQ-AC^BC + AB2BC-h AB . BC . BR ^p AC^-AB^ AB.BC^.BR . AC3-AB* ' r?f^^ AT5M^T> ((AC.BC.CQ-AC2BC + AB2BC + AB.BC.BR)3\» 2(At/'-Aii)15i:'= I -^ 4 (AC«-AB«) AB.BC2.br / - (AC.BC . CQ - AC» BC + AB^ BC + AB . BC . BR). In the given case, BP = ^ [(601* - 13)] = 3*83843. 2. Draw (Fig. 2) QD perpen- dicular to BC ; make DE = QD ; qoin RE ; make CF = BF (that IS, from the mid-point of BC draw a perpendicular thereto). Draw EH parallel to BF, to meet AB in T ; then TRE wiU be a triangle with an angle T = ABF = ABC- ACB, and the circle RPE, drawn round this triangle, will cut BC in the point P required, so that Z CQP - BRP. For / SOD = SED = SRI, ZCQD-BRI =. ABC-ACB, .-. SQC-BRS = ABC-ACB; and, by construction, we have Fig. 2. I PER + PRE = ABF = ABC-ACB ; but / PER = PQS, .• . / PRE + PQS » ABC - ACB = SQC - BRS, 64 and PRE + BUS = SQC - PQS, hence / BRP « CQP. [If BC » a, CA « *, AB « «, BR == m, CQ = «, and BP « x, wo have cot BRP = ^-^/^i^ = cot CQP = «-^'^-^^ ^^« C . :r sinB (a — a:) sin C therefore — cosec B cosec C = cot B — cot C, which gives X a — x ., J .. J. wc n tfcosB — AcosC c2_32 the quadratic equation = ; — — r— ; ox a—x ab whence, putting, for shortness' sake, k=nb^rmc—b'^■\■c^y we obtain the general value x - ^ ^^_ {k ± [A;3-4fMi) X<. Ex.— Let • « 2, y = 3 ; then X< = A^ + a* + 2x^+2x^+Zx7+Z3rfi + ix^+ and X,j = x^-tx* + x^^{l -a;2)(l ^x^) X<. 67 Solution hy W. J. C. Sharp, M.A. 1. Evidently X,- is the coefl&cient of 2< in the expansion of the infinite product (1 +ar«)( I +xH)(l + sfiz) ...,so that thisisl +Xi« + X22* + X3z8 + &c. ; but, if xz ss y, the same product = (l+y)(l+ary)(l+a:«y)(l + ar3y)... « (1 +y) (l + Xiy + X;y2 + &c.) = (1 +a?«) (1 +Xia'X! + X2a;V + X3iF3«3 + &c.) ; hence, equating coefficients, Xj = (1 + Xi) a?, Xg = (X1 + X3) a*, &c., &c., Xi = (X<_i + Xi) x\ so that Xi = ~y Sec, 1 — x •"1-** *"^" (l-a;)(l-a:2)...(l-^») [a value obtained in a different way in Art. 2]. Similarly X,-,y is the coefficient of z* in the expansion of (1 +a;2)(l + xh) ... (1 +x^z), so that this product = 1 + Xij z + X2,> 2* + Xg, J 2^ ^ &c. ; and if, as before, xz — y, this product = , '^^. (I + xy){l + x-y) ... (I -^xJy) 1 + xiy " iT^ { ^ ■" ^'•■' '^ ^ ^^•' "'"' * ■^•■' ''"' ■*■*"•)' and, multiplying by 1 + 3-> * ^ 2 and equating coefficients, a?^*i + Xi,,- = (1 + Xi,y);r, a>^*iXu + Xj.j - (Xi,j +X2,i)ar5, &c., a^*iX<.i,y + X.,y =(X,-.i,, + X<,,)a;% so that Xi,y = ^Jlll^ &c., X- -^iii-I^ililix - ^ '^ (l-a:/)(l-rc^-^)...(l-a;'-<^^) *'•'"' l-a;< *"^''' (l-.a:)(l-a:2)...(l-a:«) = (l-ari)(l-a^-i) ... (l-aji-«**i) X,-. 2. Evidently X7 = a: + .i2 + a^+ a:* + &c. = — ^, and Xj = a;3 + a:4 + 2x« + 22« + 3j:7 + 3a;"+4a;9 + &c. -«.(i+.)a.2...3.«+4.«+&c.) = ^.(-j::^^. Now X3 = (a^ + .r« + a;> -^ &c.) Xj ; for, if v„ be the coefficient of a^ in Xs and /An that in Xj, y„ is made up of /tn-s, the number of solutions of j» + ^ + r = fiin which one value is 1 ( jt?, ^, and r being all unequal in- tegers) ; of/u«_6, the number of solutions in which one value is 2 and /gS a?' o none 1, and so on. Therefore X. = :; X., « -; r— — - — -;:-, and l-a;3 (1— ar)(l— a?-)(l -iT) a repetition of the same argument leads to the result above stated. 7377. (By Professor Syltester, F.R.S.) — Integrate the equation in differences Wn . i — «« + w (» — 1) «,» -i + (2« — 1) »«, 68 where «» denotes the product of n terms of the fluctuatiBg prog^ression 1,1,3,3,5,5,7 Solution by W. J. C. Sharp, M.A. The equation u„+i = m« + « (« — 1) Wn-i is remarkable, as, though it is of the second order, when solved by successive substitution it only involves one constant. The solution is «n = «,»«• This is easily verified as follows : — Let n a= 2jp — 1, then »«« + »(« — 1) «„_ia=l .1.3... 2jp— 3 . 2p—l , a + 2 (jo- 1) {2p- 1) . 1 . 1 ... 2jp - 3 . 2j»-3 . a = 1.1.3.3 ..2p-3.2p — 1 .2i?-lfl = «n+ia. Let n =B 2j», thena>na + n («~l)«n_ia « 1 . 1 . 3 ... 2j» — 1 . 2p — l . a + 2p {2p- 1) . 1 . 1 . 3 ... 2;?-l . a = 1.1.3.3 ...2p— 1 .2j3-l .2p+l ,a = «„+ia. The solution of the given equation may therefore be put in the form Un ^ o)n(t + Vn, whorc Vn IS the value of M„ obtained by putting a = «i = 0. Then Vi = 0, Vg = 1, r^ = 4, t'4 = 25, Vg = 136, Vg = 1041, Vj = 7596, &c. I have been unable to find a functional expression for v„. [The reduction in the number of constants only applies when «, &c. are integers, and seems to be due to the fact that, for such yalues, two successive equations, (« = 1) w^ = «j, (n = 0) «i = «3, are of the first order.] 7544. (By the Editor.)— Construct a triangle, having given the base, the vertical angle, and the ratio of the segments of a given chord of the circumscribed circle drawn parallel to the base, cut off between the circle and the sides of the triangle. Solution by Margaret T. Meter ; D. Biddle ; and others. On the given base AB construct a segment of a circle containing an angle equal to the given vertical angle ; and let QR, parallel to AB, be the given chord. Produce the chords QA, RB, to meet in O ; divide AB in the given ratio at S, and through S draw OC, cutting the given chord in I, and the circumference in C. Then ABC will be the required triangle. For, let AC, CB meet QR in H and K ; then, since QR is parallel to AB, we have AS : SB = QI : IE - HI : IK - QH : KR- 69 7536. (By Professor Sylvester, F.R.S.) — If 3fi— 2 points are given on a cubic curve, and through 3»— 3v— 2 of these an («— yj-ic be drawn, cutting the cubic in two additional points, and through these and the remaining 2v given points a third curve of order y + 1 be drawn, prove that its remaining intersection with the given cubic is a fixed point. Solution by W. J. 0. Sharp, M.A. This theorem is a consequence of Professor Sylvester's theory of Residuation [Reprint, Vol. 34, p. 34). Taking any 3y points, if A and B be the additional intersections of an (w — y)-ic through the other 3n— 3v — 2 points, and C that of the (v + l)-ic, is through the Zv points and A and B, the Zn—'6v — 'l other points are coresidual to C and the original Zv points. If A', B', C be corresponding points obtained by taking another («— y)-io through the same 3^ — 31^—2 points, the Zv points and are coresidual to the same Zv points and C^ and C and G' denote the same point. This point is the single point coresidual to the original 3n — 2 points, for the i«+l)-ic system composed of the first (n— v)-icand the corresponding v¥ l)-ic is such that the 3n — 2 points are residual to A, A, B, B, and 0. Also the «-ic through the 3»— 2 points and C will meet the cubic in a residual point P, which is, therefore, coresidual to A, A, B, B, and there- fore residual to G, (t.^. it is the tangential of C) ; and the 3» — 2 points are residual to P and 0, and therefore coresidual to C, which is therefore the same, however the 3n — 2 points may be taken. The theorem alluded to by Professor Sylvester is given in Salmon's Higher Curves, Art. 154, p. 131, and is shown to be a consequence of Pro- fessor Sylvester's theory of Residuation, Art. 160, p. 135 ; it is identical with Mr. J. J. Walker's Quest. 7068. If « = 3 and ^ = 0, the theorem becomes : — ** If cubic curves be drawn through seven points on a given cubic, the lines joining the two remaining intersections of any of these with the original cubic will all pass through a fixed point upon it." If « = 2 and y = 1, it becomes : — " If four points be given on a cubio and through any one of these a straight line be drawn meeting the cubic in two other points, the conic through these and the other three original points meets the cubic again in a fixed point ; and, as a particular case of this, ' If a conic osculate a given cubic at a given point A and touch it at B, it will pass through the single point coresidual to the tan- gential of B, and three coincident points at A.' " In this way innumer- able theorems may be deduced. 7512. (By Professor Townsend, F.R.S.) — An ellipsoid and any in- scribed polyhedron of maximum volume, or circumscribed polyhedron of minimum volume, being supposed to bound two solids of uniform density in their common space ; show that both solids have the same principal axes at their common centre of- inertia. Solution by the Proposer ; C. G-raham, M.A. ; and others. The polyhedron, whether inscribed or circumscribed, being always regular in the particular case of a sphere, therefore, for both solids, 'X(yzdm) = % l,\zxdfn) ^ Q, :&(xydmj^Of for every triad of rect- VOL. XLI. I 70 Myular pliines p«ssin« through the common centre, in that particnlar case. Ami every triad of such planes for any mass or syfltem of masses, tranarorming into a trind fnr which 2 (y'l' rfm) - P, I (zyd;n) = 0, S.{.fj^dm) " 0, in every transformation for which y — \x, y" — iiy, z' — yi, whore k, /i, rare constants ; therefore, &c., for the general case. 7230. (By the Editob.)— On a square (A.) of a cheas-hoard, a Imight is jJaced at random : find the probability that it can march (1) from that squiire (A) lo a given square (B), as, for ojtample, to one of the comer- squaiea, within a moves ; and (2) over b squares in leas than c moves, for instance, over the four corQCr-squares of the board. Sol ul ion by D. Biodlb. Ins'ilvingthis problem, it is well to remember three things ;— (o) That, according to his position, the knight's command of Bquares varies. When on one of the 16 ct^ntral squares, he commands 8 squares; when on the 16 which flank the bordeia of the 4' central set, he commands 6 squares ; whi^n at the comers of the 6' set, he commands 4 squares, and when on any of the i middle squares of each side also, he commands 4 squares ; when on a border square adjoining the corner on either side, he commands 3 squares ; and when on a comet- square, he commands only 2. Con- sequently therp are— 16 squares on which his range is 8, 16 „ „ 6, 20 „ ,, i. and his average range is o\. (A) That, as be moves from his original square 1, 2, 3 moves, his range dndorgoes a branching process; and his command of squares from one position overlaps that from another, being often partially similar, though never entirely the same. Thus, from either of the i central squares there are 8 once removed, but, instead of S times S twice removed, only 26. (c) That the choaa-hoard is so far symmetrical that an eiamination of the knight's progrcsa from 10 out of the 64 squares is sufficient to give us data for the whole board. The 10 squares referred to are those wBch, roughly speaking, are included within the rigbt- angfed triangle whose hypotenuse is half a dia- gonal of the board starting from either comer. Each of those on the balf-diagooal represents 4 similar ones, including itsolf; each of the other 6 represents S similar ooi', Iniluding itself. Or, dividing the board into quarter blocks of 16fquares, E = B, I = C, N = D, K"G. = H, and P = M. The only squares that have none corresponding in the same block are A, V, L, Q. ^ B 1 N B F K C G D TI BI 1 Q 71 The following table gives a precise account of the num^ er of squares that can he reached by the knight from each of the 16 in 1, 2, 3, &c. moves : — Original Square. Reached in 1 move. Reached in 2 moves. Beached in 3 moves. Beached in 4 moves. Beached in 5 moves. Beached in 6 moves. A 8 26 24 5 B 8 22 24 9 C 6 20 26 11 D 4 16 24 16 4 E( = B) 8 22 24 9 F 8 19 24 12 G 6 17 25 14 1 H 4 14 23 17 5 I(-C) 6 20 26 11 K(=Q) 6 17 25 14 1 L 4 13 26 18 2 M 3 12 23 19 N(-D) 4 16 24 15 4 0(-H) 4 14 23 17 5 P(=M) 3 12 23 19 6 Q 2 9 20 21 10 1 Sum \ Totals i 84 269 384 226 41 1 .. Average 4 H out of 63 16i^ out of 63 24 out of 63 out of 63 2i out of 63 out of 63 le fclfeftfoi 1. From the foVl^oing table we find that the probability of (A) and (B), io< A taken at random, being within one move is TiTUff = T-7» within 2 moves, ,'\,'\j»y ; within 3 moves, -^^Vff J within 4 moves, -^\ ; and within 72 6 moves, -^M}. The opposite extremities of either diagonal are the only positions wmch take the knight six moves to march hetween. The chance of his being placed on a comer-square is ^ = -^V > ^"^ ^^® chance of his liaving to march to the opposite comer, JW. Consequently, -— — — = —— 16 . 63 lOOo is the probability of his having to take 6 moves in marching from one position to another, and this is the remainder left by 1 - ^^^{%i already found. But, in the question as stated, (B) ia giiven, and (A) alone taken at random. Moreover, one of the comer-squares is specially selected for (B). Now in bur table Q, is the comer-square, and we are able to state that if a = 1 move, the probability is ^^ ; if « = 2 moves, ^^ ; if « = 3 moves, |^ ; if a= 4 moves, Jl ; if «=5 moves, f| ; and if « = 6 moves, then f ^ = 1, or certainty. The sum of the figures in the table, opposite the letter denoting the square selected for (B) (according to the question), up to and including those in the column devoted to the number of moves selected for a, will always be the numerator, and 63 always the denominator, of the probability re- quired. 2. The second part of the question is more complicated, since it is im- possible to tell which of the b squares may be nearest to (A) ; and the number of moves will vary not only according to the distance of (A) from the series b, but also according to the order in which the b squares are taken, unless, as in the instance given in the question, they are symmetri- cally placed. We can, however, discover the average moves taken by the knight in crossing from one position to another, both taken at random, by multiplying the sum totals given in the table by the numbers given at the head of the several columns ; then taking the sum, and dividing by 1008 (their original grand total). Thus T5Wd (8* + 538 + 1152 + 904 + 220 + 6) = 2*881 nearly, or yg^. And this multiplied by b will give the average number of moves taken by the knight in marching from (A; to the series b and through it ; because, though d + 1 = the number of squares marking the several positions of the knight from (A) onwards, yet the number of intervals « b only. Of course, c must always equal or exceed b. In the instance given, of the 4 comer- squares, we know that c must equal or exceed 15, to come within the range of probability, because 5 is the lowest number of moves taken by the knight in moving from one comer to another ; and we will even allow that (A) may be itself a corner- square. In any case, the distance of (A) from a comer-square never ex- ceeds 3 moves. Consequently, if c be 18 or more, the probability =» 1 or certainty ; and if c be 14 or under, the probability = 0. There are 16 squares from which the knight can reach one or other comer in 3 moves ; 36 from which he takes 2 moves ; 8 from which he takes 1 move ; and 4 •which are corner- squares, and in which the move must be reckoned 0. Therefore, if (■ pa^ ■\- qx^ -^-rx -^ k «« 0, and de- duce therefrom a solution of Quest. 7330 {Heprint, Vol. 39, p. 111). Solution by Dr. Curtis; S. Greenidqe, M.A. ; and others. If « = i (« + i + c + d) = — i^, we have (Area)* = («-«)(a-*)(a-c)(«-rf), « «*— (rt + * + c + d)«8+(ai + fl^ + &c.)«2— (ai(? + &c.) s-^-ahed ^ 2* 22 2 When A: «= 0, one side of the quadrilateral vanishes, the quadrilateral de- generates into a triangle, and we obtain the result in Quest. 7330. 7601. (By Professor Hudson, M.A.) — The lenses of a common astro- nomical telescope, whose magnifying power is 16, and length from object- glass to eye-glass 8^ inches, are arranged as a microscope to view an 77 object placed f of an inch from the object-glass ; find the magnifying power, the least distance of distinct vision being taken to be 8 inches. Solution by B. H. Rau, M. A. ; Sarah Marks ; and others. Let F and /be the focal lengths of the object- and eye-glasses of the telescope. Then, by question, F+/=8iin., and — « 16; therefore F = 8 in., / = i in. When arranged as a microscope, tiiese glasses are in- terchanged in order. Let O, E, PQ, pq^ be the centres of the j^ object-glass, eye-glass (of the telescope), the £_ P 5-'"' ' object viewed, and its image, respectively. -—J.'''' ^ Then Op = F = 8 in. ; PE = f in., ^ also :rrn + ?r — -7-5 whence =^ = -^— «= 4. Now the magnifying EP EjD / EP ^~t pq PQ »^ 8 8 j»E . power = ^^-^ -* — = ^-^ . — = — .^ =4, Op distance of distinct vision Op PQ 8 PE 7605. (By J. J. Walker, M. A. , F. R.S.) — Referring to Question 1686, show that (1) the circles drawn on the common chords of three mutually orthotomic circles as diameters have not a common radical axis (as erroneously stated in that Question), but have the same radical centre as those circles ; and (2) their common chords are equal to one another, and (3) respectively parallel to the radii of the circle through the centres of the orthotomic triad, drawn to those centres. Solution by As^^tosh MukhopIdhyat. My solution to Quest. 1686, is incorrect; the a:-coordinate of C being in reality — b cos A cos B, so that the i3rst term of equation (3) should be {x-^b cos A cos B)2. Equation (4) is correct. But, subtracting (3) as cor^ rectedirom (1), we get, not the same equation as (4), but — arcosA + ysinA = i(ccosC + ^ cos B- a cos A) (6). Hence, the circles have not the same radical axis. Solving for x and y from (4) and (6), we get for the coordinates of the radical centre of the circles (1), (2), (3), a; = 0, y = i cosec A {c cos C + * cos B — a cos A). But these are well known to be the coordinates of O. Hence, the three circles on the common chords of the orthotomic triad as diameters, instead of having a common radical axis^ have the same radical centre as the three original mutually orthotomic circles. [This can otherwise be proved by reasoning similar to what is followed in Townsbnd*s Modern Geometry, Vol. i.. Art. 183, Cor. 3.] Again, if P be the circumcentre of the triangle ABC, its coordinates VOL. XLI. k ^8 are easily seen to be —\a-\-b cos 0, \a cot A ; but C is (2 cos C, 0) ; hence the line PC is =s ■■ , or a?cosA + ysmA — ^ cos G cos A (6). y — iacotA ' This is obviously parallel to (4), which is the radical axis of (1) and (2). Hence, we inftT that the radical axes of the circles on the common chords of the mutually orthotomic circles as diameters are parallel to the radii of the circle passing through the centres of the orthogonal triad, drawn to these centres. Again, we see that (4) and (5) diJSer only in this, that the intercepts they make on the axis of x are on opposite sides of the origin ; hence, it follows, from elementary jfeometry, that the portions of these lines which form chords of (1) are equal, — as, indeed, can be shown by direct calcula- tion, since (chord)^ = 4 be cos B cos C — (c cos G + b cos B-a cos A)*, in both cases ; this interprets that the common chords of the three circles are equal. 7593. (By R. Knowles, B.A., L.C.P.)— A circle passes through the ends of a chord PQ of the parabola y^ = 4aa? and its pole {hk) ; prove that (1) its equation is x^ +y' '" x (a— A)y + h (2a— A) = ; a a VII Vf (2) if PQ is perpendicular to the axis, the focus is the centre ; (3) if the circle cuts the parabola again in OB, the middle point of the line joining the poles of PQ and CD, with respect to the parabola, is the focus. Solution by Maroaket T. Meyer ; B. H. Rau, M.A. ; and others. The equation to PQ, the polar of (A, k) with respect to the parabola y2 _ \ax is ky—Qax — ^ah = 0. The equation to CD must be of the form ky + 2ax + ^ = (since PQ and CD are equally inclined to the axis of the parabola), and that to any conic through P, Q, 0, D is y2 _ ^ax -^ \ (Ary — 2ax — lah) {ky + 2ax + /) « 0, and if this represents a circle, we have (1 + Kk'^) + A . 4a« = 0, whence A = - (A;' + 4a«) -^ ; hence the equation to the circle is (fc2 + 4«2) (y3 - ^ax) - A: V + ^a^^' + 2aA {ky + 2ax) ^liky- 2ax - 2ah) » 0, and /=2a(2a— A), because the circle passes through (M). Thus the ^2 4- 'hi" If circle becomes x^+y^- ^^ x-'±{a^h) y + h (2a-h) =0. If PQ is a a perpendicular to the axis, X; = 0, and the centre is the point (a, 0), i,e, the focus. The pole of CD, i.e. A:y + 2rtaj + 2a(2a— A) = 0, is the point (2a — A, — fe). The coordinates of the mid-point of the line joining the poles of PQ and CD with respect to the parabola are i (A + 2a— A) and i(fc— ft), i.e. (a, 0), i.e. those of the focus. 79 7294. (By A. McMuRCHT, B.A.J — ^Without knowing the angles of a triangular prism, show that its retractive index can be determined by observing the minimum deviations of rays passing in the neighbourhood of the three angles ; and if these deviations be denoted by 2a, 2/3, 27, then fi is g^ ven by /i' — n^ (cos a + cos fi + cos •>•) + /i[co8(i3 + 7) + cos(7 + o)+cos(o + /8][J — cos(o + i3 + 7) « 0. Solution by D. Edwakdes ; Professor Nash, M.A. ; and others. If 0, (py iff be the angles of the prism, and 2a the minimum deviation at one angle, then it is known that sin (a + i0) *^ fi sin \0, and similarly for 2fi and 27. Also, since e + + tan i^ tan ii^ + tan ^^ tan i0 ~ 1, sin a and tan {$ «. , &c. &c., whence, substituting and reducing, we fi — cos a have the result in the question. 7373. (By R. Russell, B.A.) — Determine (x) and (f> (x) where they are of the form — , so that, by putting y — (x) or (ar), the Cx i- D quartic (abcde)(x, 1)* = and its Hessian may turn into the quartic {abcde)(i/f \)* — and its Hessian. — («) The determination of 0{x) and ^ (x) depends on the solution of a cubic, {b) The roots of the quartio may be represented in the form a, 6 (a),

— tttt + ^t;^ + 77;^ « 1 •••••• (lf2% AD BE CF AD' BE' CF' ^ [In the second result, the lines are considered as signless mag^tudes ; if regard were had to the signs, the > should be omitted.] Solution by Bellb Easton ; B. H. Rau, M.A. ; and others, 1 This is true of any three straight lines AD, ^ BE, CF passing through a common point P. PD aBPD aCPD aBPO For AD aBAD aCAD aABO' PE „ PF - BE * CF ' hence ^ PD PE PF aBPC + aCPA + aAPB ^ aABC ^ . AD BE"*"CF'" AABC ""aABC* ' 2. Since P' is on the side of BC remote from A, - aPBC + aPCA + aPBA = AABC ; hence we have + — — + --— = 1. AD BE CF 7541. (By Professor WoLSTENHOLME, M. A., Sc.D.) —The coordinates of a point being x = a (w2 + w-2), y ^ a (w— w*), where wis the parameter, according to tho usual rule the locus should be a quartic, since we get four values of m for determining the points in which tho locus meets • any proposed straight line. Nevertheless, the locus is the parabola y'^ — a (a*— 2«). Account for the discrepancy. Also, with the same values of (^, y), the equation of the tangent is m*a;— 2m (w^— l)y + a(i»*— 4m2 + l) B 0, which would make the class number 4. Solution by W. J. C. Sharp, M.A. ; R. Knowles, B.A. ; and others. The explanation is that m^ + w-2 = (m— m -*)2 + 2 = ^* + 2, suppose, and bo the equations may be written x = a (^" + 2), y = at, which of course represents a unicursal curve of the second order ; also the equation to the tangent may be put in the form ar— 2 (m— m-i)y + a(m2— 4 + i»-2) =0 or x—2fy + a (t' — 'l) = 0, so that the curve is of the second class. If a be the inclination of the tangent at {x, y) to the axis, m = —tan ^o or cot |ra. [The proper resultant of the two equations is {y^ — ax ■*■ 2a^Y ~ 0> "o that the parabola should be considered as doubled, every straight line meet- ing it in two pairs of coincident points ; and every tangent counting as 2. Every point on the curve and every tangent is given twice (for the values kf - fc-* of m) : this implies that every point of the locus is a node and every tangent a bitangent, which can only happen when a curve is double. 81 Of course the change of notation adopted above obviates all difficulty. Hence, writing down both resultants as determinants, it follows that a, 0, —a?, 0, a, 0, 0, a, 0, —X, 0, a, 0, 0, a, 0, —a;, 0, a fty — y, -a, 0, 0, 0, 0» «» — y» -«> 0, 0, 0, 0, a, -y, -a, 0, 0, 0, 0, a, — y, -a, 0, 0, 0, 0, a, -y, -« «, 0, 2a— X, 0, ff, 0, 2a— X 0, a, -y, 0, 0, «, -y a<(j^3_flpa; + 2fl')«]. 7493. (By W. J. C. Sharp, M.A.) — Show that at an inflexion on the curve U =* 0, «12» W22» ''2 Ui, Ws, 0. [This is an application of the form of the Hessian suggested at the end of the Solution of Question 5762.] Solution ^y G. B. Mathews^ B. A. ; J. 0*IIeoan; and others. At an inflexion we have by Euler's theorem, = Wji, W12, tti3 ^21> *'2J> *'23 ''siJ ^'kj ''sa ««ii. •■ tl. 2l» Wii, «,2, a;Mii + yMjg + 2^13 «2l> **22> XU2i+1/U^i + ZU.y^ 1/31, M32, a;W3i+yW32 + 2M33 W12, (w-l)wi «*22, (n-l)t<2 (w-l)wi, (w-l)w2> («— 1) [iPMi+ywa + ^Ms] •'n* «'i2» (w-I)mi «21» «22» (»-1)M2 (w — l)Wj, (w--l)W2, «(« — !)« or, since m = at all points on the curve, the stated result follows. similarly 7523. (By S. Tebay, B.A.) — Find the mean value of the radius of curvature for all points of an ellipse. Solutioti by B. H. Ba.u, M.A. ; A. Mukhopadhyay ; and others, — • X t/ The radius of curvature at the point a?y of the ellipse — 5- + -75- ■■ 1 i* p= . — ( 1— tf'-— ). Let be the complement of the eccentric angle of the point xy ; then a; » a sin ^, and y = a cos <^ ; therefore _2 f> = — - (l-c*sm-<^)i. 82 \ pds fig 1 The mean value of p is ; and -— « « (1 — *• sin* 1. where » is the cubical dilatation -^^ + — + — , and Va the familiar dx dif dz symbol of operation ( — ) + f^J + ( — 1 ,at the point a:, y, z of the solid. ^'^^f ^^yf ^d^f Supposing now these equations to be satisfied for a system of plane waves propagated as in the question, and represented in consequence by the equations, 4 = fc cos a.f{lx + mt/ + nz—vt) = h cos a 'f{p)t »; = A; cos B ./(<►), f = ^ C0S7 ./(«p), where a, /3, 7 are the direction angles and k a small constant representing the absolute magnitude of vibration, I, m, n, and v the direction cosines and the velocity of propagation, and / any arbitrary periodic function oscil- lating between extreme limits of finite magnitude, and representing in consequence vibratory motion ; we get, by substitution in the equations of propagation, the three following equations of connection between the several magnitudes involved, viz., p V' cos o = (fi + Jv) / (I cos a + m cos jB + « COS7) + y cos o, p v^ cos fi = (fi + ^u)m (/cos a + m cos $ + n cos y) +y cos $, p v^ cos 7 = (/A + Aj/) n (/ cos a + m cos jS + w cos 7) + 1' cos 7, which give a, jS, 7 and v, when determinable, in terms of /, m, n, /i, vy and p, which are supposed to be all given or known. By elimination of v'^ between these latter equations in pairs, we get immediately the three following equations of connection between a, jS, 7 and ly m, «, viz., {m cos y—n cos /8) (/cos a+m cos jB + « cos 7) = 0, (n cos a — I cos 7) (/ cos a + m cos /3 + « cos 7) = 0, (/ cos /3 — w cos a) (/ cos a + m cos ^ + n cos 7) = 0, which show at once that, either coso : C08/3 : COS7 = / : m : «, or /cosa + mcosi3 + ncos7 — 0, and establish in consequence the first part of the question. Solving for t' in the two caaee respectivdy, and denotine the cor- responding velocitieB by v.and p, as correspondiiig to normal and to trans- versal vibratJona reapectively, we get, in aiuwer to the secood part of the qneation, that ('a)^ — - — " and that («■}' -^ — ; which diow that the former depends on the two coefficients /i and r, and is always greater than the 1att«r, which depends only on the coefficient y of the Bubslacce. That Vi should depend only on r, and that v„ ehould on the con- trary depend on boUi n and c, would appear also a priori from the obvious consideration Uiat, while tianevereal vibrations can from their nature produce only change of Form, normal vibrations, when other than those of the entire mass as a rigid whole in its space, must on the con- trary produce at once changes of volume and of form ia tbe molecules of the BDbstaDcs, 7638. (B7 the BDiTOB.)-~If from a given point O, in the prolongation through C of tbe baas BG of a given triangle ABC, a straight line OPQ be drawn, cutting the sides AC, AB in P, Q ; show that, R being any point in the base, the triangle PQR will be a mniimum when a parallel QS to AC through Q cuts BC in a point S, such that 08 ia a mean proportional between OB and OC. SoliHiimt ig {!) A. H. Cubtis, LL.D., D.Sc. ; (2) O. Hbppbl, M.A. 1. Let S be the point anch • thatO8>-0B.OC; then, if SQ he drawn parallel t^ CA, 8P will be parallel to BA, dnce OQ:OP - 08 : 00 - OB : OS ; and, if OFQ" be any other cutting line, we have ABQP-SQ'P>8Q'P'; hence the triangle P8Q is the m mon vertex at 8, their base ang through ; moreover, aP'RQ' : FSQ" - diatance of R from Fft' : dis- tance ol S from P'Qf- OR : OS - a known tatio, hence those two tri- 2. Draw PT parallel to CB, and let it meet QS, the parallel throogh Q ioAC, inT; then, patting OB -i, 0C-<, AC~^, OS-f, themaximnm value of aPQR depends upon that of QS . OR-PC . OR, or on that of QS- PC, or QT. Now we have , andQT-M*-^)fr-'), 88 hence the maximum value depends upon that 'o{ h + e — ar, or of (A»-c*)«-(Ayx'*-a:*)^ and this evidently occurs when x = i*6*, or x^ = be, that is to say, when OS is a mean proportional hetween OB and OC. 7644. (By W. S. McCat, M. A.)--Prove thai the three lines that join the mid-point of each side of a triangle to the mid-point of the corres- ponding perpendicular meet in a point. Solutions by (1) A. H. Curtis, LL.D., D.Sc. ; (2) Harold Harley, B.A. 1. The trilinear coordinates of the middle point of the side c, the axes heing the sides of the triangle, are \csixi B, ^csin A, 0, while those of the middle point of the corresponding perpendicular p^ are i/73COsB, i/^scos A, i/73, and the equation of the line joining these points is resin A-y sinB— zsin (A— B) = (1), while those of the two corresponding lines are ysinB— 2sinC— arsin(B — C) = 0, asinC— a?8in A— ysin(C— A)=0...(2, 3). If we multiply (1) by sin 2C, (2) by sin 2A, (3) by sin 2B, and add, the coefficients of x, y, z vanish identically ; hence the lines meet in a point. Again, if we add (2) and (3), we obtain arsin B — ysinA = (4) as the equation of a line passing through the intersection of (2) and (3), and obviously through the vertex C, while ysinC— JssinB = and « sin A — a: sin = (5, 6) are the equations of the two corresponding lines through the common point of (1), (2), (3), and the vertices A and B. The form of the equation (4) shows that the line which it represents makes with the side a the same angle which the bisector of the side c makes with b ; and hence it follows that the point of intersection of (1), (2), (3) is the scrond focus of the ellipse [Brocard's ellipse] inscribed in the triangle, and having for its first focus the point of intersection of the three lines joining the three vertices to the middle points of the opposite sides. • 2. Let D, E, F be the mid-points of the sides of the triangle ABO, and G, H, K the mid -points of the perpendiculars from A, B, C on the sides of the triangle DEF ; then, since ED is parallel to AB, therefore the right-angled triangles CEK, BFH are similar ; hence we have EK^EC^£ FG^^ DH_ g . FH BF c ' DK a' EG b' therefore DH . FG.EK = DK.EG. FH; whence, by Cbva's theorem, DG, EH, FK meet in, a point. [Mr. Tucker remarks that the theorem in the Question is given in §11, p. 7, of NiiUBERG*s paper **Sur le Centre des Mcdianes antiparaUeles," the point of intersection being what is known as the Point de Grebe, or Sym- 89 median point, of the triangle, — ^a fact unknown to the Proposer, who ob- tained the theorem from the three rectangles in Question 7612 having a com- mon centre (the Symmedian point). The property in the question may be more generally enunciated as follows : — ** If the mid-points of the portions intercepted on any three concurrent lines from the vertices of a triangle, between these and the opposite sides, be joined to the mid-points of the corresponding sides, the three connectors will pass through the same point," — a theorem which may be proved thus : — If D, E, F be the mid-points of the sides and AG, BH, C K any three concurrent lines meeting the sides in G, H, K, and ^, A, ^ be the mid -points of AG, BH, CK ; then ^, A, k lie on EF, FD, DE respectively, and F^ = iBG, .^E = iGC, Ek « iA/-, JtD = iKB, &c. But BG.CH.AK = GC.HA.KB; hence Yg.Bh.K/e ^ g^,hF . ArD, therefore &c.] 4516. (By the late T. Cottbrill, M.A.) — In a spherical triangle, of the five products cos a cos A, cos b cos B, cos e cos C, cos a cos b cos c, — cos A cos B cos G, one is negative, the other four being positive. In the solution of such tri- angles, what parts must be given that the affections of the remaining three can be determined by this theorem P Solution by J. J. Walker, M.A., F.R.S. (1) If cos a, cos b, cos c are all positive, and a>b>e\ then cos A alone may be negative, since both cos b, cos e > cos a and therefore a fortiori > cos e cos 0, cos b cos a. But — cos A cos B cos is opposite in sign to cos A. Hence either the first or last of the five products alone will be negative. (2) If cos a alone is negative, then cos B cos are both positive, but cos A is negative. Hence of the five products cos a cos b cos e alone will be negative. (3) If cos a, cos b are both negative, but cos c is positive, a>b; then cos A must be, and cos B may be, negative, cos must be positive. Hence, of the five products, either cos* cosB or —cos A cosB cosC alone will be negative. (4) If cos a, cos bf cos c are all negative, then cos A, cos B, cos C are all necessarily negative. In this case, of the five products, cos a cos b cos c alone can be negative. It follows from this theorem that, a, by c being the given parts, if all are < iir, then one only of the three angles can be > iir ; but if a, alone, > ^ir, then a must be > , B, C < iir ; if two only of the given parts, as a, ^, > iir, then one of the two angles A, B must, both may be, > iir ; if all three of the given parts are >iir, then all three of the angles A, B, C must also be > iir. The same things may be predicated vice versa of angles and sides, save that, if two only of the given angles are obtuse, the opposite sides must also be obtuse, otherwise three of the five products would be negative. 90 In the other cases of solution, in which the theorem gives some clue to the affections of the parts to be foand, the reservations are too numerous to make its application useful. , . . ** His saltem accumulem donis. ..." >» 7550. (By J. Griffiths, M. A.) — If < = 5 + Jsnw . sn (K -«) and modulus — ■i^/3, K « I ; show that '^ Jo (l-isin^e)* dt [(^-2)(U-3)(U-4)(^-6)]* s du. Solution by G. B. Mathews, B.A. ; D. Edwardes ; and others. ^ , • /xr \ a 8n« cn« -5 -. . .- „ su'm (1 — Sn^tf), t-i - }8n«Bn(K-«) -I -j^^, ^'-T^ + ^-^-j-i^-—^. (<-2)(<-5)=(»-7<+10-fj^^ir^-f A^-^^'. therefore [(<-2)(<-3)(«-4)(<-5)]» - ^\ ^2-^j2-3»^ . . . ^ „ , (cii't«-sn'e<) dn'« + j!;'8n'i« [(y-«)(y-i3)(y-7)(y-«)]* [(«-7)(i8-«)]* ' (l-fc^Bin^^)*' a-7.i8-8 J 91 7511. (By Professor Wolstknholmej M.A., Sc.D.) — A, B are the g^ven centres of two circles; Pp, Vp' the external common tangents, Q/ly QV ^^6 internal common tangents, P, Q being on the same side of the axis ; Pp, Q!q' intersect at right angles in V, and P'j»', Q^ at right angles in V' : prove that (1) P, Q, q\ p' lie on one straight line, P', Q', q^ p on another straight line, whose directions are fixed, and these two straight lines and W meet in one point O ; (2) the common tangents Pp, Py are equal to the sum of the radii, and Q^, Q'q' to the difference ; (3) the points of contact lie on four fixed circles, and the com- mon tangents pass through two fixed points ; (4) PQ', PQ, pq\ p'q all intersect in one fixed point C bisecting AB ; (6) PQ, P'Q', J9g, p'ql are all of equal length, and the ratio Pjo' : Qg' is the duplicate ratio of Pi' i Qg J (6) the ratios OP : j^'O, OQ : Og' are equal, and are equal to the ratio of the radii of the two circles ; (7) the common tangents and the two straight lines through the eight points of contact all touch the same parabola, focus 0, and directrix VV. Solution by W. J. C. Shasp, M.A. Let AB meet Vp and Q^ in K and K' respectively, and let Q^ meet Pp in W ; then the points P', p', Q', /, V, W are the refiexions of P, p, Q, y, V, W with respect to AB, — for the one - half of each circle is the reflexion of the other with respect to the same line ; and, if C be the middle point of AB, the circle with centre 0, passing through A and B, will also pass through V and W and their reflexions V and W, for VB bisects the angle q'Yp, and VA bisects ^'VA, and AVB is a right angle ; similarly AWiB is a right angle. Hence WW passes through 0, and it and PF, pp\ QQ', qq \ an d V V' are aU perpendicular to the line of centres. Now let VK and VVV' be the positive directions of the rectangular Cartesian coordinates, and a,b, ^ be the radii of the circles with centres A, B, C, so that (2c)2 =(«-*)« + (« + *)« = 2 (a2+ 42) or2c««a' + *«, then the equations to the circles are a- + y2 + 2aji;-2rty + a2-. = A, a?' + y«-2*a;-24y + 42 - s B, and fl;2 + y3+(a-i)a;-(a-hd)y= Os C, and therefore W is the point — (« — i), 0, and W is 0, a + 4, and the polar of W with respect to Ai^bx — ay-k-ab = 0, which is also that of W with respect to B, and PQ^V is a straight line, and therefore its reflexion V'Qfqp, the polars of W with respect to B and of W with respect to A, ax + by—ab '^ and these meet on (a—b)y^ (a + ^) ^> ^e perpendicular VOV from the origin on y -4 = — '^-^ (*—*)> ^^ ^^ o^ centres and at a + b the point O where these meet. This proves (1), and (2) follows at once from the values of the coordinates of the points of contact. Evidently the circle through any two of the points P, Q, y', jp', and with its centre in AB, passes through the reflexions of the two points, and there «re six such circles, the two given circles being two of them (3). VB cuts the polar of V with respect to B {pq') at right angles, and, as they are the 92 diagonals of the square YpBq'f bisects it ; then pq* passes through G, and similarly so do PQ', P'Q, ^nd p'q (4). Again, AW « BW - €^2, therefore PQ - 2 ^^- ^^ ^ 2 ^ and j^g ^ 2^^ '^^ ^ 2 ^, therefore PQ = pq, and these are equal to their reflexions P'Q', p'/. And OP : yo - PK :i>'K « « : «, OQ : q'O « QK' : K'y =« » : *, which proves (6), and P/ : PO = a 4- i : a and 0/ : QO =» a-* : a, therefore P/ : Q/ = (a + *) PO : (a - *) QO, and PO : QO = Pj? : Q7 because the triangles PpO, Q^O are similjEur, therefore P// : Q 4, or to a single broken chain in the case of a square whose side has an odd number of places > 6. A knight could not move in a single endless chain in a square whose side has an odd number of places, because on a chequered board the colour of the square changes at each move. In a square of 49 places, the 49 would be of the same colour as 1, and could not therefore be a knight's move from it. For the square of 49 places, starting from a comer as before, I get the square (Fig. 5) filled up with a broken chain 1 to 45, a link 46, 47, and another link 48, 49. The link 46, 47 can be connected with the link 32, 33, and the link 48, 49 with the link 36, 37, whence we obtain the single broken chain (Fig. 6). For the square of 81 places, I first obtain Fig. 7, containing 3 broken chains, viz., from 1 to 73, from 74 to 77, and from 78 to 81. The ex- tremities 81, 78 of the third chain may be connected with the first chain 95 "by means of the link 16, 17, and the second chain with the first chain by* means of the link 69, 70, thus obtaining Fig. 8, which is a single un-^ broken chain. 26 23 38 9 36 21 48 7 39 10 25 22 63 8 35 20 24 27 12 37 60 49 6 47 11 40 63 52 57 54 19 34 28 13 56 59 50 61 46 5 41 64 51 62 55 58 33 18 14 29 2 43 16 31 4 45 1 42 15 30 3 44 17 32 Fig. 1. 48 57 22 27 50 65 20 29 23 26 49 56 21 28 51 54 68 47 24 17 60 53 30 19 25 16 59 46 31 18 61 62 36 45 2 15 62 43 32 9 3 14 37 44 1 8 63 42 38 35 12 5 40 33 10 7 13 4 39 34 11 6 41 64 Fia. 2. 18 15 20 29 36 13 21 30 17 14 7 28 16 19 22 23 2 31 32 9 4 3 24 33 10 Fig. 4. 1 12 35 8 27 6 25 34 11 6 26 ^i^.J^\ 62 63 With Figs. 2 and 3. 38 29 22 69 36 31 20 61 23 26 37 30 21 60 35 32 28 39 24 17 68 33 62 19 25 16 27 40 63 18 67 34 50 41 2 15 56 43 64 9 3 14 49 42 1 8 55 44 48 51 12 5 46 53 10 7 13 4 47 52 11 6 45 54 Fig. 3. 9 30 19 38 7 28 17 20 39 8 29 18 37 6 31 10 43 49 ' 47 16 27 40 21 32 45 42 5 36 11 44 41 46 48 26 Id 22 33 2 13 24 35 4 1 12 23 34 3 14 25 Fig. 5. 9 30 19 42 7 28 17 20 43 8 29 18 41 6 31 10 47 40 33 16 27 44 21 32 49 46 5 38 11 48 45 34 39 26 15 22 35 2 13 24 37 4 1 12 23 36 3 14 25 Fig. 6. With Figs. 5 and 6. I. 45 t 37 44 -49 3&-»48 ^ 1*48 The loop marks the starting- point of a chain. 96 18 28 41 54 11 26 42 66 12 27 40 63 29 14 67 68 81 76 66 43 16 74 64 73 16 30 67 68 69 62 44 17 70 66 60 79 31 66 69 78 71 77 18 46 2 33 20 47 1 32 19 46 3 34 Fig. 7. 13 32 46 69 33 14 60 47 16 34 48 21 36 70 22 49 1 36 39 62 9 10 26 38 63 8 61 80 37 24 76 60 7 72 23 36 61 6 49 4 36 22 21 48 6 With Fig. 7. I 73 * ei 4 80 K^ 8 F " bs ' . U J 46 68 12 31 61 72 16 77 71 62 78 69 63 20 2 37 23 60 11 30 44 67 17 76 68 81 73 66 64 19 79 74 24 61 3 38 The loop marks the starting- point of a chain. 43 66 9 10 29 42 67 8 66 18 41 28 76 64 7 1*10. 8. 80 27 40 66 6 63 4 39 26 26 62 6 29 46 3 14 27 48 1 4 16 28 47 2 13 26 17 62 63 60 26 64 49 6 44 69 66 63 12 37 24 18 31 62 61 64 67 60 11 43 6 66 68 61 38 23 36 32 19 8 41 34 21 10 39 7 42 33 20 9 40 35 22 II. Solution by D. Biddle. (1) If A represent the column in 16 46 30 which the loiight stands, and B the roWy then A ±1, B±2, or A±2, B±l will represent the position at the next move. The annexed diagram (Fig. 9) shows how on an ordinary chess-board the knight may proceed to every square once, and return to the square from which he started, which may be in any position, the cycle being com- plete. If we denote the columns by the first figures and the. rows by the second figures in a series of numbers, we obtain the following : — 11, 23, 16, 27, 48, 67, 88, 76, 84, 72, 61, 43, 22, 14, 26, 18, 37, 68, 77, 86, 73, 81, 62, 41, 33, 21, 13, 25, 17, 38, 57, 78, 86, 74, 82, 61, 42, 63, 71, 83, 76, 87, 68, 47, 28, 16, 24, 12, 31, 62, 64, 66, 36, 54, 66, 45, 53, 65, 46, 34, 55, 36, 44, 32, in which each figure up to 8 occurs 8 times as first and 8 times as second, and in wluch the successive numbers are formed by the addition of ±(10±2) or db(20dbl). Now (10 + 2) is added 9 times and deducted 9 times; (10 — 2) is added 9 times also and deducted 9 times; (20 + 1) is added 8 times and deducted 8 times (if the knight completes tiio cycle by Fig. 9. 97 returning to its original square) ; and (20 — 1) is added 6 times and de- ducted 6 times. The figures denoting change of column amount to 46—46, those denoting change of row to 50— 50. The balance is perfect. But, since the arrangement is unsymmetrical, it is evident that a difference in the course can be effected by simply starting from each of the 4 corners in rotation, by taking either of the two directions which load out of the comers, and also in each of these 8 cases by reversing the course. Thus from one primary arrangement we obtain 16 distinct routes by which the knight can complete the round of the board and return to the square from which he started. We need not here consider the number of primary arrangements that could be made of this kind. But we may point out that to fulfil the requirements of the problem, as regards the ordinary chess-board, it is not necessary to be able to return to the original square. In Fig. 1 it is easy to see that by going forwards from 1 to 27 and then backwards from 64 to 28, we could finish on a remote square and yet traverse the whole board as required. Similarly, we could finish on 12, 43, or 68 ; and it is not improbable that, by modifica- tion of some one of the several primary arrangements (each with its 16 distinct routes), we could begin and end on any two specified squares of different colours. (2) Where the number of squares on the board is odd, as in the given instances, 49 or 81, a complete cycle seems impracticable; that is, the knight cannot return to the square from which he started. The balance between the outgoing and return moves is necessarily imperfect, where there cannot be an equal number of each. But it is quite possible to comply with the requirements of the problem in regard to a 7'^ board, so far at least as 26 out of the 49 starting-points are concerned. The following diagrams (Figs. 10, 11) give two arrangements from which the 25 tours mentioned can easily be mapped out : — 27 16 5 46 25 14 3 6 47 26 15 4 45 24 17 28 35 40 37 2 13 48 7 38 1 34 23 44 29 18 41 36 39 12 33 8 49 20 31 10 43 22 19 30 9 42 21 32 11 Fig. 10. Routes. 1—49 (Fig. 10) 1—49 (Fig. 11) 49—1 (Fig. 10) 49—1 (Fig. 11) 27—49, 26—1 (Fig. 11) 43—1, 44—49 (Fig. 11) 19 4 29 6 21 8 11 28 37 20 39 10 31 22 3 18 5 30 7 12 9 36 27 38 17 40 23 32 45 2 47 26 43 16 13 48 35 44 41 14 33 24 1 46 49 34 25 42 15 Fig. 11. No. of similar starting-points. • • • • t • A 4 • t • • ■ • A 4 • • • • • • A • • t • • • o 4 • • • • • • ^ 4 25 Treating the squares in the two arrangements as we treated Fig. 1> wo find that, in Fig. 10, (10 + 2) is added 6 times and deducted 6 times, also (20 -I- 1) is added 6 times and deducted 6 times ; but (10—2) is added 7 98 times and deducted only 5 times, and (20 — 1) is added 5 times and deduc- ted 7 times: balance = — 22 =« 22 — 44 (the terminal squares in the chain). In Fig. 3(10 + 2) is added 7 times and deducted 4 times; (10 — 2) is added 10 times and deducted 7 times; (20 i- 1) is added 4 times and deducted 5 times ; (20-1) is added 5 times and deducted 6 times: balance = + 20 :« 31-11. About this latter there seems no regularity, which leads one to imagine that there is no real reason why the knight should be unable to complete his course fiom the remaining 24 squares of the 7* board. But these tours are certainly attended with greater diffi- culty. On turning to the 9' board, we find that we can divide it into two por- tions, a central sot of 5^ squares, and an outer £ringe two squares deep, and that these two portions can each be entirely traversed by the knight^ without crossing the boundary line between the two, provided he start from a comer square. In Fig. 12, the simplest arrangement is laid down, and this would serve for a great number of distinct routes, with but slight modification. In Fig. 13 the two portions are used conjointly. 13 26 39 62 11 24 37 60 9 40 63 12 25 38 61 10 23 36 27 14 59 76 71 66 61 8 49 64 41 70 66 60 77 72 35 22 15 28 75 58 81 62 67 48 7 Fig. 12. 42 55 80 69 64 73 78 21 34 29 16 67 74 79 68 63 6 47 56 43 2 31 18 46 4 33 20 1 30 17 44 3 32 19 46 5 71 64 27 40 69 62 25 38 67 28 41 70 53 26 39 68 61 24 65 72 63 4 9 14 61 66 37 42 29 10 15 62 65 8 23 60 73 66 6 64 3 60 13 36 81 Fig. 13. 30 43 16 11 58 7 2 49 22 17 74 67 6 1 12 69 80 36 44 31 76 19 46 33 78 21 48 75 18 45 32 77 20 47 34 79 7040. (By Rev. T. R. Terry, F.R.A.S.)— If i? and q be two positive iDtcgors such thati? >3', and if r bo any positive integer, or any negative 99 integer numerically greater than p, show that p~-q^\ p^-r-l 09-y+l)(p-$^ + 2) ' (^ + r-l)(i? + r-2) '* P ' p — g + r' [This identity has been suggested by Professor Sylvester's Quest. 6978, but a proof may be given independent of the theorem in that Question.] Solution by W. J. C. Sharp, M.A. It is easy to see that the equation holds for all values of §- if r « or 1, and for all values of r if $- = or 1. Suppose it to hold when $-—1 and r — 1 are written for q and r. ,..1 y-1 r-1 ^ (q^l)(g.2) (r-l)(r-2) ^,^ p-q + 2'p + r-2 (j»-g + 2}(p~^ + 3) (p + r-2){p + r—3) p — q+ 1 jp-f r~l p ' p—q+r ^— ^+I*i9 + r-l (i^-^ + l)(i?-^ + 2)* (j» + r-l)(j5 + r— 2) ^ 1_ __£ r__ ^ P-g+^ P-^-r-l _p-q p + r , p — q + l p + r— I p p — q + r p p—q + r* and therefore by induction it holds for all positive values of q and r so long as q j» if g = or ^ = 1. Suppose it to hold for q—\ and » + 1, put for q and 8 ; therefore 1 y-i ^^-l , (y-l)(y-2) (..+ 1X^ + 2^ ^^ p-q + t 8-P + 2 (jo-?+l)(i?-? + 2) («-^ + 2;(«-;? + 3) __ p — q + 1 g— ;? + 1 . therefore ~ p '8 + q —p ' i_ q ^ .^. qjq-^) sjs+i) ^,^ (jo-^ + l)*s-i? + l (p-y+l)(i?-5 + 2) {8-p + l)[s-p + 2) ^ !____? a__ ^ p-q + l ^ 8-p+l ^ p-q s-p p — q+\ 8—p+l p ' s + q—p p s + q—p^ and the formula holds for all values of s > j» ; therefore, &c. 6878, 7422, 7653. (By B. H. Rau, M.A.)-— Given a concave spherical mirror, a luminous point, and the position of an eye perceiving one of the reflected rays j find the point of incidence and reflection on the mirror. 100 Solution by D. Biddlb ; Bellb Easton ; and other $. Let A be the luminous point, B the point through which the reflected ray is to pass, CD the concave spherical mirror, and E the centre of its curve. Then Z APE-BPE, and AE, BE are similar chords of the circles EPA, EPB of which EP is a common chord. Draw AF, BF at right angles to AE, BE, and GH tan- gential to the reflecting surface of the mirror at P. Then, since EPG, EAG in the one circle, and EPH, EBH in the other circle, are right angles, the centres of the circles must be at the mid- points of EG, EH. Moreover, since A E, EB are similar chords, the diameters of the circles must bear the same ratio, EG : EH = AE : BE, and the complementary chords also must bear the same ratio, AG:BH=AE:BE. We also have, given, Z AEB ; and AB, with the perpendiculars drawn to it, EI, FN; also AN and BN. Moreover, / ALE = APE, and ZBE^=BPE. Consequently, EKL is an isosceles triangle and EI bisects KL. Again, ELG and EKH are right angles; and ZEGL-EAB, and zEHK=EBA. Wherefore, the triangles ELG, EEH are similar to EIA and EIB, the sides being severally as EL : EI ; and EM, the prolongation of EI, cuts off equal portions of each, and joins the apices of two triangles EGH, MGH, which have the same base GH, the height of one, EGH, being the radius EP of the curve of the reflect- ing surface. Now EI . EM=EL«. Consequently, if EM be found, and a circle be drawn on EM as diameter, L and K will be its points of inter- section with AB. Then G and H can be readily found, and GH will touch the reflecting surface in P, the required point. Let EM = Xy then, since Z lEL = AEG, therefore we have EA : EG = EI : EL =EL : a? and EA^ : EA2 + AG^ = EIx : a;2-EI : x, _ EI(EA^-t-AG2) . EA2 ' hence thus, putting AG «==?/, EA = 1, EB = flr, then, by drawing perpendiculars to AB from G and H, we find GH-=|AB-(^— + — ji/^^ ^(aF-BfJ ^• - an^bn« „ , „ f = -r^-^ 'FT^i and ^ = Let AF "*" BF .AF FN FN« AF BF then GH-2=(AB-/^)2+^V. Moreover, EG2=1 +y-andEH2= (1 +y)2a3. Now, the area of the triangle EGH = i (2EG2EH2 + 2EH-GH2+ 2GH2EG2-GH4-EG4-EH*)*. 101 But the area of EGH » i (GH . EP), also. Therefore GH . EP = i (2EG2EH' + 2EH2GH2+ 2GH2EG3-GH<-EG*-EH*)*, and 4EP2[(AB-/y)2+i^V] = 2a2 (1 +y2)2+2a3 (1 +y*)[(AB-/y)>+i^V] + 2[(l+a2)(/^ + ^=)-(l-a2)3+(l+a«)AB«-(^ + ^2)AB2 -2(/'+ir2)EP-]y« -4AB/[(1 + aS) - (AB2 + 2EP2)] y + { AB« [2 (1 + a^) - (AB^ + 4EP2)] -(l-aV}-0. This equation enables us to find G in any given instance, and then, if we draw a circle on EG as diameter, the reflecting surface is cut in the required point P. Thus, to give an example, let AE = 1, EB = -6429, AB = 1-4143, and EP=-6367; then AF = 1'6143, BF = 1'7286, FN=:l-4429, AN«-4429, andBN = -97l4, whence /= -5976, ^ = -4780, and a = EB = -6429. Our equation then yields the following result, after reduction : — S^-3-6468y3--2o29y2 + 6-397ly-2-3881 = 0; whence y =-416= AG, and BH = aAG= -2263. Find G and H by these, join GH, and the perpendicular EP will give P. [A solution by Dr. Curtis is given on p. 69 of Vol. 39 of Reprints, "] 7669. (By Professor Townsend, F.R.S.) — A thin uniform spherical shell being supposed to attract, according to the law of the inverse fifth power of the distance, a material particle moving freely in either region of its space external or internal to its mass ; if, in either case, the cur- rent velocity of the particle be that from infinity under the action of the force, show that its trajectory will be an arc of a circle orthogonal to the surface of the shell. Solutions by (1) A. H. Curtis, LL.D., D.Sc. ; (2) the Proposer. 1. If a denote the radius of the sphere, p the distance of the centre of the sphere from the attracted point, and r the distance of this point &om any particle of the shell, and if V denote the function which, differentiated with regard to a;, y, «, will give the corres- ponding components of attraction. — 4. ff it^ » _i f /i2Fg^singrfg JJ ^ " *Jo[a^ + p^-2a/>co8«P (lira I 1 __ 1 \ _ inrd^ 4p I {a-pY (a + rt^ / " («2-p2) /iira _ As this expression involves only p and constants, it shows, as also appears VOL. XLI. N 102 d priori, ihai the total attraction passes through the centre of the sphere, and for the orbit we must have no constant being brought in by integration, as the velocity is that from infinity ; or the equation of the orbit may be written p^ mta* = kp. This curve will lie in a plane passing through the centre of the sphere and the line of initial velocity of the particle, and is plainly a circle as radius of curvature ^-^ » k, and, when j? s 0, p » + 2u;— 1 ^0. These equations give «=y2, v=i(5-2y2), w«=i(3-2v/2); and cos A, cosB, cosC are the three roots of ic8-y2a;2 + ^(6-2v/2)a;-i(3-.2v'2) = 0. Here 0?—^ is a factor, and, solving the remaining quadratic, the three cosinosare -5, -8080488, -1061648. The angles are 60% 36°6'39"-4, 83° 54' 20"- 6. It will be found that the given triangle has its sides in the ratio of the sines of these angles. 103 J The angles of the triangle are 60°, 60° ± o, where cos a = y/2—^ ; the ii of the several circles of the triangle are R - 28970-66, r « 12000, rj = 63789-81, r^ = 40970-35, r^ - 23121-88, and the distance between the orthocentre and the circumoentre is 23479-64.] 7603. (By the Editor.) — If on a rectangle AOBZ two random points (P, Q) be taken, P on the base OB, and Q on the surfEu^e OZ, show, by a general solution, that, OA remaining constant, (I) as OB increases in- definitely from zero to infinity, the probability that the triangle OPQ is acute-angled decreases from i to ; and (2) in the cases when OB = OA, OB = |0A, 0B = 20A, OB = 40 A, the probabiUty will faU short of \ by the approximate values -^j y^y, -^^, -^, respectively. Solutions by (I.) D. Biddle ; (II.) the P&oposeb. I. The point Q, in order to form with OP an acute-angled triangle, must be between the parallels AO, KP, and outside the semicircle OSP. The average length of OP = iOB, and the chance of Q being in the variable space AP (OB being fixed) is also \. Consequently the chance of Q being in position to form with OP an acute- angled triangle is i, when the space enclosed by the semicircle vanishes as it does when OB has diminished to zero. When OB is infinitely greater than OA, the semicircle absorbs on the average the whole of the space AP, and leaves no room for Q in the requisite position. Conse- quently the chance is then 0. The actual chance at any limit of OB and any position of P is repre- sented by the ratio subsisting between that portion of AP not included by the semicircle, and the entire rectangle AOBZ. And when the whole of the semicircle lies within the rectangle at all positions of P, that is, when OB does not exceed 2A0, the mean area of the semicircle can be deducted from the mean area of AP, and the remainder, in the ratio it bears to the entire rectangle, gives the required chance. In these instances OA. OP— ^ir . OP^* gives the area of the space in question for any single position of P, and the moan area for all positions of P = OB . OA ^i-| . i»^), so that the probabiHty of Q being in this space falls short of i by -1309 = ^i^ when OB = OA, by -19635 = ^^'^^^^ when OB = |OA, and by -2618= i^^ when OB -20 A. 168 ^ y J jgy In the fourth instance given, viz., when OB ^ 40A, the case is materially altered, since the semicircle (on OP when OP > 20A) extends beyond the boundaries of the rectangle. However, we have already ob- tained the mean area of the said space whilst the semicircle is within the rectangle, that is, in the present instance, when OP 40B. 104 Thia area, for any single position of P, is OP . OA— (J^ir . OP' - segment). The segment of the semicircle when OP « OB « 40A is easily found, since its height — OA. The sector is accordingly jast | of the semicircle and the segment -yW . OP^- OA (0P«-40A*)*. From this it diminishes to zero when 0P = iuB. To the same point, the mean area of the rect- angle is |OB . OA, and of the semicircle f . ^w . OB'. Consequently, the mean required space for that portion of the series is (OB' } — f . Jt . — - ) + mean segment. And the mean required space for the other portion of the series will be 0B.0A(i-A.|,.Q2). The mean between these two wUl bear the same ratio to OB . OA that the required chance bears to unity. The mean segment can be found from a series ranging from ^ diameter to zero in height, and each multiplied by 4 / (1 — 2^)"', since we have h = (iOP- OA) / OP, and OP^ : OA' = 40A2 : OA' (1 - 2hy. This gives a result of *6596 OA' as the mean of the segments extending beyond the rectangle when OP > ^OB. Hence we have OB . OA ( J-f . ^ir . 55\ + -6696 = -2876, OB.OA Ji-yX,.^x.g|) = -4764, i (-2876 + '4764) = -382 = mean space required for Q. Now 4 . -382 « 168 : 16-044 ; thus the chance is 16044 ^ J 67-956 168 2 168 II. Otherwise .—The triangle OPQ will be acute-angled, if Q fall any- where on an area (S, say) contained between the convex circumference of the semicircle OSP, the two tangents OA, PR, and the side AZ of the rectangle. Hence, for every position of P, the probability of an acute-angled triangle will be the ratio of the area (S) to the entire area of the rectangle on which Q must fall. Also the probability of P's falling on any portion of the side OB will be the ratio of that portion to the whole length of OB. Let the breadth OA of the rectangle « unity, its length 0B = 2a, and 0P = 2a;; then, putting Sp S^ for the respective values of S when x 1 , we have Si = 2a;-iirar2, Sj = 2a;-ir2cosec-' x- (a:'-l)*. Hence, putting p for the probability that the triangle APQ will be acute- angled, and, supposing the length of the rectangle not less than twice its breadth (x not < 1), we shall have But, if (2) the length does not exceed twice the breadth (\ not > 1), we have only to consider S^ ; and then 2a j? =1 Sidx {&)- 105 Now we readily find I S^ e^a; » x^^^ira^, I Sj rfa? = 1 — Jir ; 3Sjdip« 3a;>-2a:(a»-l)*-^cosec-ia; + log.[a?+(a;3-l)*]; Hence, when the length is not less than twice the breadth (a not > 1), («) gives p - i-(^^i=IlL^Acosec-iA + g-Lb^^ (a'), an expression which, by patting a » cosec a, may be written p = i— -^cosa-^coseca + ^sin^a.logecotia (o"). When the length is four times the breadth (A •■2, a»^ir), then, from equation (o') or (o")> "we have j5 «= ^ (3 - y3) - iV + jV 1o8^« (2 + -/3), or j» = -09166 = ^ nearly. When the length is not greater than twice the breadth (A not > 1), equation (/3) gives p « i--^Air (i8'). When the length is equal to twice the breadth (a» 1), then, from equa- tion (i8'), we have p « i— t^^it, ot p =: ^ nearly. When the rectangle is a square (A — i), then we have P = i- A'T, or p - Ij^, nearly. If we suppose the side OB (or 2a) to increase without limit, the side OD remaining constant, then p decreases without limit, and becomes zero when A is infinite ; and, as OB decreases without limit, j9 increases up to i, which is its limit when A is zero. 7676. (By J. J. Walker, M.A., F.RS.)— If P{«y«) = is the equation to any surface referred to rectangular axes, show that the equa- tion to the curve in which it is cut by the plane x cos a + y cos $ + z cos 7 =p, referred to the foot of j9 as origin, and the line in which the plane is cut by that containing the line p and the axis of 2, and a line at right angles thereto, as axes, is obtained by substituting for Xy y, z, in F (xyz) » 0, p cos a + (y cos j8 — « cos y cos o) cosec 7, pcoB$ — (i/ cos o + « cos i8 cos 7) cosec7, i?cos7 + «sin7. [It may readily bo verified that these formulae give, e.ff.y as the equation to the section of a;> i- y* + «« - r^ = 0, y' + z^-r'^+p^ = 0.] Solution by W. J. Curran Sharp, M.A. In a paper ** On the Plane Sections of Surfaces, &c.,** read before the London Mathematical Society, in December, 1883, I have shown that, i^ (a"i, yi, «i), (flJai Vii «2)> (^8> ys» «3) ^0 any three points, and if 106 Ax^ + fix^ — y£3^ ^^^ 1^ substituted for x, &c. in the equation to a surface. A + ZLt + f the resulting equation in \tA¥ is the equation, in areal coordinates, to the section of the surface by the plane throuf^h {xi, yi, z^) , (x^, y,, 2j), (^, ys, z^), these being the vertices of the triangle of reference. Mr. Walker's origin of plane coordinates (A) is the point (p cos a, p cosiS, pcosy), and his axes are the lines = -^L- « ■ ^'". — 2 a p, cos a cos/3 8in^7 j?cosa — g ^ y — jp cos $ z— pcosy __ C08/3 C08O ~ ' and let the two points B and C, which with A determine the plane, be chosen in these lines, in the negative direction, so that for B p = r, and for C 3=l + 2^ + E(ir) + [3y + E(ir) + 3(^-l) + E(ir) + ...3(0)+E(ir)] + [3fe-I)+Ei(r + 3) + 3{^-2)+Ei(r + 3) + ...3(0) + Ei(r + 3) + E|(r— 3), the last term being taken only if r > 3, whence it may be denoted by E J (r + 1)], .-. J^, -l + 2^+E(ir) + 3iy(^ + l)+(l + ^)E(ir) + 3i^(^-l) + ^Ei(r + 3)+E^(r+l) = l + 2g' + 3g'8 + ^[E(ir) + Ei(r + 3)]+E(ir) + E(ir) + Ej(r + l). Now E (if) + E i (r + 3) = r + 1 ; and, by trial of all cases from r = to r =6, wemay substitute r + E^ (6- r) for 1 +E(ir) + E(ir) +E J(r+1), therefore nPi= 2q-^Sq^ + q(r+l) + r + Bi{6-r) »(3^ + r)(l+y)+Ei(6-r), which proves the theorem, since 33^ + r =* i (6j^ + 2r) « i [« + R (J«)]. 7506. (By S. Tebay, B. a.)— Find (1) the form of a when x' + a and «^— a are rational squares ; also (2) deduce the simple values 108 aad (20 give a naat method (with examples) of cidciilating corresponding numeriaJ valuee <^ s and a when « only is integral. Solution by G. B. Mathbwb, B.A. ; the F&oposbr; and others. Let x^-^a = (a? + m)': then x — -- — , and x^—a « — . Let a«-6af«« + m* - (a-^m^^ ; then m^- M^^, Take a = i.'f—n **» * heing any arbitrary quantity ; then Lm=^i, and ..<±dtt*Pi. Take * = 2 (A;2-/2), then x = (A;-/)2 + 4P, and a = 8 (fc-S/) (Ar^-^). Again, let ^^ ~ ^ « ^ he a fraction in its lowest terms ; thus Take i « ?- : then a « pq, and a; = ? ""^ . « ' ^** 28t Example I.— Let fc - 7, / = 2 ; then a = ^2 « ijL??*a. Thus/? - 1, y = 5, » — 2, < = 3; therefore a s^pg ^ 6, and ip « ^ ; thus the two squares are (f ^)2, (^)2. n.— Let Jk =- 6, / « 1 ; then a « ^' = |-^ i» ; and wefind, as before, 6.1' a s 6, a; = j, and the squares (i)', (^)*. III.— Let fc « 9, / = 1 ; then a « f^^ **,* as before, « - 15, « « J^ and the squares (J)«, (V)'- [If we assume ar^+rt = (m + «)2, a^^a ^ (m—n)^, we have a *= 2mfi, a:3 »B f/t^+n^ = q ; hence we may take m = (Ar^— ^, « = 2W; then If ? « 2/', Jk = (]k'-0» ^e ^vo ^ = (fc'-^)2 + 4/'3, a= 8r(Jk'-0(>^ + 0(*5'-30 = 8r(fc'-3^(;fc'*-r). As an example, if a; « 101 = 10^+ 1^, we may take A: =^ 10, / = 1, « = 40 . 99 = 3960. See also solution of Quest. 7468, on p. 119 of Vol. 40 of Reprints.'] 7598. (By Professor "Wolstenholmb, M.A., Sc.D.) — 1. Circles are drawn with their centres on a given ellipse, and touchiug (o) the major axis, (jS) the minor axis ; prove that, if 2a be the major axis, and e the eccentricity, the whole length of the arc of the curve envelope of these eccentricity, the whole length of the arc of the curve envelope of these 109 circles is 4a A + 1^' log 1^], ia f{l - e')^ + 2 ^^^^ (a, /3). 2. Circles are drawn with their centres on the arc of a g^ven cycloid, and touching (a) the base, (jS) the tangent at the vertex ; prove that the curve envelope of these circles is (a) an involute of the cycloid which is the envelope of that diameter of the generating circle of the given cycloid which passes through the generating point ; (j8) a cycloid generated by a circle of radius ia rolling on the straight line which is the locus of the centre of the generating circle (radius a) of the g^ven cycloid. 3. Circles are drawn with centres on a given curve and touching the axis of X ; prove that the arc of their curve envelope is a;— 2 f y dd, where dfi X. y are the coordinates of the centre of the circle, and — = tan B. dx Solution by A.'K. Curtis, LL.D., D.Sc. ; Professor Rau, M.A.; and others. The curve, envelope of circles described as in each of the cases included in this question, is by Quetelet's construction (see Salmon's Higher Curves) an involute of the caustic by reflexion of the curve due to rays perpendicular to the fixed line which the moving circle touches. In (I), (a) and (jS), it is required to find the length of a certain involute of the curve which is the caustic of the ellipse due to rays parallel to an axis. 1. Let the figure represent the ellipse, let DE be any incident ray codirectional with the ordinate y, EF the normal and EG the reflected ray, making respectively with the axis major the angles

g [*' ^^sin4>rf]* = ia^l-\^U {j^\ log ( l^). h tlie formula VOL. XLI. 110 The discussion of (2) is similar, the difference in the form of the result from case (I) being due to the fact that the formula now employed is dz I = sin-^«. 2. This depends upon the fcict that the caustic by reflexion of a cycloid due to rays perpendicular to its base consists of two cycloids as in figure. 1 ^ /- "'V,..;/\l WK each touching the given cycloid, and having for base one half of its base, and that this caustic is the envelope of the diameter referred to. This may be shown geometrically thus : — Let ABA' be the g^ven cycloid, DE any incident ray, GEK the generating circle in the corresponding position, and G its centre, draw GH perpendicular to ECE', and on GO as diameter describe a circle GHCH', then Z DEG = L CGE = Z CEG, and therefore CE is the re- flected ray and the radius through E of the generating circle in the corres- ponding position, and arc GH, subtending Z GCH at circumference of circle GllG = arc GE, subtending same angle at the centre of the circle GEK, therefore — line GA ; hence the locus of the point H is a cycloid to which HG is a normal, GE a tangent, and AO the base ; thus (2, a) is proved. (2, j3) appears from the fact that B is obviously a point on the locus, and it is therefore evident from the figure that the particular in- volute, in this case, is the cycloid stated in the question. 7699. (By R. Knowles, B.A., L.C.P.) —Prove that in any triangle cos A cosB cosC 1 ^-» (TsinB asinC AsinA R Solution by Maubice Pontonnier ; J. Brill, B.A. ; and others. On a d sm A = -— , c sm B = — , « sm = ^=r ; 2R* 2U' 2R * done (1) vient d'oii : 2RcosA . 2R cos B ^ 2Rco8C be ac ab 1 a cos A + b cos B + c cos C abe , 2^' mais ft2 + c2_^2 C08A= , — , cosB = : 2bc ' Ill onaalors: -a^-b*'-c* + 2aW + 2a^c^+2b^c^ = R2 ' ou 4*2^2 sin* A - ^^, ou -At = 2R, R* sin A formale connue, done (1) est verifiee. [ cos A . . __ gco8A + ... __ a cos A -t- 5 cos B + g cos C _ 1 1 (TsinB 2A a.RcosA + i.RcosB + c. RcosC R ' J 7602. (By Professor Hudson, M.A.) — A ray proceeding from a point P, and incident on a plane surface at O, is paxtly reflected to Q and partly refracted to R : if the angles POQ, POR, QOR be in arithmetical progression, show that the angle of incidence is cot-^ f ^~^ ) . Solution by C. Morgan, B.A. ; J. A. Owen, B.Sc. ; and others. Let a, a' be the angles of incidence and refraction ; then P0Q = 2o, POR = ir-o + a', QOR = ir-o-a', and, since these angles are in arithmetical progression, we have 2o'= 3o— a'— t, and o— a'= ^t. But, if (jl be the index of refraction, sin a = /a sin a' =^ sin (a— 60°), whence the stated result follows. 4904 & 6884. (By Dr. Hart.)— Find the equation of the Cayleyan of the cubic ar*y + y*a + «2a?+2»/wry« = 0, and compute the invariants of this cubic. Solution by W. J. Curran Sharp, M.A. For this form of the equation to a cubic, the Hessian, the discriminant of the polar conic yy2 + jjy'i + xz'^ + 2 (mx + y) y'z^ + 2 {my + z) z'x' + 2 (wa + x) a// = is m^ [x^y + yh + z^x + 2mxyz] —{a^+y^ + :i^-' Zxyz) « 0, and therefore, when this meets the curve, {x + y + z)(x^ + y'^ + z'^'-yz—zx—xy) = 0, an equation which represents one set of inflexional axes, of which one is real and the other two imaginary, and, since the discriminant ^(3^— 6?M + 4m-)2 of the binary cubic determining the real inflexions is positive, the cubic cannot be cuspidal. In other cases the reduction to this form may be effected by identifying the inflexional axes with the above lines, and the curve is referred to such axes that (y, z), («, x), (a?, .v) lie upon the curve, and each is the tangential of the one after it in cyclical 112 order, and therefore its own third tangential. The Cay ley an is the con- dition that ax + fiy + yz = should cut the conies U, = z' + 2/;/yz + 2xy, Us = J:* + 2t/z + 2mzXf U3 = y' + 2zx + Imxyy in points in involution, t.«., The invariants are S=-(w»* + 3m) and T = - (8in« + 36m» + 27) (and therefore the cuhic cannot he cuspidal) ; the discriminant is 27 (8m3 + 27). 80 that, if the cuhic he nodal, w^ = — ^. and T is positive, f.#. {Quarterly Journaly Vol. xvi., p. 192) it is cronodal. 7427. (By Professor Townsend, F.R.S.) — A lamina, setting out from any arbitrary position and moving in any arbitrary manner, being sup- posed to return to its original position after any number of complete revolutions in its plane ; show that — (a All systems of points of the lamina which describe curves of equal area in the plane lie on circles fixed in the lamina ; (b) All systems of lines of the lamiua which envelope curves of equal perimeter in the plane are tangents to circles fixed in the lamina ; (c) The two systems of circles, for di£ferent values of the area in the former case and of the perimeter in the latter case, are concentric, and have a common centre in the lamina. Solution by G. B. Mathews, B.A. The motion is determined by the rolling of a curve in the lamina upon a curve in the plane ; thus the results at once follow from Kempe's and McCay's theorem [given on pp. 82 to 86 and p. 101 of Minchin's Uniplanar Kinematics]. Ibl^, (By Professor Wolstenholme, M.A,, Sc.D.) — ^If we denote by F (a:, «), the determinant of the «th order X, 1, 0, 0, 0, 1, X, 1, 0, 0, ... 0, 1, Xy 1, 0, ... • • • • • I 0, 0, 1, X, 1 0, 0, 0, 1, X prove that F(ar, 2r+ 1) = j:F(a2--2,r), F(a:, 2r) = F(a;2-2,r) + F(j;2~2, r-1), ¥ (x, n) = [ x-2 cos — ^ 1(^—2 cos — — ] X (a;— 2 cos — ^ )... ( a;— 2cos -^^ |. Solution by B. Hanumanta Rau, M.A. ; Prof. Nash, M.A. ; and others. By expanding the determinant, it is easily seen that ¥{x, n)-'xY{x, «-l)+F(ar, w-2) = (1)' 113 and, therefore, F (x, n) is the coefficient of ^** in the expansion of (1— ary + y^)"* in ascending powers of y. Let x = 2 cosO and = o. n + l ' then (l-2cose,y + y2)-i = {l-ye^)-^(l-ye-^)-^ (e^-e-»<'){l-tfe^){l-ye-^) sinO"- "* .-. F(ar,fi) =, !lEi^±l)i =, 2»sin(0 + a)sin(0 + 2a)... Bin(0 + «a)...(2). But sin (0 + o) sin (0 + na) = sin (0 + o) sin (o — 0) =s sin^o— sin'O = cos^O— cos^o = (cos6— C08o)(co8 6— cosfio), and so on. Therefore F {Xj n) = 2" (cos 0— cos o) (cos 0— cos 2o) ... (cos 0— cos na) = (ar — 2 cos o) {x— 2 cos 2o) . . . {x—2 cos na) . If fi = 2r + 1, then cos (r + 1) o = cos iir = 0, and (a;-2coso)(a;— 2cos«o) — (»2— icos^o) =■ (a;'— 2 — 2 cos 2o), r,¥{x, 2r+l) =a;(a;«-2-2coB2o)(«3_2-2cos4o)... = icfar«-2-2cos-^Vit«-2-.2cos — ^ ...= a:F (a?2-2, r). \ r+l/V n+lj Again, from (1), F (ar, 2r+ l)-irF (a;, 2r) +F (a:, 2r-l) = 0, .•.a:F(a',2r)=»F(a:, 2r+l) + F(a;, 2r-l)«a;F(a;»-2,r)+aF(a-2-2,r-l), therefore F (», 2r) = F {x^- 2, r) + F {x^- 2, r- 1) . 7410. (By W. J. C. Sharp, M.A.)— If N : Dxbe a fraction in its lowest terms, and D = 2*. 6* . «*. A*", c** ..., where a, b, c, &c. are prime numbers, the equivalent decimal will consist of h or k d on -recurring figures (according as A or A; is greatest), and of a recurring period, the number of figures in which is a measure of a^-^ (a— 1) . A"*-^ (d— 1) . c"-^ {e— 1) ... Solution by G-eoroe Heppel, M.A. Let the equivalent decimal have p non-recurring and q recurring figures. Then N : D = K : 2p. S**. 10*-^ Hence, obviously, p must be equal to the highest index of either 2 or 6 in the factors of D. Also, sup- posing D to have but one other prime factor a, then, from Fermat's theorem, the maximum value q can have is a— 1. If 5 has any smaller value, then, since in actual division we have remainder after every q nines used, and we have remainder after using (a — 1) nines, therefore q measures a— I. Now, if D contains a factor a', suppose that the result- ing period is one of e digits, and let 10'' = C. Then I measures C— 1, therefore it also measures the following : (a-l)(c-l)+fl! or (a-l)C + l, (a-2)(C^-C) + (a-l)C+l or (a-2)C« + C+l, (rt-3)(C3-C-) + (fl-2)C'^ + C+l or («-3) OHC2+ C + 1. 114 Proceeding in this way, we see that / measures 0-* + C*~' + .. . + C + 1 ; therefore it measures (>— 1 or 10'*'— 1, or q repeated ae times. Hence, the maximum period for a being (a— 1), for a' it is a (a— 1) ; for a' it is 0' (a — 1 ), and so on. Consequently the greatest possible period for D, as given in the question, must be a'-* (a— 1) . b**-^ (b— 1) . c"~* (e— 1) ..., and from the reasoning given above, the actual period always measures the maximum period. 7657. (By J. Crocker.)— If an ellipse be described under a force/ to focus S ana/2 to focus H, and SF » r, HP » rj ; prove that dr^ dr \r r j / Solution by D. Edwardes ; W. G. Lax, B. A. ; and others. Resolving along the tangent, we have Again, i:!«C/'+/')8inSPT -(/+/)* that is, v' = (/+/') — . a Differentiating this with respect to r, and remembering that dr' -i^dr = 0, -•^-^ -'Hi- f^y-^ '"-''■ by (1). Putting r + r' for 2a and reducing, we have the stated result. [In a paper by Dr. Curtis " On Free Motion under the action of several Central Forces" (Messenger of Mathematics, New Series, No. 109, May 1880), this question is discussed, as a subordinate case, and the condition -— — (Fr) — - — T (Fr^) — 0, deduced, which is equivalent to the above.! r^ dr^ ' r^'^dr^^ / » » ^ j 7547. (By R. Tucxer, M.A.)— PFR, QFS, are two orthogonal focal chords of a parabola, and circles about PFQ, QFK, RFS, SFP cut the axis in points the ordinates to which meet the curve in P, Q', R', S' : prove (1) locus of centres of mean position of P, Q, R, S is a parabola, (latus rectum ^L) ; (2) 2 (FP') + 2L = 22 (FP) ; and (3) if also normals at three of the points P, Q, R, S cointersect, then y^^ 5' y "*) — — 24L-''. 115 Solutions by the Proposer. Let am^y 2ami, &c., be the coordinateB of P, Q, R, S ; then equa- tion to PQ is (♦nj + W2)y-2a? — 2am^m^ (1), with condition (1-Wi2) (l-fW2«) +4miW3 « 0...(2). ^/ The equation to PR is \/^ (mi + ma)^— 2a? = 2am^in^ (3), «^ and it is a focal chord, .*. m^m^ *" — 1> similarly m^m^ = — 1 (4) ; ^^ (1) 2^ » a (mj + iTij + fns + m^, 4F ■» aSm^, and 2(»nrm,)=-6 (6). therefore 4p = 4a («— 3«}, that is, y^ ^ a (i— 3fl); hence locus of mean centres is a parabola whose latus rectum « ^L. (2) The equation to circle round PFQ is aj* + y*— a {m{^ + m2^) a? — 2a (m, + Wj) y + t^mim^ (4 + m jiWj) « 0, therefore abscissa of F= am^m,^ (4 + miWj) and FF= a {tn^-¥m^ by (2) ; therefore 2 (FF) + 2L « 2a [4 + 2m«] - 22 (FP). (3) The equation to the normal through any point (d?, y) is am8+ (2a-a:) m-y -= 0, therefore 2 (m) » for P, Q, R, and — Sj(-) - 2j -^ - Sj -V- - A l>y (5) 24L-». Or, (1) thus : — The equation to PR is y — A; (a;— a), therefore arj + ar8 " 2a f 1 +-t- )> ^i + ^8 "" -r* Similarly ajj + 3:4 =« 2a (1 — W)^ y^^y^ — — ^ak, therefore 4j — 2y ■» 4a f — - -kYi.e., y « a f — - — A: j , F=.a(l+A;2+-^); therefore 2 « A:2+ J- -2 = — -3, aS A;3 a ' ».^. , y^ " a (i— 3a) . [This is also the locus of the intersection of two orthogonal normals (see Smith's ConieSf p. 104).] 7658. 1^7 S. Constable.) — The vertex of a triangle is fixed, the vortical angle given, and the base angles move on two parallel straight lines ; construct the triangle when the base passes through a fixed point. 116 Solution by Arthur Hill Curtis, LL.D., D.Sc. For greater generality, let the lines not be parallel, let A be the fixed vertex and OBE, OCF the fixed straight line on which B and C, the extremities of base, move, D being the fixed point on base; draw AE, AF, AG perpendictdar, re- spectively, to OB, OC, BC, then Z EGP - AGE + AGF = ABE + ACF « BOC + BAG, and therefore known ; hence one locas of G is a seg- ment of a circle on EF containing this known angle, while another locus is a circle on AD as diameter ; the intersection of the two loci gives point G, and determines the required base BC. 7476. (By D. Edwabdes.)— If xyz » {1-x) (2-y) (2-«), show that ,» 5 xyzdxdy = --——. 1 = f Solution by G. B. Mathews, B.A. ; Sarah Marks ; and others. From the equation, and then by putting 1 —a: for a?, we have •'o Jo _./_j L_+_i 1-4. ^ -*\^X2.3=^ 22. 4*-« 3^.52 4«.6= "J' ~ U' 3=« 1.3/ \ 2- 42 2.4/ \ 32 53 3.5; ie_ 6 2 ^ 4 117 '■•-x. 7666. (By Professor H aughton, F. R.S.) — Prove the following formula for finding the Moon's parallax in altitude in terms of her true zenith distance, viz., sin^ = sin P sin « + i sin^ P sin 2» + i sin-'* P sin 3« + &c. Solution by D. Edwarobs; J. A. Owen, B.Sc. ; and others. If Z be the observed zenith distance, sin/? = sin P sin Z or sin p ^ sin P sin (2 + jo), therefore e^P - l-sinPg--^' l-8inPe^» Taking the logarithms, and expanding the right-hand member, we have p = sin P sin 2 4- 1 sin- P sin 2z + ^ sin^ P sin 3« + &c. Now, lip be very sm^l, sinp ^ p approximately. 7664. (By D. Edwardes.) — If the sides, taken in order, of a quadri- lateral inscribed in one circle, and circumscribed about another, are a, b, e. d ; prove that the angle between its diagontds is cos-* . ac-\- Od Solution by J. A. Owen, B.Sc. ; E. Knowles, B.A. ; and others. Let A, k be the diagonals, A the angle between b and e, and the required angle ; then the area of the quadrilateral is \hk sin Q — i (ad + bc) sin A, .•. sm => — sm A ; but cos A = — — — ; — ; ac + bd 2(be-t^ ad) hence, remembering that c + a ^ b + d, we have sin2 A - -^*f -^; hence cosa - { 1- -i^, } *, cos a - i'^-^. {ad + bey I (ac + bd)'^) aci-bd 7575. (By Professor Wolstenholmb, M.A., Sc.D.) — Two normals at right angles to each other are drawn respectively to the two (confocal) parabolas y^ = 4« (aj + w), y^ = 4^ {x + b) ; prove that the locus of their common point is the quartic 2y = {ai+bi) [x^2 {ab)h]k + (a\-b^) [a? + 2 («3)i]i, which may be constructed as follows : — draw the two parabolas y2 = (a-\-b)X'iab ± 2{ab)^ {x-a-b), and let a common ordinate perpendicular to the axis meet these parabolas in P, Pf Q, qy respectively, then the quartic bisects PQ, P^, ^Q, pq. Also the area included bewoen the quartic and its one real bitangent is fa%2 {tn+l) {m — l)\ whore a = biH*y and a > b. These results will only VOL. XLI. p 118 be real when ab is positive, or 'when the two confocals have their con- cavities in the same sense, but in all cases the rational equation of the quartic is fy« - air + 2a*) (y> - *x + 2ab) + ab (a -*)' = 0. [The qaartic is unicursal, but has only one node at a finite distance (x'^a + bf p'^O); there is sing^ularity at ao , equivalent to two cusps. The class number is 4, and the deficiency 0, so that 28 + 3ic = 8, $ + ic = 3, or « - 1, « « 2.] Solution by B. H. Rau, M.A. ; Professor Nash^ M.A. ; and others. The equations to the normals to the confocal p&rabolas f/^=4a {x + a)f f/ix=.ib (x + b), are respectively y « tnx^a {n^ + m), y =* m^x^b (wi' + Wj) (1, 2). If these are at right angles to each other, m| » 1 and therefore (2) becomea . ^._i^j(±,^X) (3,. From(l)-(3), . („^ ±) -«„»(«+ ±) _ ^ («+ ±) = 0; therefore am^ + — =» a? : therefore a*m+ — « [»+2 (ad)*]*, and e^m = [a;— 2 (aft)*]*. m m Eliminating x from (1) and (3), we have y » am, tn .-. 2y - 2 (— -am) = (^ +3*) [a;-2(a*)i]* + (a*-*i) [a^ + 2(a*)4]4, nnce the radicals may be taken with either the positive or negative sign. The quartic may be constructed with the help of the curves y = (a* + li) [x-2 (a*)J]4 and y « (a* - *i) {x + 2 (aA)4]i, which are the parabolas given by the equation y2 = {a + b)x'-iab ± 2 (a*)* (x-a-b). Squaring the equation to the quartic, we have 4y2« (a + ^ + 2flUi)(a:- 2aJ*i) + (a + *- 2ai*i)(ar+ 2aiM) + 2(a- *)(a:2-4a*)i ; or [2y2_(fl + J)a; + 4a*]2 ^ 4 (a-b)^x^-l6ab {a-b)\ or [y^-ax + 2ab) (jf^-bx ■k'2ab) + abia-bf = 0, which is the rational equation to the quartic. 7287 & 7353. (ByProfessorWoL8TENHOLME,M.A.,D.Sc.)— (7278.) Two circles have radii a, b, the distance between their centees is c, and a>b + e; prove that, (1) if any straight line be drawn cutting botii circles, the ratio of the squares of the segments made by the circles has the minimum value fl{[(« + 3)2_c2-|4 + |-(a.^)2_^2]4} ; ^{[(« + 3)2«^2]4_[(«_J)?« * + (? are the roots of the quadratic u^b^—u (a^ + ^— AE + EB>AP + PB, hence AP + PB is a minimum. [See Reprint, Vol. 39, p. 59, and Vol. 41.] 5522, (By Professor Asaph Hall, M.A.) — If a planet bo spherical and ^ be the angle at the planet between the Earth and the Sun, and a the radius of the sphere ; prove that the distance of the centroid of the planet's apparent disk from its true centre will be — sin^i^ when the o 3ir planet is gibbous, and — cos^ ^

Sv'~Y 3ir * 2 ia X = iira' + iira"-* cos ^ When the planet is crescent, we obtain _ 4a l--cos^ a, and there- fore H is always between D and A'. Applying the result just obtained to the segment cut off by DA' from the nine-point circle, whose radius is IE, we have sin 9- tDA'-i- JE - ^' - ^£ - ^"'I^a'"'"^ """ '^~*^' hence B = B-C; bIw A- i(i-c); li -~^ -~^'-i-i~^; II. Solution iy MoKOAK Jenxihs, U.A. FrofeBsoT Gkhibe's reBult may be put in the fbUowing geometrical (1) If a circle be irscribed in a given segment, the diameter of this circle ig equal to the product of the segments of the chotd divided by the height of the supplementary segment. If I be the centre of the in-circle tonching the h segment at E and the chord AB at C ; and H, M, L, the mid-points of the arc of the given segment, the chord, and the supplementary arc, respectively ; then EC passes throuKh L, and EI through 0, Uie centre of the circle AUBL ; and i^ ^ EC-CL AC ■ CB . OL " LE ° LE . LC ~ LM . LH ' IC - I AC^_CB ' ML therefore EAX the diameter perpendicular BO, A' being the mid-point of BC, IH, AD perpendiculare on BC, AN parallel to BO, da pHTBllel to BC cutting EL in t>, AD in h. Then In . lo : AIP = lA . IL : LA' -KL.r:LN.LK, I« . I« : LN . KN = KN .r : LN . KN ; therefore lu . If — EN . r. Now, the ceotroid of the triangle ABO being the internal centre of similitude of the circum- '- rcle and the nine- points circle, NK - 2«*, where ni is the height of the ,o nino.„n:,,(= „i~,i„ ™ the opposite sidoofA'Dto 1. segment A'iD of the nine-points circle, oi -, and therefore the in-circle touches the n ■inteU hy C. P. Hodgson ( MATHEMATICAL WORKS PUBLISHED Br FRANCIS HODGSON, 89 FARRINGDON STREET, E.C. p In 8yo, cloth, lettered. ROCEEDINGS of the LONDON MATHEMATICAL SOCIETY. Vol. I., from January 1865 to November 1866, price lOs. Vol. II., from November 1866 to November 1869, price 16s. Vol. III., from November 1869 to November 1871, price 208. 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Section I. — Mathematical Tables 2 „ II.— Algebra 2 6 ,, m. — Theory of Equations and Determinants 2 ,, rV. & V. together. — Plane and Spherical Trigonometry 2 „ VI. — Elementary Geometry 2 6 „ Vrr. — Geometrical Conies 2 Part II. of Volume I., which is in the Press, will contain — Section VEIL — Differential Calculus ., {ready) 2 ,, IX. — Integral Calculus (readi/) 3 6 „ X. — Calculus of Variations. ,, XI. — Differential Equations. ,, XII. — Plane Coordinate Geometry. „ Xin. — Solid Coordinate Geometry. Vol. II. is in preparation, and will be devoted to Applied Mathematics and other branches of Pure Mathematics. The work is desifoied for the use of University and other Candidates who may be reading for examination. It forms a dig:est of the contents of ordinary treatises, and is arranged so as to enable the student rapidly to revise his subjects. To this end, all the important propositions in each branch of Mathematics are presented within the compass of a few pages. This has been accomplished, firstly, by the omission of all extraneous matter and redundant explanations, and secondly, by carefully compressing the demon- strations in such a manner as to place only the leading steps of each prominently before the reader. Great pains, however, have been taken to secure clearness with conciseness. Enunciations, Rules, and Formulae are printed in a large type (Pica), the FormuUe being also exhibited in black letter specially chosen for the purpose of arresting the attention. The whole is intended to form, when completed, a permanent work of reference for mathematical readers generally. OPINIONS OF THE PRESS. " The book before us is a first section of a work which, when complete, is to be a Synopsis of the whole range of Mathematics. It comprises a short but well-chosen collection of Physical Constants, a table of factors up to 99,000, fh)m Burckhardt, &c. &c. . . . We may signalize the chapter on Geometrical Conies as a model of compressed brevity. . . . The book will be valuable to a student in revision for examination purposes, and the completeness of the collection of theorems will make it a useful book of reference to the mathematician. The publishers merit commendation for the appearance of the book. The paper is good, the type large and excellent.*' — Journal of Education, ** Having carefully read the whole of the text, we can say that Mr. Carr has embodied in his book all the most useful propositions in the subjects treated of, and besides has given many others which do not so frequently turn up in the course of study. The work is prmted in a good bold type on good paper, and the figures are admirably drawn. ' *— Nature. " Mr. Carr has made a very judicious selection, so that it would be hard to find any- thing in the ordinary text-books which he has not labelled and put in its own place m his collection. The Geometrical portion, on account of the clear figures and compreaaed proofs, calls for a special word of praise. The type is exceedingly clear, and the printing well done.*'—EduccUional Times. •'The compilation will prove very useful to advanced students.*'— 2%tf Journal of Science, T Demy 8vo, price 5s. each. RACTS relating to the MODERN HIGHER MATHE- MATICS. By the Rev. W. J. Wright, M.A. Tract No. 1.— DETERMINANTS. „ No. 2.— TRILINEAR COORDINATES. „ No. 3.—INVARIANTS. The object of this series is to afford to the young student an easy introduc- tion to the study of the higher branches of modem Mathematics. It is pro- posed to follow the above with Tracts on Theory of Surfaces, Elliptic Jn- tegrals and Quaternions. Fcap. Svo, 176 pp., price 2b. AN INTEODUCTION TO GEOMETBY. -^ FOR TEE USE OF BEQI^NEES. 00N8I8TINO OF EUCLID'S ELEMENTS, BOOK L Accompanied bt Numerous Explanations, Questions, and Exerciser. By JOHN WALMSLEY, B.A. This work is characterised by its abmidant materials suitable for the train- ing of pupils in the performance of original work. These materials are so graduated and arranged as to be specially suited for class-work. They fur- nish a copious store of useful examples from which the teacher may readily draw more or less, according to the special needs of his class, and so as iu help his own method of instruction. OPINIONS OF THE PBESS. ** We cordially recommend this book. The plan adopted is founded upon a proper appreciation of the soundest principles of teaching. We have not space to give it in derail, but Mr. Walmsley is fully justified in saying that it provides for ' a natural and continuous training to pupils taken in classes.' " — Athenedum. ** The book has been carefully written, and will be cordially welcomed by all those who are interested in the best methods of teaching Qeometry."— School Guardian. ** Mr. Walmsley has made an addition of a novel kind to the many recent works intended to simplify the teaching of the elements of Geometry. . . . The system will undoubtedly help the pupil to a thorough comprehension of his 8\xh}ect."— School Board Chronicle. ''whenweconsider how many teachers of Euclid teach it without intelligence, and then lay the blame on the stupidity of the pupils, we could wish that every voung teacher of Eucli