THE EVAN WILHELM EVANS MATHEMATICAL SEMINARY LIBRARY THE GIFT OF LUCIEN AUGUSTUS WAIT ^h ±\m\i.o... \ 7348-1 3 1924 031 192 838 olin.anx Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31 924031 1 92838 SOLUTIONS WEEKLY PEOBLEM PAPEES BY THE EEV. JOHN J. MILNE, M.A. bonbon : MAOMILLAN AND CO. 1885. Y. ^3 Richard Clay & Sons, BREAD STREET HILL, LONDON, E.fc. and Bungay, SuffoUi, fVJ?^ PEEFACE. Mt purpose in bringing out the present ^ volume is to fulfil the promise made in the Preface of the Weekly Problem Papers, and to place before students the various methods which are serviceable iu solving problems in Elementary Mathematics, many of which methods are not to be found in the ordinary text-books. My object being therefore both to increase a student's stock of mathematical knowledge, and to teach him to put it to a practical use, I have in many cases given two different solutions of a problem when I thought it would be to the advantage of the reader to do so. In almost every case 1 have given a preference to the methods of Elementary Geometry, as I think there is a tendency at the present time to allow them to be to some extent supplanted by those of Modern Geometry, which, although more fascinating, are scarcely as valuable a training to a student previous to his entering the regions of higher mathematics. With regard to the arrangement adopted in the following pages, a little explanation is necessary. Some years ago I began to form Problem Papers for the use of my pupils, without any vi PllEFACE. view to publication, and I selected the most suitable questions which I met with from time to time. Amongst other sources I drew largely from the Tripos Papers of 1875 and 1878, which are generally acknowledged to contain some of the most interesting of the problems which have appeared of late years, and solutions of these questions have been brought out by the examiners. When I decided to print my collection of Problem Papers, with solutions, two courses were open to me, viz. either to insert the questions selected from these two Triposes, and merely to indicate their origin, or to omit them altogether. Owing to their instructive character I was unwilling to adopt the latter course, and I felt that I should be doing good service both to teachers and to students in bringing these questions before their notice. For the use of those who do not possess the solutions of these two Triposes, I have added in an Appendix an equal number of alternative questions which are to a great extent similar in character to the corresponding problems. I have also added two notes, one on Geometrical Maxima and Minima, and the other on the geometrical method of inves- tigating the envelope of a line which moves subject to certain conditions. The problems bearing upon these two subjects are extremely interesting, although hints as to their treatment are seldom to be met with. My thanks are due to my former pupils and to my other friends who have rendered me great assistance, but especially Pheface. vii to Mr. E. F. Davis, M.A., late of Queens' College, Cambridge, who supplied me with many very neat solutions, particularly in geometrical problems. Those marked with an asterisk are all due to him, and it is a source of great regret to me that I was not able to avail myself still further of his aid. Owing to the nature of the present work it is impossible for me to expect that it is quite free from errors due to transcription. It is hoped, however, that the context will allow them to be corrected without difficulty. At the same time I should be glad to receive notice of any errors, or of suggested improvements in any of the methods employed. On the following page will be found a list of the errors which occur in the "Weekly Problem: Papers, most of which were due to the papers from which they were taken, and were detected on working out the solutions. JOHN J. MILNE. Chestnut House, Hevbesham, Milnthoepk. EEEATA IN WEEKLY PEOBLEM PAPEES, p. 4, last line, for Uq + a^ read Oq + ai. P. 9, last line but two, for 03 read Oj. P. 15, last line, for a = ^ read u. — fi. P. 16, 2nd line, for (- |) read (- ^X- Vapek II. VIII. XII. XIV. XVI. XVII. XXV. XXVII. XXIX. XXXIX. XLI. XLII. No. 6, instead of the word ' in,' substitute ' con- secutive terms of a fixed.' „ 7, in the last factor, for + a^ read — a^. „ 2, in first line, for x read a;". „ 7, in last line, for a sin (a - a') read a sin (a - o'). „ 4, for cos 76 read cos 55. „ 3, log 3 = -47712. See note at end of solution. „ 2, in the last line but one, for a; = — , read a; = +. 5, last line, for ' intersect ' read ' intersects.' 6, before the word 'Shew' insert 'The paper is on the point of falling over.' „ 2, for - — c^ read - — A m 2 7, the last line should be —,^ + t— -j = 1 (aa)2 ' (*^)2 ^ <^' 2, the first line should be If p, q, r are all unequal positive integers, and x is positive and not equal to unity. 6, for ' axes ' read ' semi-axes.' 1, for 2«a:(l + .r)«-i read <2.nx{\ + icf<—K Errata. Paper XLIV. XL VII. LII. LV. LVII. LVIII. LXI. LXXIII. LXXIV. LXXXVII. XOV. XCIX. J) 1J No. 6, , h , 6, , 1, , 1, , 1, , 1- , 2, , 3, , 6, , 6, , 5, , 1, , 1, , 2, , 1. after ' of ' insert ' the intersection of.' last line, for J read J. third line, for 2a read 2a. in last line, io-cl = m, = n = ±^ read ' The real values of /, m, n are given by for 576 read 2304. fourth line, for ' index ' read ' radix.' (3), for 63 read 61. for^%ead^. Q,in Qin for nir read imr. at the end add 'which also touches the chord.' first linp, for ' conjugate ' read ' any two.' at the end add ' and if KM and QN intersect in^', shew that QE = QN.' for 7 read 5. first line, for b read a -\- b. insert 2 in the L of the 2nd fraction. (3), this series should be T^ a2 e2 T^ KB 172 ~ 12.32 52.72 .92 + , second line, for sec - read sin-. „ 2,i „ 5, last line, for ' an ellipse ' read ' a conic' EEKATA IN SOLUTIONS. pAPEK II. No. 2. (1), the denominator of the second fraction in each line should be ex + a. Page 9, last line but one, all the + signs should be — .. „ „ last line, for + write + (- 1)". „ 11, last line but three, for 309 read 306, and make the same correction in the last line. Papee VIII. No. 3, second line, for AB read AC. „ X. „ 2, first line, the numbers should be 1', 2' . . . „ XIII. „ 2. (l),lastlinebut one,for^+y read 4+y(3i + l)- „ XIV. „ 4, first line, for sin 2fl - sin 6 read sin 16 — sin 0. , . v , . „ sin2"fl J sin2"5 „ ,, „ last line but one, for — ^ — — read " " " " ' sine 2sin5 Page 38, last line but three, for BB read BC. Papek XVIII. No. 4, for « - a read 2(s — «). „ XXI. „ 7, first line, dele §C. „ XXV. „ 5, in the 6th line, C is the intersection of ^5 and iE,S. ,, XXVI. „ 2, after ^yzw insert + xyz{x + y + z). „ XXVII. „ 2, fourteenth line, for aV^ read ay\ „ „ „ „ in the next line read a' = y = a^d — ad'. Papeb LXXIV. No. 1, last line, for 7 read 5. Page 286,.LXXIX. 3. (2), this should be cos 6 cos - 2 ^ 1 , 1 cosg , 1_ 2__ -4- C0S2 - C0S2 - COSZ -^ COS^ - COS^ - COS^ _ Page 292, XVIII. 1, seventh line, for 'quicker' read 'longer.' „ 294) „ 7, at the end insert ' For geometrical proof see Besant, Beet. Hyp. Props. 9, 10.' , „ „ XIX. 2. (1), for 3 read 6 throughout. SOLUTIONS WEEKLY PROBLEM PAPERS. Paper L 1. The value of the bill is £506 5s. .•. value of assets is £1423 Os. 4A .-. £2jl34 10s. 6d : £1,423 Os. 4d. :: 20s. : 13s. Ad. Ans. 2. as+bc={a-\-b)(a+c), 6s-]rca={6-{-c){b+a), cs-\-a// = (c+a){c+li), .: (us + be) (fo + ca) (cs + ab) = (a + by {b + c)^ (c + af. 3. 2*:r+2*+l |2j;'-6;r+5|2V''-2*(2*+l>+2'-2*+l 2.^^+(2'+2V^ - 2^(2^+l)a:^- 2(2*+l>- 2^2^+1).^; 2V-2*+l>+5 2*(2'-2*+l>+2'''-2V2*+ 2^-2^1. 4. 1 + 2cosO - 7)cos(7 - a)cos(a - /3) = 1 + cos 03 - 7) (cos (7 - ^) + cos (7 + /3 - 2a)} = cos ''((S - 7) + 1 + i{cos 2(/3 - a) + cos 2(a - 7)} = cos 2(/3 - 7) + 1 + cos \^ - a) - 4 + cos 2(a - 7) - -J = cos=(^ - 7) + COSX7 - «) + cos^(a - 0). S B SOLUTIONS OF 5. Let the bisector of A meet BG in D. . A sin - Then ■p sin^' ^ sinC'"i> 2 (. sin ^ sin C J ' 0/ 2 \b ^ J 1 /sin ^ sin ^\ 2aVsini< sin i B C ""2 1/l^M '°'2 1/1 , IN -^ = 2(« + J'-7- = 2W + y' .■. by addition the required result follows. 6. Let DG cut the circles in Q and H. Then BTL' = Z)© . DC, i)i?^ = BG . DH . . . . (1) Since ^7jC5 = AC'G = A6G, .: the triangles DGB, DAO are similar .-. BG.DG = Aa.DC=- AG.DG (2) SoAC.J)H=BE.DG=BG.I)G (3) .-. by (2), (3) BG.DH = DC' ; and by (1) BE . DF = DGK 7. Let the tangent at P meet the major axis in T'. Then CA'^ : GS^ :: CN : CO :: GN . CT' : CG . GT' ^' = '?? = :^ CS CG C&' " CG CS :. ^Zfis parallel to. PJ". Now GT is parallel to SP, PT', and .-. parallel to PO. . are equal, .\RT= HP. CO . GT' = CSS , Sr ^CH ' SP ~ GR : CT = GA; .: HT is perpendicular to RHTP is a rectangle, /, its diagonals 1. .r" — nx + ; Paper II. - 1 I >z.r«+i - (n -\- V) .v^ + I \ nx - (n + 1) K.ri+l — n^x'' + rfix — nx — {ji-\-V)x^-\- A^ — rfx + M.r + 1 - (« + l).'e't + M(« + l)a'-^(« + l)+w + l . x'' - ar + 1 WEEKLY PROBLEM PAPERS. 3 Now x"' - nx + li - 1 = x" — 1 — n {x - 1) = (x - 1) /^-i + a^-^ +- + 1 - «} and if we put » = 1, tlie exp. in the 2nd. bracket = .•. (a; — 1)" is a factor oi. a!^ — nx -\- n - 1 .-. (a - 1)2 is the G.O.M. ■ 2 (i) "^ + & , i-^ + a ^ (a + i) {^ + 2) cx-\-h-\- X {a — c) cx + a-\-x(b — c) 2(cx-\-a-i-b)-{-x{a-\-h — 2c) ex + 6 ex -{■ b ex -\- a -\- b •• '^ ex + b -^ '■'^ cx + b - ^^ ex + a + b ^ X {a — c) X (b - e) _ x (a + b - 2c) " ex -\- h ex -^ b ex ■\- a -\- b .: either x = a — e b — c a -\- b — 2o " ex + b ^ cx + b cx + a + b .... a' + b^-o {a^ + b^ a simple equation, which gives x = - ^ .^2 ^ p _ e (a + b) \' (2) >^¥^r^ + '\fx^ + b^ = ^-^' and (x^ + a^) - {x^ + 62) = ^^ - b^ .: \/x' + a? - V^H^' = 2i» e" - U^ 2V*' + a^ = 2x + 2x .: 4x^ + 4«2 = 4x^ + 2 («== - 4^) + — 1^ f^z — ^2^2 ... 2(«2 + 42) = ^__^ &c. (3) By subtraction we get B 2 SOLUTIONS OF - = 1 evidently satisfies the equation, and the product of the roots = the last term .•. the other root is - = ; • y a - h J It X — 1/, we have x^ = y^ = — . , _. • Tc " - '^ „ d {fi - rf a — b"' a [cr -\- b' -{- c' - U - ca — ab) ab-riT^ ''° s-a's-b (s-c){ab-s.(s-c)] S , 3. = , — = 5 = - = (SO. r^ _S_ S s s — e *■ ""^ ^ = 2d> ' and A = -2- . a — c , b ha — Bo .: cos 6 = + — = — . be — 3a So cos(^ = —^- — .*. 4 (1 — cos 6) (1 — cos (j)) = cos 6 + cos (j>. 5. Let ABC be a triangle whose sides are parallel to the three given straight lines, the centre of the circle, OG the radius perpendicular to BC. Make the angle GON equal to the angle BAG. Draw NM parallel to BO, Ml' parallel to AC. Then MNP is the triangle required. For arc 3fG = arc GN, .: angle MOG = angle GON. .: angle MPN = angle BAG &c. 6. Let P, Q, B be three points on a parnbola whose ordinates are y„ yj, ys, and let the inclinations of the tangents at P, Q, It to the axis of X be 61, 62, 63- Then cot 6,, + cot 6^ = 2cot d^ ■'• 2«"^ 23 "" 27 •■■^i+^3 = ^y2- •^3-^2=^2 -^1 = com. dif. = constant. .•. § is the vertex of the diameter Q/^ whose ordinate is PVR. :. TQ = \TV; and if iQJf the tangent at Q meet TR and TP, the tangents at R and P, in L and J/, the triangles LMT, RPT are similar. Also area LMT = i iJPT = i PQR = I QV .VR sin QVR. WEEKLY PROBLEM PAPEES. But RV^ = Qr. -^^^, and RV. sin QVR = ^-^-^ ALMT 160 .'. the A has a constant area as long as ^3 — y^ is constant, which is the case in the question. 7. Let ABGD be the square, A resting against the wall and B attached by a string to the point E. BK, CN', JDH are perpendiculars on the wall. 6 is the centre of the square, Tthe tension of the string, W the weight of the square, M the reaction at j4. CL and CM are perpendiculars on the lines of action of T and E. If the vertical through G cuts CW in N and AM in F, the line of action of T must pass through F. Let a be the angle which AB makes with the wall. Then BEA = a. Then T : ^: iE :: 1 : cosa : sina. Taking moments round C, T.CL= W.CN -^-B. CM Now CL = a cos 2a, a being a side of the square CM = CA cos [- -\- a) = a (cos a - sin a) CN = Cff sin ( J + a) = a (cos a + sin a) .•. cos 2a = i cos a (cos a + sin a) + sin a (cos a - sin a) TT .•. cos^a = 3 sin a cos a, and a =^ ^ i •'• cos a ^ cos a sin a . •■• ~3"' ^ "T" ' ,•. CN' : T>H : BK :: sin a -j- cos a : cos a : sin a ::4: 3 : 1 SOLUTIONS OF Paper III. 1 /I 1. (i-.rL 1+1.+...+ dLtiliii^ilLil^ ---'+•. " r - 11 <• n r-ll ) .". required sum of coefficients = coef. x in (1 — a;) n' - + r - 1 .•. required ratio is = 1 -\- n (r — 1) : 1 n Let a; = =- or "2. o 5 (-2)2 (-2)3 = •2+-02+-0026+-0004+-000064+-00001084- . = -2231412 . . . •"• JoSio J = /» X -2231412 and | = ^^ .'. 8 logia 5 - 2 = -43429 x -22314. = -096907 . . . .-. login 5 = 3)2-096907 = -698969 . . . = -69897 correct to five places of decimals. WEEKLY PEOBLEM PAPERS. sm d sin 6 sin." 6 °o "■ = a •'• ^n = sin 6 cos 6 COB 6 J J cos'^g sin^ ^ 1 1 •'. m -f- ?t = a I s~ ^^ "' I a" ^ a' sin'e cos* 5 sin* e cos* 5 {mnf 4. For cos a write c. r . ^ ,1 °°^'' I cos2a cos (ji — 1) a LetC = l+-^ + -^ +... + ^^„_i Bin a sin 2a sin (n — 1) a S = ~- + -^r- + • • . + — ^s:::! .C+Si=\+- +-^+... + . gas C .•. (0 + iSi) {l — " (cos a + J sina) } = 1 - - (cosw a + i sin wa) , equating unreal parts sin n a and c = cos a .: C = — - — . cot a = if ma = ir. cos" a By equating real parts we find _ . ^ cosua — Ssiaa = 1 — • cos" a 5. Let be the centre, B the radius of the circle round ABCB, and let G be the other point of intersection of the circles round FLC, BCE. Then the angle FGO = ADC = CUM. .: EGG + FGG = EGG + EBC = two right angles. .*. EG and GF are in one straight line. S SOLUTIONS OF Bisect EF in Z, and draw the tangents FH, EK, LM to the circle round ABCB. Then FG.EF = FC.EB = Fm and EG.EF = EC. ED = EE? .: EP^ = FH^ + EK'^ .: AFL^ + 2iJ2 = FH' + li- + EK' -\- E^ = FCP + ^-02 = WL' + iFU .: FL^ + iJ2 = 0£2 = £jf 2 + Je2 .-..FL = iilf.. .•. the circle, centre L and radius LE or ii'' passes through If. And the radii LM, MO are at right angles, .•. the circles cut orthogonally. 6. Let the forces be denoted by -j^' ■^• Then resolving along the normal at 3 we have sin (7 sin ^ , . , fi . — fi . ) which = 0. .*. the resultant must act along the tangent. 7. Let the diameter through Q' meet the tangent at P in T. Join EQ' cutting the curve in L and the diameter at £ in M. Then BL . EQ' = RMK Now BM : BQ' :: BP : BT :: I : 3 .: BM = J BQ'. .:BL. BQ' = i RQ'^ .■.BL = i BQ'. .: BL : Zq :: 1 : 8. Similarly B'L' : L'Q :: 1 : 8. 1. 2a - Paper IV. II - xf y ' 2a - e y)^2 .: '^.ay -3/ = ^2 + a:2 - 2zx %az - %z' = / + o;^ - "i^y Again 3 (j/ - z) = WEEKLY PEOBLEM PAPERS. 9 .-. (y - «) {2a - 3 (y + ^)} = (y - a) |2x - (y + z)} and y- z^feO.*. «+y + = «. ('i' - 2/f _ (^ - ^)^ s y _ y (x^ + 1/^ - 2gy) - ^ (z2 + gg - 2ra) ~ y« gg (y - z) + (y - z) (y'' + ^^ + y.^) - 2a; (y^ - z') yz .•. 3y0 = a;^ + y^ + 2^ + ya — 2.Ty — 2a;z .: (j - ^f = ^ (2y + 22 - a;) = a; (2a — 3x) (y - •^)'' /. 2a — 3x = • x 2. The given expression 5a - 20 - 7 f 4/3 + 3y - 3a , 50 - 7y + a} = 2 cos j^^ I cos ^ + cos f 1 284-2y+2a 8a-6/3-4y , 6a+40-8y , 4a-804-67 = 003 ^ 1-C08. J^ l-COS J + COS -^ = required result, since a -{■ + y = tt. 3. Let ABC be the given triangle, Og, O3 the centres of the escribed circles opposite B and C. OJ), O^E are perpendiculars on BA, BC and OJP, Ojfif are perpendiculars on GA, CB. ED, GF are produced to meet in A'. Then the angle BDE = BO2E = 90° - ^ .-. ADA' = 90° + 1- So AFA' = 90° + |- .-. A' = 360° - (180° + ^-Y^ + A) = dQP -^ So A" = 90» - I' = 45° + j; 5" = 45° + | ; C" = 45° + J J," = 90= - ^" = 674° + ^, -B"' = 67r + f ; C" = 67J° + | ^« = i) + - , & = -D + 2„. C« = D + 2„ • 10 SOLUTIONS OF A "B C Now when n is indefinitely increased, each of the angles :^ . ir ■ ^ ' => 2" 2" 2" ultimately vanishes, and J„, Bn, 0% each become = D, which is .-. 60°. .•. the triangles are ultimately equilateral. 4. Let the forces be denoted by P, Q, R. Their directions pass through the centre of the circum-circle .: P : Q : B :: BG : CA : AB :: sin^ : smB : sin C :: sinEOF : sin ^OD : sin DOS .'. the forces are in equilibrium. 5. It is true that one of the values of e2n7iV^ jg cos 2njr + n' - 1 sin 2mr and that one of the values of each of these expressions is unity, .". one of the values of eSirV-i — Qjjg of the values of e^f'*^^. To find the other values of the expression e2»wv^-i, introduce the common factor ic, and let the arithmetic logarithm of x be a. I log a; = a + imr^— 1. Thus we see that a quantity has an infinite number of logs, one being real and the rest imaginar y, and the latter are found by adding to the real log the quantity 2mr v'- i. The suitable value to be given to n is to be determined from the circumstances of each case. Thus to be absolutely correct, in the question the equality should be written .: raising each to power V- 1 and this is evidently true when n = — 1. Similariy it can be shewn that there is a value of each of the other expressions equal to one of the values of e^-V- 1_ WEEKLY PROBLEM PAPERS. 11 6. Let A and B be the centres of the circles which intersect in C and G, and let the tangents at C and the common chord cut A3 in D, E, F respectively, so that A and B are on the same side of the common chord CFQ. Then a = DOE = BCQ + FCG = CBA + CAB = -t - AOB BO _ Bin CAB CF ^ 1 ■'• AB ~ Bin ACS ~ CA ' sin a ab sin a ah sin a .: CF ■■ AB ija^ + b'^ + 2ab cos o 7. Draw mif perpendicular to SH. Since CK = CS = CH .: SBH is a right angle. Since GS- = CR .: CSR = CBS = EST .: the triangle SUM = triangle Sifr .-. SM = /Sy. Similarly it can be shewn that HM = HZ. Sinoe RM^ = SM . MH = SF.H^Z = BG^ .-. on the tangent at £. Again SF : SP :: SY : BZ :: SM : HM :• .Vlf . HM : HM^ :: BM^ : HM^ :: SR-' : HR\ Paper V. 1. 1037 (10-12 1 301 30000 37000 2- 301 . 30301 30301 3032 3090300 6064 6699000 30963 54 BM = EC, .: R lies 12 SOLUTIONS OF 2. Suppose that the expressions are equal when a and h are interchanged. be — a^ ca — b"^ \ Then = a + ^^qr^7 + ^ - (* + „. + ^2 + ^) bo — ca — {efi — i^) . M / 1 a + b + e \ = («-*) I 1 -«^ + A^ + .»| a + i + c 1 - -^—, — j^—. — •„ = since a — b i= Q). a^ + 6'' + r Now we will prove that when this condition holds, the espressiona are equal when a and c are interchanged. For . be — a^ _ , . ab - e'^ \ " "^ a^ + b'^ + (? ^" "*■ a' + b^^ cV a^ + b^ + ir' ^ '\ a^ + b^ + c^i = 0. Also, if a + + c = 1, we have a^ + b'' + c' = 1, .: {h + of = (1 - ay- ... 42 + c2 + 2ifl = 1 - 2a + a^ .-. 1 + 2i(; - a^" = 1 - 2a + a«, .". ie + a — «^ = 0, . be - a^ , , , r, •• " + 2 , ;■, I — J = a + is - a= = 0. a' + b^ + tf 3. Assume y = x^, so that when a; has any particular value, y = the square of that value. .•. x = kJj/, and the required equation is oy + b Vy + c = 0, which reduces to ay + (2ae - b-yj + (? = Q. Again, assume 1/ — iJx, so that when x has any particular value, y = the square root of that value. .•. x = y^, and the required equation is ay* -\- bf' + = Q. WEEKLY PEOBLEM PAPERS. 13 4. cos 2d - cos 46 = 2 cos 26 cos 46 — cos 25 + cos 66, .: cos 56 = — cos 46 = cos(7r - 46) .: 66 = 2n7r ± (it - 46) .: either 105 = (2» + 1)tt, or 26 = (2« - 1)it. 6. If h denote the height of the room K" cot =a = AC'' = »= + BC" = a^ + h'' cot^ft cot^a — cot^|3 Now if o = 18°, cot^a = 5 + 2 VS, and if ^ = 30°, cot=^ = 3, - '" ^^' ^^^ = l|'(2V5-2) = 12=(2V5-2) 5 + 2V5-3 2(1 i- V5) 42 = 122 X 2-47212 .-. ^ = 12 X 1-572 ft. = 18 ft. 10 in. -368, &c. 6. Let 6 be the eccentric angle of P. The equation of any one of the series of paraholas satisfying the given conditions is {x — a cos 5)2 + (y — * sin 6Y = {x cos a + y sin a - p)\ li y = 0, X = ±\la^ - b\ This gives us .-. a cos 6 = p cos a and («2 - 52)gin2„ = p^ ~ a^cos''6 - 6' sin =5, .". eliminating p we have («= sin 'a + 6^ cos *a) (sin =5 - sin ^a) = 0, .: a = njr ± 6. By giving different values to », it will be found that there are only two different parabolas. Their equations are {x - a cos 5)2 + (y - 6 sin 6y = (a: cos 5 + y sin 5 — a)" (tc - a cos 5)2 + (y - & sin 5)2 = (.t cos 5 - ^ sin 5 — a)2. The equations to the directrices are y — — X cot 5 -j- « cosec 6, y — X cot 5 + « oosec 5, and these make with the axis of x angles ^ + 5 and 5 "" ^' .-. the angle, between them is 25. 14 SOLUTIONS OP 7. Ciy" = ACAD ^ AC(AC - CD), .: CD'' + AC.CD- AC' = 0. ,.0D = ^14^AC. .; taking the positive sign we get BE = BD ^ BC + CD = AG + '^^~ ^ AG = ^^ ^^ BE BE V5 + 1 .'. AGE = 72° = ^ of four right angles. Paper VI. 1. Substitute fory in terms of x from (1) in (2) and we get x^ (ly'e + aH) — Ixad + (? — 4^ = 0, If this has equal roots aH''' = (Wc + (fid) {d — b'^), which reduces to a? 4^ a?d + y-c = cdjOT ~ + J =1. ad ad a Now X = ri — i — ~j = T = ~ Vc + a-d do c 1 c — cfi cd — a'A b'o b y = 1 ^^-"^ = —bT = ~iw = 6^ = 5* 1 2. 8o = 9 .-. 23" = 32 ; 36 = 5 .-. 3 = 5^ (1) Let logi„l = xj, .: 10^1 = 1 .-. % = logi„2 = iTj .-. 10a;2 = 2. £ £ /10\, Then 2 = S^t = 53a6 - V"^/ 2 2 2 .*. 2' "^Saii = lOSai .•. 2 = 102+306 -i- 2 •. lOs'j = lOS+So!" .-. a;, *2 2 + 3ab WEEKLY PROBLEM PAPERS. 15 (3) logid? = 2^3 _ 3a ~ 2 + Sab (4) logio4 = Xi So 3a 10^4 = 4 = 22 = 102+3a» .-. a;. = ■ ^ ~ 2 + 3a5" 3. sinB = sin(0 + ^) .-. sin^ J5 + sm2 - sin^^ = siniJ . sin (0 + ^) + sin (C + ^) sin (0 - A) = sin£ {sin {G + A) + sin (C - 4)} = 2 Bin£ sin C cos A. 4. Let P be the orthooentre of ABG. Then PDF = PBF = | - ^ PZ)^ = PCB =: - - A .: PD bisects the angle FDF .: P is the centre of the circle inscribed in DEF. Let / denote its radius. sin 4 a Then FF! = AE . -: — t, = c cos ^ . - = a cos J. sm O c BEF EFB\ .: FE = r' (cot — 2 1- cot — g— j- Todh. Trig. Art. 249. , , ^ , ^^ , sin (B + 0) , sin^ .*. a cos J. = r (tan B + tan 0) = r 5 ^^ = / 55 ^ ■ ^ ^ ■' cos^cosO cos B cos .•. / = -; — 7 cos A cos B cos C sin.4. = 2 iJ cos .4 cos B cos (7. 5. Let AG hei the vertical wall, ABCD the rod, being its middle point and B the position of the rail. Draw BO and A F perpendicular to thtj wall, and OjP vertical. ThenP, the reaction atB passes through F. Let W be the weight of the rod, and 6 the angle it makes with the horizon, and a its length. Taking moments about A we have P.AB= W.AF AB= —a = Tfi . -^; AF = AGcosd = | cos5 coso lo cost^ z 16 SOLUTIONS OP .: p = w . s C0S2 e: Also P : W :: sin AFC : Bin AFP :: 1 : cos 5. .: P = W . „ = W . 8 cos' e cosfl .-. cos' 5 = I .: cos(9 = i .-. e = 60°. 6. Join DB, EO, FB, FB, EG. Then AS : EB :: AG : GD .: GE is parallel to DB. Similarly FH is parallel to DS .". GE is parallel to FH .'. they are in the same plane .•. Gff and EF are in the same plane with them. 7. Tripos 1878. Monday morning. No. 11. Paper VII. 1. Clear of fractions, and we get x>{a + b-c- a)+2x''{ab-cd)+s; {ah{c + d) -cd{a + b)] =0. Then a; = is one root. Since two of the roots are equal, suppose (1) the equal roots to be = 0. .*. ai (c -{- d) — cd {a + 6) ,1+1 = 1+1. ' a ^ 6 c ^ d (2) If the equal roots be not = ii;^a + b-c-d)+ 2x (ab - cd) + ab (c + d) - cd (« + A) = 0. Since this equation has equal roots .-. {ab - cd)'' ~ {a ■\- h - c - d) {ab {c + d) - cd {a + b)} = 0. Now the expression on the left vanishes when u. = c, a = d, b = c, or b = d, and it is of the fourth degree, .•. we may put it ■=i {a — c) {ii — d){b — c) (b — d) where i is some numerical constant. Let a = 5 = 2, c = (f = 1 .-. ,{: = 9 - 2 (8 - 4) = 1 .-. A = 1 .*. one of the quantities a or 4 is equal to one of the quantities c or d. WEEKLY PEOBLEM PAPERS. 1 li b == d, X ab - cd d ~ ab - ch , ■ = - 0, a - c - 6, 0. a + b — c — lia = c, X ab - ed 1 ^ ~ ah — ad - «. 0. a + b — c — 2. (1 + ])'' = 1+1 + 1(1- -') 1 « (I - 1) (t - 2) 31 + • 21 + " " - + I+ 3.1 22.21 3.1.1 23.31^ 3.1.1.3 3.1.1.3.5 2''.41 25.51 + .. ■ • - + k 3 1 2-.21 r 1 1.3 1.3.5 "^12^.31 2un 2».5r- -1 •'• 2 1 1.3 13.31 2-'.41 13.5 ■^2=.5i -.... -3 1^ + 2 +2^.21- ^ / 1 /23 ,-'\ 23 2 ,_ 3. X — ff cos li — 2 cos Q = 01 X cos R — y ■\- z cos P — J .'. since P + § + i2 = (2» + l)jr, we get ^ y ■" „ s"hrp = s"Rr^ = ^h^=-^^"pp°^« .'. arcos §+ycosi' — z = ZlsinPcos § + cosPsin § - sinijj- = A'(sin(P+0-sma} = 0. To find cosP, multiply (1) by x, (2) byy, (.3) by z, and subtract tho first result from the sum of the two latter. We thus obtain y^ + ^^ — is^ = lyz cos P. 4. Let P be the orthocentre, and § the centres of the circum and nine-point circles. Then Q bisects OP. From 0, Q, P draw OM QB, PN perpendiculars to BC. Then FN — PB cos C = -^—?. cos BcosG=2It cos .B cos C, Sin o and OM = i2 cos il. 18 SOLUTIONS OF .-. QR = i{PI/'+ OM) = ;j(2cos^cosC+ coa^) = 2Cos(C - B). 5. Tripos 1875, Blonday morning. No. 1. 6. Substitute and clear of fractions. .-. ax-'rb + J-la;/= {ex + d)^ - ct/r; + ij - l\{c.v + d)ri + y^]. Equate real and unreal parts. .-. aia-c^)+crijy = di~li\ ,-. x[(a - c|)^+cVj = {"- c^id^ -i)- cdr,\ y{(« - c|)^+cV) = cri'^dt - l)+dr,{a - e^), +dY{c^-ekf, = {(« - cif+oV\{{dl: - bf+d'^r,'], .-. (:J^+/){(«-0|)'+^V} = (^^-i)^+«V- Now if X, y describe a circle, we must have 1^ = x"^ -\- y^ Ar px -\- qy, where k is constant, .-. !c^[[a - c^f + c"ri^ = {{a - c^,^ + cY\[^^ + y^ + px + qy], = {d^ - by + dW + p{a - c|) (;') (cV^" - {y - kf + Jia\y - kf - {h0 - h? - ha^) (y - Vf + k{¥ + ;i= + dF) {x -h){y- k) = .-. hk[{x - hf - (y - ky] = {ff -B- a") {x - h){y - k). Paper IX. 1. Let ex^ — «» + 5 be one factor of ax^ — ix^ + c, and let kx •{■ m be the other, .■. aafl — 6x^ -jr e — {cx^ — ax + b) {kx + »')> = kcx^ + x^{cm — ak) + j;(44 — am) + i»;. .■. equating coefficients of like powers of x, k m cm — ak — b kc a \ a'^b'' bV^ce- ^ be - a?- ^ ac ^ ac ^ 1 .: - ic = be - a , WEEKLY PROBLEM PAPERS. 21 .*. «2 = Uc, * = -, m = 7' .'. {c3^ - ax-\-b\{--\--\-=-^-\-x'{ 1 + 4 ^)+— ' ^ '\m, m) m ' \m ml \m m / m = bx^ — ex + a. 2. The (r - l)tii term is r2 |P + 22 + . . . . + (r - If} (12 + 22 + .... + r^} r2 (»•- - l)r(2r - - 1) T{r + 1) (2r + 1) 6 6 36 1 (r - 1) (r + 1) (2r - 1) (2r + 1) _ -^ , ^_ + _£_ + ^_ r-l^r + 1^2r-1^2r+l ^ 7~^ ~ r + 1 " 2?- - 1 + 2»- + 1 .•. if Sn denote the sum of n terms, 6 6 6_ 6 24 24 , by partial fractions. 2^^ ^3-1 n + l~ n + 2 4-1^ 2» + 3 6 = ^ - (» + 1) (« + 2) (2» + 3)' 3. psinS + qcosB = r(sin5 + cos5), .•. {p — r) sin 5 = (»■ — 2) cos 5, sin 5 cos B 1 • • r - 2 i) - r V(i' - J-)^ + (2 - '')'' Now 1 = sin 2^ + cosM = /sin 2^ + 2' cos =5, ... (g _ r)» + (p - rf =f{(- X = z . r ■\- X. Again r -i- x = 2z = ix + i, .: r = 3x + 2, .: if we give to .r in suooe"ssiun the values 1, 2, 3, .... we obtain for r the values 5, 8, 11, ... . .•. neglecting the scales of notation in which the radix is 1 and 2, we see that the above number can only be found in one scale out of three. 2. The A.M. of 1, 2, 3, 3m is known to be > the G.M. 3mi3m + 1) . „ ^ „m/ r .-. ■ ^ i-3m > V(3»0I 24 SOLUTIONS OF .", it is sufEoient to prove that J/Smi^M + 1)2 3m + 1 ,, , „ ^ 3m + 1 /y — 5^-^-! — ^ > — 2 — ' **'■ ™'''* ^"^ -^ — 2 — or 3ra > 1, ■which is true Urn > 0. 3. Todh. Trig. Art. 272 gives an expression for tan n6 in terms of tan^. If we put tan 5 = -/S or -^' = | or ^- .-. «5 = a multiple of n or 277, since re is a multiple of 6. .-. tan nd = 0, and the required results are obtained by equating to zero the numerators of the expressions. 4. Let SB' cut AA' between A and A'. Then iCBC = iir - {BCB' + BC'B'), = 2i7 - {2(7r - BAB') + 2(7r - 5^'S')} 5 .-. C5C' = 5^5' + BA'B' - TT, = 27r - (ABA' + ^5'^') - n-, = jr - (^00' + y/0'0) = OAff. If 55' cut ^J' produced, CBC will equal the supplement of OAO'. 5. Let the normal at P meet the major axis in 0. Through G draw a line LMGL'M' perpendicular to the normal, and meeting SP and PS' produced in M and M'. Draw SL, S'L' perpendicular to MM'. Then SP : S'P :: SO : S'O :: SL : S'L' :: |^ = ^' SM S'M' :: ^ ■■ ^7^ :: S3f : S'M', since PM = PM' :: SP - PM : PM - PS'. 6. To make the highest card project as far as possible from the table, we must move it until it is on the point of falling. Then move forward the one immediately below it until it is on the point of falling, and so on. When the lowest card projects as far as possible from the table, it is evident that the highest card will then project a maximum distance from the table, and that each card will be on the point of tailing independently of the rest. 7. Let iS be the vertex of the triangle, S'P the base. If S'SP = 6, S'PS = 26, cos 26 = ^^ ~ \ cos 6 = '"'^ ^ ^ - WEEKLY PROBLEM PAPEES. 25 Let SS' = e, and e = the eccentricity. Then 2AC = SP -\- PS' = c + 2e cos 20 = c(l + 2 cos 26) = 2e cos 6, c .'. -^ = 08=6. CA = e . ccos 6, 1 2 V5 - 1 2 cos ^ \/5 + 1 2 ' .-. e{e 4- 1) = 1. If SB be semi latus rectum, SB = e{SA + AX) = e(e + V)AX = AX. Paper XI. 1. 1000 lbs. = 453-59 kilog. 1000 miles = 1609-27 kilom. .-. the question may be stated thus : If 453-69 kilog. can be carried 1609-27 kilom. for 25-2 francs, how- many kilog. can be carried 100 kilom. for 20 francs ? 100 : 1609-27 ,^„ ^„ , 25-2 : 20 "• *^^'^^ ' ^"'• _ 1609-27 X 453-59 _ 729948-7793 •■• ^"^- ~ 126 " ~ ■" 126 = 5793-24. 2. l(x + s/) (a + 6 - - ^ = i"^ - o-h xy{,a -{■ b - c — d) = ah{e ■\- d) - cd{a + h), ■■■{^S{a-\-b-c-df = {cd-aby-{a-^b-c-d){c(b{c+d)-cd{a-irb)]. Now the expression on the right vanishes when a = c, or a = d, ot b = e, or b = d, and .'. = k(a — c) {a — d) (b — c) (b — d), where k is some numerical con- stant. By equating coefficients of any term, as aH'^ we find k = 1. ■'■ ^"^Y"^ + b - c- d) = J(a - e){a - d) {b - e) (b - d). I l\-i 1/1 1.3 1 , 1.3.5 1 ) 3-V1-4J =1 + 412 + 274 ■4 + 2":47e"4=i + - •■7' .-. Given expression = 4 | (j j — 11 = 4^—^ — l| = g(2- ^?>) >J3. 26 SOLUTIONS OF 4. sm(A + V j = sin^'jT + ^ + |j = - sin(^4 + |J' sin(^^ + i) + sin(-^ - sin/ i"! . /■ »r\ ' .•. Expression = '—. — ^ J 2 sin y^ cos ■= sin ^ . ,, . „5r Bin''^ — sm^- 1 4sin^ _ 3 sin ^ ~ 4 sin ^^ — 3 ~ 3 sin ^ — 4 sin '■'A 3 5. tiiny = sin 3^ cosec 0^. Vl + .r^' + Vl - a-2 Vi + :j:2 - Vl - 1 + tany 1 — tany l+.r2 ^2 •■• Vl + .^2, Vl - .-v^ sec ^y + 2 tan y * ' 1 — .r'^ sec ^y — 2 tan y 2 tan y - . .•. ar^ = s-'- = 2 sin y cos v = sin 2;/. sec *y J J J ' 6. Tripos 1875. Tuesday morning. No. 6. 7. Let W, TQ be two of the tangents meeting the axes in f, V. Join S'T meeting BG in B, and draw the ordinate TN. Then the angle STt = S'T*, and TtS = Tt'R, .: TSt = t'RT =-- S'lW, .: the triangles STN, S'TN are similar, .-. TN^^ SN.NS'. .: the locus of T is a rectangular hyperbola, having S, S as vertices. It is evident that iV cannot fall between S and S'. WEEKLY PROBLEM PAPERS. 27 Paper XIT. 1. Multiply tlie three equations together •'• (2 + •t) (^ + ^) (y + a;) = ± ale, .-. bydivisioiiy + 3 = + ^~, z + x = ± -„ x -{■ y = ±~. a c By addition we have !r + y + z = hi ± - + ^ + f.^]-, t « b e J which at once gives us .v, y and ^. 2. Let — = 1 -. [x — a) [x — 0) X — a X - b .:x = A{x - b) + B{x - a), ,\ A + B = 1, AO + Ba = 0, .: A = — ^ ,B= ^, a — b a — '0 \ ( b a \ .: expression = -{ V > a — b Kb — X a — X) =^{('-r-('-r}' e f ^ 1 n 1 i a'^ - b'n 1 .•. coef. oi af^ = 7 s~ > = . • a — I? ( o» a^i a — b an . b'^ 3. Consider the identity {a+b + c) (a2 + i2 + c2 _ Jc _ <;« _ ai) = c3 + js + ^.3 _ 3^3^, Since this is always true whatever be the values of a, 5, c, let a = e»S b = e^*, c = e^^, where i = *«'— 1, _•. ^gai -\- e?i + en) {e^ai -J- gS^i -j- g^i _ 4fi+T)i — e(.-/+ii)i — d''+W \ = e3a« + e3/3i ^- gSyi _ 3elc+ l3+.y)i .: putting for e"* its value cos a + i sin a, &c. {cos a + COS j3 + cosy + i (sin a -j- . . .)\ {cos 2a + . . . — cos (/3 + 7) + ' (sin 2a + . . .) } = cos 3n + cos 3/3 + cos 3y — 3 cos (a + (3 -f- y) + i (sin 3a + . . .) ; .■. equating real parts we obtain the required equality. Another relation can at once be written down by equating unreal parts. 28 SOLUTIONS OF 4. Let be the centre of the circum-circle, P the ortho-centre. Let ^P produced meet 5(7 in M. Through draw ON, OQ perpen- dicular to BC and PM respectively. Then PM = PB cos C = \- °°^ "^ cos C = 2i2 cosB cos C, 2 sin C PM = P£ sin C = ^'^' ^°^^ sin (7 = 2i2 cos£ sin 0, 2 sin (7 ON = R cos -^, £A^ = ie sin ^, .-. 0P2 = (piif - OiV)2 + (BM - BN)\ ^ R"^ [{2 cos B cos - cos ^)2 + (2 cos B sax C - sin^)^}, = 52 {(cos B - C - 2 cos AY -f sin« (C - B)], = 722 {1 _ 4 cos^ (cos^^^ + cos£ + G)], = 7^2 {1 - 8 cos ^ cos B cos 0} . 6. Let the circles round JDE, CDF meet again in 0. Join OD. The angle ^0£ = ALE = Ci)F = CO J, and FGO = FDO = sup. of ODA = OEA, .: OA : OF :: OE : 00 .: OA . 00 = OE. OF. 6. Let P be the vertex of the diameter PO which bisects AB, and let PT, the tangent at P, make an angle 6 with the axis. Then if a;', y be the coordinates of any point Q on the parabola referred to PO and PT as axes, e,m^6 ia > r r 7. Tripos 1878. Tuesday morning. No. 2. Paper XIIL (a + J + c) (a + J - c) (a - & + c) (a - 6 _ c) = [a-^ + 62 _ c^ + 2«5} {«=> -I- i^ _ ^2 _ 2«6|, = (a^ + 6^ - cf - 4a^b\ = a* + i* + ^•^ - 26V - 2cV - 2a'i». WEEKLY PROBLEM PAPERS. 29 2. (1) This equation may be written {^^7-^~{^^\-^ + 1)3_ {h + lf. Put a; + - - 2 = 16y, «■» + -i - 2 = 166, :. y = t is one solution. The others are found on division to be given by the quadratic Vy^ — h -^ y. Then the three values of X -\ — , and .•. the sis values of x can be found. X (2) From (1) and (2) by elim. z we get h {a - c) X -y a {}> — c) y = a-h -\- aV' - 2alc' So from (1) and (3) {a - c) x + {h - c) y = ab — c^ a{h - cf .*. X = ^ . (c — a) {a ~ b) The values of y and z can at once be written down by making symmetrical changes of the letters «, b, c, 3. The given expression = i {cos 2 (^ + -B) + cos 2 0} {cos2 {A - B) + cos 2 0} + i {cos2 0- cos2{A+B)} {cos2(^-£) - cos 2 0} = i {cos2 0cos2 (A +B) + cos 2 C cos 2 (4 - B)}, = cos 2 cos 2 A cos 2 £. * ?-2?-3 >i— s — cl ^ s . s — a _^^_ 2s-{ b + c) /{s -b){s- c) ^ ' {s — b) (s - c) ^ s . (s — a) = 2s - (6 + c) = «. 5. The angle EAC = i ABO = ABD = ACD, ,: EA is parallel to CD. 30 SOLUTIONS OF The angle EOA = ^ AGB = CBD = GAD, .: EOia parallel to AD. 6. Let QNjR be the chord cutting the major axis in N. Let A'Q meet the tangent at A in L. Then the angle QAL = QA'A ; .-. QAj^ + QA'N = QAN + QAL = one right angle. .•. QAE + QA'B = two right angles. 7. Let G be the centre. Then it is evident that for any given weight to have the greatest efEect in upsetting the table, it must be placed at one of the corners, A suppose. Let JE), P be the middle points of the adjacent sides. Let W be the weight of the table, and let P be the greatest weight which can be placed at A without destroying equilibrium. Let AQ and EP intersect in H. Taking moments round U, we have P.AH = W.EG, .■.P= W. .: no weight less than the table when placed upon it can upset iL Papkk XIV. 1. {ax - hi/Y' + iftx - hi/) {at/ + hx + It/) 4 {ay + hx + lifp = fl%'2 — iuixi/ + h-'if + a^xi/ -\- ahx^ + ahxt/ — ahy^ — Iflxij — b-//^ + a'^y' + iV + Uy + labxtj + IVxtj + lalf, = (j?^ + d'^xy + d^y'^ + aia;^ + dhxy + ahy''' + W'x'^ + li-xy + b'^y'^, = (^2 + ^y + 2/^) { m'c — n'b . I'x + m'y + w'z _ ^ ~ //' + mm' + ??/»' . W -\- mm' + ^2'^' / Similarly it may be shewn that ^//^r , i ^ is equal to each of the other expressions. ^■2+2+2+2-'^' •■• t^°i 2 + 2-' = ^^" 1'^ - 2 - 2/ =_,- *^"l 2 + 2J ' ••• = t^^ii + 2J + ^^''il + 2> J -S C , ^ D tan ^ + tan 2 tan - + tan g J B ' CI) 1 — tan Z) tan 5 1 — tan - tan ^ (tan-n+tan^^^l-tan^tan-gj+^tan^+tan^j^l^tan^tan^l 5 32 SOLUTIONS OF (A B\ C D , ( G D\ J B .: ^^tan - + tan -) tan -tan - + \^tan - + tan ^) tan - tan ^ = tan - + tan 2 + tan - + tan - • A r Now tan -z = -, &c. ^cc? + crfi + dub -Y abc. ~ a -{- 6 + c -^ d 4. sin 25 - sinS = 2 sin 5 cos 5, sin id - sin 26 = 2 sin 5 cos 35, sin 65 — sin 45 = 2 sin 5 cos 55, sinS^S - sin (2" - 2)5 = 2 sin 5 cos (2" - 1)5, .•. by addition we find sin 2"5 cos 5 + cos 35 + cos 55 + .... + cos (2" - 1)5 = . - > ^ ' sm5 = 2""^ cos 5 cos 25 cos 45 cos 2""^ 5. Todh. Ti-ig. Art. 129. 5. Let one force act at A and the other two at B, and let each force he denoted by P. Then the resultant of the two forces at B must be a force P parallel to the force at A. .". the forces at B must each make a.i angle 60" with the resultant, and .". must make an angle 120° with each other. 6. We will first prove the following useful proposition. ' The perpendicular from the focus on any tangent and the diameter through the point of contact intersect on the directrix.' Draw iSr perpendicular to the tangent at F and produce ST to meet the directrix in I). Draw the tangents SQ, DQ[. Then since the directrix is the polar of the focus, QC^ passes through S, and is per- pendicular to SB. Bes. Con. p. 10. .■. QQ is parallel to the tangent at P. .". the diameter through P bisects QQ and passes through D, the point of intersection of tangents at Q, and §'. Prom this the question in the text follows at once. WEEKLY PROBLEM PAPERS. S3 7, Draw C^ perpendicular to AB. Then CD^ = DE^ - EC = {DA - AEf + {AC^ - AE^) = LA^ + AE'' - 2DA . AE + AB^ - AE' = DA" - AD . 2AE + AB^ = 9^i;a - 5AB^ + ^^a = lAB^. Paper XV, 1 . Let P denote the population at the beginning of the 1st year, and let Pi, P2, P3 denote it at the end of the 1st, 2nd, 3rd .... years. Then P^ = P + {l-l^)p^p + Z. = l|p. 181 /181V So P2 = lao ^'^ ^ \ 180/ "^ ' ®° °°' .'. if .T denote the required number of years ••, x (log 181 - log 180) = log 2. •301030 .„,„„ , ''■ * " -002407 ^ ^^^^^ nearly. 2, Employing the method given in the notes 1, (1) we have ic" — l/c ca' — da ab' — afh^ .: (Ic' - b'c) {ah' - a'd) = {ca' - daf. 3. Take logarithms of both sides, .-. {X + 2) (1 - log 3) = 2 (2^ - 1) log 3, 2 1 •■• " ^ 5 logs- 1 " -69280325 " ^'^^^ * * * 4. The angle BOP =~ + | =900-1, IlP = PC=tseod, 34 SOLUTIONS OP radius of circle round BOP = i BP cosec BOP = " ^^ = " ^^^ - ^cosicos^ i ^/^Ho - a) {s - e) 2 2 _ a6 \/ca {s — i)_ .*. product of radii of the first three circles is _ g'^V. dbc J{s - a){s - 6) {s - c) 43 . r^ . / 4' • »^ -s^ A3 r3 () {u — a) = uv, (1) (a + i;) (» - ^) = M», (2) since n — a = v — ^. From (1) ^2 = „ (a _|. „) ; From (2) c^ = /3 (a + «) ; .-. a" : »2 :: a : ;3. 2. Clearing the second equation of fractions we get a [p? + /) -^^ xy (x -\- y) = a^ -\- a?- (a; + y) + axy, .: 2a3 4- ary (x + y) = a^ + a' (x + y) + axy, •■• (a: + y - 1') (^y - «^) = 0. .*. either x -{- y = a; (1) or a;y = a^. (2). From (1)^ = fl - :r, .-. x^ + {a - xf = 2aS .-. a; = | (1 ± \/3). .-. y = a - ir = I (1 T V3). From (2) {n — w)^ = 0, .'. .r = y = ± a. The negative sign is seen by the 2nd equation to be inadmissible, D 2 36 SOLUTIONS OF 3. Let B denote the foot of the tree, BA its height, G and D the first and second positions of the observer. Then BCA = 61° BDA = 46°, 1)AC= 5°, BAC= 39°, DC = 30 feet . _ „. sin 46° AC = 30 . — — --> sino" .'. log AC = log 30 + i sin 46° - L sin 5° = 1-417712 + 9-856934 - 8-940296 = 2-334350. BC = AG . Bin 39°, .: log BG = log AG+ L sin 39° - 10 = 2-334350 + 9-798872 - 10 = 2-133222, log 135-90 = 2-133219, log 135-91 = 2-133251, .-. BG = 135-90093 . . . Note. — The question is given as it was originally stated. The value of log 3 is incorrect, and the value of AB would be found to be about 190 feet, which is a good height for a tree. 4. /(2fl) = (I - tan= 6) /(fl) = ^^ ./(ff), , /(26) ^ /(g) ' ■ 29 cot 2d ~ e cot 6 "*' /. /{6) = m . cote. 5. Let A and B be the given points, CD the length of the given straight line. In the given circle place a straight line equal to CD. The angle in the segment which this straight line cuts off is known. On AB describe a circle containing an angle equal to this angle, and let the two circles intersect in 6 and H. Let AG, GB cut the circle in E, F, and let AS, HB cut the circle in K, L. Then the triangles EFQ, HKL satisfy the given conditions. For the angles EGF, KHL are each equal to the angle in a segment cut off by a straight line equal to CD. .-. EF and EL are each equal to CD. 6. Let ABCD be the tetrahedron. At J, B, C, D place weights respectively proportional to the lengths of the tangents from these points to the sphere. Let the sphere touch the edges AD, BC in P, Q. Then P is the C. of G. of the weights a.tA,D; Q the C. of G. of those WEEKLY PROBLEM PAPERS. 87 at B, G. '.•. the C. of 6. of the four weights lies in PQ. Similarly the C. of G. of the system may be shewn to lie in each of the other two lines joining the points of contact of the sphere with the opposite edges AB, CD and AG, BD. .: these three lines are ooUinear. 7. Let xy = i" he the equation to the hyperbola referred to the asymptotes as axes. The equation to the ellipse will be ^y = X= (- + f - l)', where - + f = 1, is the equation to the chord of contact. Then the equations to PP' and QQ' are evidently - + I -. 1 = ± f , a A and .". these two lines are parallel to the chord of contact. PArER XVII. 1. By ordinary division we find the quotient is \J^x + x^ + x'^ + !C*' + a^+!fi + !i' + a^+ •r''. 2. We may consider a, h, c as roots of the equation if , y J I 1 '— = 1, which is of the 3rd degree in le. Assume k + a = \; .-. A + /3 = X + /3 - a. &c- .: a + a, a + i, a + c are the roots of £ I jy 1 ' = 1 \ '^ \ + fi - a "^ \ + y - a .-. by multiplication, we have X' + ^iX2 + A^\ + Ai = 0, where A3 = - a; (/3 - a) (y - a). Also, since the product of the roots is equal to the last term with its sign changed, a; (a - ^) (a - y) = Xi . Xj . X3 = (« + a) (i + a) (c + a). 38 SOLTTTIONS OF 3. Since cos a + cos j3 + cos y = 0, and sin a + sin ^ + sin -y = 0, .•. e"* + e^* + er* = 0. Now whatever be the values of a, J, c, if a + i + c = 0, a3 ^ j3 ^ c3 _ Zabo = {a + b + c) {a^ + V + (^ - bo - ca - ab) = 0. .•. writing e»» for a, &c. e;3ai _}- e3/3t + e37» = 3e(i+i8+7)» .-. equating real and unreal parts, we obtain the required relations. 4. Let BEF be the triangle, and let DP produced meet the circle in C. Then DP = -^ i sin- 2 and the angle CPE = PBE + FED = CTB + PEF = CEP, /. CP = CE = 2B sin ? .-. PA.PB = PD.PC= 2Br: 5. Let ABCB be the quadrilateral ; ^, J", +i^) 5ab{a+by(^a^+b*)=5ab{a^+2ab-\-P)(a.*+b*) = 5a'b+l0a.^''+5a^^+5aV+ 10a!'b<^+5aF= B, l5aH%a+b){a^ i-b^) = 15a^''+l5aW+l5a^^+15a^^= G, .: A+B+C+3&aHHa''+b^)+70a^b^ =a»+Sa'b+28a^^+56a^b^+10a*b*+5&a^^+28a^«+8ab''+i^ = {a+bf. 2. Tripos 1875. Monday afternoon. No. 5. 3. a + b + c = 0; .: (a + bf = - ,fi; a^ + 6^ + (fi _ {a + bf - a^ - b'> • • 5 6 = abla^ + ^a^b + 2ab-i + b^}. WEEKLY PROBLEM PAPERS. 41 So = = a6(a -\- b\ and (a + bf = c\ a' + b'^-ir (? ^ (a + a)^ + g' + ^i^ 2 2 = ffl2 + oA + i- = (a + 4)2 - ab, a^ + i« + (^ , .'• 5 = - «*{(« + ^)' - «i(« + 5)1, _ a3 + i3 + (,3 a2 ^ 62 + (;2 3 2 From the given conditionsj we have e<"« + e^« + e^» = 0. .•. in the above result writing e** for a, &o. we have 6 ^' "3 ' ■ 2 .*. putting for e^ai its value cos 5a + jsin 5a, &c., and equating real parts, we obtain the required relation. If we equate unreal parts, we have sin 5a + sin 5^ + sin 5y 5 cos 3a + cos 33 + cos 3y sin 2 a + sin 2/3 + sin 2y = 3 2 sin 3a + sin 3|3 + sin 3y cos 2a + cos 2/3 + cos 2y + 3 2 4. Tripos 1875. Monday afternoon. No. 9. 5. Tripos 1875. Monday morning. No. 2. 6. Let B be the middle point of BC, the C. of G. of the triangle. Then P = ^W. .: by symmetry the force will just be able to lift a corner of the triangle if ajiplied at B or 0. 7. Produce ST to meet S'P in Z^. Then PBYK is a parallelogram, .'. PR = KY= SY. 42 SOLUTIONS OP Paper XX. 3 - VS 1 1. o ~5'"rp7 + 557/17"r 2 "3^3.7^3.7.47^3.7.47.2207^ = •3+-04761904762+-00101317123+-00000045907+.... = -38196601125. .-. V5 = 3 - -76393202260, = 2-2360679775. 2. 3^"+^ _ 8;j - 9 = 9"+^ _ 1 - 8» - 8, = (9- l){9''+9"-^ + .... + 9 + l-;,-l}, = 8{9"- l+9"-i -1 + .... + 9-1}. The quantity in the brackets is evidently a multiple of 9 — 1 ; .•. the given expression is a multiple of 64. Another method of solving this class of problems is as follows : — By trial we find that the expression is a multiple of 64 when n = 1 and when » = 2. Suppose that it is a multiple of 64 when n = p. Then writing p + 1 for n, and subtracting the valae of the expression when n = 2>t we have 9P+' _ 8(;, + 1) - 9 - {9" - 8;, - 9 J, which = 9''<9 - 1) - 8, = 8(9^ - 1), and 9 — 1 will evidently divide 9*" — 1. Thus we see that if the expression is a multiple of 64 when n = p, it is also a multiple of 64 when n = p -{■ 1. Now we know by trial that this is the case when 11 = 1, and when « = 2 ; .". it is the case universally. 3. Tripos 1878. Monday afternoon. No, 8. 4. Tripos 1878. Monday afternoon. No. 9 6. Tripos 1878, Tuesday morning. No. 3. 6. Tripos 1878. Monday morning. No. 1. 7. Tripos 1875. Monday morning. No. 10. WEEKLY PROBLEM PAPERS. 43 Paper XXI. 1. Tripos 1878. Monday afternoon. No. 1. 2. Tripos 1878. Monday afternoon. No. 2. 3. Tripos 1875. Monday afternoon. No. 10. 4. Tripos 1878. Monday afternoon. No. 8. 6. Tripos 1878. Tuesday morning. No. 4. 6. Tripos 1878. Monday morning. No. 4. 7. Let PQR be the triangle formed by the tangents. Draw PB, QG parallel to the axis, and make at P and Q the angles RPS, RQS equal to the supplement of BPQ. Then S is the focus. Draw SD and SE perpendicular to PQ, PR. Then DE is the tangent at the vertex, and SA, perpendicular to DE, is the axis, and the parabola can be described. Paper XXII. 1. For X write tan A, &c. .: tan A + tan B + tan C = tan A tan B tan 0; .-. A+B+G=n. 180° ; .-. 2A + 2B + 2C=2n. 180° ; .•. tan 2A + tan 2B + tan 20 = tan 2A . tan 2B . tan 20 ; , ^ „, 2 tan A 2x and tan 24 = , -, — j-r = -■, •„ &c. 1 — tan •^.4 1 — x^ 2. Tripos 1878. Monday afternoon. No. 3. „ «7 + 67 + (.7 «7 + J7 _ (a + J)7 ?l±^«I = „. + „6 + 6^ 44 SOLUTIONS OP .•. by multiplication we find aJ + W + c' _ a^+h^ + (fi a^ + &' + c" 7 ~ 5 2 Then proceeding exactly as in XIX. 3, we obtain the required relation. 4. Tripos 1875. Monday afternoon. No. 11. 5. Let ABCD be the tetrahedron. Let the resultant of the forces along DB and DC cut BC in E. Let them be replaced by this resultant. Then the resultant of this force and the force along DA will cut the plane AliC in a point in AE, and is .". neither wholly in, nor parallel to this plane. But the action of the forces along AB, BC, GA lies wholly in the plane .450. .•. the system cannot be in equilibrium. 6. Tripos 1878. Monday morning. No. 3. 7. Tripos 1878. Monday afternoon. No. 9. Paper XXIII. 1. Tripos 1878. Wednesday morning. No. 3, 2. Tripos 1878. Monday afternoon. No. 3. 3. Let AB be the rod, A the hinge, C the point to which the other end of the rod is fastened by the string, so that AC = AB. Draw AF perpendicular to BC, and let the vertical through the middle point of AB intersect BC in G. Then the action of the hinge at A must pass through G. Let T, W, B denote the tension of the string, the weight of the rod, and the reaction of the hinge. Then taking moments about A, 2a being the length of the rod, and 6 the angle which it makes with the horizon, a 6 16. fF. -cos5 = T . asin^- and T = W; .: ^ cos5 = sin^' .66 1 ,6 - 1 + x's .-. sm^^ + sm 2 - 2 = ; -•. sing = ^ Taking the positive sign we have ,2£_, _:_,« , 4-2^3 2 - 1 - sin^^ = 1 6 ■JZ . /3 . 1-= Vj- 6 /3\i. « /3\i ^=(4) '■ .•.^ = 2cos-i(j)- WEEKLY PROBLEM PAPERS. 4. (1) sin 64 + sin 65 + sin 60 = 2 sin {3A + 3S) cos (34 - 35) + 2 sin 3C cos 3(7 = 2 sin 3C {cos {3A - 3B) - cos {3A + 35)} = 2 sin 30 sin 3ji sin SB. tan 1 + tan -7 4 .'. the given expression ■ /IT A\ . (v B\ . /w C\ sin(---)sm(j--jsm(j-j) ^ cos(- -4-)- (I- £\ in C\ 1 ~ l) -^ (4 - 4 j Now 5 + 2 n A' 2 2' .5 + ,r 4 ^ •• 4-4 4'*°" ,y .S + O.C+4.4+5 .•. iV = Sin — sm — !- — sin 4 4 4 1 . B + C / B - C 2A+B+ C\ = - sin — (cos — cos !— J — ■ — I 2 4 V 4 4 / 1 / . B , . G\ If. i(A+B+G) .A\ 1 / . A , . B , . C i\ B + G + A A + B J) =: COS — — — COS — - — cos \ 4 4 4 1 B + G , B - G ,^^2A + S+C\ = ~ COS ^!^— (COS + COS • !— ) 2 4 V 4 4 / 1/ B , C\ , 1 r 2{A+B+C) . A\ \ I A , B . G\ = J I COS 2" + ""^ 2 "'"^ 2 / which proves the equality of the two given expressionsi 46 SOLUTIONS OF h h" 6. Lot C = - cos ^ + ^— ; cos 2^ + , c 2e' S "- sinA + --., sin 2A + . c Zc' .-. C +&• = -. (?^« + ^ . e2-^« + .. . c 2c'' = - log (1 ■ _ b eAil .'.1 (cos A + i sin ^A) = e ■O-Si e-c . e- Bi = e- ■0 {cobS - i ', sin S). ,: equating real and unreal parts, *-r pnR iV — " ~ b cos A a cos c B (1) c _ . „ 5 sm c A a sin 5 c ' (2) square and add (1) and (2). .:e2c = 'L:C = 'iog'-. c^ a Similarly, if .„ a cos B , ifi o 7j 1 6" = -— + -,cos25 + ... S' =- sin5+ ^, sin 25 + ... c 2c' we shall have C = log - ' b .'. the given expression — C — C = log log - = log - • aba Note. — If we take the value of <7 + C we find Jcos^+acosB , i'' cos 2 J. + «2 cos 2B , , c* Also, if we divide (2) by (1) we get tan S = tan B, b b^ .*. jB = - sin A 4- ~„ sin 2 J. -4- . . . c 2(^ WEEKLY PEOBLEM PAPERS. 47 6. The angle EFB = PCD = GQG ; and the angle FHR - CGQ. .: the triangles FHB and CGQ are similar. .-. HR : HF :: CG : GQ .'. GQ.RR = HF . CG = constant. 7. The given equation can be put in the form {y - m-^x) {y - m^x) = where mi -]- m^ = , miOTj = - • c a If these two straight lines are at right angles, 1 + (mj + mg) cos ro •{■ m^m^ = 0, b 1 + cos a = "-(J+D- Paper XXIV. 1. Since a, are the roots of the 1st equation, ... a + |3 = _ a, a3 = J (a' - b^), .: (a - 0)= = (a + j3)= - 4a^ = l\ .-. a - ;3 = IF *, .-. ia = - {a ± h), a' - ^ = ± ab, .: the equation whose roots are a + j3, a — ^ is 4;!= - 2aa: + a? - ^^ = Q, or x^ + {a ± b) X ± ab = 0. 2. Let a? denote the number of feet in a side of the one carpet, y the ■number in a side of the other. Then we have to find when ax^ + ^^ is a min. subject to the condition x +y = const. Now {ax - ^iff + aP{x + yf = (o + /3) (aa;" + /3/) ; X V , .'. cuc^ + ^y is a min. when ax - py = 0; i.e. when q ~ ~' .'. the areas are as «' to y", i.e. as /3^ to a*. 48 SOLUTIONS OF 3. The given expression = i{sinO + 7 - a)sin2(i3 - y) + sin (y + a - /3)sin2(y - a) -f- sin (a + ^ - y)sin2(a - |3)] = j{cos - 3y + a) - cos (3^ - y - a) + cos (y - 3a + ^) - C0s(3y - a - /3) + cos (a - 3/3 + y) - COS (3a - /3 - y) ] ; = 0, since cos (— A) = cos A. 4. Let B^C be the given triangle. On AB, AC describe two similar isosceles triangles BDA, AEC, having the angles at their bases each = 6. Let F be the middle point of BG. Then DF'' = F& + BD^ - 2FB . BD cos {B + 6), EF^ = FC^ + CE^ - 2F0 . CE cos (0 + 6), and JDF = EF, FB = FC; .'. BD^ - GE^ = %FB{BDcos{B + 6) - CEcQs{G+e)}, ••■ i^6 = 2^5{'"=°^(S + ^) - *<=o^(C+ 6)], a = sTTTg'.cCcosBcosfl- sinBsinfl)- 5(cosCcos5 - sinCsinfl)}, = ■x{ccosB — b cos G), since c sin B = S sin C, (?■ - V- .-. cos^^ = -; A cos 6 = -^; .'. 6 = 45°; ■^ v2 /, the vertical angles must be right angles. 5. Let the given lines be of lengths a, b, c. Then (a + 6)» - c2 = (a + 6 + c) (a + 6 - c). Construct a rectangle having Gff, EE for adjacent sides such that aH=a + h + c, 6K=- a+l - c. Produce OH to L, making EL = EK. On (?£ as diameter describe a semicircle. Produce KE to meet the circumference in M. Then the square on EM is easily seen to bo the square required. WEEKLY PROBLEM PAPERS. 49 6. GAB = 30°, DAB = 60°, BAB = 90°, FAB = 120°, ,„ sin 120° ,- sm d0° ' AD = AB sec 60° = 2^5, AE = AG = Va4B, ^J" = AB, .: horizontal component = JB + 2^Ccos30° + SADcos 60° + 44Scos90° + 5AFcos 120°, = ^b|i + 2. V3.^+3.2.i+0 + 5(- i)}' 9 = ^5-2' vertical component = 240 sin 30° + SAD sin 60° + iAE + 54 J" sin 30°, = Ab[2. V3.^+3.2.-^+4.V3 + 5. ^}' 21v'3 45 .-. resultant = -g- v'gl + 132a, = AB\/3W. 7. Let P6 be the normal, § its middle point, FN, QM the ordinates of P and Q. Then MN = ^iViy = ^S. .-. SM = AN; .: QM^ = iPN^ = ^5 . AN = AS . SM. .: the locus of § is a parabola, vertex S, and latus rectum ^ that of the original parabola. 50 SOLUTIONS OF Paper XXV. 1. Let X — ahe the common measure. Then i£ we write a for x in each of the given expressions, they will heoome = 0. .-. a= + «a + 4 = 0, a^ + a'a + 6' = 0; . "' ^ g ^ 1 . •'■ ab' - a'b " 6 - b' a' - a' .: {ab' - a'b) {a - a') + {b - b'f = 0. 2. Since 1, x, x^ are in H.P. •■• l + i = I-' .••'^'-2^' + l = 0; .'. (a; - 1) (:!;2 - ii; - 1) = 0, and a? - 1 ^t ; .: x^ - x = X . . {A) Since 1, y', y^ are in H.P. •••l + /3= J; .•./-2y+l = 0; ■•• (y - 1)(/ + y - 1) = 0, andy - 1 ^ 0; .-. / +y = 1 . . (5) Subtracting (S) from (.4) we have a;^ - y2 - (ir + y) = ; .-. (^ +y) (« - .5- - 1) = 0> and a; +^ ^ ; .-. a: - y = 1 . . . (C) •■; y + 2/" = 1 = ^ - y = «" - a' ; .*. — y2^ y, a;, a;" are in A.P. the common difference being = 1. Again from (C), x^ — xy = x, andy'^ = 1 — y; .-. a;'' - .ry + y- = a; - y + 1 ; .*. - f +y + ^ + ^ = {a: +y){x - y +\) <= {x + y) (x^ - xy + y=) = a^ -\- y^. 3. The given expression B+C S-C C + A C-J , A + B A-B = cos — 2 — °°^ ^o 1" °°^' — 2 — ""^ — 2^ +COS — 2 — ""^"o — = ^{COSJ? + COSC + cos C + cos .4 -|- COS^ 4- COS 5} = COS A + COS B + COB C. WEEKLY PEOBLEM PAPEES. 61 A ri- X o,. "OS ^6 - sin "6 sm fl cos 5 ' .•. tanO = cotfl - 2cot2e; 1 e 1 6 ^tang = .Tcotg — cot 6. So jtan- =jcotj -gCot-. •■• 2 *^"2 '*' 4 4 ^ 4 '^°'4 ~ ''°*^' By proceeding in the same manner, and dividing by 2^ 1, 5,l,fl 1 e 1 e. 1 ^ e 1 6 I e I e 25*^1125 + 2Ji^a^^e = 2^ 2'« ~ f* 2*' 22»-l**° 221-1 + 22» 221 "" 22"'^°*22tt 22^r::2°°*^ 22*^2' ,'. by addition cot e + 2 tan^ + ^.Un^^ + -■■■ + p^tan^; = 22^'=°^^ 1 a^i" 5 1,. , = - . _r . cos r;r- = -X when » is very larffe. 5. Let ^ be the centre of the circle to which the tangents are drawn, S the centre of the other circle. Let the circle, centre B, cut PQ in R and PQ produced in S. Then AP bisects the angle QPQ', or BPS. .: the arc AR = arc AS. .: RS is perpendicular to A£, and .*. fixed in direction. Join QC, CQ'. Then the angle gCA = supplement of ASP = ARP = supplement of ARQ, = supplement of ACQ. .: ACQ + ACQ! = 2 right angles. .: QC and CQ are in one straight line. .•. AR, RS, and Q(^ axe concurrent. 6. Tiipos 1876. Tuesday morning. No. 4 e2 52 SOLUTIONS OF 7. Since y = x tan a and y — (c tan are at right angles. ,'. 1 + (tan a + tan j3) cos <» + tan a tan /3 = 0, l + tan_tan_ •*•'"'''» = - , 11^ ,, 19,r tan _ + tan _ TT COS- J 19- 11 cos 24 " cos- . 19 + U ~ . 5n- sm — 4 . TT Bin - 4 ~ Va" *" 4' Paper XXVI, 1. By ordinary division we find jg = -652631578947368421. 12 18 By LIX. 1, we know that each of the decimals tj,) 7q' • ■ ■ Tq <'<"'" tains the same 18 digits in the same cyclical order, and as the right hand figure of any one of the decimals is formed by multiplying the numerator of the decimal by unity, the other digits can be written down from right to left by inspection. Thus ^g = -368421052631578947, 17 Yg = -894736842105263157. 2. The given expression = vP'(x-\-y-\-i) +2ze2(3:2+^2+2^) ■\-w{:^^ry^-\-^-\-ii^^^-V^x-\-xy) — ^xyzw-\-xyz{sc-\-y-\-z)-\-^xyzv> = - «;«+2«i=(j;2+/+02) ^^w%yz-\-zx^xy)Jrw{^-\-y^-\-^- ?.xyz) = - te*+w^2{x^+y^ ■ifz^)+yz+zx+xy^ - w\(i?+y^-Jtz' -yz-zx-xy) = -K>2[K,2_(a;+^+z)2}= 0. 3. ExpreBsion WEEKLY PEOBLEM PAPEES. 53 . 1 . (J L) 28in.rsiny Icos a? cosy/ cosy — cos a 1 I ^ , l^ ~ cosy + cos* 2sina:smy \cosa: cosy/ 2. a; + » . X — v sm — i-2 sm - — i 2 2 i m-^y X a; - «f = ■ z_ = tan — ~ tan i'. a ^-tV X — v 2 2 2 cos — -^ cos 9 2 2 4. Let 6 denote the circular measure of the angle 143° 14' 22", •ft ^0 5. ••^=4 =2' ■-- 180° ,,^810000 3.,,^5__^ 143° 14' 22" 257831 5. Let A, A' be the vertices of the given section. Then since OAL, CA!D are right angles, the sphere passes through A and A'. Now the centre of the section of the sphere made by the given plane will be the intersection of CD with AA', and will .•. coincide with the centre of the conic. .". the plane intersects the sphere in the auxiliary- circle. 6. Let be the centre of the circle, OD the perpendicular. Produce DO to meet the sphere in Z)'. Let r, R be the radii of the circle and sphere. Then OD = rjl, OB = 2R - OD == 2R - r x'2. Since DV is a diameter of the sphere, DAD' is a right angle, and DA' = DO^ + OA^, since DO is perpendicular to AO ; and VA^ = DfO"" + 0A\ .'. is,'' = Dm = DA'' + DA^ = 2r2 + r= + (2^ - ^2ry + r' = 6i-2 + 4iJ2 - 4V2Br. .-. 4 V2 . -ff . »• = 6^2 = 3 ^2r. V2>-, .-. AB = 3^2/- = 3. OD .-. i2 = J OD. 7. Let CZ) be the equi-conjugate diameter to which FM is perpen- dicular. Equation to CO is ~ = ~- .: if x'y' be coordinates of i',the a 6 equation to RM is 6 (y - y) + « (^ - ^ = 0. 54 SOLUTIONS OF The coordinates of M obtained from these equations are (4/ + ax'), -^^ {bf + «y). o'' + «2 ' ' ' " 0^ + Similarly, the coordinates of JV are ~ (if - ax'), -r-K-r. (¥ - "if). .: if I, i; be the coordinates of Q, the middle point of MN, aV Ay V = — — ' + b^' ' a' + 62 Now the equation to the tangent at P is ■5^ + -— =1. a'' 0^ „ ., normal „ (y - /) _ = (j: - a;') |j, and we see by trial that the values of ^, ij satisfy this equation. .'. the normal at P passes through Q. *This question may also be solved geometrically. Let Gc, Cy be the equiconjugate diameters, and let the tangent at P meet them in T, t, and bisect Ti in Q. Then the angle PCT = QCt. (See solution to XLVTII. No. 7). Draw FN, PM, PU perpendicular to Cx, Cy, Ti respectively. Produce CQ to L so that CQ = QL, and complete the parallelogram CTLt. Then, in the triangles PMN, CtL we have PM perpendicular to Ct, and PN perpendicular to CT, and .•. also to tL, and the angle PMN = PCT = QCt, and PNM = PCi = QCT = CLi. .: the triangles PMN and CtL are similar, the homologous sides being perpendicular. Now, in the triangle CtL, tQ bisects CL, .". in PMN, PU (which is perpendicular to iQ) bisects MN. Paper XXVII. 1. Let a = a-\-{p — V)^\ and a = y . pf-'^\ i = a + (2 - l)^l(^) i = y.ps-ll(5) « = « + (»■- l)/3) c = y.pr-l) From {J) a - h = {p - q]^, b - e=={q - r)ff, c — a = {r - p)^. .•. o»-e = •y(9-'')/3.p(p-iX9-r)/3^ and similar expressions for i'-<» and c"-'. .'.in the required product, WEEKLY PEOBLEM PAPERS. 65 index oiy = (q-r + r-p+^-q)^ = 0, index of p = {{p -l)(q-r) + (q- 1) (r - p) + (r - 1) (p - q)}^ = 0. .•. given expression = y° . p" = 1. 2. Divide both equations by ^', and write « f or - . We get y az3 + bz^ + cz + d = (1) a'»^ + i'z'' + ■ .'. 1. Z il i Again, by Todh. Trig. p. 157, No. 40 ; 1, except when A ■■ .; COS A + cos 5 + cos C :|> f . ABC 8 sin 2 sin ^sin 5< 1, except when A = B = C,in which case it = 1, WEEKLY PROBLEM PAPERS. 61 4 It Ein^AcosB - y sin ^B cos A + «(cos ^^ - cos 2^) = 0, X sin 2^ cos C + ^(cos «^ - cos ^G) - z sin ^C cos ^ = 0, .*. eliminating x we have = y { cos 5 (cos ^A ~ cos ^C) + sin "^B cos -,i} - «{sin2CcoB.4cosjB - cosC(cos-'^ - cos^^)}, = ^(cos ^ - cos 5 cos C) (cos C 4- cos ^ cos B) - IS|(C0S J. - COB 5 cos C) (COS^ + COS C COS -4), = y(cos + COS 4 COS B) - z(coaB + cos Ccos A), = 63/ - cz. Todh. Trig. p. 156. No. 24. .'. by — cz = k suppose • 2A T, .Ma^BcosA , cos'^B - cosM\ , /sin 2£ cos A , BmA + BwaA-B\ = n — b — + ; > , sin.4 , . ^ , = k (sin B cos A -\- sxaA — B), , sin4 . ^ „ = k sin A cos B, a .'. ax = k = by <== cz. 6. Let the perpendiculars from D meet the sides BC, CA, AB re- spectively in F, 0, H, and let those from E meet the same sides in F', - e") - Ukxy + !i?{^ - (?) = 0. If %, »?2 ^^ ^^ tangents of the angles which these lines make with the axis of x, ihk k^ - c2 % + Mj = If the two lines are at right angles we must have 0=1 + (% + '"2) cos Q) + m^m^ Ihk cos a Ifi — c'' _ A' + fc'' + 2^ ;^cosm - 2 e' .'. (^, K) lies 'on the circle 3;2 ^ ^2 + 2.'!;y cos «> = 2 A Paper XXIX. Now x^'^' = .r„ .-. ^i"'"^' = a:/* = 2:3; .-. a:i'"='°*"'» = x^' = 3:4, &c. . . .f 1 •'a — •'11 ,', ^r^a^a . ■ • • ^71 ^ 1. 2. o%2 + ^y + c2^2 = ; aV + iy + cV = ; 1,1 1 ~ ~ O' = l'^ =-—(* = R suppose, .'. I - a-'x = Rx; I - b''^ = Ry ; \ - c^z = Rz. .: oV + 6y + cV - {aV + by + c««3) = i2(aV + by + c^r*) = 0, .-. a«a^ + «y + c*z^ = aV + by + ^z" = 0. WEEKLY PROBLEM PAPERS. 63 Also a^x^ + by + c<2' - (aV + by + ^^) = Msi.^ + by + A3) = 0, .'. a«a^ + by + cV = af'x^ + *V* + «*«^ 3. i2 COB 2^ + 2«c COB (5 - C) + c» cos 2(7, = J'(cos*B - An^E) + 24c(cos5cosC + sin5sinC) + c\co%^G - sin'C) = (S cos5+ccosC)» - (5 sin5 - cwaCf. Now 4 = c cos A ■\- a cos C; c = a cos J? + 4 cos A, .-. 4 cos 5 + c cos C = 2a cos 5 cos C + cos A{c cos 5 + b cos C), = 2a cos B cos (7 + a cos J = a cos {B — C)\ «8in5 - csinC = ^ — . {scu^B - sinSC), sm^ ' = ^j^ sin (5 + C) sin (5 - C?) = a sin (5- C) ; .•. given exp. = a?'\^oi,\B - C) - sm\B - C)}, = o2co8 2(J-C). 4. Let jS = sin ^a; - i sin 22ar + . . . , C = cos 2:2; — i cos ^2x + • • • • .-, (7 + S = 1 - i + i - . . . . = Iog(l + 1) = log2. Let C = C — S = cos 2:;: - ^ cos 4a; + |^ cos 6x . . . . and S' = sin 2x — ^ sin 4a: + 4 sin 6a; ... . .-. C + S'i = e^ii _ I e4rf + I eSici . . . = log (1 + e2"), .•. 1 + e'lt = gC'.gS'ij .•. 1 + cos 2x -\- i . sin 2a: = eC (cos /S" + i sin S'). .•. equating real and unreal parts, eO' cos S' = 1 + cos 2a: ; eC' sin 5' = sin 2a:. Square and add. .-. e2C' = 2(l +cos2a:) = 4cos''a:.-. eC = 2cosa; .-. C = log 2 + log cos a:, ... C - /? = C = logS - logseca:, and C+ ix + yf + {x + yf, ... (a: + j,)3 _ 6 (a: + ^)2 + 12 (a; + 2,) - 8 = - I, .-. (^ + y - 2)3 = - 1, .-. a; + 2/ - 2 = - 1, .•.x + y = l^ _ p = 2 and .ry = 1 )' " -^ « = J {1 ± V^^} ( 2/ = J {1 T '^^a} By symmetric changes of the letters, we can obtain two other sets of values for x, y and «. 3. Since (x + y)^ = 4a^ + (* - 2/)^ we see that if the product of two quantities is constant the sum is a min. when the quantities are equal. Now the product 2cos<9 V3 ^ . J3 ' 2cose .'. the given exp. is a min. when 2 cos Ji .„„„/! 4. V3 . -, n J. I" V3 2cos5 2 6 4. The sides of the triangle formed by joining the centres of the circles are b -\- c, c -\- a, a -{- h. If « be the semi-perimeter, s = a -\- h -\- c, s — {h + c) = a, &c. ■ ■ ~ a2 - (a + J + c)2 ~ a + & + c' ■ ■ r^ ~ 6c "*" ca "*" a&' 5. Tripos 1878. Monday morning. No. 6. 66 SOLUTIONS OF 6. If B be one extremity of the minor axis, SB = AG. ,: A is the foot of the directrix of the parabola. .•. its vertex bisects SA. If we put SB = CA' we see that another parabola can be described whose vertex bisects SA'. 7. Let D be the point of suspension, A CUB the rod, E the position of the ring, and C the point where the vertical through D cuts the rod. Then C is the middle point of the rod, and DE is at right angles to AB. Draw AF perpendicular to DC produced. Let DAB = a. Then AC == 1 ft. AD + DE ==^ 9 ft. Since the tension of the string is the same throughout, CD bisects the angle ADE, and since DMA and DFA are right angles, a circle will go round DBF A. .: FDA = FAG =6; and DAF = ^ - ADF, .-.a + e^'^-e, .-. a = I - 25. Now 1 = AC= AFsecB = AD cos (a + 6) seed = AD tati 6, .: 9 tan 5 = {AD + AF) tan 6 = {AD + AD sin a) tan 6 = AD{1+ sin a) tan 6 1 - tan'^ 2 = 1 + sina = 1 + cos 25 = 1 + l-l-tan^fl l+tan25* .•, 9tan35 + 9tan5 = 2. If for tan 6 we substitute 3 — 3 , the left-hand side becomes = 9 (3~* - 3~*)3 + 9 (3~* - B~*) = 9 (3~^ - 3 . 3~^ + 3 . 3~" - 3~^ + 3~* - 3~^) = 9(3"^- 3~* + 3"^-3"^ + 3~*- 3~*) = 9 (3-' - 3-') = 9 (1 - i) = 3 - 1 = 2. . . 3 — 3 is a root of the equation. WEEKLY PROBLEM PAPERS. 67 Papee XXXI. 1. Tripos 1878. Wednesday morning. No. 3. 2. Tripos 1878. Monday afternoon. No. 6. 3. (1) The given expression = oosC{asmS-6smA) + cosA{ismG-cemF) + cosS{cBinA-asmC) = 0. (2) Let - — - = _ — ^ = _ — - = 2? suppose. Bin A sin S BmO .■.a =: jBsin.4, h — iJsinJB, c = BsinC. : ^^-^^ ^ jji sin'^-sin'^ ^ ^ sin (A + B) sin (A - B) " GosA+oosB cosA+oosB „ A+B A-B 2 cos — ~ — cos — - — 2 2 = 2iJ=sin:!i±:?sin^^ = 2R\Bm^~ -sin^D.&c. .-. given esp. = 2iJ2(sin2^ - sin2?+ sin"^ - sin2^+ sin^f'- sin^-) = 0. *4. Let F be the point within the regular polygon ABC . . . oi n sides whose centre is 0. From P and let fall the perpendiculars BMx, OiVi on the side AB, and draw PR-i perpendicular to ON-^, and use a similar construction and notation for each of the other sides. Let OP = 8. Then ON-^ = OJTj = . . . = radius of inscribed circle = r. Also iJj, R^. . . lie on the circumference of the circle on OP as diameter, and BrfiB,^ = B.,pB^ — . . ., each being the supplement of an angle of the polygon. .•. Br^B,^ = B^Bt = . . . .". BiB^Rg ... is a regular polygon. Now since N-^N^ . . . Nn is a regular polygon, .-. SPNj' = niONi' + OP') = TO (»-2 + 8=). See Casey, Bk. IV., Prop. IV. Similarly SPB^' = n (2.— + ^) = | SI But PMy^ + PRy^ = PiVi^ ^PM-c = ^PNi' - 2i'iJi= = TO (r= + 8=) - ^ fi2 = «,r= + ^i F 2 68 SOLUTIONS OF 6. Tripos 1875. Monday morning. No. 3. 6. Tripos 1875. Monday morning. No. 8. *7. Tlie equation (a; - of + {y - bf = {aw + fy + yf _ ,„2 , 02-, ("^ -f- /3y + y)^ '^'^ -TP) a^ ^ /32 represents a conic whose focus is (a, b) directrix ax -\- &t/ -\- y = 0, and eccentricity = Va^ + ^^. Now the necessary and suificient conditions that two conies should be identical in magnitude and form are (1) the eccentricities must be equal ; (2) the perpendicular distance of the focus from the directrix must be the same. .-. a' + ^2 = a'2 + ^'2 (1) (aa + h^ + y)^ _ {a'd + V^' + yO" a' + 0=" a'2 + H'^ or (aa + i/3 + yf = {a'd + V^' + yj .... (2) Paper XXXII. 1. Tripos 1878. Wednesday morning. No. 4. 2. Tripos 1878. Monday afternoon. No. 5. 3. (l-tan?)(l-tanf) oos| - sin-« Jg-f-j^-f (l+tan-)(H-tan-) cos-+Bm^ __cos-+-sm- '°' 2 ~ ^'° 2 ^^° (4 ~ i) ^ 2 (cos- -sm-)sm- cos- + sm- cos(j-|j 2(cos-+sm-)cos- sino — 2 sin^-j; 2 _ sin a + cos a — 1 _ sin a -}- sin )3 — 1 sina + 2cos2^ ~ sina + cosa + 1 ~ sina + sin0 + l' WEEKLY PROBLEM PAPERS. 69 .'. Pq'=I'C^+CQ^-2FC.GQcosG '2ai So §72==«=(2i=+2c=-a=)-3;(362-c2+a=')+6=. BP^=x\2e^+2a^ - b') - .t(3c2 - a^+h^)+,^. = (3x^ - 3jy+l) (^2+62+^2) = i{a^+b''+c^)+3{^-iy(a^+b^+c'). 5. Tripos 1875. Monday morning. No. 6. 6. Let P, Q be two points on the conic, and let SP, SQ make angles ^, (j) -\- a with the axis of a:, a being constant. Let the tangents at P and Q intersect in R. The equation to PP is - = ecos 6 + cos (6 — (}>) ; r „ „ RQ „ — — e cos d -j- cos (0 — ^ — a). .: at R, e cos 6 + cos — tji = e cos 5 + cos 6 — (f> - a, .: d - (t> = 2jr - e + ((> + a, .-. e _ ^ = ,r + ^ ; ,'. the locus of i2 is the conic - — e cos 6 — cos - • r 2 This may also be solved by reciprocating the theorem, 'The envelope of the chord of contact of tangents to a circle which cut at a constant angle is a concentric circle.' 7. Let the plane of the paper represent the section of the cone made by a plane containing the axis. Let BO, DE be the diameters of the base and top of the frustum. Produce BB, CE to meet in A. Let the axis of the cone cut BG in F, and DE in G. Let H, K, L be the C. of G. of the cone ABC, the cone ADE, and the frustum BDEG respectively. Let LF = X, AG = OF = a, and the angle BAG = 2a. 70 SOLUTIONS OF Then EL = t-^,HK=^-^. Then volume ABE = 3AG . circle DJf = San-anan^a; .-. volume £DEG = ABG - ADE = 6a JT 4ffl2 tan'' a — Saira^tan^a a _ 2 ^ _ aa^irtan'a ^ 1 . . ^ ^^ •'• 3^ ~ 21fl.3 7rtan2a 7' '"^ 28^' 4 .-. i(? : LE :: 17 : 11. Paper XXXIII. 1. Tripos 1875. Monday afternoon. No. 6. 2. Let & = 1 + 2 (1 - «) + 3 (1 - «) (1 - 2ffi) + . . + k(1 -«)...(!- ffl- 1»> + -(1-«)...(1-k4 :. -Sji+i = 1 + 2 (1 - c) + 3 (I - ffl) (1 - 2a) + . . . + (n+l)(l-«)...(l-M)+i.(l-a)...(l-^«), .-. -3(a+/3+c). Now BPC+ CPA+APB = ABC; ••• (»-2+'-3)«+(''3+''i)/3+(»'i+»'2)y = (»--J-i)a+(r-»-2)^+(»--r3)c. 5. Tripos 1875. Wednesday morning. No. 1. 6. (SP . S'P = GD^ = CP=. 7. Let I, j3 be the eccentric angles of ^ and F, . .: EP = ffl2 (cos a - cos §y + ^2 (sin a - sin /S)^ = 4 sinH (a - ^) {«' sin^i ^ (a + 0) + ^2 gos^ J (a + /3)} = 4ap2sin2i(a-0). ■*■*■ 2CF^^ 2sinH(a-i3) = l-cos(a-^) = 1 - cosa cos/3 -sinasinjS __ 1 _ •^1*2 _ sl3£2 72 SOLUTIONS OF Paper XXXIV. 1. Tripos 1878. Monday afternoon. No. 4. 2. Tripos 1875. Wednesday morning. No. 5. 3. Expression = 2(cos''|+cos*i'^) = ^j /l+oos|y+[l+cos^)n = H('+vV+(.-iin-H^+l) = i- 4. Tripos 1875. Thursday morning. No. 2. 5. Let ABCD be the quadrilateral, the centre of the circle, E, F, G, H, the feet of the perpendiculars from on AB, SO, CD, DA. These points are the middle points of the respective sides. Join J)£, Then AHE ia i ABB, and CGFis. i CDB. .: ARE + GQF = i ABQB. Similarly, by joining AG we may shew that BEF + DBG = i ABGD. .: the sum of these four triangles is ^ ABGD. .: the remainder, the quadrilateral EFGH is also half ABGD. 6. Let Tt, T'if, the tangents at P and Q meet the axis major in T, T. Then P&C = SPT + STB, and QSG = SQT' + ST'Q. .: subtracting, QSP = SQl" - SPT, since STP = ST'Q. Similariy QS'P = S'Qi! - S'Pt. And SQT' = S'Qt', SPT = S'Pt. .: QSP = QS'P. 7. If (jTiyi), (iTjyg) be the coordinates of the points where the chord of contact intersects the curve, we must substitute the value of •^ from — -j- II. = 1 in the equation - + IT = 1. Thus y =t (1 - -1^) . .•.i=-f+'-7i-^'r 3-2 A , bVti\ „ «2^ , ^2 WKEKLT PROBLEM PAPEES. 73 ... ^, + ^, = 2 ^ - i (tlAll\ =2h^ (It + ^-) .,., = (_ - 1) ^ -, (— ^) = («^ --W)^K1>+ b'O The values of y^ +^2 and yi^j ""^y be written down by inter- changing a and b, h and A. = l{(.'!'i+'!'2)'-4.'i^i«'2+(yi+y2)'-4yiy2} = 1 ' +^ -('^ +* H«^+pJ+(-F+^)i«^+^^) I i^^'+Z-O Papee XXXV. 1. The number of combinations of 15 things taken 3 at a time is ^ 15.14.13 ^ 1.2.3 .•. the number of days on which the girls could walk out having at . 5 7 13 least one different companion each day is ■ — = 7. 13. Now suppose that on the same day a and b walk in the same row. If there were no restriction, they could do this 13 times, viz. once with each of the other girls ; and similarly for each couple. But by the conditions of the question, they are only to go once, .". the number n, 7.13 „ of days = -~^~ = 7. The question may also be considered thus. Suppose we consider a. She can only walk with 14 different girls, and she must always walk with 2, but not with either of the same 2 again, and she can therefore only walk with a couple-of different girls for 7 days ; and similarlj' for the others. The oraer for these 7 days is 74 SOLUTIONS OF 1st day. 2nd day. 3rd day. 4th day. 5th day. 6th day. 7th day. a, b, c a, e, g c, d, i d, h,j e, i. h c,h,o j, n, a d,e,f h,f,h 6,g,n a,i,m h,l,n d, I, a f,k,c 9, h, i i,j, o k,a,h f, g, I j, c, g e,j, 6 o, g, d j, k, I I, m, c m,f,j h, o, h m, b, d g, k, m h,m,e m, n, o n, d, k o, e, I n, c, e o, a,/ i, n,f I, b, i For further information respecting this interesting problem see the Ladies' and Gentlemen's Diary for 1862 and 1863, and the Proceedings ef the London Mathematical Society for 1881. 2. Tripos 1878. Monday afternoon. No. 5. 3. Tripos 1875. Thursday morning. No. 2. *4. Let OEi, OB^ .... be the straight lines drawn from parallel to the sides of a regular polygon of k sides, and on them let fall the per- pendiculars PRi, PB2 .... from the point P. Then R-^, ^2 .... lie on the circumference of the circle on OP as diameter, and R-JJR^ = R^ORg = . . . . each being the supplement of an angle of the polygon. , •. ^1^2 = ^2^3 = . . , . .•. R^R^Rs .... is a regular polygon, ,•, by Casey Bk. IV. Prop. 4, Cor. 1 SPR' = 2n . (^j" = I OP'- This may also be deduced from XXXI. No. 4, by supposing the polygon, whilst retaining its regular form, to become indefinitely diminished. It may then be regarded as a point coincident with 0, the centre of figure, and the sides all pass through dividing the angle 27r into s equal angles. .-. SPil/i" = w^ + jS" = |8^ since r = 0. 5. Let the plane of the ellipse APA' be perpendicular to the plane of the paper. Let be the vertex of one of the cones from which it can be cut. Let the cone be cut by the plane of the paper which is supposed to contain the axis. In the triangle OAA' inscribe the circle ESF, and on the other side of AA' draw the escribed circle HS'G, so that 0, E, A, G, are in one straight line. Then A'0-AO=-{ffO-ffA')-(OG-Ae)=SO-OG+AO-HA' = AS'-A'S'= constant. .•, the locBs of is a hyperbola which has A and A' for its foci, and 8, S' for its vertices. WEEKLY PROBLEM PAPEES. 75 6. Take the centre of the circle as origin, and let^Cff = 6. Then th e coordi nates of A are (a, 0) ; JB (- a,0) ; E (a cos 6, a sin d) ; K (a cos d + 2a, a sin 5 + 2a). Equation to ^^ is ^ = - -^J-A^^ (^ _ ^) =_(,._ ^^^otf' . 6 cos - sin - 2 2 (1) ■V Tjxr a sin (5 + 2a) , , , , . ^ (6 \ fl 1 1 + cos (5 + 2a) I \^ / .*. y 005(2 + "} = (i^ + «) ^(2 "^ ")' e . 9 . e e .', ^(coBHCOSa — sin^ sina) = (ar + 'i')(8in2C0Sa + cos^sina) ; (2) a Eliminating jr between (1) and (2), ^{y cosa + (a; — c)sina} =(:?; + ff)|ysina - {^x — a) cos a}, .•. x^ ■\- y^ — lay tan a = i^, .: x^ -\- (y — a tan aY = a%l + tan ^a) = a' sec V 7. Using the same figure and letters as in XXXII. No. 7, let us take the case where L, the C. of G. of the frustum is vertically above H, CE being supposed in contact with the plane. Then CE = CF cos ICA -f FL cos FAO, .'. a sec a = 2a tan a sin a + na"-- <^os a, 11 . „ 17 .•. 1 = 2sm''a + —a cos ''a, .". sm^a = j,- And we see that if the vertical angle of the cone be decreased the perpendicular from L will fall within the base LE\ ,: the frustum will not topple over if the vertical angle of the cone is less than 2sin-i\/2- 76 SOLUTIONS OF Paper XXXVI. 1. i- = -142857142857 .-. ~ = 14-28671428 ... |=^-14=-285714.... Again ^ = 1-428571 .... ...|=3L°-1 = -428571.... Similarly for -, -, -• 1 ■ • 142857 Now Tr = -142857 = 7 999999 7 X U2857 _ 7x1 _ •'• 999999 ~ 7 ~ ' .-. 142857 when multiplied by 7 must give a series of nines. 2 3 From the way in which =, ■=,.... are formed, it is obvious that no new integers are introduced, and that the cyclical order is not changed. See also XXVI. No. 1, and LIX. No. 1. 2. Let n = '7m -\- p, .'. K — I = 7m -^- p — 1, • . ij = integer H ^ By trial we find p = — 3, or 4 are the only ones which give an odd remainder, and thesse are practically the same. Let « = 7ra + 4. We have to find the sum of all terms of the series whose general term is (7ot + 3) (7ro + 4), or 49ra» + 49m + 12. .-. By Art. 26 the series to m terms = 49 '^^ + ^ n ^'"+l} + 49 -(^) + i2m. Now if we put !M = 0, « = 4, .-. n(n — 1) = 12, which is the first term, . -. to include this value of m, we must add the constant 12. WEEKLY PROBLEM PAPEES. 77 .-. the sum = i9m{m + 1) { — ^^ + 1| + 12(«» + 1) = i9m(m + 1) . ^^^i-? + 12(m + 1), and m = "-'—' = -^ |(«-4).-^^ +12} = 21 ('^ + ^) (''' + 6;j - 4). a /3 B 3 ^ _l_ ^ tan g + tan g- tan^^ + tan^ }. tan — t: — = = 2 a 8 8 8 1 - tan - tan 5 1 - tan^ ^ tan^ tan^, (l+tan^gj tan^ ^ ~ = |- tan = tan^. 1 -tan*^ 1 - tan^g .-. >27r + ^ = ^^-^- If « = 0, a + (3 = 2(^. 4. Let 2^-56' be a triangle, 7), ^, F the feet of the perpendiculars. Then by geometry we know that A, B, C are the centres of the escribed circles of DBF. Let p, r' be the radii of the circum- and in-scribed circles of DBF, r^r^r-i the radii of the escribed circles, centres A, £, C. Then ^^ = p^ + 2pri; f = p2 + 2pr2; ^^ = P^ + ^pr^ip^ = p" - 2pr' ; p = f Now n + r^ + r^- i" = ip, XXII, 4. ... ^2 4- yi + ^2 + ^2 = 4p2 + 2p . 4p = 12p2 = SR". 5. Since the angles at A and I) are right angles, a circle will go round ABDK .: CE . CA = CD . CB = ^BCK 6. A varying geometrical quantity has a maximum or minimum value when it has the same value for two consecutive positions. Let BAC be the maximum triangle, BA'G a consecutive position, indefinitely near to the first. Then the triangle BAG = BA'G. .: AA', which is ultimately the tangent at A, is parallel to BC. 78 SOLUTIONS OF Similarly it may be shewn that the tangents at B and C are parallel to AG, AB. .: the maximum triangle is equilateral. .*. area = 3B0C, where is the centre of the circle, = l.BO. OCsin 120° = ^ iJ^. 2 4 For all questions of this kind, see Theory of Maximum and Minimum treated without the aid of Differential Calculus by the present writer. 7. Let PG be the normal, Q any point in it. Draw the ordinates FN, QM. On SP let fall the perpendiculars GK, QL, and draw QL' perpendicular to PJf. Let ASP = a, ASQ = 6, so that (r, 6) are the coordinates of Q. Then QL : GK :: PQ : PG :: PL' : PN, .: QL : PL' :: GK : PN :: SG : SP :: SA : AX. .: QL = e . PL'. .: r . sin(5 - a) = e{PN - QM) = e(SPsma - r sinfl); . „ . ,„ , SP . esina .•. e sm fl + sin (fl — a) = r 1 + ecosa Paper XXXVII. 1. At the first observation let x and y be the angles made with the vertical by the hour and minute hands respectively. Then at the 1st observation it is 30 + y min. past 4, 30 + w .". the hour hand has gone — r^j-^ min. divisions, 30 + » .-. a; = 10 ~^, .-. 12 iT +y = 90 . . . (A) At the 2°'5 observation it is 30 — a; min. past 7, .'. the hour hand has gone — rs — min. divisions, 30 - ir ••• y = 5 + -j^ , .: 12y + ar = 90 . . . {E) From A and B by subtraction, we get x = y. WEEKLY PROBLEM PAPERS. 79 2. (i) \/"^{bx - a^) - V - (ftm -W) = a-h, and evidently = («-*)(^- ^^ )' .•. by division, we get \J ^{bx- a") + V -(«* - *^) = •» —^ .'. 2 \ r{b!e - a") = a; + a - b — ^ a^ + 2ab^ + I ab a^\ 2ab + «3 _ ab 1 / a^\ 2ab + «3 = -{ax — -r ) ; .-. r«.r -- 1^) - 2a fy ax -j- {2ab + «2) = 0. .-. sj ax -~^=a± Va^ + 2a4 -f /J^ = a ± (a + i) = 2fl + «or - i. If vre square both sides, we at once get the values of x. (2) ir +y + V?^^ = ff (1) y^lx"^ - f = 26 (2) From (1), Va;^ - / = a - a; - y (3) .•. w^ — y^ = a^ ■\- x"^ ■\- ^ ■\- ixy — 2ax — 2«y, .•. 2/ + 2xff - 2ay = 2ax - «». From (2) and (3), ay - xy - f =• 2b (4) «' - 45 ' 2ax - a'' + 46 = 0, .'. x = — g^ ■ 80 SOLUTIONS OF Substituting this value of x in (4) we get which gives the value of y. 3. ^ + 5 + C=»Mr, .-. sin(^ + 5)=+ sinC, cos(^ + 5)= + cosC, sin 2C sin 1A sin 25 sin 1A + sin 25 sin 1A - sin IB 6 ~ 4 9 ] 2 2 gsin(^ + 5)cos(^ - B) = p 2 2 "k A Qsin(^ + 5)cos(^ — B) = 2 sin Ccos c) 2 2 sin {A - 5) cos (-^ + 5) = g sin Ccos c\ .: cos {A - B) = ± S cos C, sin (^ - 5) = :p ^ sin 0, .: 1 = 9cos2C + g sin 2(7, .-. sec^C = 9 + g tan^C, .-. Stan'^C = 72, .-. tan^C = 9, .-. tanC = ± 3, . „,, 2tan(7 ±6 3 .•. Sm 2C' = :i — ;— : tt, = z — r-p; = + j- > 1 + tan-'C 1 + 9 n" . raff ?r .■. sin 2A — ±1, .". 24 = »wr ± ni •'• A = -^ ± j, ■'• tan 4 = + 1. Again 2 Bin(^ + (7) cos (2^ - C) 28in5cos5'\ • /, , „ ^ ■ „ ^ ■ ^ ^ = J I sm (A + C) = ± sin B 2mn{A-G)c0B(A + O 2sin.gcos5 Ln«(^ + /7) = +cos5 .-. cos (4 - C) = ± 2 cos B, Bin{A - 0) = + ^ sin A .-. 1 = 4oos2B + J Bin25, .-. sec^B = 4 + j tan^^, .-. 3 tan =5 = 12, .-. tan^B = 4, .-. tanS = ± 2. 4. Denote the two expressions on the left by X and F. By the ordinary formula, X = „ . „ • ' 2sinj3 WEEKLY PROBLEM PAPERS. 81 In finding Y we will consider (1) n odd, (2) n even. (1) when n is odd ; r=cos(n- 1)0 + cos (n- 1 - 2)/3+... + cos{m-l - (n - 3)}^ w, -3 to — ^ — + 1 terms. f 1 /n-1 A) . n - 1 = cos|»- 1 -^— 2 -^j (/3sm — 2 — ^cosec^; » + 1 „ . w - 1 1 = cos -g-psm — ^^ — 0cosec/3= 5-(sinn,/3 - sin0)cosec0; 1 1 + (- l)"-! .: X — T = ^ = J , n being odd. (2) when n is even ; F=cos(n - l);3 + cos(n-l - 2)/3 + ... +cos{n-l -{n- 2))/3 ra- 2 to — = — + 1 terms. = cos j » - 1 — (^ ~^)\ sin -2' cosec |3 nfi n^ 1 = cos-^ sin -„ coseojS = g- sm?i/3cosec/3; 1 + (— 1)™-1 .: X — Y = = J , n being even. 5. Let PS, QR intersect in Z. By symmetry L lies on the diameter through A. It also lies on the polar of A, which is a fixed straight line. .•. i is a fixed point. 6. By the same method which was employed in XXXVI. No. 6, we can shew that when a triangle inscribed in an ellipse has its maximum value, the tangent at any vertex is parallel to the opposite side, and .". the diameter which joins any vertex to the centre bisects the opposite side, being conjugate to it. .•. the centre is the intersection of the bisectors of the sides. We know that corresponding areas in the ellipse and auxiliary circle are in the ratio of b : a. Now by XXXVI. No. 6, the area of the 3 a/3 maximum triangle in the auxiliary circle = —r— cfi. .: the area of the maximum triangle which can be inscribed in the ellipse is — j— ai Obviously, as in the circle, an infinite number of such triangles can be inscribed, all having the same area. G 82 SOLUTIONS OF 7. Let the coordinates of P be {x'^'). /. Then the coordinates of U are -{a;' — ae), j- ,: Equation to ^Cis j—, = y i(x' - ae) - a „ . „ . ff x — ae Equation to SP is -, = ■ f X' — ae ,'. to determine the coordinates of Q we have i/{x' — ae — 2a) = i/'(x — a)\ y(a;' - ae) = f(x -a))' , X — ae^ , V ••■ x' = , y = -^— . 1 — e 1 — e /. if the equation to the original hyperbola is -; — "- = 1, the a equation to the locus of Q is {x -^ ae)' _y2 «2(1 - e)2 ~ 6\1 - ep- x^ The lines parallel to the asymptotes are evidently -jj — 7-2 = 0, showing that the two hyperbolas have parallel asymptotes, and are .•. similar. Also the transverse axis = a{e — \) = C8 — CA = AS. Paper XXXVIII. 32 1. B, the last dealer, has to pay away to the others — X 4 = 64 counters. And as the total is 32 X 5 = 160, J' has 160 - 64 = 96, which gives the last table. The others can be obtained by working backwards. A has O „ -0 ., E „ When At the A deals. B G J) F end 81 2 4 8 16 32 41 82 4 8 16 32 21 42 84 8 16 32 11 22 44 88 16 32 6 12 24 48 96 32 160 160 160 160 160 160 WEEKLT PR6BLEM PAPERS. 83 L 1 2. Consider two consecutive numbers m" and (n + 1)" . Eaise each to the «(» + l)th power, and divide by «."■. This gives us n and [} -\ — ) , the latter of which is equal to 2 + a proper fraction. .•. if f IN" « > 2, « > I Z + - ) ; .•. the quantities gradually decrease. Now consider IJs and ^2, or 3'' and 2'. We see that VS > V2, .•. V3 is the greatest of the numbers. 3. Since the tangents are in A.P. and A = 45°, tan A + tan B .-. 2tanB = tan^ + tan C = 1 + tan (7 = 1 - i _ tan^tani? 1 + tfin ;? 2 tan B = 1 - 1 - tan jy ~ 1 - tan £ .: either tan B = 0, which is inadmissible, or tan5 - 1 = 1, .•. tan B = 2, .: tan C = 3- 2 . 3 .•. sinB = —r, sm C = -7=' 1 smAHinO ,„ „ „ 2 _ x'lO „ , „ /t ••• 3 = area=2 ^^,-^5- -•• ^^ = 2.3 ^^2 -^=8, ••• ^=2^2, ^sin^ „ ,_ 1 "^5 '- bsinC ,_ 3 ">'6 4. a' = 2r cos n 5 . <7 ^ J9 «'iV 8.3 A B C _ '^ ^'" 2 ''" 2 '°' 2 ""^ 2 =°' 2 , — r- = -7- cos n COS 3 cos.r = 8)- . • -, abc aba 2 2 2 A abc cos- r^ 2sin_g 2sinC r^ 2' 6~ '~T~ "'W g2 84 SOLUTIONS OF 5. Let abed be the figure formed, so that B, u, h, ff, are in this order in a straight line, and similarly Dcdiy. Then since AC is equal and parallel to CA', .'. AA' is parallel to CC. Similarly BS is parallel to DV. .: the figure abed is a parallelogram. Join AC. Then since Jff is the middle point of AB, and I/d is parallel to Ba, .'. Ad = da. .: abed = 2Aed = AdD. Similarly it may be shewn that abed is equal to each of the triangles AaB, BbC, CcD. But these 5 figures make up ABCD. .'. abed = rABOD. 6. Bisect AB in C. Draw through a straight line perpendicular to AB. The point P will always lie on this straight line. Now the angle QPA = PAB. .: QP is parallel to AB, and .•. perpendicular to PC; and AQ = QP. .: the locus of § is a parabola, focus A, and directrix OP. 7. Take as origin, 0(7 the axis oi x, and a perpendicular to OC as the axis of y. Let the coordinates of the fixed point be (A, 0). The equation to any straight line CPP' is of the form y = m(x — A) . . . (1), .•. y — mx — — mi. The equation to the conic, which passes through the origin and touches the axis of x is of the form ax^ -\- by^ -\- cxy ■\- ey = Q (2) If we make (2) homogeneous by means of (1) we get ax" + bf + cxy - ey . ^ ~ ^^ =0 .... (3) This, being a homogeneous equation of the 2i"i degree denotes two straight lines passing through the origin 0, and since it is satisfied by the coordinates of the points which satisfy (1) and (2), it is evident that (3) represents OP, OF- (3) may be written x^ah + mxy{ch -\- e) -\- y^(mbh — e) = 0. If this splits up into (x — miy) {x — m^y^ = 0, Ml = cot COP, m^ = cot COP. n{ch -\- e) ck -\- e cot COP + cot COP' = mi + m2= - = constant. mak ah WEEKLY PROBLEM PAPERS, 85 Paper XXXIX, 1. Let X, y, z denote the number of direct routes from London to Cambridge, London to Oxford, and Cambridge to Oxford respectively. Then 11 = ;;: + y^ ; and 13 = y + zx. .: By subtraction (.r — y) (2 — 1) = 2. Since a, y, z are integers, the only solutions are from 7 5 either « — 1 = 2, a; — y = 1, /. a = 3, a; = „, y = „, inadmissible, or 2 — \ — \, X ~ y = 1, .'. z = % X = h,y ■= %. 2. Consider jt) quantities each equal to xl-'', q quantities each = i^-r, -and r quantities each = a*-?. ™, . . ,, pxi-^ A- qx''-v A- raS-i Their A.M. = ^- -^ f^ g(g-»-)+gfr-g)+rOp-g) Their G.M. = x i'+9+'' = ^° = 1, Since the A.M. > the G.M. pxi-r + qxr-p -|- rafP~i - , p + q + r except when k = 1, when the A.M. = the G.M. which is also the case when J) = 2 = r. 3. Produce OG to meet the fixed circle in E. Then OB . OG = OP . OP' = OA . OE = constant. And OG is constant. .*. OD is constant. .". D is a fixed point. Join PA, PD, CP, GP', EP'. Then since PDOP' is a quadrilateral in a circle, .-. the angle OPD = OOP'. Similarly.OPJL = OEP. And OOP' at the centre = twice OEP' at the circumference. .-. OPD = twice OPA. 4. Since dBOj and OiAO^ are right angles, a circle will go round O^BAO,, .: OA . OOi = OB . 00^. Similarly it may be shewn that each of them = OG . OO3. 86 SOLUTIONS OF r OB r . r Again, 0A.00,=—^. —j, = ^ . B . sin - sin - sin - . sm - sm - ^ ^ ^ Z ^ . B . C r . fl sm — sm -^ 2 2 a . a = 4r.o^v— j-=4ri?. .^ 4.B.C~'^* A ^~' '2 sin 4 sm -n cos 2 sm ^j sin 5 sin jj cos -5 5. Let P be the position of the point within the triangle ABG. Let the perpendiculars from P on BC, CA, AB be denoted by x, y, z respectively. Then evidently (?;!; + 6y + cz = 2A (1) We have to find when the expression a;'' + ^^ + 1? is a minimum subject to the condition (1). Through P draw a straight line PI) parallel to AB, and suppose that z remains constant whilst x and y vary. In other words, suppose P to move along the line PD. Then we have to find the point on this line for which a;^ + y'' is a minimum subject to the condition that ax + hy is constant. Now (ax + hyf + (})x - ai/Y = (a^ + 6") {x^ + /), X y .'. x^ + y" is a minimum when hx — ay = 0, i.e. when ~ = ^* Similarly, by supposing x to remain constant whilst y and a vary, we shall get r = -■ .'. ^ x, y, and z all vaiy, the expression x^ -{-^^ + ^ has its minimum value when X y z ax ■^- ly -\- cz 2A a h c d^ + 6^ + c^ (i2 + 62 4. ca If u^ be the minimum value of x"^ + y" + ■z^, ^ = («» + 62 + c2)2 ('^^ + i^ + <^) = ^a^ 6" + 62' 6. Square and add. .-. j5^ + / = 42 + £2 + 44B sin 6 cos A Multiply. .'. xy = AB + {A^ + B^) gin 5 cos 5, Eliminate 6. iAB A' + B' : U^ + £2) (r* +^2) - 44Bry = (4" - B^^^ WEEKLY PROBLEM PAPERS. 87 7. Let LMN be an equilateral triangle. Let M be on the axis of x, N on the axis of y. Let {x,^/) be the coordinates of L, and let NMO = 6. Then x = LNsinLNy = 2a sin (150° - 6) = 2asin(fl + 30°) = aVS sin d + a cos B, y = iMsiniil/a; = 2a sin (fl + 60°) = «ViJ cos fl + a sin 6, .'. in No. 6 writing a\/3 for ^, and a for 5, we have Mx" + /) ± 4a3'j3 = a\ We have the double sign because L may be on either side of MN. Paper XL, 1. Since the number is less than a million, its cube root has only two figures. Now (lOp + qY = 1000p3 + 300^2 + 30pq^ + f. .'. q can at once be determined by considering the units' digit of the number, and we find p by taking the number whose cube is the greatest integer which is not greater than the number given by the three left- hand figures of the given number. 2. Let &x denote the price per head of the 10 sheep. .'. &(x — J) denotes the price per head of the 5 sheep. Let lOy -\- 2 = total price of the 10 sheep, .-. 10y + z = 10a: and lOz -{- y = 5x - f , .-. 8y - 19z = 5; .-. y ^.19r + S, z = 8r + l. Now y and z must be positive integers legs than 10. .'. the only admissible values are obtained by putting r = 0; .'.y = 3; « = 1 ; .-. lOi = 30 + 1 = 31 ; .-. a: = £3^^ = £3 2s. ; x-i = £2 12s. 88 SOLUTIONS OF ABC ^^°1 , 3. tan -^ + tan 2 + tan 2 = + . . . . cos- 2 .A B G . sin -cos -cos- +.. . . "" A B C cos -cos 5 COS - C . A + B A B . C COS -r Sm ^ \- COS -: COS ^ Bin - ^'l C} AA- B , A - B\ 2 cos 2(cos -^— + COS — 2— j = 2. = 2 = 4. = 4 COS'' -^ + cos -^ cos ^ sm -r A - B . A + h COS -^ sm Q- + cos — -; — sm — „ — - ^ , . Cf A B . G\ 1 + smr^lcos -^cos-x — sm ^1 n sin C + «• (sin A + sin B) , , . Of A B A+B^ 1 + sm -( cos - COS -n — cos — „- — 1 sin A + sin Ji + sin U . A . B . 1 + sm n sm ^ sm ^ sin^ + sini^ + sinC 4. Let FAE, FBD be the tangents at A and B ; BCE the tangent at G. Take the centre, and draw GOD cutting BG in ff. Similarly draw GEE, OKF. These lines are respectively perpendicular to BC, GA, AB. Then AF = B oot AFO, AH = M cot AEB, GJJ = R cot OBO. .-.perimeter of BFF = 25(cot AFO + cot ABE + cot GDO) = IB.f^s.-a.FAB + tan ^^(7 -|- tan D0£) = 22E(tan C+ tanS + tan^) = 2iJ tan^ tan B tan G, since 4 + jB + C = 180°. WEEKLY PROBLEM PAPERS. 89 „ (? . ^ 2S ^ ahc. Now fr-75 = sm^ = -r-; ,'. R = -rrr' 2iJ he ' 4:8 .: perimeter of DEF = ^UnAtaaS tan C. 5. Draw DQ parallel to CEF, meeting AB in G. AE AF 2AF Then -^ = pa = YB' ™°® "^^ ^ ^"^'^' ^ ^^^"^ *^® middle point of BG. 6. Let jS be the common focus, H, H' the 2°^ foci of the variable and fixed ellipses. Let the ellipses touch at P, and on the tangent at P let fall the perpendiculars SY, HZ, E'Z'. Let (a, b), (a', b') be the axes of the variable and fixed ellipses. Then since SP makes with PY the same angle which PH and PH' do, .•. PH'H'm a straight line. (1) SP + PH' = 2a' ; SP + PH = 2a. .: HE' = 2(a - a') = const. . •. the locus of IT is a circle. H'Z' H'P (2) SY. H'Z' = V ; SY.HZ= b\ .: -^^ = const. .-. -gg - const. JTlff •'• fJTo ~ const. .'. the locus of H is an ellipse similar to the fixed ellipse. 7. Let (xu yi), («o,y2), {m^jy^i be the feet of the normals from the point 11 P(a, ^). The equation to the normal atP isp - y = — -{a ~x) . (1) Eliminating x between (1) and y^ = 2px we get y^ + 2p(p - a)y = 2fp. •'• Vl) Vit 2/3 ^emg ^^^ roots of this equation, 2/1 + 2/2 + 2/3 = 0. 2/22/3 + 2/3S/1 + 2/12/2 = ^P{P - o). Square the first of these equations, and subtract twice the second, and we get x^^ x^-^- x^= - 2{p - a). From (1), ^ . pip - yii = - yiO - a^i), ••• yi(i3 - yi) = - 2a^t + 2*1*, and y^ = - 2p.Ti + 2yi\ .-. adding, 2(^1" +^1^) - 2{p + a)^i - ^yi = . . . . (2) 90 SOLUTIONS OF ••• Hv + Vi^ - (/> + «)^i - g-^i = Similarly x^ + yi - {p + ajx^ - ^^2 = (4) and ffj^ + y^ - (p + a)x^ - g^a = Now /.2=(3;,-a)2+(y.-0)2+(^2-a)2+(^2-/3)2+(^3- a)H(^3-;8)s, .•. by means of (4) we have /.= = B{a? + /32) - (a - ;,) (oTi + iTj + .rj) - (2^ + |)(yi + y^ + y^), = SCa'i + ^2) _ 2(a - ^)^ .". the required locus is the ellipse x'' H- 3/ 4- ^px - if = P. If P be a point on the line such that the sum of the squares on the three normals is a minimum, and = k% then the sum is also a minimum for a point P' on the line indefinitely near to P. Now P and P' both lie on the ellipse x'^ + 3y^ + 4P^ ~ ^P' = ^^- ■'• "^^ must igive i such a value that the ellipse may touch the given line. Now the general equation to the tangent at (^V) to the ellipse referred to its centre, which is at ( — 2p, 0), is y — y" = — ^7-, (X — X) = — r-) (X — X). .•. m = - o -> or the point of contact lies on the line 3my + a? = 0. . ■. changing the origin back to the point (2p, 0) we have bmy + a; + 2^ = 0. Paper XLI. 1. 6 men do -6006 of the work in 2'12 hrs. .-. 1 man does — „ „ „ 1 hr. 61 ■. 3 men do 100 }j )i 1) 3„ •. 7 boys do 49 100 1) » I) 3„ • 6 „ „ 7 60 » )) )i 1 „ WEEKLY PROBLEM PAPERS. 1198 There remains -— — of the work for 6 boys to do. 0\J\J\) 1198 50, oi, ni • ■m>^l li'^«-=2hrs.51^mm. 2. The expression on the left = ar(l-:r)-'-»3(l-a;3)-l+a;-'(l-;r5)-l__. 91 they will take Add the columns vertically. = .i:(l-»'+;r«-a:6+...)+x'(l-a'<+a^-^l''+...)+^'(l-««+.T"--)+- 1+x^ ' 1+x^ ' 1+x^ ' ■■■ 3. 2 sin (a + 6) sin (/3 + c^) = 2 sin + 6) sin (a + ), .: coa{a - fi + 6 - G. M. .-. -^ (Sm + 1)2 > [[3m\™ See Appendix. 3. By hypothesis — -^ — , j — , — ~ — are m A. P. A B 180° - C 180° - A 180° - B . , „ •• —c—' —^—' —5— "'■' '° ^- ^• 180° , 180° , 180° , . , _ .'. -^ - 1,-2 1, -g- - 1 are in A. P. 94 SOLUTIOKS OF ■•0 2':b^^*'"^-^- .-. C, J, S are in H. P. Let = G0° 4- !c, ^ = 60° + y, if possible, where x and y are positive. Then B = 180° - 60° - a; - 60° - y = 60 - a; - ^. Since C, A, B are in H. P. .-. 0:B::C - A: A - B, .: 60 + a; : 60 — a; — y :: a; — y : a; + 2y. Now the 1st term is > 2nrl, and 3rd term is < 4th, which is impossible. 4. A is any point on the circumference, D the centre. Let the chord OQPR which is parallel to Ali meet AC in P, and the circum- ference in Q, R. Then the angle OPG = BAG = OBG. .: P is on the circle round OBG. Buti) is also on this circle, since OBD + OCD = 2 right angles. .-. BPO = BCO = a right angle. .•. QB is bisected in P. 5. Let FN, RQ be the ordinates at P and R. PN^ = RN.NA = iAS.AN, .: BN = iAS. .: PR^ = RN.BA = AAS.AR = QR\ 6. Let B be the fixed point in the axis. Draw BR perpendicular to the tangent at P meeting SP in Q. Draw PK perpendicular to the directrix. Join SK meeting PR in Y, Then SY is perpendicular to PY. Since BQ is parallel to SK, the triangles BQ,S, PSK are similar. .-. HQ -.SB-.-.SP: PK, .: SQ = SB = const. 7. Draw CM, ON perpendicular to PQ. The equation to PQ is ^^ + -^ - 1 = 0. .: GM^ - '^^' and PO^ = 4 (W + «'P - M'^) ifi'V + a'V) ^^^IV 7 ^"'^ ^<^-* («2^= + ^.^.)2 XXXIV., 7. .-. area CPq = i Clf.P« = '^'^^ (^ + a^F - g'^^)* WEEKLY PEOBLEM PAPEES. 95 .-. area CqpP = CVq + OPg aV/' (i'^s + g'F - «"&')* (^"^3 + fliifc' - g"^"') (Vh^ + a^/^» - a°5')* Paper XLIII. 1. Let a{by-\-cz — ar) = b{cz-\-ax—by) = c{(ix-\-by-cz) = R suppose. .-. — = by-\-cz — ax; — = cz+ax-by; — = ax-\-by—cz. a b c Adding the 2nd and 3rd of these results we get ^b c) bo aba Similarly y = -S . '-~-^, z = R abe aba .-. :^+y + ^ = ^(« + * + «) = o. aba 2. (« + lj3 = „ (;i _ 1) („ _ 2) + 6« (re - 1) + 7;z + 1, ... the »th term of the series = ^ + ^ + -^ + 1, .•. sum to CO = ff -|- 6« + 7e -f- e = 15e. 3. Using the formula of Todh. Trig. § 288 cos 55 = 5 cos (9- 20 cos'^+lG cos'5 ; sin he = sin 6(1 - 12 cos^^+lG cos«6) = sin 6(5-20 8in26+16 sin<6) ; 96 SOLUTIONS OF .•. sm55-cos55 = 5(sm^-cosfl)-20(sin'd-coB35)+16(8in«5-cos55) = (sin fl - cos 6) (1 - 2 sin 25 - 4 sin^ 18) : sin55+oos55 = 5(sin5+cos5)-20(sin3|9+cos35j+16(8in55+cos'e) = (sin5+oos S) (1+2 sin 2i9-4sin2 26) ; and tan (^-D= tanfl — 1 sm6—cos6 tan (9+1 sin 6+cos 5 .". by division we obtain the required result. 4. CP = ^^^IA2 (U sin/3 ' ^p ^ a sin PJjQ ^ _ &sin(P^C'+a + 2^) sin/3 sin0 .-. a sin P40 = b {sin P^C cos (a + 2/3) + cos P^C sin (a + 2/3)} ; .-. taiiP^C= - ^""(" + 2/3) a + ^ cos (a + 20) . • T) y^ J- i sin (a + 20) .•. sinP^<7= ± '^'' V«^ + A^ + 2ffl4 cos (a + 2/3) By substituting this value of sin PJC in (1) we obtain the value of CP. 5. (1) Tripos 1875. Monday morning. No. 4. (2) Let CE be the common chord of the two circles. Then the angle CBD = BBC = CBA = ACE. .: ABO = ABC, and ACB = ACE, .: LAC = EAC. .: CE subtends "at the circumference an angle which is one-fifth of two right angles. .'. CE subtends at the centre an angle which is one- fifth of four right angles. .-. CE is one-fifth of the circumference of the small circle. 6. Tripos 1875. Monday morning. No. 7. 7. This question is partly solved in V., No. 6. There we see that two parabolas can be described, and that the equations to the directrices are a; cos ^ + y sin ^ = a. The lat. rect. are evidently twice the distances of P (a cos 0, b sin (f) from these lines, and = 2 (a - (a cos^ cj} ± b sin^ 0)} = 2 (a ^ *) sin^ cj>. WEEKLY PBOBLEM PAPERS. 97 * Papek XLIV. 1. Tripos 1875. Monday afternoon. No. 4. 2. Let ax -\- b -\ — = y, .•. ax'- — (y — b) X -\- c == Q, _ y — b ± '\i {y — by — 4ao '2a Since x is real, the expression under the radical cannot be negative. ••• (y - 6f 4 Uo, .•. y <[: * ± aVcwT AX^ + Bx + C _ Again, let j!x' + £'x + 6" ■" .-. (4 - A'y) a;2 + (£ - B'y) x + - C'y = 0, .-. 2 {A- Ay) x=-{B- B'y) ± V(^ - B'yf - 4 (^ - Ay) (C - 0'^). Since x is real, the expression under the radical cannot be negative, and y will have its limiting values when, this expression = 0. When this is the case, 2a; (A - A'y) = - (B - B'y), ,.2Ax + B=^i2A'x + B')y = i2A'x + B').^±^±^ .: (AB' - A'B) a^-2 {CA' - C'A)x + BO' - B'C - 0. 3. 2 sin^ = sin C + cos C, .: cos 24 = 1-2 sin^^ = 1-4(1 + 2sinC'cosC). = i(l-2sinCcosC), = 4(cos C-sin C)^ = f-^cos C- -i-sin C)^ = cos» f^+C). V2 V2 ^* sin^B = sin Ccos C ,: -^cos 2B = i-sin''B= J-sin Ccos C=cos2(|+C) = cos 24. 4. Tripos 1875. Monday morning. No. 5. 5. Tripos 1875. Monday morning. No. 11. 98 SOLUTIONS OF 6. The equations to the two normals can evidently be written y = m {x — 2a) — aiiP (1) y = _ (a; - 2a) r. (2) m m' , , ,. »i^ — 1 , n . m^ — I ,: subtracting, . h; — 2a) = a . = — , .: X - a = a. ^ ■ — '~ (3) ni- Again, eliminating x — 2a from (1) and (2), yiX - m^) = a ( n?), .: y = a. ™ "*" ... (4) \m m ,". from (3) and (4), if — a {x — a). 7. Let the tangent at T meet the major axis in R, and the minor axis in §, and let (a cos ^| 6 sin ^) be the coordinates of S. The equation to the tangent at P is - cos d> 4- i. sin d> = 1 ad .: CR = a sec ^, CQ = b cosec = a' + b'' + a^tun^cj) + b^cot'cf,, .: QB^ will be a min. when a'' tan" ^ + i^ cot^ ^ is a min. Let «^ tan^ + 4^ cot^ cp = t^, .'. aHan'^ - aHan^^ + ^2 = 0, ^ 2a2 The min. value is given by u* = 4a^i^, .". u^ = 2ah. Also see XL V. 2. .-. qR"" = a' + b' + 2ab = (a + bf, .'. QB = a + b. The singularity is evidently a min. for there is no max. limit *We add a purely geometrical proof of the above. Let the tangent at P meet the major and minor axes in T and t. Then when Tt is a min. Tt = T'H, where T't is the consecutive position of Tt, i.e. Tt moves at this stage aa if it were a line of constant length WEEKLY PROBLEM PAPERS. 99 sliding between two rectangular axes. Draw TB, tB parallel to the axes meeting in R, the instantaneous centre. Then if EP be joined, HP is the normal at P. .-. PB? => PT.Pt = cm. .', PR = CD, and is measured outwards along the normal. .-. OR = a + k See XLI., 6. Since CTRt is a rectangle, Tt = CB = a + b. Paper XLV. 1. Lets =l+2x + 3i;^ + ... +{n - 1) a;»-2 + na^-\ .-. Sx = X + 2!c^ + ... + (a - 1) a;«-i + nx^, .-.8(1 -a) = 1 + K + a;2 + + a;»-i - nx'^ 1 — X 1 — X „ _ 1 — (» + 1) .r" + tix^+^ •'• '^ (1 - xf 2. Since the product of the two expressions ae^" and 6e-^^ is const, and = all, .: the sum is a min. when the two expressions are equal. Each of them is then = Vproduct = >\/a6. .: the sum = 2\/a6. 3. (1) Let /S = oosfl.sinfl + cos«^.sin2e + cos3fl.sin36) + . .. C = cos5.cos(9 + cos2^.cos2e + cos3fl.cos3^+ . .. /. C-h Si = co8e.e"+cos'e.e"' + cqs^e.^+ . . . cos 6 . e" cos 6 (cos fl + i sin S) ~ 1 - cos^.e'' ~ 1 - cos 5 (cos d + i sin 6) _ cos 6 (cos d + isin6) _ . . ^ cos ^ + »' sin 6 ~ sin fl (sin 8 - i cos ^) ' i sin 6 + cos 6' .•. S = cotd. Evidently C = 0. (2) The given series = the sum of the two series log,2 + »^%... + 2Iog.2+(%?r + ... = ^"^'^ - 1 + e''"^"' - 1 = 2 - 1 + 22 - 1 = 4. H 5J 100 SOLUTIONS OF 4. Let G = cos a; sin {m ■{■■^ ) +moos (a; + -] sm{a;+-!: — -J-\-\- ... it ' it' il (S = sin a; cos (a: + — I + » sin (a; + - I cos [a: + ■= — 5-^/ + • • • .■.C + + a. Let the tangents at P and Q intersect in T. The equation to P^is - = e cos 5 + cos {6 - 0), r „ „ QT „ - = e cos 6 + COB (0 — (j> - a), .: at T, e cos 5 -f- cos {6 — (f) = e cos 6 + cos {6 -

) = COB {e - ^ - a), .: e - ^ = ± {0 - (j> - a). Taking the negative sign, 5 = + -:, .*. the locus of r is - = cos 2. + e cos (4> + -), r 2 2/ Isec-^ or = 1 + e sec - cos (0 + _ 1 . This is the equation to a conic, whose axis is inclined to the former at an angle ", and whose eccentricity is e sec -• Paper XLVI, 1. Put a? = a -f- - + -, .'. bx^ — ahx - a = (1) 1 'l If a, =^ J + i + i, .: cx^ - bcx - 6 = (2) If X = c ■{ 1 — , ,: a.v' — acx — e = (3) a X 102 SOLUTIONS OF Multiplying (1), (2), (3) together, and dividing by abc, we get «« - fa+i+c);r5 -{'Lj^ij^l-bc-ca-abXi^ \b c a J V a b c ' jdtj^t^'L-bc-ca- aiy - («+4+c)a; -1=0, *c a o / or a;«- \-{a-\-h^-c)x{3^-\-\)-\-{ab-\-bc-\-ca)^t^x^-r)-\-{a-\-l-\-c-abc)^ .*. the three continued fractions satisfy the given equation. _ ai± Jo?Tfl + \ab From (1) 1 1 The positive sign gives us o + ... d + a + The negative sign gives us _ • • • • + a -|- + From (2) and (3) we obtain similar values. . the other roots of the given equation are _11 _\ \ _11 6 + a+..., C + 6+..., a + c+ 2. (H -=- ? = 2, \i) 9 ' . loge2 = log (i)' - log| = 2 log (l + 5) - log (1 - i) V3 2 Sii'^a'P •• V +\32'^2 3*'''3" 38^""T 3. The common denominator of the three fractions being — sin (a — 13) sin (3 — y) sin (y — a), the numerator is sin (5-/3) sin {6-y) sin (|3-y) + sin ifi-y) sin (tf-a) sin(y-a) + sin (5 - a) sin (6 - j3) sin (a - 13), WEEKLY PROBLEM PAPERS. 103 = i sin - y) |cos O _ y) - cos {26 - fi - y)} + i sin (y - a) [cos (y - a) - cos {26 - y - a)\ + i sin (a - 0) |cos (a - 3) - cos (25 - a - 0)} = J {sin 2 O - y) + sin 2 (5 - /3) - sin 2 (5 - y)} + i {sin 2 (y - a) + sin 2 (6 - y) - sin 2 {6 - a)} + i {sin 2 (a - /3) + sin 2 ((9 - a) - sin 2 {6 - 0)} = i {sin 2 (a - ^) + sin 2 O - y) + sin 2 (y - a)} = I {sin (a - (3) cos (a - /3) + sin ((3 - a) cos (a + |3 - 2y)} = J sin (a - /3) {cos (a - /3) - cos (a + ^ _ 2y)} = sin (a - j3) sin (a - y) sin (j3 - y) = — sin (a - /3) sin (/3 - y) sin (y — a). .•. given expression = 1. 4. Let B, C denote the bases of the hill and tower respectively. A and D their summits. Then ACB =^ a, ADB = ft CD = a. Let AB = X, DAG = 6. Then CDB +^ + 5 + | - a = ir. .: CDB = J - + (9 - o) ; CDA = ^ + a - 6, a sin 6 AC 1 . ■ „ • / /,v ; - — = ~ — , .'. !i)Svo.6 = a sin a cos (a — 6), .^C cos {a— 6) X sin a .•. sin 6 {x - a sin^ a) = a sin a cos a cos 6, . a sin a cos a ir — osin^a sin 6 cos 5 Now CB = x cot a, DB==a cosec O + (9 - a), OD" + CB^ = Z)£2, .-. a2 + x^coi^a = a^ {1 + cot^ + 5 - a)j, .*. a; cot a = a cot (/3 + 5 - a) .", X cos a sin {& ■\- 6 — a) = a sin a cos (|3 + 5 — a), .*. sin 6 ^ cos a cos O — a) + a sin a sin (/3 — a)} = cos 6 {o sin a cos (/3 - a) — .i; cos a sin (^ — a)\. (1) 104 SOLUTIONS OF Eliminate 6 by means of (1), and arrange the result in descending powers of x. We have Aosasin(/3 - a)+cu;Bina |cos'''aCos(/3 - a) - cos(;3 - a) - sina cosasin(j3 - a)} + a' sin'a Icosa sin (|3 - a) + sina cos (^ - a)} = 0. .*. x' cos a sin (/3 - a) - ax sin^ a sin j3 + a' sin^ a sin ^ = 0. This equation is unaltered by writing jr - a, n- - /3 for u, j3. Now j3 may be > ^, but a must be < -. .". only one value of x satisfies the given conditions. The other value gives the solution of some other question. 5. Let ABCD be the square, P the point on AC through which the four circles pass. Let EFOH be the quadrilateral, E, F, 6, H being points on the circles round PAB, PBC, FCD, PDA respectively. The angle POD = PCD = 4 right angle. The angle PRD = PAD = i right angle. .-. PCD = PHD, .: PG = PH. Also PD = PB, .: the circles round APD and APB are equal. .-. the angle PHA = PEA, .: PH = PE. Similarly PFC = POC, .: PF = PG. .: the circle described with centre P and radius any one of the lines PE, PF, PG, PH will pass through the extremities of the other three. Also PHG = PGH = PCD = ^ right angle. .-. HPG is a right angle. And PEF = PFB = FOB = i right angle. .-. EPF is a right angle. .-. HPG = EPF.' Add to each FPG. .: EPG = FPH, and the sides which include them are equal. .'. the base EG = base FH. Since the triangles EPG, SPF are equal in all respects, and PE is perpendicular to PF and PG to PH, ,: the base EG is at right angles to FH. 6. Let S he the given focus, C the centre of the auxiliary circle. Produce SC to S' so that CS' =■■ CS. Then S' is the other focus, and if 8S' intersect the circle in A and A', these points are the vertices. In CS take a point X such that CX : CA :: CA : CS. Then X is the foot of the directrix. .'. since we know the focus, directrix nnd eccentricity, we can describe the curve. WEEKLY PEOBLEM PAPERS. 105 7. The equation to a chord PF' is - = e cos ^ + secfl cos (6 - a). r In the rectangular hyperbola,e =V2,andseo)3 = \/2, since iP/S(P'= -. 4 J_ .: equation to PP' is — = cos 5 + cos {6 - d), which is the equation r of the tangent to a confocal and coaxial parabola. This may also be proved by reciprocating the theorem, ' The locus of the intersection of tangents to a circle which cut at right angles is a concentric circle,' the centre of reciprocation being taken on the circumference of the inner circle. Paper XLVII. 1. ^/^^ + x{Jb-c)+d? + ^/x'+l:lfi-a)-i-b■' = -i^x' -{■a!{a-b)+cK Square and transpose. .-. ifi+2xg}-a)-^a^+W-.c^ = -2 \'{sr'+x(^b - c)+a^} {x'+x{c - a)+b'}, .: a^+ix%b - «)+2a?ja2+4= - c2+2(i - a)2|+4a;(« - a) (a«+6» - c^) + a^+b* + ci + 2a^b^ - 2aV - 2JV = 4:fa:'^ + x^{b-a)+a;^ (a^ + b^ + be - ba - c" + ac) + X {a^ (fi - a) + b^ {b - c)} + aV'], .-. 3x*-2x%a'+b''+c'-2bc-2ca-2ab)+ix{a-b) (ac+bc-ab-c^) - (a* + b* +c* - 2b'^c' - 2cV - 2aW) = 0, ,•. ar« + 2fx>' -\-ix{a-b-){b - c){fi - a)-\-r = 0, .". K* + 2pa;' + 2^ + r = 0, where y, g, and r satisfy the required conditions. 2. From the r things there can be formed [r_ permutations. „ n- r „ „ ,, n- r „ .:/{r) = \r \n - r. 106 SOLUTIONS OP .■. the expression on the left hand [n ^ \u - 1 ^ |2 \ro - 2 ^ ^ |n - 1 ^ [?i = 1 (1 + If = ?:. 3. (1) sin 18° = ^^\ .-. cosnS = 1 - tll:^^ 1M^2V6 ^ ^ 4 ' 16 16 ' 10 + 2 \'6 tan36== 2J^f ° =ll^.=M±45)tanl8°= V5tanl8». l-tan218° ^_ 3- Vs 2+2^5 6-t-V5 ^/5 - 1 (2) tango = _sinl8^ ^ 4 ^ ^ ^ - I 1 + cos 18° ^/lO + 2 v^5 4 + VlO + 2 x/5 ^+ 4 n/5 |4 - VlO + 2V5|' And 16 - (10 + 2 x/6) V5 - 1 _ v^5 - 1 1 V5 + 1 6 - 2V5 (\/5 - 1)2 v'5 - 1 4 ' .-. tan 9° = y^^jtik - Vl0 + 2V6|- 4. Let ^5C be the triangle required. From the vertex A draw AD perpendicular to the base BC. Then we have given the values of BC, Ah, and AS'' ^ AC'. Then AB'^ - AC^ = {BV + AV) ~ (42)=' -}- DC^) = Biy ~ Z)C2 = BG (BD r-DC). .: BD ~ DG is known. And BD + DC is known. .•. BD and 2)C are known. /. the position of U is known. And since we know the value of DA we can construct the triangle. WEEKLY PROBLEM PAPERS. 107 5. Let the tangents OQ, OF meet the directrix in the points F F. Draw the tangents FP, F'Q. Let OQ, FP intersect in A, and FP', FQ in B. Then SA bisects the angle P8Q, and SB bisects the angle P'SQ'. Since F and F are on the directrix, .'. PP' and QQ' pass through S. ,: JB passes through S. Again, since OSQ = OSF, and QSA = F'SB, .: OSA = 0S3. .: OS is at right angles to A£. This may also be treated as an instructive example in reciprocal polars. Take EF, any chord of a circle, and draw the diameters EG, FH. Corresponding to the lines EF, FG, EH we have the points 0, A, B. Since FQ and EH are parallel, and .•. meet at infinity, we see that AB passes through S. Since FG is at right angles to EF, we see that AO subtends a right angle at S. 6. Let AB be the circular arc, its centre, and let the particles be situated at the points A, P, Q, . . . B. Let AGP — 6, so that (« — 1) 6 = 2a. Take as origin, OA as axis of x, and a line through at right angles to OA as axis of y. Then if (xi y-^, {tc^ y^j . - . be the coordinates oi A, P, , . . and if (^, y) be the coordinates of 6, ^ = -(•J'l + ^2 + •••• + »«) = - (1 + cos 6 + cos 25 + • . • . + cos n — 16), r n — \ ^ . ti6 6 r . na a = -cos — s"— flsin -^cosecET =- cos asm r cosec 2 2» n - 1 n — 1 1 + ^2 + +yn), - (sin 6 -\- sin 26 -{■.... -\- sin n - iff), since yi = 0, r.« — 1„.«5 6 r . . na = -sm— 5 — e sm TT cosec K = -sin a sin ^ cosec 2 2 K n — \ ffi-1 r . ;2a a ••• 06 = n/*= +^^-= - sin ^^^j cosec^^^- 108 SOLUTIONS OF Writing this in the form -sm 11 — T]a . a n — i n — 1 a sin- r or - . sin - 1 n - 1 we see that if the number of particles be indefinitely increased, their C. of G. will ultimately coincide with that of the circular arc. Now when n is indefinitely great, — ^^^ and - may be considered as in- n — 1 definitely small compared with unity, and the value of « - 1 is ultimately equal to unity. V .'. if 6' be the C. of G. of the circular arc, OG' = -sin a. a 7. Let be the origin, P the point (h, k) PA, PB perpendiculars on the axes of x and y respectively ; PN, PM are parallel to 1 iiese axes, and PQ is perpendicular to AB. Then OA=OM+MA = h-\- kcoaa; OB = ON + NB = k + kcosw. The equation to AB is 1 = 77-7 + -prr, = TT~j 1* / — n • ^ OA OB A-\- i; cos a ' k-\- h cos a' or x{h cos 0) + A) + y(^cos m + A) = (^ + icosw) {k + ^cos -\-lc]h-'r{h-\-hcosa>)k-{h-\-kcosai){h-^hcosa>)\ . (Ac0Sa)-f A)=4-(/i:+Acosa>)^— ii(A+Ac0Sa))(i5:+Ac0S(B) Bk\\ -cos'6>ysin'' Suppose the equation to PQ iohe Ax -\- By -\- C = 0. (1). WEEKLY PROBLEM PAPERS. 109 Since this is perpendicular to AB, whose equation is (A cos 0) + % + (icoso) + % - (A + iJcosffl) (kooato + A) = 0, /. by /SaZni. Con. Art. 26, Cor. 2, we must have A{icoaa> + k) + B{/i:coBa + A)=[A{k + &cosa) + B{&oos(o + h)}ooS(o, .-. Ak BA, .-. 4 = - ^• A k Since P§ passes through {Ji, k), (1) may be written A{x -h)^B{y-k) = % or h{x - ^) - % - i) = 0, .-. hx - hy =h^ - k\ c\ b^ + ^ (c - a)', a^ Paper XL VIII. 1. The given expression = 2(a - «c) { (i - c)\ac+ab-bc-ia^)+c\ab + bo- ca) + b\ac-\-be - ab) } = 4(a - *c){i2(;2 _ ^2(^ -<;)=} = 4(a — Ic) {be + ab - ac) (be + ca — ab) = 16(« — 4c) {u — ca) (« — ab). 2. Clear of fractions, and arrange in descending powers of x. .: x^ja?ib-c)+^'(c-a)+y\a-b)}-x{a%b'-c')+^{c^~tl^)+y\a'^-b^] + a^c(b - c)+ fi''ca{e - a)+ y'abifi, - i) = 0. Since this equation has equal roots, . •. { x\a^ - (?) +^\(? - a") +y\a^ - i^) } 2 =4 {d?{b- c)+^\c - a) +/(a - b) } {aHc{b-c)^^ka{c-a)+y''ah{a - b)] 110 SOLUTIONS OF Arrange the result in terms of the powers and coefficients of a, |3, y. .-. a\b ~ cy + p\c - ay + y\a - 6)* - 2/3V(c - af {a - bf - 2-fa%a - hf (6 - cf - ia'd^b - cf (c - of = 0, .-. ± a{h - c) ± /3(c - a)± y{a - J) = 0. See XIII. 1. 3. Let S = a sin 6 + a^ sin25 + «' sin 35 + . . . . = ccos 5 + «« cos 25 + o' cos 35 + . , . . .-. 0+ 8i= ae" •}- flV + . . . . = (1 - ae")-'^ - 1, .'. (C+1 + Si) (1 - ae") = 1, .-. (C + 1 + Si) (1 - a cos 5 - i . «sin5) = 1. Equating real and unreal parts (0+l)(l - acos5) +;5'.fflsin5 = 1, (0+1). asin5 - S(l - acos6) = 0, .'. S(a^ sin 25 + 1 - 2fl cos 6 + a^ cos =5) = a sin 0, asm 6 S = 1 - 2a cos 5 + fl2 52 4. sin5 = 5(l-^-,)(l-^J.... ...e«-e-« = 24 --,)(! --)... Let 58 = It, .-. e'-e-'=2:r(l + l)(l + y(l + J-,). Let 6i = " 2 ,*. ^ - e" 5 17 37 '^ • 4 ■ 16 ■ 36 •^• (1) ■(i+y(^+y (2) WEEKLY PROBLEM PAPERS. Ill Divide (1) by (2). .-. e?+e-^-=2(t + l)(l + J-^).... „ 2 10 26 = 2t-9--25---- = 2.^' .-. 4a;2 - / = (e^ + i^f _ (/ _ e^f = 4. 5. Let ^5<7 be the right-angled triangle. On the hypotenuse BO describe the square BPQC. Let BO, CP intersect in E. Draw ED and EH perpendicular to AG, AB respectively. Suppose AG > AB. Since the angles at A, M, D are right angles, .'. HED is a right angle. The angle BEG is a right angle, .'. BEH = DEG. .: from the right- angled triangles EBH, EGI), since EB = EG, .: EH = EI). * 6. Let OSaS' be a square, and OM, ffM', SY, ST perpendiculars from its angular points upon the variable straight line. Then Casei/, p. 24, OM^ + ffM'^ - iSY . ST = area of square. .-. if OM^ + O'M'^ constant = B, SY . S'Y' is also constant, and the variable straight line always touches an ellipse, foci S and S'. If a, b be the semi-axes, SY . S'Y' = i^ = ^^s^i _ g-j^^ and area of square = 'iCS^ = i^aV', Also, and 0' are on the line through the centre at right angles to the major axis, i.e. are on the conjugate axis, and CO = CG — CS. ' * 7. Let Ca, Gfi be the equioonjugates,. and let the tangent at P meet, Ca in Q and <7/3 in B. Draw PX, Pji parallel to Ca and Q3. Bisect JiQ in D, and draw PL parallel to Ca. Then Cn . CQ == Ca^ = C/S^ = C\ . C% .: Cji : C\ :: CB : Cq :: BL : LD :: GL : ID, .'. Cji : Pji ■.: CL : LD. And the angle PjiC = GLD. .: the triangles PC/i, DCL are similar, .-. the angle PCfi. = DGL. i.e. CP and CD are equally inclined to the equioonjugates, and .•. also to the major axis. Now if we produce QC to Q' so that CQ' — CQ, RQ' is a tangent to the ellipse, and is also parallel to CD. .: CP and BQ^ are equally inclined to the major axis, as are also OP and MN. . : RQ is parallel to MN. Similarly it may be shewn that the tangent from Q is parallel to MN, 112 SOLUTIONS OF Paper XLIX. 1. (l + icy = l+nx+ ^^^~ ' x"^ + Is- (1 + ;,;)» = ^ + .... + l^-z^^Jg'^-^ + • • • . •'• /iW = "O^f- ^"■'' in (1 + •■J')^" = coef. »"+S ,-. (1 + :r)2» = 1 + . . . . +/i(0>" +/i(l)^+i + . . • . (1 + a;)" = a;" + . . . . + j ^ _ ^ 1^ '^"-'" + • • • . •■• /2('') =" coef. a;2i-f in (1 + sc)^ = coef. ir''+'', .-. (1 + ;r)3» = 1 + . . . . +/2(0)a;'' +/j(l) . a^^+i + . . • . ^ ' ' ' ' \n — r \r •'• /sW = coef. a;2n-'- in (1 + a:)*» =■ coef. ;!;2''+''. So/4(r) = coef. a;3»-r in (1 + xfn = coef. ar^n+i-, /^(r) = coef. a:8»-r in (i + a;)6» = coef. a;37i+r. • • • • • .". if ra be even, and = 2p, 2(p+i)» /»(r) = coef. X ^ " in (1 + ^f^'^^^ = coef. aJ'"+'- l(2y + l> •. fm{n) = ooe£ a?!"" or coef. it(p+i)» in (1 + a!)(2?+l)» = .(^ .(■ lyipn If »? be odd, and = 2p + 1, /m(r) = coef. ajCp+i^+r in (i + a;)2(p+i);;, I 2(ff + l)re .•./™(n) = coef. :r(P+2)''in (1 + a:)2(i,+l)^ = ^'^^-^-^^- ^— WEEKLY PROBLEM PAPERS. 113 2. Consider a quantities each = -, 4 quantities each = -, c quantities a each = -• c Their A.M. = - " * " ^ Their G.M. = (i • ^. ■ ^J' 1 +6+C. Since the A.M. > the G.M. 1 3 / I \a+6+c- a ■\- b -\- e This may be shewn to hold in a similar manner for n quantities. 3. Let the angles referred to their respective units be Ix, x, Sx. 100 , „ 10000 Then 2x + -^x + 3. -^^x = 180, .•. X = 15, and the angles are 30°, 25°, and 125°. i. Since the product of tan''^ and cot ^5 is constant, and = 1, .•, the sum is a minimum when tan^fl = oof^^ = V(product) = 1, .-. tan e = ± 1, .-. = (n ± jJtt. 5. Let BC, BB, BE be the three given straight lines, BB being between BC and BE. Let the given point A be on the side of BE remote from the other two lines. In BC take any point 0, and draw CH perpendicular to £Z), and produce it to K so that CH is to HK in the given ratio. Through K draw KE parallel to BB, cutting BB in E, and join CE, meeting BB in D. Through A draw ANML parallel to CDE, meeting BE, BD, BO in N, M, L respectively. Then ^i is the line required. Tor LM : MN :: CD : DE :: CH : HK, .: LM : MN in the given ratio. 114 SOLUTIONS OF *6. Let H, K be corresponding points in AB, AD, so that A3 : MB :: AK : KD or HB : KD :: AB : AD. Let EH, CE meet in 0, and draw OM parallel to AB to meet BC in M, and let E lie between B and G. Then OM : CM :: CD : KD, OM : EM :: SB : ££", .-. OMhCM.EM: : CD.HB-.BE.KD: : CD.AB-.BE.AD: -.ABhBE.BC. .: the locus of is a hyperbola having CE for diameter, and line bisecting CE and parallel to AB for the conjugate diameter. If E lies on CB produced, M lies between C and jB, and the locus is an ellipse. 7. When two conies have double contact, their equations can be put in the form S = 0, and ii — Kc? = 0, and their difference is evidently a perfect square. . •. in the question we must have afi{a — a) + 2xy(c — y) + y\b — |8) + 2x(a' — a') a perfect square. .•. all the terms involving x andy must be of the 2oii degree. This gives us the two conditions a' - a' = 0; (a - a) (& - |3) = (ff - yy. Papeb L. *1. $ = Ix + my -\- )iz ; (1) x = l^ -{■ m-q -{■ «f ; (4) r) = nx + !y + mz; (2) y = n^ + Ir, + mC; (5) C = mx -\-ny + h; (3) z = m^ + iir] + l^ ; (6) From (1), (2), (3), (4) we have (c = l{lx -\- my -]- nz) + m(nx + fy + mz) + n(mx + tty + h) = (P + 2mn)x + (m2 + 2;ra)y + (mi + %ln)z. Since this is true for all values of x, y, z, .: P + imn = 1; (J) m^ + 2»/ = 0; (5) «2 + 2/m = ; (C) and the same three conditions are obtained by expressing y oi z in terms of x, y, z. From (B) and {(T) m^ - n^ - 2l{m - ti) = 0, .: {m - n) {m -{• n - 21) = 0. WEEKLY PEOBLEM PAPERS. 115 Prom (B) and (C) we deduce that when m and » are real, they both have their sign opposite to that of /. .: la + It - 21 =^ 0. .'. m — n = 0. .'. excluding zero values, we have 21 ■{- m = 0, and P + 2iii' = 1. •'■ ^ = ± i- .•.»» = »=- 2^ =+ *. See Errata. 2. Assume that i^ + px + q A + ;7— .+ {x — a) {x — b) {x — c) X — a' X - h where A, B, and C are constants. Then 3c2 +P35 + 2 ^'A{x — 6) (a; - c) + B(x - c){x— a) + C{x - a){x - 6). Since this is always true whatever be the value of x, we may put X = a. .: a" -{■ pa •\- q = A{a — b) {a — c). Similarly by putting x = b, and u, = e, we get ^2 ^pj) +• 2 = B(h - c) (6 - a); c^ + jpc + q = C{c - a) (c - b). Thus A, B, and C are determined. ^ ^A x\-'^ Now a a\ a' A .'. the term involving x^ in this expression is — ^ , ^ ■ .*. the general term required is 1 a^+pa + q 1 5^ + Sp + g 1_ c^+jJc+g . a»+i ■ (a - 6) (c - a) "*" bn+i * (a - 6) (6 - e) "^ c»+i ' (6 - c) (e - a) 3. The given expression cos .rf oos^ . cos (7 sin 5 sin C sin Csin ^ ^ sin ^ sin5 sin ^ cos ^ -|- si" S c°s B + sin <7cos C sin A sin ^ sin C ,in2A + BiniB -\- sm20 and ^ + 5 + £7 = 180°, 2sin^sin^sm6' ' ' t^ ' 4 sin A sin 5 sin C 2 sin ^ sin 5 sin C "~ I 2 116 SOLUTIONS OP 4. Let E, F be the points of contact on CJB, CA of the escribed circle opposite C; E', F' corresponding points on BC, BA for the escribed circle opposite B. Prom A draw AE perpendicular to BC, meeting EF produced in 6, and E'F' produced in 6". We will first prove that C B {s — b cos C) cot n = (» — c COS B) cot p' {S — I) cot g = = («-(•) cot -g . . t . (1) C C B 3 bsxaC = c sin B ; .■. 24 . sin^ 5- cot _ = 2c . sin^ sr cot -^ ; O B .'. J(l — cos G) cot ,7 = c(l — cos jB) cot n • . . (2) .•. adding (1) and (2) we have G B (« — J cos C) cot s' = {s — c cos jB) cot ~ • Now GS = EE tan HEG = {GE - CE) cot DEF Q = (s — 6 cos (7) cot ^ , since DEGF can be inscribed in a circle, = {s — c COS B) cot K = G'E. .: G' coincides withG. .*. EF and E'F' intersect AE: in the same point G. B Again A6 = GE — AE = {s — c cos B) cot ,, — esmB B\ B cot 2 = (* — ccosB — c . 2sin^-^) 7? = {s - c) cot 2 = ''« • 5. Join FB. Then the angle EAB = C^2), and ^£5 = ACD, ,•. the triangles ABE, ACD are similar, .-. AB : ^i? :; ^2) : AC, .: AB . AC = AE . AD, .: AB.AC+ AD' = AD{AE + AD) = AD . ED = DC . DB WEEKLY PROBLEM PAPERS. 117 6. Let CP, CD be the equiconjugate diameters, and QCQ the diameter perpendicular to CP. Let §?g be the chord through Q parallel to CD. Then Qq is the double ordinate of OB. Let the tangent at q meet CP produced in T. Then Gr.CT= CF', .: CV .VT = CP'' - CV^ = QV^ = QF . Fq. .: a circle will go round QCqT, and the angle QqT = QGT = a right angle. .•. Qq is a normal at g. Similarly it maybe shewn that the chord Q'q' is normal at g'. 7. Let a, ^, y be the eccentric angles of I, M, N, 8 that of P. Then by Sahn. Con. Art. 244, Ex. 5, a = 240° _ ^8, /3 = 120° - ^8, y = - iS. And by Art. 231, Ex. 5, the area of the triangle formed by a, ft y = 2ab sin J(a — /3) sin J(/3 — y) sin i(a — y) 3a/3 = 2ab . sin 120° sin 120° sin 240° = ~ ah = const. 4 And XXXVII. 6, —r— ab — the area of the maximum triangle which can be inscribed in the ellipse. Paper LI. 1, (1) From (1) and (2), X y {c ■>ra)(a-b)- {c - a) {a + b) (« + b) {b - c) - {a - b) (b + c) z "" {b + c) (c - a) - (i - c) (c + a) .'. 7 = = ',, = —, -Ji = ^ suppose. be — a? ca — b^ ab — r From (3) by simplifying we get abx . bey , caz ,, . .7% ; + - — ^ + ~, -„ = n(bc + ca -f ab), be ~ a^^ ca - b^ ab - o' ' " /. B(fie -\- ca + ab) = «(ic + ca + ab), .: B = n, .: X = it{bc - «°) ; ^ = }i{ca - b")) z = n{ab - e% 118 SOLUTIONS OF (2) Put y = b — a -\- X. Then the equation becomes V(y + A) (y - a) + V(y -b){y + a) = \/{g-{- a){y + b) - 'kaT, (1) and (y + h) {y - a) - {y - b) (y + ff) = 2(4 - ajy, Adding (1) and (2) , , 2(4 - a-)y 2V(y + 4)(y - «) = 7(y + «)(y + 4) - 4.4+;^^^^|Jp== (y - a) {y + 3^) . VCy + a) (y + 4) - 4ff6 .'. either y = a, and a: = 2a — 4, or 4(y + 4) {/ + (fl + 4)y - 3a4} = (y - a) (y + 34)2, .-. 3/ + (5« + 24)/ - (2a4 + 54^)y - 3«42 = 0, ••• (y - «) (3/ + 5(« + % + 3«4} = 0. This gives us the 3 other values of y, and solving, we get finally X = la - b, a, l{a - 116) ± \/25(a2 + 4^) + 14«4. (3) a:y(:r* - /) = a'' = xYix^ + /)2 (^/ _ rf, .-. a;= - y2 = (;r= + /) (a:^/ _ %xy^ + 1), neglecting the factors x'^y\x^ + y^), which do not give admissible values of x and y when equated to zero. .-. a;^ - / = ^y^ 4 a;y _ ^^^ _ 2.^^6 + a:^ + /, .•. sd^f + a;>6 _ ijSy _ 2a:y3 = - 2, .-. xy{xy^ - 2) (ar"'' +/) = - 21 From (1) xy{xy3 - 1) (a:^ + /) = ffl / ^ ' .: a{xy^ - 2) = - 2(ay5 - 1), and ••■ V ' ;+2' ••-/ i-„+2' from {A) by substituting the value of xy^ — 1 Now WEEKLY PEOBLEM PAPERS. 119 ••• (r^ - f) (« + 2)^ = {a:' +■ /)«^ .-. 2x''a + 1) = f(a' + 2/8 + 2). (« + 2)^ ~^ "^ ■ 2(« + l) ' 8(^ + if {a + 2)" {a' + 2« + 2) , («2 + 2« + 2)< («2 + 2« + 2)' • 16(« + 1)' 2(a + 1) (» + 2)2 2. Let 4/ + a, 4ot + 6, 4k + c be 3 nmnbers, in which a, b, c are all niimbei'S < 4. The sum of their squares is 10(/^ + »K^ + »«2) + 8(al + bm + en) + 0" + ¥ + c" =^p + a" + b^ + c\ Now if this is of the form ^x + 7, we must have tt' + V + c^ = iy + 7. For this to be possible, we must have either one or three of the numbers a, b, c, odd. Now all the sets of 1 odd and 2 even, or 3 odd numbers less than 4 are 1,1,1; 3,3,3; 1,1,3; 1,2,2; 1,2,0; 1,0,0; 3,2,2; 3,2,0; 3, 0, ; 1, 3, 3. And we see that in none of these cases is a^ -{- V -{- c^ of the form 83, + 7. 3. Tripos 1875. 1^^ Tuesday morning. No. 2. 4. Take the common centre as origin, and a line through it paral'el to ABC as the axis of x. Let the coordinates of A, B, C be (a, p), (/3, p), (•y, p), and let a, b, c be the radii of the circles. Then a" = d^ -\- p^\ b^ = 0^ +p'; c^ = -f + p\ The equations to the tangents at A, B, C are ax -\- py = a'; ^x -\- py = b^; yx + py = c'.. .•. by Salm. Con. Art. 39, the area contained by these lines is \a{- c^p + Wp) + /3(- a'p + c2p) + y(- Vv + <>^P)Y ' 2(«p - tip) (/3p - yp) (yp - op) ' ' [ai^ - /) + &{y-^ - a=) + y{a^ - ^X ipia - ;3) (0 - y) (y - a) (g - P)(P - y) (y - g) ^ AB . BC . CA 2p 2p 120 SOLUTIONS OF 5. Tripos 1875. Monday morning. No. 9. 6. Let tlie triangle DBF be inscribed in ABC so tbat D is opposite A, &o. Then the triangle FDE = FAE, each being one fourth otABC. Draw AO, i^fl" perpendicular to BF, and let FF and AT) intersect in K. Then since AEF = DEF, and they are on equal bases, .•. their altitudes are equal. .•. AO = DF. .: from the right-angled triangles AGK, DHK it follows that AK = KB. 7. Since the line joining two of the points mentioned is at right angles to the line joining the other two x.^ - Xi x^- ^3 ••• (y2 - yd (j'4 - yi) + (-^2 - ^i) (.^i - ^s) = o- But yi = ~ &c. c^ c s •*1'^2 **'3''^4 .•. a:ia:2*3-'^4 + c* = 0. Papee LII. *1, The given expression = 3271 . 52» ^ 54ii _ 35 . 62u _ 32« -f. 34 = 32.1(5271 _ 1) -I- 5471 _ 1 _ 35(5271 _ 1) = (521 _ 1) (3271 -I- 5271 _ 34). Now 521 - 1 = 25" - 1 = (1 + 24)1 _ i = M{2i) = 11(2'' x 3). Also 32» + 52" - 34 = 32« + M{24) + 1-34 = 32'» + Jf(24) - 33 = M(3). And 32» + 52» - 34 = (4 - l)2m + (4 + l)27i _ 34 = 2|l + \ g ' . 16 + M(256) } - 34 = 32»i(2jt - 1) + i!/(256) - 32 = 2/(32) = M(2^). ,., 327. ,^. 52» - 34 = M{3 X 2'). '. the given expression = M(2<' X 3=) = if(2304). See Errata. WEEKLY PEOBLEM PAPERS. 121 IT 2. ADE = -^—A ; AE = eoosA^,p^=— r'= — 2R coaA cosBcosC. Todh. Trig. cap. xvi. Ex. 27 ; A = 2R^ sin A sin B sin G Pi = AEsmB = ccosAmiB = 2iJ sin B sin (7cos ^, •'• P'i- + Pi +^3 = 22J(sinBsinCcos4 + sinGsin^cosB + sin^sinBcosC) = 2ij{siuCsin(J. + B) + ^(oosA-B- cosX+B)cosC| = 222^1 - oos''C+ i(cos A^^-oosA+B) cos 0} = 2ij|l + i cos (7(cos Z^^ + cos A+B)] = 2iJ(l + cos A cos B cos G) ; P\Pi +^"2^3 + I'sPi = ^^ ^'° ^ sin jB sin (cos ^ cos B sin C + . . •) = 4iJ2 sin ^ sin B sin (cos ^ sin B -)- C + sin ^ cos B cos 6'| = 4iJ2 sin^ ^ sin B sin C {cos B cos C - cos B + G] = 4iJ^ sin" 4 sin" B sin^ C = ^, ; and Pi (p^ + P2+P3)= - r' (2iJ + /) = - 2^/ - r'^ A" .•. sum of products 2 together = — — 2B/ — r'% P1P2P3 — ^^^ ^'^ -^ ^^°^ -^ ®™^ ^ • ^-'^ ""^ ^ ''°^ -^ ''"^ ^ A" , A^ Pi iPiPi + APa + P3P1) =-»•'■ ^3. ,•. sum of products 3 together = 0. A" PiP2psPi = - ^'-J^ •'• PiPiPaPi ''■^^ ^^^ roots of the equation ^ - 2Ex' + d' - 2iJ/ - r-') «''-%■'" = 0- 122 SOLUTIONS OF 3. [Tt + 1 + 2 is divisible by 2 ; \n -i- 1 + 3 is divisible by 3 ; [» + 1 + 4 is divisible by 4 . . . (» + 1 -{- n -\- 1 is divisible by m + 1. Now the numbers on the left are consecutive numbers, each of which hfis a factor, and .". is not a prime. 4. Let FT be the tangent to the other circle. Then PQ.Pq = PT^ Now the chord of intersection is the radical axis. .*. the result follows by Casey, p. 113, Cor. 1. 5. Let .4, ^ be the centres of the circles, C and F their points of intersection. Let ECD be the position of the chord required such that reel EO . CD is a max., and let E'CV be a consecutive chord through C. Then since the rate of change of the rectangle is zero .-. rect. EC.CD = rect. E'C . CV. .". a circle will go through BVEE'. And since D, If are consecutive points on the one circle, as are also E, E' on the other, we see that the line required is the chord of contact of the circle which touches the two given circles, the circle being such that its chord of contact passes through G. To draw this chord, we have only to notice that it makes equal angles with the tangents at E and I), and .". also with the tangents at G. .: the required chord bisects the angle between the tangents at the point of intersection. *6. Let the diameters through Q, B meet PS in M and N respectively. Produce PP to meet SK in T, and NR to meet PK in F. Then SN. NP : SM . MP :: BN : QM, and MP:NP:: QM : VN, .: SN : SM :: BN : VN, .: SB : SL :: ST : SK. .: PS is parallel to KL. This problem can also be solved by Pascal's hexagon. Thus, if e, be the point at infinity in which the axis again meets the curve, and ffiffi denote the tangent at oo, i.e. the line at <», the theorem tells us that the points of intersection of the following pairs of lines are collinear, viz. {PQ, Sa) or K; (Qa, SB) or L, and (sm, PB) or the point at infinity on PB. .'. KL is parallel to PB. 7. Let the tangent at P meet the tangents at A and A' in T and T" Let ST', S'T intersect in Q. Join PQ. WEEKLY PROBLEM PAPEES. 123 Since TS bisects PSA, and T'S bisects PSA', .: TST' is a right angle. Similarly TS'T is a right angle. .-. a circle will go round TSST. Again SQT = 8TT + S'TT = ST'T + SST = ST'T + TSP = SPT. .: a circle will go round SQPT. And since TSQ is a right angle, ,•. TPQ is a right angle. .'. PQ is normal at P. Paper LIU. 1. (1) Employing the method of p. 19, No. 11., we have S =1-2-3-2 + 1 + 6 + ... S,= -3-1 + 1 + 3 + 5 + ... S, = 2 + 2+2 + 2 + ... .". the n*^ term is of the 2nd degree in n, and may be written in either of the forms Jn^ + £n + C. . . (1) or J'n {« + !) + S'ti + 6" . . . (2) If we take (1), and put ?i = 1, 2, 3 in succession, we have j^B + C =1; 4:A + 2P+ 0= - 2; 9^ + 35+C=-3. From these equations we get A ■= 1, B = — 6, C=6, .•. the n^ term is »' — 6» + 6. .'. by p. 18, Art. 26, the sum of n terms ^ n(n + l) i2n + 1) _ g ^ 9i(n + l) _^ g^ 6 2 If we take (2), we have iA' + B'+C' = l; 6A' + 2J' + C'=-2; 124' + 35' + C'= -3. From these we find A' = 1, F = - 7, C = 6, .•. the Ji*'' term is Ji (ra + 1) — 7ra + 6, w (>t + 1) (w + 2 ) „ n{n + l) , .: sum of n terms = — ^ — ' — ^- — ' • „ + o«. Of course the two results thus formed are the same, and merely difEer in form. 124 SOLTTTIONS OF (2) S = 2 + 5 + 12 + 31 + 86 + 249 + . , . 8^= 3 + 7 + 19 + 55 + 163 + . . . S^= 4 + 12 + 86 + 108 + . . . Here the second difference series gives us a geometrical progression, common ratio 3. .-. p. 20, No. III., the Tith term is of the form 4 + 5» + <7.3»-i. Putting M = 1, 2, 3 in succession A+£+C=2; J + 2B + SC=6; ^ + 35+90=12. These equations give A = 0, B = 1, 0=1. .: the ffitii term is ti + 3"— i.. » . ra (»i + 1) , B» - 1 . , sum of n terms = — ^ — - — -' -I • 2 ^3-1 = i{«(»+l) + 3«-l}. ^' 1.4.7 ■'' 4.7.10 "^7.10.13 '^•" mi. tv. i 3ii — I 3re — 2 + 1 The atii term = — ' (3n - 2) (3» + 1) (3» + 4) (3» - 2) (3» + 1) {in +4:) + (3ra + 1) (Sn + 4) ^ (3m - 2) (3w + 1) (3ra + 4) To write down the sum of » terms of this series, we use the following rule. Strike out the first factor of the n^^ term. Write the result negatively and divide by the product of the coef. of n and the number of factors which remain. Then add a constant which can be determined by giving a suitable value to n. •■• ^""^ °^" *^"°' = - 1-3^4 - O (3>. + l)V + 4) + ^- To determine C, we observe that when « = the sum = 0, 2. (2jt)» = 4ji2 = 4 {1 + 3 + 5 + . . . + (2ji - 1)} = 4 + 12 + 20 + • ■ . + 8m - 4 = an A.P. with common difEerence 8. WEEKLY PROBLEM PAPERS. 125 (2« + 1)2 = 47i2 + 4^ + 1 = l + 4+12 + 20 + ... + 8re -4 + 4+ 4+ 4 + .. .+4 = 1 + 8 + 16 + 24 -f . . . + 8» = 1 + an A.P. with common diflference 8. 3. Let A be the summit, B the foot of the hill, D the point in JB such that AD — Z.DB, C the point in the plane. Then by the question, CD bisects the angle ACB. .1 BD BO sin (5-30°) V3 1 ,. , . • • o = ^r-y = vr; = '^^ — :; — - = -r^ ^ cot 6. .-. tan 6 B DA CA sin fl 2 2 . ' ' ' 3 V3 - 2' 4. Let ABC be a triangle. Produce BA to E, and bisect the angle CAE by AD, meeting the circum-circle in D. Join DB, DC. Then the angle DGB = MAD = DAG = DBC. .: DB = DC. Again, let Oj, Ojbe the centres of the escribed circles opposite B and C. Then O^BO^ O^CO^ are right angles, and DB = DC, .: the circle on 0^0^ as diameter passes through B and C, and D is the centre. .•. D is equidistant from B, C, 0^, O3. *5. Let Oi, O2 be the middle points of AC, BC. Join EOi, ED, FO^, ID. Since AD= DB= O^A + O^B, .: 0.,D = O^C = O^F; and 030= OjC = Oi^, and D^ = DF, .'. the triangles O^DE, O^DF have corresponding sides equal; .-. the angle DO^E = DO^F. .: O-^E, O^F are parallel. .•. EF passes through G, the centre of similitude of the two circles. 6. Let be the circum-centre, P the orthocentre of ABC. Draw FY, OY' perpendicular to BC. Then since GAB = PAC, &c. .-. and P are the foci of the inscribed ellipse. .-. PY. OY' = (J minor axis)^. ,'. the product is the same for all the sides. This can also be proved as follows : PY = rCtanPOr= ^C cos C cots = ^^= . cosJBcosC sm.B = 2ffi cosjB cosO, OY' = B cos A. ,\ PY. OY = 2R^ cos A cos B cos 0, a symmetrical expression. 126 SOLUTIONS OF 7. Let ABC, PQR be the circum- and inscribed triangles. Let^, 5, r be the points of contact of BC, CA, AB. Through F, Q, H, q, r draw diameters meeting BC in L, M, N, m, n, respectively. The diameters at p, q, r will bisect the chords QB, BP, PQ. Then QB = Mli = ML + LN = ILn + Um = 2mn = 2^2 = 4:BC. Similarly it may be shewn that RP = 4C4, PQ = iAB. Paper LIV. 1. (1) S = 4 + 18 + 48 + 100 + 180 + 294 + . . . Si = 14 + 30 + 52 + 80 + 114 + . . . 8^= 16 + 22 + 28 + 34 + . . . ,53= 6+6+6 + ... Here the 3rd difference series gives us a series of equal terms. .•. the nti term is of the 3rd degree in n, and is of the form A + Bti + Cn{n+1)+I)n(n + 1) (« + 2). Writing » = 1, 2, 3, 4 in succession, we have A+ B+ 2C+ 62)= 4^ A + 2B+ 60+ 24Z)= 18 A + 3B +12C+ 602) = 48 A + iB + 200+ 12023 = lOoJ Attending to the remark given in No. II. on p. 20, we find A = 0, B = 0, G = - 1, D = 1, .: the mtli term is re (?i + 1) (»i + 2) — n (ji + 1). When the n^^ term is of the above form, to obtain the sum of it terms we have only to introduce an additional factor at the end of the n^ term and divide by the number of factors thus obtained. .-. sum of n terms = » (^ + 1) (r. + 2) > + 3) _ n(n + !)(» + 2) 4 3 = i K (» + 1) (« + 2) (3n + 5). WEEKLY PROBLEM PAPERS. 127 (2) a = 5 + 11 + 22 + 41 + 74 + 133 + /Si = 6 + 11 + 19 + 33 + 59 + -Sa = 5 + 8 + 14 + 26 + -83= 3+6+12 + Here the 3id difference series gives us a geometrical progression, common ratio 2. .-. p. 20, No. III., the n^'^ term is of the form A + JSn + Cit {tt -\- 1) + D . 2»-i. Putting K in succession = 1, 2, 3, 4 we have A+ £ + 2C+ D = 5i A + 2B + 6C+2D = 11 ^ + 35 + 12c i- 4i) = 22 ^ + 45 + 20c + 8D = 41 . From these we find .4 = 1, £ = - 1, = 1, D = 3. .-. the Jitii term is re (»i + 1) - » + 1 + 3 . 2«-i /. the sum of re terms = '- — ^ — ■ — ; ^-~ — -' + a + 3 (2" — 1). (3) Consider the series formed by the numerators S = 6 + 8 + 16 + 34 + 89 + 252 + . . . S^= 3 + 7 + 19 + 55 + 163 + . . . ^2 = 4 + 12 + 36 + 108 + . . . /. the n^ term is of the form A -\- Bn -\- C. 3"-i. Putting X = 1, 2, 3 in succession we have ^ + £+0 = 5; A + 2B + 3C = 8; ^ + 3£ + 9C=15. From these equations we find A = 3, B = 1, C = 1, ,: the wtii term of the original series is 3«-i +n + 3 1 _ 1 ^ « + 3 1 n(«+l)(re+2)'3"-i »(»+l) («+2) ■*■ ji(re+l) (n+2) " 3»-i 128 SOLUTIONS OF As on p. 23, No. 5, assume ' » + 3 1 — An-\-S _1 A(n-1)+B 1 ji(n+l) (n-f-2) ■ S^-i — (?i+l) («+2) ' 3»-i ra (re + 1) ' 3»-2' .-. ra + 3 = - 2ul?i2 - (3^ + 25) ra + 6 (4 - 5). Equating coefs. of like powers of n on both sides, we see that A = 0, B = — ^ satisfy this identity. .*. the nf'^ term of the series 1 1 1 1_ 1 1 1 ~ n (re+1) (k+2) 2 ' (m+1) (w+2) ' 3'»-i "^ 2 ' m(M+l) ' 3n-2' = C- .: the sum of n terms 1 2 (Ji + 1) (re + 2) 2 (re + 1) (re + 2) 3»-i To determine O, we observe that when re = 0, the sum = 0, .•.0 = -J- + i.J_. 3 = 1=1. 2.1.2 ^ 2 1.2 4 , sum of n terms = 1 — (' + !i.)- 2 (re + 1) (re + 2) 2, m"-! — 1 = mult, of ra ; re™-i — 1 = mult, of m, .". (m™-i — 1) (re™-i — 1) = mult, of mn. ,: Jot"-' . re™— 1 + 1 — ('»"-■' + re™-i)| -i- rare = an integer. .•. (ra"-! + re™-i — 1) -j- rare = an integer. .•. ra"-i + re™-' when divided by mn leaves a remainder 1. 3. Since the ships are moving uniformly, we may imagine the first to remain at rest, and the other to be moving with a velocity equal to the difference between its own and that of the first. Let A denote this stationary position of the first ship, £, 0, B the positions of the second at the different observations. Then -^ = — = ""(a- /3) . ^ ^ :^ _ sin {.& - y) AC~ AC ~ sin (8 - a)' q ~ OD ~ sin (0 - y)' . ^ _ sin (a — 0) sin (5 — y) " q sin (/3 — y) sin (6 — d) WEEKLY PROBLEM PAPERS. 123 .•. p sin(;3 - 7) (sinS cosa - cos5 sina) = gsiii (a - j3) (sinfl cosy - cos5 siny), .*. tan 6 \p cos a sin (/3 — y) — g cos y sin (a - ^;} = ^ sin a sin (iS — y) — g sin y sin (a — /3). 4. The angle BBC = ^Z)C = i^i)^ = FBI, and ^/"^ = BW, .: BEG = BAF, and the triangles ABF, CBB are similar. .-. the angle BCE = BAF = CUB = BEG, .; the triangle BEG is isosceles, and .-. also AFB. 5. In the ellipse C& : CN :: CS^ : GA^ Cg -.FN:: GS^ : CB\ .-. CO.Ggc^ CN. FN. Now PIP : AG'^ - CN^ :: BG^ : CA^, FN^ .GN^ = -^ (AC' - CN^) . GN^ a {AG^ - CIP) . CNK Now the sum of these two factors is const. .•. the product is a mai?. ■when CN^ = AC^ - GJf", .: CN^ = iAC\ .: Pm = hBG^, .: GP^ = GSP + PJV2 = J (AG^ + BC^), .'. GP is one of the equioonjugate diameters, and .'. 4 triangles can be drawn. 6. Let ABG be the triangle, the circum-centre, P the orthoeentre. Let B, V be the middle points of BG, AP. Then if BV and OP intersect in N, DV is the diameter and N the centre of the 9 point circle of ABG, and N bisects both OP and BV. .: OBPB is a parallelogram. Then resultant fA PB, PG is represented by 2P2) ; and PA = IPD', .: resultant of PA, PB, PC = twice resultant of PB, PV = 2 OP. 7. Let (h, k) be the point from which the tangents are drawn. Then by Paper XXXV., No. 7, W am fjfi ifi \ «2 + ^2 (length of chord)2 = 4 (^- + -, - IJ j^ ^ = 4j3^, since {h, li) is on the given curve 130 SOLUTIONS OF Paper LV. 1. Let X, y, z denote the digits, r the radix of the scale. Then the muiibers are r^x ■\- ry ■\- z, r^y -{■ rz -{- x, ■fiz -\- rx -{- y. Since these arc in A. P. r'(x + z) + r (x +y) +y + ^ = 2 (r^y + rz + x), .: .T (!-2 + r - 2) + y (- 2?-2 + r + 1) + « (r^ - 2r + 1) = 0, .-. .V (/• + 2) (r - 1) -y{2r+l)(r - I) + z{r - 1)^ = 0, and r - 1 ^ 0, .-. x{r + 2)-y(2r+l) + z{r-l) = 0, 3 .'. z -\- X — 2y = (v — x) = f suppose. »■ — 1 Then since y — x < r — 1, .: is a positive or negative proper r — \ fraction. .•. t is integral and lies between + 3 and — 3, and .•. = ± 1 or ± 2. .-. 3 (y - .r) = (r - 1) «, .-. r = 1 + 3 . ^-^ = 1+3?. .•. the radix of the scale exceeds by unity a multiple of 3. Now y — X = pt, and z ■{■ x — 2y = t, .'. z = x -\- (fip -\- V) t. Now z and x are positive, and each < Zp + 1, .'. t =1= ^ 2. .:( =f ± 1. If i = -\- l, X may have any value between 1 and p — 1 inclusive. „ t = - l,x „ „ „ „ 2^ + 2 and 3p „ .'. there are 2 Q) — 1) solutions. Let any set of digits be y=a, x=a+pt, z=a-{p-\-l)t, where l, = ±l. The com. dif. = {ip-\-lfx-\-{?,p+\)y+z-{?,p-itlYy- {Zp+l)z-x = {Bp + \f {x-y) + {Bp + \){y- z)+z- X = {Zp + 1)2 . pt + (3p + 1) (p H 1) ^ - {2p + 1) t = ± 3;> (3^)2 + 3^ + 1) = constant, since p is constant. WEEKLY PROBLEM PAPERS. 131 2. Denote the left hand member of the last line of the question by 2, Multiply the first 3 equations by Xj, /ii, vi respectively. .•. Ai«i — X^\-iCl^ = p (XiJ«2 - ^Vi), «Vi*i - ^yy-ih = P (ft''2 - Mi'^z)! AiCi - xyv^e^ = p (1/1X2 - i'i/i2)- .'. adding, we get pS = J'a- - xy (Xi^j + ^11^2 + "i^a) = o- (p^ + /) - scy (Xi(?2 + Mi*2 + ^x'^il = ap^ + y- \axy - x^ (Xi«2 + /*i62 + ''i<^2)}- 3?ow a^^ — a^^y = p (fii — Vi), V^ - ii.ry = p (vi - Xi), V^ - "iJy = p (Xi - pi). .•. a-2 (Xi«2 + Pl52 + »'i«2) - "^^^ = p (Xipi - XiPi + piwi - piXi + j/jXi - vipi) = 0, .'. p2 = pV, .'. S = pa. 3. Tiip'os 1875. Monday afternoon. No. 12. 4. Let be the centre, and § the middle point of the arc SO. Join d(^ meeting CP in i?", and let AO produced bisect the arc CD in Q. Then the arcs AP and CQ together = the arc P§. The angle PJ.l'' = sum of angles subtended at. the circumference by the arcs AP and C§ = the angle PFA. .: PA = PF. Again, since the arc AP = CO, .: the angle PCJ = CAO. .". PG is parallel to J.O. Similarly it may be shewn that AQ is parallel to AO. .'. AOCF is a parallelogram, and GF = AO. .: PC = PP +F0 = AP + AO. This proves that 2 sin 5i° = 1 + 2 sin 18°. 5. Tripos 1875. Wednesday morning. No. 1. k2 132 SOLUTIONS OF 6. Let tbe equations o£ the ellipse, director circle and tangent be -I + -S = 1 ; ^' + y' = «' + ^'; - <=os5 + ^ sin5 = 1. The equation representing CP and CQ is evidently ^2 + / = („2 + 1^2) ^f cos 5 + I sin ^y, which on reduction becomes , , 2i sin 6 cos 5 6^ a n Writing this in the form - ^'i'^) (y - Wj.?) = 0, J2 we see that ranW, = — -, which is the condition that the lines should a2 be conjugate diameters. 7. Suppose one of the distances, AP, to remain constant, i.e. suppose P to move along the circumference of a circle centre A. Then we have to find the point P on this circle for which BP + PG is a min. Then P is such that BP and CP make equal angles with the tangent at P, and .•. also with the normal AP. (See Theory of Maximum and Minimum by present writer). .•. the angle APB = APC. Similar!)', by supposing PB to remain constant whilst PA and PC vary we should find the angle APB = BPO. .: when AP, BP and CP all vary, the point P is such that APB = BPC = CP^ = 120°. .-. on the sides describe arcs of circles containing 120°. The point of intersection of these arcs is the point required. Note.— If one angle be greater than 120°, the point P will evidently coincide with C. Paper LVI. 1. (1) lY + !^/' = 27 (2.1-2^2 - * - y), a-2y + a-/ = 3 (4.ry - .r - y), ■•• (^ + y) (^y + 27) = 54^2^2^ (■i^+y)(^y + 3) = 12.ry, WEEKLY PROBLEM PAPERS. 133 .\ dividing, .t^ _ 3xy + 9 = fxy, .-. 2xy - 15zy + 18 = 0, .-. (2xif - 3) (ay - 6) = 0, .-. ay = I or 6. either (a: + y) (6 + 3) = 12 X 6 .-.a; +^ = 8i ijy = 6 ) .: X - y = ± 2 VlO or (a:+y)(t + 3) = 12xf .-. a; + y = 4 1 ay = f i .-. .r = 4 ± VlO; y = 4 =F VlO ; or a: = i (4 ± VlO) ; y = J (4 qi VlO). Also ;r = 0, y = obviously satisfy the equations. (2) Multiply up, and we get d^z + c^y = iry« = A + A = a^y + ^% .". (4^ — c^) a; + ff^y — A = A — c^y + (rt!^ — i") 2 = 0, * _ y _ ■^ _7?p Substitute in - + - = ^. .-. iJ2 iP + C' - »2) (c2 + «2 - b^) {cfi + i^ _ c2) = 2. This determines if, and then x, y, « can be found. f3) (a;2 - «2) «/4«2 - :j:2 = 2a3, .-. ra;4 - 2a2a;2 + a*) (4^2 - a:^) = iifi, .-. - > + 6«2a:* - ^cfix^ + 4fle = 4fl«, .-. a« - 6 A* + 2a^x'^ = 0, ... ^2 (a:2 _ 3a2j2 = 0, .-. a; = or ± a \/3. 2. 1 + ^/-3 = 2g + V— 1 . ^) = 2 (cos^+ ^A^sin^) ... 4 (1 + x^-^e—l = 4 . 2«»-i (cos ^ + V^sin 5^"' = 2«»+^[cos(6»-])!I + V^^l sin (6«-l)^} 134 SOLUTIONS OF = 2«"+l (cos I - V^ri sin |\ \6b+1 (1 _ V^«"+l = 2^"+! (cos I - V^Tsin Ij = 2""+^ (cos (6« + 1) ^ - V^siii(6«+1)^{ = 2^"+^(cosE- ^/3lsing, ... 4 (1 + Vr'ajsn-i - (1 - V^^°"+^ = 0. 3. = • = tan^ X + sec 2x V3 1 — cos 2a: , n sec 2x - 1 , „„„ n^ = + sec 2x = + sec zx 1 + cos 2x sec 2^7+1 2 = 1 + sec 2.r 1 + sec 2x 7 « _ 10 .'. (1 + sec 2x)^ - ■ "' , . (1 + sec 2x) - 2 = 0, V3 2 Va 2 v/3 2 V3 2 Vs 2 — /. sec 2x = -T- or 5 — 4 v/S. V3 Now — '^' is negative, and tan^ a; is positive. .*. sec 2x is negative. .•. we must take the second value. ...cos2. = _l_ = -i + l^. 5 - 4 ^3 23 *4. Let A-i, A„...Aj'bQ the angular points of a regular heptagon inscribed in a circle, and let be any point on the circumference between A-i and A-,. Let A1A2 = A^A^ = . . . = a; A^A^ = A^^A^ — . . . = d. WEEKLY PROBLEM PAPERS. 1G5 Then from the quadrilateral OAiA^A^ we have by Euc. VI., D, (,0A■^ + 0^3) a= OA^d; So {OA^ + OA^ a = OA^ /; {OA^ + 0^s)o = OA^d; (OAi + O^j)^ = OA^d; {pA^ + 0^,) a = 0^6 1^; (0^8 - OA^ a = OAjd; (OA^ - OA,) a = OAi d. ,'. by addition, we have {OA, + OAi + OA, - {OA^ + OA, + OA, + OA,)]d = 2a {OJi + OAs + OA^ + OA., - {fiA -f- OA^ + 0^^)}, .-. {0^2 + 0.^4 + 0^6 - (0^ + 0^3 + OA^ + O^j)} (<; + 2«) = 0. Now rf f 2a ifc 0, .-. 0^1 + Oy^3 + 0^5 + 0^7 = OX + OA^ + O^o- The above method is perfectly general, and may he extended to tli« case of any regular figure of an odd number of sides. 5. Lot be the fixed point, AC, BB the chords tlirough at right angles. Let :E, F, a, a" be the poles of AB, BC, CD, J)A. Let IC, GH meet in K, and let EF, DG meet in L. Then AEB = 7r - 2EAB = n - 2ADB = ZBAC = 2CBG = ir- COB. .: AEB + CGB = 2 right angles. ^ .'. a circle can be described about EFGH, Again, F is the pole of BC, and H is the pole of AD. .: FH is the polar of the point of intersection of AD and BC, which evidently passes through O. Similarly EG passes through 0. Again, BED = DHO, being in the same segment, and EBO = ir - IBO = tt - EDO = EDO, .: FOB = HOD = FOB. .: EOF is bisected by OB, and .-. also EOIIis bisected by OA. 6. SI^ + S'P'^ = 2{S0^ + OP') = 2{S0^+BQ^ = 2(80^ + 0B^+ OQF) = 2 {SB^ + oq^) = 2 {OA^ + oq^) = AQ^ + A'q\ and SP + 5'P = ^^' = Aq + A'q, .: SB = Aq-, S'P = A'Q. [36 SOLUTIONS OF 7. Through M draw MS at right angles to MJf, meeting the xis in If. 1'hen JIA . AN = AM'^ = PW^ = iAS . AN. .: AH = 4AS. .: fi" is a fixed point, and A.V is a fixed line, and MN at right angles 1o MH. .: the envelope of MN is a parabola having H for focus and AM for tangent at the vertex. Paper LVII. 1. (1) ^y = 6 4- 7 + 18 + 45 + 94 -I- 171 -f , . . S,= 1 + 11 + 27 + 49 + 77 + . . . S.,= 10 + 16 + 22 + 28 + . . . ,S'3= 6+6+6 + ... Here the .Brd difference series gives us a series of equal terms. . the n*'' term is of the 3rd degree in n, and is of the form A+£>i + Cn{n + l) + Bn {n + !)(« + 2). Putting ffl = 1, 2, 3, 4 in succession, we have A+ B-\- 2C+ 6Z) = 6> A+2B+ 6C+ 240= 7 ^ + 3.B + 12C + mo = 18 ^ + 45 + 20C + 120Z) = 45J From these equations, we get A = 'i, B = — \, C = - 4, D = 1. .-. the nti" term is » (» + 1) (« + 2) - 4». (k + 1) - « + 9. .•. the sum of n terms = J«(« + l) (» + 2)(« + 3) -!»(»+ 1) (« + 2) - i« (» + !) + 9» = ^ {« (» + 3) (3k - 7) + 88}. (2) 5 = 7 + 13 - 7 - 181 - 1149 - 6111 - . . . /Si = 6 - 20 - 174 - 968 - 4962 - . . . S^= - 26 - 154 - 794 - 3994 - . . . ^3 = - 128 - 640 - 3200 - . . . WEEKLY PEOBLEM PAPERS. 137 x'leve the 3rd difference series is a G.P. common ratio 5. .-. the »tii term is of the form A + Bn + Cn {it + 1) +D. 5n-i. Determine A, B, C, D as before by giving » the successive values 1, 2, 3, 4, and we find ^ = 1/5 = 2, C = 3, D = - 2. .-. the rath term is 1 + 2m + 3m (« + 1) - 2 . S™-!. .-. the sum of » terms = » + « (m+ 1) + « (« + 1) (k+ 2) - J (5" - 1) = ;z (« + 2)2 - J (5» - 1). (3) By tlie method of III., p. 20, we find the «tii term of tlie series formed by the numerators is 3 . 2"-! + «2 - 12. .•. the «*•' term of the given series 3.2™-i + m2 - 12 1 ii(n + l)(n + 2) 2"-! 3 ^22 - 12 1 ti {>i + 1) {n + 2) ' n(?i + l){n + 2) 2™-! As in v., p. 23, assume «2 - 12 1 _ An-\- B J A(n-1)+B 1 »H«+1) (m+2) ■ 2"-i — (m+l) (a+2) ■ 2n-i » (ra + 1) ' 2"-2' .-. »i2 - 12 = - Ajv^ - (2^ +B)n + i(A - B). .: equating coefScients of like powers of », we find that A = — 1, B = 2 satisfy this identity. the »*•» term = _ (;» - 1) - 2 1 = C- {n + 1) 2»-2 (« + !)(« + 2) 2»-i = Un-l - Un. : the sum of ti terms of the given series 3 1 « - 2 1 2 ■ 7^ 1) (» + 2) (» + 1) (« + 2) ■ 2»-i If « = 0, the sum = 0, .% C = - f. 2. Tripos 1875. Thursday morning. No. 1. 3. From (2) 2 cos2(|) + cosZ^ = 1, and sm^ e + cos^ e = I. .: 2cos' = «/2sin5. 138 SOLUTIONS OF From (1) sin ^ (2 cos ^ — 1) = 1 - 2 cos (p, .-. (sm

can be written down. 4. (sin A + sin B) (cos 5 + cos C) (cos C + cos A) o ^-5 B-C C-A.A + B B + C C+A = 8 cos cos cos — Sm ^ cos -r-^ COS — ^ — , a A - B B-C C - A G . A . B = COS COS COS — - — COS- sin- sm- . 2 2 2 2 2.2 XT . A . B C , . B . C A . . G . A B JMowsm- sm- cos- + sm- sm- cos- + sm- sm- cos- £i Z Z Vt Z A it it ^ . B . A+ G . . G . A . A4-G = sm _ . sm — -^ — + sm - sm - sin - sm 2 2 ' 2 2 2 A+ C f A+ G , . G . A\ — i — I cos — -I f- sin - sin - 1 2 V 2^22/ . A+ G A G = sm_^— cos- cos 2 . A+ G . B+ G . A + B ~ sm — ~ — sin — J — sm — i — , 2 2 2 ' .". the given expression a . A+B A-B . B+G B-G . G+A G-A = 8 sin — ! — cos . sm — ' — cos . sm — - — cos 2 2 2 2 2 2 = (sin A + sin B) (sin B + sin C) (sin C + sin A). 5. Let AD be the diameter of the circle, and let ABGD be the figure formed by half the hexagon, so that AB = a, BG = b, CD = c. Join AG, Then ACD is £i right angle. .-. d^ = (AC + CD^) d = (ai+i^ - 2abcoBB + c^)d = (a* -|- i^ + c^ + 2ai cos D) d, and cos D =• -, (I = (a2 + 42 + c2) (? + lahc. WEEKLY PROBLEM PAPERS. 139 Now the square of the area of the triangle, having a sl-2, I v2, c v2 for sides = \(ji ■\- h ■\- c) {a -\- b — c) (Jb •\- c — a) {c ■{■ a — h). Square of area of hexagon = 4 times the square of ABCB = \{a-\-b-\-c-^{fi-\-h - c-\-d){h-^c- a-\-d){p-\-a-l->r«[) = i {(« + W -{c- d)^}{(c + ^2 -{a- b)^} = i{{a+ by (C +d)^+{a- bf {C - df - (fl2 - 42)2 _ ((,2 _ (?2)2| = I i^a^^ + 242(;2 + laH^ -|- 1d^ {a^ + V^ + c^) - fl* - /S* -c* - (a2 + 42 + c2) (^2 + Gabcd) = \ S2ah^ + 242^2 + 2fl242 _ a* - 41 - c* + ^abcd + (?*} = i (» + * + «)(« + *- c) (a + c - «) (c + a - *) + abed + J (Z* = (area of A)^ + abed + i dK 6. Let ^ be the centre of the fixed circle, B the centre of the other, C the fixed point, LB the common chord. From C draw CG- perpendicular to EF, and CH, a tangent to the fixed circle. Then by Casey, p. 113, Cor. 1. Cff2 = AB . ca. Now Cff and AB are fixed, .'. CG is fixed. 7. Tripos 1878 Monday morning. No. 8. Paper LVIII. 1.(1) t + l^'t ^''l = a + b, a b X y ^^^^(^-^)^_%_ni) . . . . (1) aba: , .... x(.T 4- a) v(y + b) .-. by division ^ \ = ^-^^ — ', .-. 42.12 _ aY 4. ai2y _ a^y = 0, .". {bx — a!/){bx + ay -{- ab) = 0. If bx - ay = 0, then writing (1) ab{x — a) , abij/ — b) _ f. bx ay 140 SOLUTIONS OF we have x ■\- y = a -^ b, .'. x = a, 1/ = b , . . . id). If bx -{- ay + ab = 0, (a + x)b . b^ _ ab *'' y — > • • "~ I » a y a + X , , a^ , ¥ d^ ah X y X a -\- X .-. (a + b)x^ + 2abx - «' = 0, - ab ± «Va2 ^ ab + b^ ••• - = ,+T ' °'_^ " a a a+b (2) Write the equations in the homogeneous form 3.^(a -{■ b) — bxy — ay^ = a — c (1). bx^ + axy - (,a + b) = b - e (2). Eliminate the absolute terms by multiplying (1) hy b - c, and (2) by a — c and subtracting. .-. x%b^ - ac) - xy{b^ - ac + a^ - be) + f{a^ - be) = 0. Divide by y\ and for -write z. .: z\b^ - ac) - z{b^ - ac + a^ - be) + a^ - be = . . . (3). Evidently z = 1 satisfies this equation. Tliia leads to the result X = y, which we see is inadmissible. The product of the roots of (3) is -z , and since one root is unity, b^ — ae the other is — . This leads to b^ — ae 11 suppose (4). a^ —be b'^ — ae Now subtract (2) from (1), .'. ax"^ — (« + h)xy ■\- by^ = a — I, .". {x — y'){ax — by) = a — b. Substitute in this equation the values of x and y from (4), .'. iJ2(«2 - b'^Jf-ac - bc){a^ - b^) = a - b, .: B{a + b+c){a^ - i'') = 1 WEEKLY PROBLEM PAPERS. 141 This gives the value of B, and .*. se and y are known. 2. {e'! - !)'»■ = fx -\- ^r- 4- . . . \ = x" ■{■ terms involving highfer powers of x. Sow {e" - 1)» = e"" - »e(«-i'» + "^^ ~ ^^ e(»-2)ir - ... by the Binomial Theorem. .•. expanding each of the terms on the right, we find the eoef. of x^ is 1 {;*3 _ „(„ _ 1)3 + ?(^) . (« - 2)3 - . . . j . But no power of x lower than x" occurs in (e^ - 1)". .-. Ti? -■ n{n - 1)3 + ^^^ " '^ l{n - 2)' - . . . = . . . (^, r. and by p. 6, Art. 7, of Problem Papers, rfi + n{n - 1)3 + "^^ ~ ^^ (m - 2)3 + . . . = »2(« + 3)2«-3 . . . (^). li .•. adding (^) and {B), and dividing by 2 we get the required result. 3. Clear of fractions. sin(a + 5)sin(/3 + (^)+sin(a + (^)sin(/3 + e) = 2 sin (a + ^) sin + (jb), cos(a + 5)cos(y3 + (^) + cos(a+^)cos(/3 + 5) = 2 cos (a + ^) cos (/3 + (^). .". adding these two lines together we have cos (a - /3 + 5 -<;()) + cos (a - ^ - 5 + ^) = 2 cos (a - &), .'. 2 cos (a - ^) cos (5 - ^) = 2 cos (a — /3), .-. cos (a - /3) = 0, or cos (5 - 0) = 1, /. a - /3 = (2» + 1) I or ^ ~ = 2K7r. 4. tan ^ > 6, .'. tan ^, > ^, which is > j^-,- Now sin 5 < d, and 1 - .5 < cos 6, 142 SOLUTIONS OF 1 ^ . t.„ _1_ ^ 1 + •-'' tan 6 < ^, .•. tan - — ; 5 < 1 +, X' \ --L - -^ " 1 _ 2 2(1+:e2- .•. tan - — ; — -„ < 1 -\- X^ 1 — X + T 1 + a;2 1 — X -\- x^' 2(1 + x^y^ 2(1 + x^) 1 . '2(1 + .^2)2 - 1 1 - a; + i!;2' if 2(1 - ;r + 2x^ - afi + x^) < 2(1 + 2:i;2 + a;*) - 1 ; if 1< 2.t(1 + a:2) wliich is the case if a; > - • 5. Tripos 1878. Wednesday morning. No. 1. *C. Let y^ = iax be the equation to the parabola, (x^y^, (x2i/^ the coordinates of the extremities of the chord FQ. Then the equation to PQ is y-y^= y^JZ.L^ ix - X,) = ^^^UZll) (x - xd X-i — Xi X2 — Xx or y(yi + y.) - i.ax = yiy2 (^). If (I, I)) be the coordinates of R, the middle point of FQ, and y, — ^2 = 2(7, we have U = xx + X2= i (yi^ + ^^2^ ; 2, = y, + y^. Now (yi + y,)2 + (y, - y,)2 = 2(yi2 + y^'), .-. V + 4c2 = 10a|; .-. ,,2 = 4a| - c^. (5). To find the envelope of PQ, we have (yi + yi)^ = (yi - y^)^ + 4yiy2 = Ac^ + 4yiyj, WEEKLY PEOBLEiil PAPERS. 143 .•. the equation {J) to PQ may be written (yi + y,Y - 4y(yi + y^) + ida^ - ic^ = 0. .". expressing the condition tliat this equation should have equal roots, the equation of the envelope is ffi = 4a.r - <^, wliieh coincides with (B), the locus of B. This question may also be solved geometrically. Let QFQ! be a double ordinate of the diameter PP^, and QB per- pendicular to PK Then QD is constant, being half the difEerenoe of the ordinates of Q and Q, and QS^ = iAS.PF, .: PF is constant. Also QV<^ is parallel to the tangent at P. .: the locus of F, and the envelope of Q§' coincide with the p-irabola obtained by moving the given parabola a constant distance PF in the direction of its axis. See Errata. 7. Let (ffl cos a, b sin a) be the coordinates of P. Then those of p are (a cos a, a sin a). The equation of C? is y = - x tan a. „ auxiliary circle is x^ + f = «^ '. by substitution x'^ + -.■c^ tan '' la = a\ : coordinates of g are «2 cos a y = db sin a 1 Va3 cos 2a + 42 sin ^a V^ cos ^a + 52 sin 2a : coordinates of Q are a' cos n y = (5= sin a Vfl2 cos -a + i^ sin ^a v/«^ ' cos ^a + ^2 sin 2a : equation of QT, the tangent at §is «• cos a + y sin a = v a'- cos 2a + b"^ sin 2a. This is perpendicular to Cjp, whose equation is a; sin a — y cos a = 0. If Cp and QT intersect in T^ CT is the length of the perpendiouLir from C on QT. .'. CT = V«2cos2a + 42 sin 2a = CP. 144 SOLUTIONS OF Paper LIX. 1. By ordinary division, we find -1 = -0588235294117647. 17 The explanation required is exactly the same as that given in Papers XXVI., No. 1, XXXVI., No. 1. 2. . . X - a _ b _ a _ x — h b X — a X — b a' . {X - of - If^ _ _ (X - i)2 - fl2 b{x — a) a{x — b) ' . (x - a — b){x — a -\- b) _ _ ( x — i - a)(x - b + a) b(x — a) a{x - b) .'. either x — a — b — Q, .'. x = a -\- h, or a{x — b){x — a-\- b) -\- b{x — a)(x — b -\- a) = 0. .: x\a +b) - x[cfi + i2) = 0, (2) Square both sides, and transpose •■• n/* + iJx - 1 - «/7+T = - iVW^^. .: x + x - l+:r+l +2«yj;2-A-- 2 V^r^^a; - 2 «/a:2_i = 4 ^/.r' - ,t. .-. 3j: - 2 V.,:2 - 1 = 2(\/.i^2qr^ ^ 'Jx^^^. .-. 9;r2 + 4(a-2 - 1) - 12,« Vi^^n = 4(.r2 + .r + .x^ - .r + 2:r v'^~l). .-. hx^ - 4 = 20,r V.r2 - 1. .-. 25.1'* - 40.r2+ 16 = 400a;2(,j2 _ ])_ .-. 375a;4 - 360:!;2 - 16=0. .•. .375.1-2 = 180 ± >/32400 + 6000 = 180 ± 80 -^67 •■"' = 'J^5^>27±12x/0). WEEKLY PROBLEM PAPERS. 145 2 •'• '""" '''r5V27±i2\/6. 3 cos ^ = ('^ ~ <')(^ - ") ^ 6d-\- ao - ai - cd {d + a){b + c) bd -\- ac •{■ ab -\- cd . , 2 -^ _ 1 — cos ^ _ fl5 + Ci? ^~ 1 -\- tiosA ~ ae + bd Similarly 2 ia + c» 2 ci + at^ Now _ + - + -2 = '^' .•.0 = tan(^ + |+f) = tan - + tan - + tan - — tan - tan - tan _> it ii A it £i Jl .: (tan - + tan - + tan - ^ = tan^ - tan^! - tan^ - _ ai -\- cd be -{- ad ca -\- bd ac -f bd ba -\- cd cb + ad = 1. ,'. tan - + tan - + tan - = ± 1. ^ M ^ 4. Let All produced meet CQ in 2) and BP in E, and let BB and CQ meet in F. Let AQ produced meet PC in G and JSB in H, and let BE and CP meet in K. Then the angle ^i)P = BBB = ^ C5, and the angle SFD = EQD = ABC, .: the triangles DBF, ABC are similar. Also the angle GHK = BFF, and 6XR = J'iJi'', for HB is parallel to BF, and i/ff to -EF, .". the triangle OffK is similar to DBF, and .-. also to ABC, and the triangles OHK, DBF evidently have their homologous sides parallel. Let the angle ABB he denoted by 6. Then the sum of the homologous sides of 3BF and 6EK : the homologous side of ABC :: DF + HK : BC. 146 SOLTJTIONS OF jj HK _ HS 4B,^_ sin 6 sin C , sin (6 + C) °^ ~Wa ~ [AB ' BG"^ BC ~ sinJS ' sin^ "*" sin C ' DF _ sin e sin B sin {6 - B) BG ~ sin G sin ^ sm B .: HK + BF : BG :: sin 5(sin2 C + sin^ B) + [sin {8 + G) sin 5 + sin {d - B) sin C} sin ^ : sin ^ sin B sin C. :: sin ^(sin^ ^ + sin^ 5 + sin^ G) : sin ^ sin B sin G, for sin (5 + C) sin 5 + sin (5 - j5) sin C = sin 5 (sin 6 cos (7 + cos 6 sin C) + sin C(sin 6 cos 5 — cos 6 sin 5) = sin 6 sin (5 + C) = sin 6 sin ^. 5. Tripos 1878. 2nd Tuesday morning. No. 1. 6. Tripos 1875. Wednesday morning. No. 8. 7. Tripos 1878. 2nd Monday afternoon. No. 4. Paper LX. 1. Let w, T, y, z be the number of Liberals returned by England, Scotland, Ireland, and Wales, respectively. W, X^Y, Z . . . . . , . Conservatives .■.(l)w + !C-^-y + !!= r+15; (2) W + X + T + Z ^ 2m + 15 ; (3) X = «; (4) ;r - X = 2^ = I (y - JT). . . . (5). (6) r- iff = r + y + 10; (7) W + X+r+Z+w + x+y + z = 652; (8) a' + X = 60. By (4) a; - X = 2Z; by (8) .r + X = 60. .-. x = SO -tr Z; and X = 30 - Z. But X = z, .-. Z + z = 30. .-. TT + r + 2« + y + 90 = 652, .-. by (6) W+ to -{- r - w + 80 = 652, WEEKLY PROBLEM PAPERS. 147 .-. 2^=572. .-. r=286 From (1), (2), (8) r + 2a> = 632, .-. 2w = 346. .-. w = 173 From (2) 286 + ^ + r + ^ = 351 r= 35 .: z+ r+Z^e5. y = 68 And ^ + 2 = 30 .-. r = 35 ; and .-. by (6) ^ = 68. ^ = 11 From (5) 3^ = ^ - r = 33 .-. Z = 11, and Z + z = 30. .-. ^ = 19 X = 2. .-. X = 19 X + a; = 60. .: X = 41 2. Tripos 1878. 2nd Monday afternoon. No. 1. 3. tan2 ? = (^ - "X^ - i) , 2 «(« -- c) .: tan« ^ tan2 2 = (^ - «)(^ - g) . (^ - a)(s - i) _ {s - «)" 2 2 «(s - i) s(s - c) ~ «2 ' . . the given expression = '- — !— i — „ ^ — . Now this fraction is < 1 if («-fl)2 + (s-J)2 + (>_c)2V + y2+5!,?;^cosC, •'• y^ + •^'^ + 2y2 cos^ + ■s;^ + *^ + 2^a; cos B — x^ — y^ — 2ay cos <7 = - 2 V(/ + 02 + 2_y2 cos ^)(02 + a;2 + 2^:? cos B). Divide by 2, square both sides and transpose. .•. yV(l — cos^^) + 2;r5'22(cos B + cos ^ cos G), + a;^y^(l - cos^jS) + 2.ry«2(cos (7 + cos ^ cos B), + xH\ 1 — cos2(7) + Ixhjzicos A + cos B cosC) = 0. .•. yV sin^^ + z'x^ sm^B + x^i/^ sin^C + 2.r2y0 siri 5 sin C + 2.ry% sin (7sin ^ + 2xgz'^ sin ^ sin B = 0. ,: {yz sin ^ + ^a; sin 5 + xy sin C)^ = 0. 3. Let ABCB be the quadrilateral. Produce AB, DC to meet in F, and Z).4, CB to meet in ^. Let AB = a, BC = b, CD = c, DA = a. Then ^M^+M = cos i) + ^ cos G sin G sm C e^+d^-a^-b^ (w + ii? J2 + ^2 _ 2 + (^2 _ ^2^2 (c2 _ „2J(^2 _ l,2)!^ab + C(?) J =(«S+(;(?){c2(i2 - (?2)2+z (w + 1) 2(2» + 1)(2k + 3)' (3) 5 = - 3 + 2 + 13 + 28 + 39 + 26 - 55 - 296 . . ;Si = 5 + 11 + 15 + 11 - 13 - 81 - 241 . . S^^ 6 + 4 - 4 - 24 - 68 - 160 . . Ss= - 2 - 8 - 20 - 44 - 92 . . -S". - 6 - 12 - 24 - 48 . Here the 4th difference series is a geometric progression, common ratio 2, .•. the »ti» term is of the 3rd degree in n, and is of the form A-\- Bn+ Cn {w + 1) + J»2 (;? + 1) (» + 2) + E. 2»-i. Giving to n in suocessiori the values 1, 2, 3, 4, 5 we obtain 5 equations to determine A, B, 0, D, E. From these we find A = 0, B = ~- 1, C=0, Z) = f , J' = - 6, .-. the «tt term is |« (« + 1) (« + 2) - « - 6 . 2»'-i, .". the sum of « terms = f.i.K(»+l)(» + 2)(« + 3) - J«(«+l) - 6(2«- 1). 2. a^ + P + o^ + 3a6o = {a + b + e) (a^ + l^ + c"^ - be- ea- ai) + Caic > 2c {(« + 5)2 4- c2 - Sab -c{a + b)} + Gabc > 2c {{a + 5)2 + c3 _ c (a + b)} > 2c {2o (a + b) -o{a± b)] for a:^ +f > Sry >2c2.(a + «).. , , 154 SOLUTIONS OF 3. Let ABChe the first triangle, D, E, Fthe points of contact of the inscribed circle, A'B'C the second triangle. Then «' = AS = s — a, b' = s - b, c' = s — e. S' _, «W 4;S' Now />= :r; P =-7i5r. 4. sin a; = «cos (x + a), • • 2i I ^ 2 ^ > .: 6^» (1 — nie") = c-^ (1 + me-<'f]. f2xi l+. 1 — nici ' 2xi = log (1 -{- nU-'"') — log (1 - Mze»«) = «!£-<•* — J . B^ . i^ . e- 2ai ^ ^ _ ^3 j3 _ J— 3ai — . . , = i hi (f<»* + e-"*) - ^rfi (e3»» + e-3««) + . . .} - Il k2 (j2ai _ e-2at) - i «4 (c4i« _ e-4a«) + . . A = 2j ■{» cos a — ^ifi. cos 3a + . . .| - 2J {l »2 sin 2a - J ffl* sin 4a + . . .], .', X = n cos a — ^ »' cos 3a + . . . — |j »^ sin 2a - J K* sin 4a + . . .1. Let -^ "" 8' " ^ 4' " = ^' "8 V2 V 3 5 7 "^ ■ • 7 Vl 6 ^" 10 ~ 14 "^ ■ ■ V V2I ^a 5 7 ^- V 2-4' WEEKLY PROBLEM PAPERS. 155 2^/2 '367 5. Let the tangents at B and C meet in D. Join Z)4, and produce it to meet BC in M. Then since D is the pole of BC, and ^ the pole of I'T, .: P is the pole of AD. .•. BPCEis a harmonic range, and .". also TPQA is a harmonic range. 6. Let E and K be the middle points of AB and BC, ■ and F the position of the peg on which AB rests. Let B denote the reaction at B. Let the direction of B meet the verticals through B and S in G and fi". Then from the equilibrium of AB, B: W-.-.am EGF : sin BGF :: sin o : cos (a + /3). From the equilibrium of BC, B: W:: sinKHC: BinBHC:: cos/3 : sin 2/3 :: 1 : 2sin^, .". sin a : cos (a + /3) :: 1 : 2 sin /3, .'. 2 sin a sin /3 = cos (a + /3) = cos a cos /3 — sin a sin /3, .•. 3 tana tan/3 = 1. 7. The given equation can be written \ — sm - cos •- - 7- ( cos2 -. - sm^ - 1 \ \ a 2 2e\2 2/ J + 2 j^-(cos^| -sin^|) + |sin|cos|| (cos^|+sin^|) - 2 (sin2 1 + cos2 |Y = 0. Divide by cos* -, and for tan - write 2, ., j 2^2 -J(l-22)j' + 2f(l-2<)+4| (2 + ^3) -2(1+22)2 = 0, ■••'•S-?-)+?C-+')"+Kf-^^)'- 156 SOLUTIONS OF To find the envelope we must express tlie condition that the above expression may be a perfect square. Now {f + (?0 + e)2 = 2* + 2(fe + (2e + d^) z^ + leiz + eK .: absolute term = sq. of coef. of z -7- sq. of coef. of z^, I? + ?f _ 2 ( ^ f 1 - ^1 •■■(' + ;TS + |-)-('-jJ(i-?-) «2 + i2 - ^' Paper LXIII. 1. Let 30 Va + 37 = ( ^1^+ xf ; .-. 30 Vs" - 37 = ( \(7 - ir)^, .•. multiplying, 1331 = {y — a;^)', .•. 11 = y — x'', and 30 v^S + 37 = ifi + 3.ry + (S.i^ + ^) a/^ .-. 37 = a:3 + Zx^ = a;' + 3.r (.r^ + 11), .'. 4a;' + 33.1; — 37 = 0. x = 1 evidently satisfies this equation. .-. {x - 1) (ix^ + ix + 37) = 0, ,-. x=l, y = 12, .-. v^SO V3 + 37 = 2 Vs + 1. Let a + v/i~+ v^« + \/5 = (a; + Vy"+ Vi)3 = .^3 + 3a; (y + z) + Vy (3a;2 + y + 3z) + \/'7(3.«2 + z + 3y) + 6.t x'^. Put « = a'3 + 3.r (y + z) ; v'r= A//(3a;3 + y + 3?) ; V7= VJ(3a;2 + z + 3i/); 'JJ = 6x slyz. On substituting, we see at once that a/ -^, \/ - ,1b -d' V ^' V - rational. WEEKLY PROBLEM PAPERS. 157 Now h = y ^^ = ^^ (3^2 + y + 3z) (3..2 + ^ + 3_5,), ,-. 6a;/i = 9»4 + 12*2 (y + z) + 3 (y + 0)2 + ikyz .-. 54;<;% = 45ifi' + 36«3 4. 3 (^2 _ 2«,«3 + ^e) + ^, or 48if2 + 6 (5« - 9y5:) ^ + Sa^ + Biny smz tan - (cosy + cos z) " £1 — sin X tan - (cosy + cos z) + sin'y cos z + cosy sin^ z siny cosy sin z + siny sin z cos z _ — (1 — cos a;) (cosy + cosz) + (1 — cos^y) cosz4- (1 — cos^z)coBy siny sinz (cosy + cosz) _ (cosa: — cosy cosz) (cosy + cosz) cosa; — cosy cosz sin y sin z (cos y + cos z) siny sin z ' •nil • ii. ■ _f cosar — cosy cosz ^ , , /. if xy'i! are in the given proportion, -. ^ = B, and by sm y sinz adding numerator and denominator, each is evidently = A. 3. l-i + J-i + ---= log. (1 + 1) = logs 2 = -6931471 . . . 4. The points A, B are supposed to he fixed. Draw the tangent BH. Then the angle BFG = EOD = EAB, and .-. the triangles FBO, EBA are similar. .-. BG : BF :: BE : BA. ,: BO.BA = BH\ .: ff is a fixed point. 5. Let P he the point on the ellipse such that FP is a minimum. Take a point P' on the curve indefinitely near to P. Then PP = FP'. ,: the angle FPP' = PFP, and each ultimately = 1 right angle. But PP" is the direction of the tangent at P. .: FP is the normal. .-. SF : SP :: SA : AX, .: SP = 1. SF e 6. Suppose the weight of the triangle to he represented by « + J + e. Then \ of this is supported at each angular point, and at these points we have remaining the pressures represented by a, b, c. Divide AB WEEKLY PROBLEM PAPERS. 159 at D so that AB : BB :: b : a :: AC : CB. Then the point where the weight is placed must be in CB, which bisects the angle C. ,: the required point is the centre of the inscribed circle. 7. Let PSP' be the focal chord, QQ the parallel diameter. Then qqi = PP' . AA'. Let j5, f', g, g' be points on the auxiliary circle corresponding to P, P', Q, Q. Then q^ is evidently a diameter of the circle. .-. pp' : qq' :: PF : qq •.: Qq : A A'. But q^ = AA'. •■■ qq: = PP'- Paper LXIV. 1. Let «2 - .1 = «2 - 1 = (,.2 - i = k, .: a^ = 1;+ 1, &c. X y z X and (1) aV + h^f + eV = 0, (2) aV + iy + cV = 0, .-. (3) «««3 + iy + c*«3 = aV . a2 + jy . js + ^a^s . ^8 = flV (a + -) + «y (^ + -) + cV^^ + 1^ = /t (»V + jy + c223) + aV + «y + eV = 0. . from (2) and (3) A3 _ gy _ cV "^T^ ~ c2 - a2 ~ a2 _ g2» jy /53«3 " « (^2 - <;2) - 4 (c2 - «2) c («2 _ ^2)' , substituting in (1) we have a* {b^ - c^f + b^ (c3 - «2)* + c* (fl2 - ^2)* = 0. 2. Tlie required sum = - + - + -^ + . . . + 32 + a3 + 34 + • • • 160 SOLUTIONS OF = 1 22 + .--r- • ■ = ^.^■■. 1 2.3 + 3.4 + - < • • = 1. 3. If a = 50, b = 41, e = 21, then s = 56, s — a — 6, s — b = 15, s — c = 35. Let .r be the length to be added to b, y the length to be added to c. Then x -\- y = — 1. Kow S^ = s{s - d){s - b){g - c) = s{s - cf) {s - V) (s - c'). .-. 56 . 6 . 15 . 35 = 56 . 5 (15 - x) (35 - y), .: xy — S5x — 15y = 105; and y = — (ar + 1), .-. x^- + 21a; + 90 = 0, .-. ar = - 15 or - 6, y = 14 or 5. .•. the sides are either 61, 26, 35 ; or 51, 35, 26. 4. 7. cos^ = -3-— (a i'' + a /) . ^-^ = -4t^ «^ r + 0-) {V + c^ - «2, 2a''5V = — (a -j- a b 0^ — a b -t a ¥c + a e — a c). 2a' b a T3y symmetric changes of the letters we can write down the other two terms of the given expression. .•. by addition we iind it = — L_, (2«2jy + 2ahh^ + 2«Vtf2) * j_ A* _L * = a +6 + « . ■WEEKLY PROBLEM PAPERS. 161 5. By tte Theory of Maximum and Minimum if we suppose MN to remain fixed we know thatiVX and iilif make equal angles ynthBC, and similarly for each of the other pairs of sides of the triangle LMN. .: when LMW has a min. perimeter, any two sides make equal angles with the corresponding side of ABO. Now we know that this is the case with the triangle whose vertices are the feet of the perpendiculars. .■. this triangle gives the required position for LMN. G. Let AB represent the boy, OB the chair, and let ABC — 6, AB = I. Let R, Bl denote the reactions at A and B. Then resolving vertically iJ = W. I ' W - Taking moments about A, B' . I cob 8 = T^- ^ • ^'^ ^> -'• B' = — tan9. ^ 2 The boy will begin to slide when the friction at ^ is a max. and = liB = t>.W. W Eesolving horizontally, we have the friction = R'. = _ tan£), W .: nW= -^ tan 5, .-. tan 5 = 2/i, Ji ,'. the boy will begin to slide when 6 = tan— i 2ai. The chair will begin to slide when R' = fiW, or tan 6 = '2,(1. —. The greatest possible value of 6 which is consistent with equilibrium is the smaller of these two. 7. Let the asymptotes meet the directrix in B and D', and draw SF parallel to CD' meeting the directrix in P. Let SB be the semi-latus rectum. Then from the similar triangles SFX, CDX, SF : SX :: CD : CX :: CA : CX :: SA -.AX:: SB: SX. .: SF = SR. Let SF meet the curve in P, and draw the ordinate FN, and PX perpendicular to the directrix. Then SP : PK:: SA : AX:: SR : SX :: SF : SX :: SP : SN, .'. SN = PK = NX. .: F is the middle point of SF. IX 162 SOLUTIONS OF Paper LXV. 1. (1) ^ = - 1 - 3 + 3 + 23 + 63 + 129 -[- . . . S,= - 2 + 6 + 20 + 40 + 66 + . . . S,= 8 + 14 + 20 + 26 + . . . ^3 = 6 + 6 + 6 + . . . Here the 3rd difference series gives us a series of equal terms, .-. the »* term is of tlie 3rd degree in ti, and is of the form J + M+ C>i{n + l.) + Dn{n+ 1) (» + 2). .•. putting n in succession = 1, 2, 3, 4 we have four equations to determine A, H, G, 1), from which we find ^ = + 3, 5 = 0, C = - 5, 5 = 1, .". the tfi^ term is n (» + 1) (» + 2) - 5)1 {a + 1) + 3. .•. the sum of » terms = 1 » (» -h 1) (n + 2) (;j + 3) - I- » (« + 1) (» + 2) + Ss. In — 3 (2) The «th term of the series = ~ —-^ — , .,, ,, — i—^, ^ ' (5ii — 2) {5n + 3) {On + 8) _ |-(5«-2)-:^ ^ I i (ojz - 2) {bii+3) (5>i+6) (6«+3) {5n+8) (5?2-2) (5»+3)(5»+8)' .■. the sum of » terms, [see LXIL, 1, (2)] o ■ 6 ■ £« + 8 "^ 5 ■ 6 . 2 ■ (5« + 3) (5/j + 8)' Now when a = 0, the sum = 0, 7 1 41 25.8 60.3.8 1200' the sum of n terms 50 ( 24 5« + 8 "*" (5» + 3) (5» + 8) J WEEKLY PROBLEM PAPERS. 163 (3) The given series = Si + S^ , „ 1.23 , 2.24 3.26 , .«.2«+2 where^,= _ + _H-_+...+ ^^^-,- j c 1 I 2 , 3 , , K ■ Now ^, ^ (-^ -3^) ^^ + ^fl^ + ...+ <- + f^-4 • ^""^ 23 2* , 25 2"+2 ^ ,T, + i-Q + n "f" • ■ ■ I 2* 2^ 2''+2 2»+3 ~ [3 ~ ji ~ ■ • ■ ~ \n+l ~ |^ + 2 23 2«+3 2-1,3-1,4-1, ,;«H H [o- H T— + . . • + !?_ lL il |» + 1 = 1+1+1 +...+1 LL i^ !_? & _ 1 _ 1 _ _ 1 _ 1 |2 ii_ ' ' ' i» |« + 1 1 = 1 - |« + 1 .■. the sum of a terms of the given series ^ g _ 2'«+3 1 : + 2 1?? + 1 (l_e^)™ =r^+|+| +...\'^(_i)'. (^) I ^ Equate the right-hand members of (//) and (5), and multiply both sides by e«. M 2 164! SOLUTIONS OF ,-. e' - K62I + 11 e3x = (- irf+g_+. ..}"(! + ^ + •), equating coeffioients of a:" we have 1 2» »(«- 1) .3" = (- 1)' :. P - -«.2'^ + ^ D.S"-. , . . = (- ])" -\i- 3. Let C = sin" (j) cos kS + % sin«^^ i^ cos (« - 1)5 sin (6 - J. at right angles to the axis of a; is ar+ycos5= l-fl + I) (1) and the line through j 0, - f- + - j !■ at right angles to the axis of y is , + ,,0.6=1(1 + 1) =.,... (2) and these two lines evidently pass through the centre of the circle. Multiply (1) by x, and (2) by y, and subtract, and we have which is the equation to the rectangular hyperbola passing through the 9 points, and which .•. also passes through the centre of the circle. 6. Let one bullet be fired from A in direction of another which is being Igt fall from B, and let the first strike the vertical through B after a time t. It will then be at a distance J gfi below B. But this is the distance through which the second bullet has fallen in the same time. .•. the two will meet. If they coalesce, the horizontal momentum is unaltered, and since the mass is doubled, the horizontal velocity is reduced to one-half its former value. Now (horizontal velocity)'' = 2y . — '— •'• /. the latus rectum of the joint path is \ that of the former. 7. d = aeos^e + bsaflB + 2hAnecose, U =^ a sin2 6 + b cos? Q - 1h sin 6 cosd, .-. d - b' = {a - h) cos 26 + ih sin 25 = U sin 25 (l + ^^ cot 16] *■ 2A = 1h sin 16 {l + cot? 25} = 2A cosec 25, 166 SOLUTIONS OP and since 26 is < tt, cosec 25 is essentially positive, .*. the sign of a' — b' is the same as the sign of h. 13a;2 + 2xg + 13/ - 22ar + 50y - 23 = 0. The coordinates of the centre are given by 26a; + 2y = 22 ; 2;!: + 26y = - 50. .: s = 1, ^f = - 2. /. transferring to (1, — 2) the equation becomes 13:i;2 + 2xy + 13/ = 84. To determine the lengths of the axes we have a + j3 = 26, afl = 169 - 1 = 168. .-. a = 14, = 12, .-. U:,^ + 12/ = 84, .-. ^' + f' = 1, o 7 which is the equation to an ellipse of which the semi-axes are ^G, ^7. Papee LXVI. 1. The equations may be written I (x: - z) + ay + alxy == (1) m (x - z) + iy -\- bmxi/ = (2) » (a: - z) + cy + c« ay = (3) From (2) and (3) we obtain be {m — n) mn{c - b) In - en ' ' ' ' ' If (1) is also true, we have by substitution in (1) from (4) = Icl {m - «) + «»» («-*) + <*l {bn — cm) ^be(i-l) + Uo-i) + 4'~-.^) \it m/ I ^ \m n) = {b-c)± + {e-a)i + {a-b)^. i ma From (4) I = "^^ mn (e — b) bn - cm WEEKLY PEOBLEM PAPERS. 167 .'. either y = 0, in which case also x - z = [from (4)] £ - * n_ _ cm — in n m mn [b — c) b — c and from the equation (1 + Ix) (1 + «y) = 1 + Ix we see that y = , z = — -■ are solutions, and two other sets can a I be obtained from the two other equations, viz. y =--,«=- -; y = , z = - _. m en 2 ^2 = c°Zi^ cos^ {6 - <}) ) _ a^b^ (cos g 00-^ + sin g sin 4,)^ cfi sin^ (j> -{- b'^ cos^ <^ a^ sin^ (p -\- b^ cos^ (^ 0252 (cosfl - ~„ sin5 cotd)')= ^ c'^g(cosg + sing tan 0)^ ^ a^ ^ c* ^ _ (a' cos g sin 0' — i' sin g c os 'f a^ sin^ 0' 4- 42 Qos^ 0' Also .2 = "^^'"o^'(^-'^' ) «2sin3 0' + 42 0082 0" .22 — c'^^(cosgcop0' + sing sin 0')2 + (g2cosgRin0'- i2f.;n^cog^'^2 «2 gin2 0' ^ ^2 (,Qg2 ^' _ (gg cos2 g + 42 gin2 (9) (a2 sin' 0' + 42 cos2 0') «2 Biu2,0' + ^2 gQg2 ^' = «2cos2g + 52sin2g. 3, Write the first equation «3 sin (y + i sin0) = — P cos (.r + a cos 0) . . {J) and aa; see (j> — by cosec = a^ — I?, \cos0 / \sm0 /' .'. « sin (;r — a cos 0) = ^ cos (y — 6 sin 0) . . (2?) 16S SOLUTIONS OF From (J) and (S) we have a^lt sin(j> cos (,y^ — b'^ sin^cf)) = — ab^ siiK^ cos^ (x^ — (fi cos''0, .•. d'^y^ — M^ sin- ^ = — i%2 -|- aH"^ cos^ 0, .-. ffV + *^^^ = aH"^ (sin2 + cos2 0) = a%". 4. Let 5ff meet AG in i, and let CE meet ^5 in X, and let the perpendicular from A meet BO in S". Then from the similar triangles ARC, BKE, we have AC : AK :: BE : BK, and BE = AB, .: AB : AC :: BE : AK. From ABL, CGL, AB : AC :: AL : LG. From ACE, ABC, AC : BG :: GH : AC. From ABE, ABC, BC : AB :: AB : BE. .: compounding; these four ratios we have AB^. AG'- . BG.AK. BE. GL = AB^ . AC' . BG. AL . CE. BKf .-. AK. BE. CL = AL. GE.BK, ,: the three lines AE, BL, CK are concurrent. 5. Let the tangent parallel to PQ meet TP rap and TQ in q. Then p, q are the middle points of TP and TQ^. .: Op, Oq are at right angles to TP and TQ. .'. TO is the diameter of the circle which passes through T, p, q. But this circle also passes through S. .: TSO is a right angle. 6. Let Op, Oq, Or represent the forces. Let the transversal cut rp produced in K. , OT, _ OU qp_ _ p_K + OM _ Kr - pr+ OM L^~ pK' " OL OM ~ OM ' Op , Oq^ ^,Z>- ^ ■, ,Nr ^ Or " OL'^OM ~ "^ OM ~ "^ 0N~ ON' • -^4- « = ^ " liL^ ~6m on ' 7. Let the hall A moving with velocity V strike the ball B, and let its directions before and after collision make angles , and V = F" sin (j), for the momentum of ^ perpendicular to the line of c&ntres is unaltered by the collision. '^ + J. " * .". :; tan rf> = _ = tan a, 1 — e u .'. tan = ^ {I — e) tan a. Now when the deviation is a max. the angle between the old and 2' .-. (^ - I = TT - a. .-. a = 27r - ^1 + ^\ .". tana = — tan f- -\- '6 — e a- a l\ — e 7. V-0-— ,- Paper LXVII. 1. Let a' + c be the common measure. Then a;''' + ax + b = ix + c) [{x + -^ = ar + x(c + ^ + b x" + a'x + V = (iT + c) (;i: + Q = a;2 + :r (c + ^)+ b'. ci - = a; c + - - ■ ■ "-" c c .:c+-_ = a;c + -=a'; .•._—=«-«'; .. _ = ^_.^„ , b + b' , b' ,V{a-d) ab - a'b ' and c+—^- = a+-=a+-j--j, ^—^r, aib -b') , a'b - ab' and c^ = ac - b == r - b = r- a — a a- a 170 SOLUTIONS OF , , „ , 66' , , „ , U'(a - «') a'W - alf^ .: L.C.M. = (.r + e)(x+ -)(.r + -) = .^ + .^(. + ^') + .<6 + .+ 5) + ^ _ 3 _L 2 oIj — <^h' y_ n^b^ — "i'^ bVifl — a') -'^ +^ • b-b' ^''- a'6-ab' + b-V ' 2. Let A start at the rate of a miles, and B at the rate of h miles an hour, and suppose that each increases his speed by c miles an hour. Then at the end of 1st hour A has gone a -\- — miles, and £ has gone it c 3 I A — miles. .'. the distance between them is5 — c — c= — ..(1) ^2 16 ^ ' At the end of 2nd hour A has gone « + -^ + a H — x- , or 2« + 1c. B „ 6 + | + 6 + |,or2i + 3c. .'. the distance between them is2b — 2a — io = ^ , , . . (2) From (1) and (2) c = ^, b - a = J. .•. S starts at the rate of i mile an hour quicker than A. .". if a be supposed given, since A goes at an average rate of s + J| miles an hour, the distance from P to § is 4a + ^ miles. 3. By Todh. Trig. cap. xxiv. Ex. 17 S^ = (s —a){s — 6)(i — e)(s — d) — abed cos^- where 6 is the sum of two opposite angles. .'. ^5^ is a max. when cos - is a min., i.e. when cos - = 0. .•.-- = -, .'. 6 = t, .: the max. quadrilateral with given sides is that which can be inscribed in a circle, and its area is V(« -«)(«- b){s - c)(s - d). Again, since a circle can be inscribed in the quadrilateral, .•. a + c = 3 +(?, .'. 2s = a +5 + c + (? = 2(« + c) or = 2(6 + d\ WEEKLY PROBLEM PAPERS. 171 .'. s — a = c, s— h = d, s — e = a, s — d = b. . S = \'(« - a)(s - b){s - c)(s - d) = •^ahed. Now if r be radius of inscribed circle, r{a + 5 + c + (?)= 2. area 2 iJabcd "J abed fJ abed • .»*=. = . , or = » a -\- b -\- e -'f d a -\- e b •\- d . o abed {a-^c){b-\-d) Tlie following is a geometrical proof of the theorem that the maximum quadrilateral which can be formed from four given straight lines is that about which a circle can be described. Let ABCD be the position of max. area. Take ABC If a consecutive position keeping AB iixed. Let AD, BC meet in 0. Then since AD = Ajy, and the angle DAly is small, .•. the angle ADD! is ultimately a right angle. ,*. the angle ODBf is a right angle. And since the angle DOjy is small, .•. the angle OlfD is a right angle. .-. OD = OH; similarly OG = OC". And the base CD = base CD, ,: the angle DOO = D'OC, and .'. the triangle OCD = OC'D'. .: the angle DOD = COC, and the triangles DOD^, COG' are similar. .-. CC -.DD':: 00: OD. Now since the triangle OCD = OC'D", and the area. ABCD = ABC'Df, .: the triangle OAB = area ODABC'. Take away the common part OABG', .-. the remainder OAD = OBG', .-. OA. DD = 05 . CG', OA CC DC " OB ~ DD ~ OD .: OA. OD = OB. OG, .'. a circle can be described round ABCD. 4. The tops will describe circles the centres of which are at the roots. It is evident that after collision the line joining the tops will pass through the centre of one of these circles, which it could not do before. 172 SOLUTIONS OF 5. For the first part of this question see Besant's Conies, cap. xi. Prop. 1. Let A, S be the vertex and focus of the given parabola. Describe a triangle B'G'II having its sides parallel to the three given straight lines. Then describe a parabola passing through the points B\ G', If, and having its axis parallel to AH. Let A, S' be the vertex and focus of this parabola. Join S'B'. From 8 draw SB making the angle BSA = B'S'A!, meeting the given parabola in B. Draw BG, BD parallel to B'G', B'lf, and join CO. Then by symmetry CD is parallel to C'lf, and the triangle BCD is described as required. 6. The only forces acting on the rod .42) are the reactions at ^ and 7), and when there is equilibrium these must act along AD. Let the force at A along AT) be denoted by R. Then the rod AB is acted on by B and B, at A, and the forces at B. .', taking moments about B, we have V . sin BAG = iJ . sin BAD. Similar from GD we get Q . sin BDG = 22 . sin CDA . P Bm BDG Bin ABD Bin BAD ein A GD Q sin ACD sin BAG sin ABD sin ODA _0G OA _ BD AD " GD ' OB' ad' AG' P .AC _ AG . DC " q.BD ~ BO . ud' 7. Tripos 1878. Wednesday morning. No. 13. Paper LXVIIL 1. We see that the given expression is symmetrical, and will not be altered if we put x for a, and a for x, and similarly for the other letters. Consider each term of the first part {a^ -\- ¥ + fi')xyz. Since the given expression is symmetrical, the factors of cfixyz must be ax . ay . az. Similarly the factors of Ifixyz are hx . by . bz, and for (?xyz are ex .cy . cz. Now from the symmetry of the expression we cannot have ax and bx in the same factor. .-. the required expression is {ax + by + ez){ay -{- bz -\- ex){az ■\-bx-\- ey). 2. By Fermat („^)m - 1 _ 1 = Mim) ; {pm)^ - l _ ] = M{a) ; {mn)p - i - 1 = M{p) WEEKLY PROBLEM PAPERS. 173 •■• [M""-! - l}{(;)m)" - 1 - 1 1 .[(»««);>- 1 - 1] = M{mnp) — (jt'n—l pm+n-Z . (jjW-l -|- mP-1 . ;2m+p-2 , jjm— 1 + (k^)"'-! + (/;w)i-l + (8»«)2'-l — 1 = M{miip). Now the first part of the expression = 3f{mr>p) by inspection .■. (kp)""-' + (pm)"-! + {mn)p-'L — 1 == M(,mnp). 3. Let ^5(7 be the given triangle, i), B, F the middle points of BC, CA, AB. Then if P be the centre of the circle touching the given circles, the points of contact are evidently on PD, PH, PF, produced. .-. B = 2PB +a = 2PE + b = IPC + e. .: 2B = 2{PE + PF) + b + c. .: PE + PF = B - 1 (2« - a) = i) - s + ^ . ,: PE+PF+FF= B - s + a, .•; if StTi be perimeter of PFF, 2 2 2 substituting in (1) and dividing both sides by — a s — b s — e "2~ ■ 2 2 V -T 174 SOLUTIONS OF we have ^ s — a ^ s - 6 ^ s - c ^ D — s 4. (1) The three lines are evidently diameters of tho nine-point circle of the tri;ingle A£0. (2) Let d, e,/be the middle points of JD, BE, CF. Then 2f«' = BG, ieF' = AE, %jE' = AF, .: for the triangle B'E'F' we have dE' . eF' . fB' = fE' . dF' . eU. .: Nd, E'e, F'/meet in a point K. For further information respecting IC, see an article by Mr. Tucker in the Quarteili/ Journal of Bute and Applied Mathematics, Vol. xx., No. 78. It can also be proved that K is the point determined in XXXIX. No. 5. 5. Let BY meet the axis in T, and draw the ordinates BN, QM. Tlien by Bes. Con. Cap. II. Prop xxi., AN . AM = A&\ From similar triangles AQM, BIN, QM'' : AM^ :: NT^ : BN\ .'. 4AS.AM : AM' :: AAN^ : iJS. AN, .-. AAS : AM :: AN : AS, .: 4AS'' = AM. AN = AB!'. .: AB = 2AS. 6. Let B, E,Fh6 the middle points of BC, CJ, AB, and let he the centre of gravity of ABC. Then the forces represented by BB and BC liave a resultant represented by 2BB ; and forces represented by 2BB and BA have a resultant represented by SBO, since AO = 20B. .: if the resultant is constant, the locus of P is a circle whose centre is 0. If there is equilibrium the resultant vanishes, and P coincides with Draw PJf perpendicular to AB. from ^ to jB is — Ifftf = uslna.t - ^fft- = Pilf. The time of flight from ^ to 5 is . .*. f = . ff 9 WEEKLY PROBLEM PAPERS. 175 Paper LXIX, 1. Let a • denote the number. Then a; 120 X z ~ a 140 - 140' I z 126 ~ a - iZJ ' a -12( 140(a - 126) X 120 ■ ) 126(a - 140) ^^ u - 120 .-. (a - 126)( - 1). .-. sec2 ^ sec2 d, = - 2 "° (^ + <^) cos (g + <|.) ^ C0s2 5 C0S2 (p .: sin 2((9 + 0) = - 1, .-. 5 +-^ = niv - ^• - 6ec2 6 = V2'{tan 6 - tan (| + ^) '■/oJi n 1+tanfl) ,- 1 + = V2itan5-^^-^^|.= - V2.,-^ tan2 5 tan 6 ■■• ;- (tan'^ - 1) = ^2"; .: X = bia-nd = b + a ^2. .'. y = a tan (j) = — a tan ( -r + ^ ') = « — = a + 5 ,^2 \i / tan o — 1 An algebraical solution will.be found at the end of Todh. Alff. (3) For solution see end of volume. 2. The number of ways is evidently, the.number of- combinations of 3m things taken 2m at a time, when there are m alike of one set, m alike of another set, and m alike of a third set. ■ 182 SOLUTIONS OF .•. Todh. Alg. Art. 811, the required number = coefiScient of ar^m in the expansion of 1 - X'n+l (1 - ^)m+l ^ (1 - g)in+l 1 — X 1— .r \ — X = coefficient of x^ in ^-— J- • (1 - xf Now this expression = (l_:r™+l)3{l+3.r+ . . . + l!!i + iK^^lll) :.» + li and coefficient of x™ is evidently i{m + l)(ra + 2). „ J, . A+ B . - D 3. cos B sm ! sin ^ — 2 2 ,. C-Df.J + iB..A-I}) = i ™ -^ |sm ^ + sin^- I J + iB - 0+D A + SB+ C+ I) cos ! ;r ' cos '- J ' A - B - C+ D A-B+C-m + C0S 2"-^-- '=°' i r By symmetric changes of the letters B, C, D we can write down the other two terms of the given expression, which on addition we find to be , <[ A + ^B-C+D A-B + 3C+D / A-B + C + 3D A + SB + C - BW = -isin ^ + -^+^ + -^ |sin(^-C) + sin(g-J) + sin(2)-.g)l „ . A+B+C+J = 2 sm — ! ^ ! — Todh. Triff. Cap. viii. Ex. 3. „ . A + B+C+B . B-C. G-B ."B-B -2sm ^ sm -^- sm —^ sm — ^ . WEEKLY PROBLEM PAPERS. 183 4. Add 4 + 5 tan^ 6 to each side of the given eqaation, .-. 12 tan2 fl + 8 \'3tan fl + 4 = 5(1 + tan^ 6) = 5 seo^ 6, .: 3 tan 25 + 2 V3 tan 5 + 1 = |- sec^e, •■• "fata.nd + 1 = ± ^ seoft 2 ••• ^sinS + JcosO = ±^; .". sin('5 + |)= ±- 2 4 \ o / From the given condition we must take the positive sign. .-. i sin ((9 + !;:)= 10 + log ^ = 10 + i log 5 - log 4 = 10 + 4 log \a - 2 log 2 = 10 + J - I- log 2 = 9 . 7474250. Dif. of 60" in angle gives dif. in log of -0001874, .-. 1874 : 507 :: 60" : 16" . 23 .: e + % = 33° . 59' . 16" . 23 6 .-. 5 = 3° . 59' . 16" . 2 nearly. 5. Let 'ABCB be the first tetrahedron. Let E, F be the midfile points of JBD, CD, and P, Q the centres of gravity of the faces ABD, ACB. Then PQ : EI:: AP : AE :: 2 : 3, .-. PQ = ^EF = i EC. .: volume of 1st tetrahedron : volume of 2nd :: EC^ : PQ^ :: 27 : 1. Let a he a side of the first cube, A', B' the middle points of two adjacent sides. Then^'^'^ = (|)Vgy=jV Now let be a vertex of the ootohedron, corresponding to A'B^, and let P, ^ be 2 corners of the 2nd cube. Then Pq : E'F' :: OP : OE :: 2 : 3, .-. PQ = ^E'F' = iB'C. 184 SOLUTIONS OF And B'G'^ = 2F1P = ?|! = «=; .-. P'q = ? . .•. volume of 1st cube : volume of 2nd :: a? : P'Q^ :: 27 : 1. Now volume of 1st cube = volume of 1st tetrahedron, .-. volume of 2nd „ = „ 2nd ,, 6. Project tbe conic into a circle, and denote corresponding points in the circle by small letters. Then oe, od are at right angles, as are also oc and de. .: eh . bi = Ic? = const. .". also ES . BD = const. .-. if the circle round EOD meet 00 in F, OB . BF = EB . BB = const. .•. Ji" is a fixed point. * 7. Let .E be the point of projection. Since the particles have equal velocities, the parabolas described for different directions of projection have all a common directrix XZ"', such that if EM be, the perpen- dicular upon it from E, the velocity of projection is equal to that due to a height EM. Let ET be any direction of projection ; S, A the focus and vertex of the parabola described. Then if 8Y be perpendicular on the tangent ET, SY passes through M, and ST = YM. Now since E and M are fixed points, the locus of Y is the circle upon EM as diameter ; and since AL = iLY, the locus of A is an ellipse having EM for its minor axis, and its axes in the ratio of 2 : 1. In the second case, let a horizontal through E the point of projection be drawn at right angles to the wall, meeting it in the point H. Let jP be the point where one of the particles strikes the wall, A' the vertex of its path after impact, and A the vertex of the path it would have described if there had iDeen no wall. Draw AF, AF' perpendicular to ES. Through F draw a line parallel to EH, meeting AF, A'F' in N, N'. Join AA' meeting GH in M. Then since the vertical motion is unaltered, AF = A'F'. .: AA is perpendicular to &S, and the time from P to A = the time from P to A' ; and since the horizontal velocity is diminished in the ratio e : 1. .•. PN' = ePN, or A'M = e . AM. Now since AA' is perpendicular to GH, and A'M = eAM, and the locus of A is an ellipse by the first case, .■. the locus of A' is also an ellipse. Papeb Lxxn. 1. (1) If 1, w, iifl denote the cube roots of unity, we know that a3 + £3 ^ ^3 _ ^ahc = (a + A + c)( a + K>5 + j(?c){a + w'-b + wc), WEEKLY PEOBLEM PAPEES. 185 .-. the given expression is the product of three factors X, T, Z, where X = /!;2 + / + ^2 4- 2y^ + Izx + 2x1/ = {x + y + zf, r = -x^ + %jz + w{f + 20.r) + w2(^2 + 'i.xy) = (x + w^y + wzf, Z = x^ + 2i/z + v? ly + Izx) + w{f + 2.ry) = {x -\- vnj + «p2^)2^ • .•. the given expression = X . Y . Z = [('^ "l" y + ^)C^ + ^^y + W2)(a; -{■ wy -\- lo^z^^^ — {x^ -^ y^ -\- z^ — %xyzf. (2) bo — a^ = — - a'. .•. if a, i, c be the roots of the equation , x^ — pj:^ ■}- q.v - r = 0, .-. q = be + ca + ai, r — abe, . and if we put y = — — « » X then sty = o^c — x' = r — a;' = ja; — /j^^, . . px -{■ y = q, .: X — ^ •') P or y^ — y^i3q — p^) + y . q{3q — p^) + p\ = 0. The roots of this equation are evidently be — a^, ea — b^, ab — e% and sum of the roots = 3q — p^, and sum of product of roots 2 at a time = q{3q — p^) = g' (sum of roots), .-. (ca - V){ab - = luTT ± ll - (4a - 1).T I .-. eitlier 8«.t = (4;z -|- 1) ^ ; or 2a.- = (4n - 1)^ . 3. Write the given equations in the form (a; cos a + y sin a) cos <^ — (a; sin a — ^ cos a) sin = » sin 20. (1) — (a; sin a — y cos o) cos — (a; cos a + ^ sin a) sin = 2a cos 20. (2) Multiply (1) by cos 0, and (2) by sin 0. and subtract. .*. X cos a + y sin a = «(sin 20 cos — 2 cos 20 sin 0) = a sin (1 - cos 20) = 2fl sin' 0. So a; sin a — y cos a = 2a cos' 0, s a jl .•. (a; cos a + y sin a) + (:;; sin a — y cos a) = (2a)''. 4. Let (h, k) be coordihates of A. Then since the equation to CM is y = - a;, the equation to AM is ^(y - k) + a{x - h) =Q, WEEKLY PROBLEM PAPERS. 187 coordinates of M are ~—„ (ah + Lk), „ „ (ah + Ik), a^ -f- b^ a^ -f- P Writing — a for a, the coordinates of N are {bk — ah), ,^ (flJc — ah a^ + b^^ " a2 + b coordinates of 0, the middle point of MN are ^2 ^ ^2' ^2 + i2' .". equation to AO is <'-«(.-n^--') = <-'>(?f-^-') or bVi(ff - k) = a%{x - A), which is evidently perpendicular to the polar of A whose equation is «2 ■*■ *2 ~ ■ ■* This may be also proved geometrically as follows. Join CA cutting the ellipse in P, and draw FM', FN' perpendicular to the equiconjugate diameters. Let the normal at P meet M'N' in L', and draw AL parallel to FL', and .*, perpendicular to the polar of A which is parallel to the tangent at P. Let AL meet MN in L. Then since circles will go round PN'GM' and ANCM, .\ FN'M' = FCM' = ACM = ANM. .: MNis parallel to M'N'. .: the triangles AMN, FM'N' are similar, and AL is parallel to PL', .: ML : M'L' :: AL : FL' :: LN : L'N'. But by XXVL No. 7, M'L' = L'N', .: ML = LN. .: AL is the diagonal of the parallelogram which has AM and AN for adjacent sides. 6. D' is the point of contact of the escribed circle opposite A. Since A is the centre of similitude of the inscribed and this escribed circle, being the intersection of common tangents, .•. by similar triangles AF : AV :: radius of inscribed circle : radius of escribed circle :: ATi : AE', where E' is point of contact of latter circle, .-. AP : FV :: AE : EW, .: AE . FV = AF . EE' = AP . EG. 188 SOLUTIONS OF 6. Let P and Q denote the positions of the partioles. Let It denote the resistance of the plane at P, and T the tension of the string. Then the particle P is acted on by three forces, viz. its weight, the tension of the string, and the resistance of the plane, and when P is on the point of moving, these three forces are in one plane. Let PD be the trace of P on the plane, PE the normal to the plane, and PF the line of greatest slope. Then PP makes the angle of friction (Xj) with PS. Let PPP = , and let 2fl be the angle between the directions of the string. Then resolving in the plane (1) downwards, (2) horizontally, and (3) resolving vertically, we have (1) W-y sin a = y cos 5 + jB sin Xi cos (p. (2) TainB = i2sinXisin^. (3) Wi cos a = i2 cos Xi. Now when the angle at the pulley is a maximum, this angle is evidently = 2 right angles. 6 = -• .-. from (1), (2), (3) we have I'2 + W-?' wfia = iJ2 sin^Xi = TTi^ tan^Xi oos^'a = FiVi^ oos^a, .: T^ = W-^ (cos^a . fi^ — sin^a) = TF^ (cos^a . fi^ — sin^a) by symmetry, .". W-^ : W^ :: sin^a — fi^ cos^a-: sin^a — fii^ cos^a :: 1 — fi2^ cot^o : 1 - hj2 cot^a. 7. Consider the' motion normal to the plane. The times of the several paths are ' 2v sin a 2ev sin o 2e''-lw sin a ~T~' ~^' "g ' .■. the whole time = (1 -|- « + c^ + . . . + «"-i), g _ 2o sin a 1 — «" g ' 1 -.« Since the horizontal component remains unalt.red, and = k cosa, the space described 2» sin a 1 - e" = V cos a. . . — ' g 1 - « _ v^ sin 2a 1-1?™ (7 ' I — e WEEKLY PROBLEM PAPERS. 189 Papeb LXXIir. 1. Put a = -, b = -. c =^ ~, a b'' c" P = Q= ^' i+£ ^ ,#1 + i + . ^ a .: PO-+a^q) = q + a. 2. Let a be the greatest, c the least side. Let b = c -\- y, a = b + P, where ;3 < tf. .: a = c + ^ -{- y. .: l{a + b + a) (cfi + b^ + d^) - {cfi i- b^ + (? + Zabc) = i{e - ^) 03y + i32) + icy2 + f yS + ^^2. Now c > B, and /3, y are both positive quantities, .'. the last expression is positive. 3. Let be the centre of the inscribed circle. Then El)F=iEOF:=i (^ - f - .j) = ^ (| + ^) = i('^ + A Similarly i)5^ = i (^ + -B) ; ^ZO = i (tt + C)- 4. Let the common tangents intersect in A, and let S, S^, S^i be the points of contact in the order A, S, Si, S^. Let the circles on which are Si, S2 intersect in K Join J.E cutting the circle S in £ and and Si in D. Then since ^ is a centre of similitude, .'. JBS and DSi are parallel, as also CS, ESi, and 2>/Si, FS^. .: FB = ^,S, .41. -^C- = /SS; .4^, AB.AC^ AS\ Ab Ao (tangent)3 from E to circle S = BB ,W = SS^. SS^ 190 SOLUTIONS OF QN Q£ SQ QN Similarly if NQ meet KM in R', \ 9K =^ = ^A^^ , qs=qN. PM PK SP PM ^ ^ 6. Let y = a be the equation to the straight line, and the equation to any equilateral hyperbola be of the form The normal at any point {x'/) is , Ax' - 2/ , ,, If this normal coincides with the line y = a, we must have Axf — 2y = ; and y = a. .: A = — Since (x', a) is a point on the hyperbola, A' .'. the equation to any of the equilateral hyperbolas may be written Ahy + A\x^ - f - «2) - 4ffl2 = 0. To find the envelope we must express the condition that two of the roots of this equation are equal. Let the roots be oi, oi, 03. Then 2ai + aj = - a:^ - / - "^ (1) xy ai2 + 2aia3 = (2) a,^a, = ^ (3) xy From (2) ai(ai + 2a3) = 0, and oj 4^ 0. .-. oj = - 203, .-. from (1) 3a3 = -^^ -f- "' ^ xy .: from (3) ( _ ) = K = ^ , \xy/ 3,ry /. x'^ - y'^ — a^ = i{axy) . WEEKLY PROBLEM PAPERS. 191 7. B = mv^m.t; F=wf=^ 2m. %. .: — = F. .: 2B = Ft. t ■^ fit 2B^==^; mFs^^J!^ = 2B^. Papee LXXIV. 1. By multiplication we have ^2^2^ = abcxi/2-\-abz^y^-\-acji:h^-{-icif^z^-\-ax^^z-\-ia;yh-\-Myz^-\-x^^^z'' Since iryz = abe, this reduces to aH\^ + abxY + icfz'^ + cazV- + iax^ + bf + cz^) abe = . {A) Square the first equation. .-. «V ^ y^2 + 2«^y0 = b^c^ ; .: aV + fz^ = i^c^ - 2(^bc, .: writing {A) in the form aH^o^ + bc{a^x>' + yV) + cdfiy^ + zV) + a5(c222 + x^) = a^i^c'- + ■bc{bH^ - 2aHc) + caifd^ - 2aVc) + «i(a2^2 _ ^^^i^i-^ ^ q^ 2. The number in the scale of 7 has no digit higher than 6, and all its digits are even by the question, .•. they are 2, 4, 6, and those in the scale of 10 are 1, 2, 3. Let x, y, z be the digits in the scale of 10, then 2a:, 2y, 2z are the corresponding digits in the scale of 7. .-. a: + lOy + 10% = 23; + 7 . 2y + 72 . 2^, .-. x = 2{z- 2y). .-. X is even, and z > 2y. Now x, y, z can only be selected from the digits 1, 2, 3. .-. ^ = 2, y = 1, 2 = 3. .-. the number is 312 in scale of 10, and 624 in the scale of 7. 3. Xcos*5 - sin^5) - j(cos*^ - sin*<^) = q, - p, .: p cos2fl — q cos 2^ = q - P> ,'. p cos^ 6 = q cos^ cf). SOLUTIONS OF q = p cos* 6 - q cos*(^ = p cos* 6 — t cos* e. 2 .: cos5 ~ \/ ,^^ ,- P (? - P) q = - cos* - g cos* 0. .•. cos <}> = P \/ ^ . ^ q-p 192 Now And 4. Let the tangents at P and § meet in U. Then iJPy^ = angle in FBA = angle in qBA = i2§y^. .•. jKP = UQ,. . ■. i2 is on the radical axis, i.e. on AB. 5. Join ST and produce it to L so that IIL = ^^'. Then Bes. M. si. jlfi = jKS, and the angle TML = TMS = lOfl". .-. the angle HML = OMO', and the triangles HML, OMO' are equal in all respects. .'. 00' = EL = AA'. 6. Let #1 be the time between projection and 1st impact, i^ the time between the 1st and 2nd impacts. Then a = vcos6 . ti. .'. t, = . • Bcosfl Hor. vel. just before 1st impact = v cos 6. Vert. „ , „ „ = Bsinfl - gt^. After impact the hor. vel. = ev cos 6, .'. a = eo cos 6 . U, .: i. ev cos 9 Now the vert. vel. is not affected by the impact, and just before the 2nd impact it = vsinB — cf (i^ + t.^), and this must = since the motion is horizontal, ., ,sinfl = ^ (I + 1). .-. sin 25 = ^ . i+J. V cos 6 \ e' v' e 7. Let ABC, A'B'C be the triangular faces, (? and &' their C. of G Then by supposing the wedge to consist of triangular laminae parallel to ABC we see that its C. of G. will be at H the middle point of &&'. Now the C. of G. of 3 equal weights at A, B, C is at G ; and the C. of G. of 3 weights equal to these placed at A'B'C is at 2 2 2i = {cos^ - - sin'' _ — sin - sin - +2 sin sm - U l2 2 2V2^ 2 2// r A+B A~B A+Bf A+B. A-B A+B\\ = . ..{cos — - — cos — cos — ■ — cos — ' — +COS cos ' |> \2 2 2\22 2// = 0. 3. Let E, F, G, II, K, L be the mid(ne points of BC, CD, LB, BA, AG, AD. Then BA^ + BD^ = 2Er^ + IDP, .: AEL^ = 2EA^ ->,- 2ED'' - 4De = AB' + AC^ - 2EC^ + 2)52 ^ j)Q2 _ ^-gc^ _ ^2)2 = AB^ + AG-' -\- BB^ + DC^ - AB'^ - BCK So 4fiy2 = BD'^ + AC^ + -BC2 + AD^ - AB^ - CB^ and 4ffX2 = AB'^ + GD^ + BG^ + AD' - BD"- - AC\ .: by addition we obtain the result required. 200 SOLUTIONS OF 4. From S draw ST perpendicular to the tangent at P, and produce Srto meet the directrix in I). Let DQ, DQ' be the tangents from D. Then since D is on the directrix, QQ' passes through the focus, and is perpendicular to SD. .". the tangent atP is parallel to QQ'. .'. DP passes through the csntre. 6. Tripos 1875. Wednesday morning. No. 12. 6. Let P be the point, and PT the direction of projection, PSP' a focal chord and PN, PK horizontal and vertical lines through P. Let TPN = a, P'PN = /3, and let » be the velocity of projection, and t the time of flight from P to P*, P' being further from the directrix than P. pp, _ 2b^ sin (a - /3) oosa ff cos^ /3 and u - /3 = FPT = TPE = ^ - .-. t _ 2v sin (a - 3) cos 3 / 2PP' sin (g - B) ^ /2PF ^ ffOOSa ^ g = time taken to fall from rest through PP'. 7. 1st Geometrical Method. Let A be the vertex of the triangle, S the fixed point in the base BC. Draw SD and SE at right angles to JD and AG. Then DE is a fixed straight line. Let any circle through A and S cut AD in i and ii''. Let IN cut ^i) in 3f, and join ;Saf, &i. Then since circles will go round ALSN, and ADSE, .: the angle SlSfM = /S^i = 8EM. .: a circle will go round SMEK .: the angle SMN = /yffiV = a right angle. Now S is a. fixed point, and the point M always lies on the fixed line DE, and SMW is a right angle, .-. iiV envelopes a parabola of which S is the focus and DE the tangent at the vertex. 2nd Analytical Method. Let the equation of DC be ax -\- by = 1 and the equation of ZN be ca; -{- fy = 1 „ „ of AS he £=■!!= r. WEEKLY PROBLEM PAPERS. 201 Then the equation to the circle round ALSN is ^■2 _^ y2 _|_ 2.iy cos w — — _ "i. = 0, ••" -'(! + ?)- «■ •■■' = ^ + f=(: + i)<- + «- .•. the equation to iJVis {ax + |3y) (;/3 + ma) - rap = 0, or a^mx + |3^(^ + ai3(& + "'^ — r) = 0. .'. the equation to the envelope is (& + wy — ?)^ = ilnuy, or V & -f- V JBy = V ?•, which is the equation to a parahola touching the axes, the distances of the points of contact from the origin being - , - • / m Paper LXXVIII. 1. 220 = 2 . 2 . 5 . 11. .-. the aliquot parts of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, and their sum = 284. 284 = 2 . 2 . 71. .-. the aliquot parts of 284 are 1, 2, 4, 71, 142, and their sum = 220. 2. Put each of the given expressions = A. Then ax + ty = X (1) ; ex + dy = \^ (2) ; ex +/y = X^ (3). From (1) and (2) x{ad - be) = \{d - b\^), y{ad - be) = \{a\^ - c) .: if for X we write vA, ;;: = —, ;- s/k ad — be 202 SOLUTIONS OF To find h, we must substitute the values of x and y in (3), .-. {ai — bc)\^ = ^^{ad — be) -{-fi/{ad - be) = ek{d - b'S?) +/A(fflX2 - e). .: either X = 0, .-. .r = 0, y = 0, or {ad - bc)\*^ -{fa- be)\^ +fe - de = 0, or {ad - bc)P - {fa - be)k +/c - de = 0. . . . {A) Let a = 7, b = - U, c = 1, d = 1, e = 1, / = 9. Substituting these values in {A) we have 18F - 74A + 8 = 0, .-. /f = 4 or i .-. a; = 5 or ^\% ; y = 3 or - -j^. Again, a: - y = X (1) ; 4.t; - 5y = Xi+l (2) ; 3j! - 2y = XSn+l (3). From (1) and (2) a: = X(5 - X"), y = X(4 - X'') . . . {B) From (3) X^x+i = 3X(5 - X-*) - 2X(4 - \"), .■. either X = 0, /. a; = 0, ^ = 0, or X2» + X« - 7 = 0, .-. X" = . ~ ^ "^ -\ from {B) we obtain the values of a: and y. 3. Let A, B, C be the centres of the three circles which touch externally, r^, r^, r^ their radii, P the centre of the circle whose radius is R. Then ^ABC=BPC+CPA + APB. . . (//) .". using the formula S = v«(« — a) {s — b) (s - c), /^ABC= ^wrKT^7+^, /\BPC= •Jr^r^R {r^ + r^ + £), A CPA = ^r,nR (?-3 + ri + 5), A ^-P^ = 'Jv^Rin +>-^ + R), .■. substituting in (^) and dividing both sides by r-^r^^, we obtain the result (1). See also LXVIIL, 3. Now let the angle BPG = a, CPA = /3, APB = y, .-. a + (3 + y = 2w. .•. cos''' a + 008^/3 + cos^y — 2 cos a cos0 cosy = 1. . {B) Now CO. a = (^'+^)'+(.r^ +By-{ ri+r,)' ^ R^+R{r,+r,)-r,r, 2{r^+R){r,+RJ {r,+R){r,+ li) WEEKLY PROBLEM PAPERS 203 _ (jg + rg)(jg + r,)-2r,>-, ^ _ ir^r, {r^ + S){r^+ R) (r^ + nAr^+J-i) and similarly we can obtain cos /3 = 1 - 2X ^^±JJ, cos y = 1 - 2\ ^^-±^ '■2 »-3 Substitute these values in {B), and denote by S {r) the sum of the expressions obtained by giving to r in succession the values '"i, '•2- '•3- ... l=3-4X2(^'-) + 4X^^^y - 2 j 1 - 2Xsf:?+r) + 4X^s(i^±!iM±ii) :, = 4X'2(^±-0' - 8X^2 (^ + 0(-g + '-2) + ^6X^ ^ r ' rir^i .•.3+2ii!(l) + 2ra(y - 2 {3 + 2ia(l) + i!>2(-L)} + 4.0, '^i '^z »^3 V ?-/3 ?-3?-i rxr^ Again, the area of the triangle BFC can be written in either of the forms .-. i(^(f2 + rj) = i^VsE^r^ + r-s + -K)^ 204 SOLUTIONS OF Similarly y^r^ + r^) = Vw^(n+^^"+^ • • ^\R~ fj ~^^+73 < 'V it V ^^ Now the expression on the right is equal to 1 if 1 + if 1 + ''2 + '"3 ^ ('"2 + '•.■i)^ _|_ 2 ^i-±^. J;^!l+3±-L3+ 1 + !Vt!>, ■S '■2»'3 Vr„ro Vr, ?•! '2': '■■ ''2''3 ^ V^ r^r^^ »-i?-2 >-i ■"• '1 '^z '^s ^ r^r^ r^r-i r^r^ which we know to be true, • ^ _ * = 2- 4. Let R, be the radius and centre of the oiroum-circle, /ffi, Oa „ „ escribed circle. Then OOa' = R" + 2£ra. Now OaP.OaQ = square of tangent from Oa to the circnm-circle. .-. OaP.OaQ = OOa" - iJ2 = 2Rra ; and OaP = ra A OaQ = 2R. +5. Let be the orthooentre, D, F, F the feet of the perpendiculars on the sides from A, B, G. On BG take BP = CD. On CA take ^Q = GE. On ^5 take AR = BP. Then PQiJ is the triangle required. For it is easily proved that the perpendiculars drawn from any point in the diagonal of a parallelogram to two adjacent sides are inversely propor- tional to the sides to which they are drawn, and conversely, any point from which perpendiculars drawn to two adjacent sides of a parallelo- gram are inversely proportional to the sides must lie on the diagonal through the intersection of these sides. Let S be the circum-centre, A, B', G' the middle points. t TMs solution is by the Rev. G. Eiohardson, Winchester. WEEKLY PROBLEM PAPERS. 205 /fr> TiTp -nri Then ±ii == "?£ = 'ttX^o'" similar triangles SFO, CEO, JL^ CE CO = f-, „ „ BOG, B' so. . •. S must lie on the diagonal of the parallelogram of which AB, AQ are adjacent sides, and .•. SA bisects the other diagonal §iE. Similarly for SB, SC. 6. Join PQ meeting the axis in N, and draw the ordinates Q,G, NE'. Then AM . AC = ANK Bes. Parab. Prop. xxi. But Pif 2 = 4^^ . ^jif^ qci = iAS . AG, B'N^ = iAS . AN, .: PM^ . qC^ = B'NK .: B:N^ = PM . MF = MB^. .'. B' coincides with JR. 7. Tripos 1875. Wednesday morning. No. 13. Papee LXXIX. 1. Let If denote the number of pounds which B must pay per month. Then the total value of B's payments to A at the end of 12 months at 3j per cent = 12N+^^N ^ + ^ +- ■ + ^^ = 1213 i^. Now the total value at the end of 12 months of the money which A has paid to his landlord = .j«+a.„.(i±|±i)} = «o,o. and one-tenth of this = £4 Is. .-. 12^gi\r = £44 11«, .-. N-= £3 12s. ll^ip. 2. a+i + e + d=0, .: (,a + 67= ~ {c + S}\ .: = (» + &)3 + (c + d)^. ^gij^lfij^^j^oP-^. Z[ab{a + b) + cd{c + i)) = ^3 + ^s + c3 + (?3 _ ^ab{c + «?) + cd{a + b)\. 206 SOLUTIONS OF .-. (a3 + (^3 + c3 _|. (^3)2 = cjShcd + cda + dab + ffic}^. Again, = (a + S + ff + (?)2 = a2 4. i2 _^ (.2 ^ (^2 ^. 2(a5 + ae + «<; 4. j^ + ^^ + cd) = a' + b"- + c^ + d^ + ^[ab + cd + {,a + b) {e + g)], .: {be — ad){ca - bd){ab — cd) = a^\c'^ + d^) + cU^aP' + V^) - abcdicfi + i^ 4. <;2 + gi^) = a'b\c^+ d^ + cM^a?' + i^) + ajc^? . 2{«i + cd + {a + b){c + (/)} = aH\c + d)'^ + cH\a + i)2 + 1abcd{a + i)(c + d) = {»4(c + (0 + «<^(« + V)^. 3. Tripos 1878. Thursday afternoon. No. 2. 4. Tripos 1875. Wednesday morning. No 15. 5. Tripos 1875. 2nd Monday morning. No. 2. 6. Let CS = m.OP. Then the coordinates of § are wa cos <^, mi sin ^. Let |, ij be the coordinates of the point of contact of one of the tangents from Q. The equation to this tangent is — ^ + ■-^^ = I. Since it passes through Q, we have _ ^ma cos (j> 1^ rjmb sin (j) = - ma cos d) + 7 m sin 6, a o .•. _ = (1 — ^ ra sin 6 I • a m cos

± cos ~ — I. 7. Let w denote the weight of each pulley. Since there is equilibrium, P = — rwi-„—, .: u> = . . . (1) %n ' V 2"/ 2" - 1 • \ I Let /, /', A, fi, • • • fa denote the accelerations of P, W, and the pulleys beginning with the highest. Then by considering the spaces moved over in the same time, we have /» =/,/«-! = 2./' . . . /i = 2»-i/,/=2V'. .". the kinetic energy of the system at the end of time t = 4(2"/' . tY . P' + 4(2'- 1/ . i)2 . M, + . . . + i (/i)2 . a, + 4 (/'i)2 W. I 92» _ n = 4/V2J22" P'+W' + w I^^ty} = i/H' {22iP' + ^' + J. (2» + 1)(2" P - r)} (2.). Now the work done bygravity upon the system in the time i = iffl . ffP' - igfl (Ato +/,,^ + ... +/,„ +/'r') = i/^V(2"i" - ^' - 2'^P + r) (3). .•. equating (2) and (3), and remembering that / = 2"'/', we obtain the result required. 208 SOLUTIONS OF Papee LXXX. 1. Put each of the given expressions = X. where A = i'K + Sa^ic + Ifle'^) + 2q{a% + ab^c - a?b - aVe)" + ?-(fl2J2 - 2^242 + a2i2) = p(a2 + bc)\ Again X = Fac±q^Mjz^l-_Bab where i)^ = pifi^e + a*c2 - aSc - abd^) + (?{2a24(; + {be - a^f + Sa^icj + r{aJ^c - ab'^c + aH - aH) = q{fl^ + bef. Also X = ^'^ - ^g^^ + ^''^ where Kg = ^(aV _ 2ffi2g2 ^ ^2^2) + ^qi^abc^ - abi? + a^e - a^e) + r(i2<;2 + ia^c + a") = ?-(a2 + bcf. ■ Pa2 + 2Qa5 + M^ Pac + §(*c - a^) - Lab r ^ ~Pc^ -2Qca + Ba^' 2. The number of ways is evidently the number of combinations of 4m things taken 1m at a time, when there are m, alike of each of 4 sets. .•. Todh. Alg. Art 811, the required number is the coefficient of iK^m in the expansion of 1 _ jm+l X — ,'f»'+l 1 — .T^+l 1 — a:™+l \ — X \ — X \ — X \ — X = coef. 01 x^"' m I I , \ \ — X I WEEKLY PROBLEM PAPERS. 209 = ooef. of .r2»» in |l - 4«'»+i +....} il + ... . _^ m{m + \){m. + 2) _^^_i _^ {2m + 1)(2;»» + 2)(2>« + 3) ^2». + . . .) ^ (2;« + l)(2»i + 2)(2ot + 3) _ . m{m + l)(w + 2) = ^+J |(2»2 + i)(2»8 + 3) - 2»;(ra + 2)} = J(m + 1)(2m2 + 42B + 3). 3. Consider the expression fl3(J2 _ (.2) J^ j3 (c2 _ a2) + cV - i'')- If we equate it to zero, it is satisfied hy a = b. .: a — i is a, factor. Similarly b — c, c — a are factors. .-. the expression ^ {b — c){c - a){a - b) [A{a^ + i^ + e^ + B{bc + ca + ab)]. Equating coefficients of corresponding terms on both sides we find ^ = 0, 5 = - 1, _ ,,3(^2_^)+g3(,2_^.)+^(^2_i2) ^ _^^^^^^^ ^^^ ^^^_ (» — b){b — c){c — a) Since this is true whatever be the values of a, b, c, let a = €'*, b = efi', e = er', .•. a — b — cos a — cos |3 + j (sin d — sin j3) = 2 smC_-_- sm— X-t: + 2t sin — 5-^ cos ^-- = 2.-sinl^(cos--±-^+.-sinl±i) = 22 sin — ^ . e-J--'- ^oa^-b^^'U sin (a - /S) . «(«+«^S 210 SOLUTIONS 01" a\Ifi - (?) 2i sin (8 - y) ■ e(3°+g+r)i (a — b)(i — C)(o -a) „. . a — ts . a — y . y — a ,,.,,. 2 2 2 ^ _ J sin «3 - y)g2" ^ . a — /3 . /3— 7 . 7 — a sm sm ^^ — •' sm '-; — 2 2 2 /. from [C) we have S {sin (g - 1 )i;2'4 . . a-iS . /3-7 . 7-0 I -^ 4 sin -^r-^ sm ^^^ — i sin '- — 2 2 2 .•. putting e2« — cos 2a + i sin 2a, &c., and equating real parts, we have 2{co8 (a + ,8) + cos (/3 + 7) + cos (7 + a)} _ ^ cos 2a sin (/3 — 7) _ cos 2a cos \ {y - ^) 2. a — tf . ^ — 7. 7 — a . a ~ B . 7 — a Sin -— -T sin "z — ' sm i — sin sin -^^ — - 2 2 2 2 2 ^ „ cos 2a cos {i(a - /3) + Vy - n)i . fi — S . 7 — a sm sm ' 2 2 = 2 COS 2a{cot i{y - a) cot i(a - /3) - l}. By transposition, we obtain the required result. The problem might be solved more directly as follows. Let s = cos 2a sin (/3 - 7) + cos 2/3 sin (y - a) + cos 27 sin (a - |S), and _y = sin (/3 - 7) + sin (7 - a) + sin (a - P) = - 4 sin iO — 7) sin i{y - a) sin i(a - /3). Todh. JV2>. viii. Es. 3. .-. ;r — ycoa2y = sin(|3— 7)(cos2a-cos2y)+sin(7-a)(cos2;8-cos27) = - 2 sin (/3 - y) sin (a - 7) sin (a + y) - 2 sin (7 - a) sin (7 - /3) sin (0 + y) = 2 sin (/3 - 7) sin (y - a) {sin (a + 7) - sin (/3 + 7)} = 4 Bin (/3 - 7) sin (7 - a) sin J(a - j3) cos J(a + 13 -f- 2y) WEEKLY PROBLEM PAPEES. 211 = - 4y cos JO - y) COS J(y - a) COS (^-t^ + v) ■ = - 2y(cos4(/3 - a) + cos(^ - y)|cos(l±-^+y) = -^{cos(/3 + 7) + cos(y + a) +cos(a+(3) + cOB2y} .: X = y {cos (/3 + y) + cos (y -|- a) + cos (a + /3)). From this the required result follows as in the previous method. 4. Let A'B'C'V be the given quadrilaternl, and let A, B, 0, D be the points of contact of the circle with J!B', A'jy, G'B, B'G>. Let AC and BD intersect in 0. Then the angle BD'C = D'BC, and ABD = BAD, .-. BAO + ABO = i(5r - ff) + J'tt - B') = n - i{B' + B') = 77 — _ since a circle can be described about A'B'C'D' •J. .' AOB is a right angle. .'. AG and BD intersect at right angles. Now let K, L, M, N be the middle points of AB, BG, GD, DA. Then KZ and MN are each of them parallel to AG, and KN, LM are each parallel to BD, .'. KLMN is a rectangle, and a circle can be described round K, L, M, N. Let 0, I be the centres of the ciroum- and inscribed circles of A'B'G'ff. Join A' I meeting the circum-circle in B. Then since A'l bisects the angle BA'V, .: E is the middle point of the arc B'jy. Produce EO to meet the circle in F. Then r = A'l sin '^', / = C7sin %■ = G'l cos -', for A' + C' = w, ^ ^ ^ .-. A'l.IG' = 2r2cosec^'. Since F is the middle point of the arc B'D', .: C'J bisects the angle B'G'jy and passes through I. .: A'l. IE = G'l. IF= B? - bK Also A! a = 2iJ sin A'B'C' = 2^ sin A'EC' = 272 ^ EI and EH = %B sin B'^H = AG' . B'B = All. IG' 8R^ . r^ _ 8R^r^ A'l. TE ~ i2- - 62 P 2 212 SOLUTIONS OF Again, let A'£' = a, SC = b, CD = c, DA' = d. Let A'C, BD' intersect in P. Then the area of A'B'C'U — l-T—j • Tu '^ "J abed — 'J abed. .: sin B' = —-— — ■ ab + ea „ ., „ a sin A'B'B' ad sin B Now AB = : — = ■ ,„ . „ sin e JO sin 6 be sin B' Bo OP A'C Bind {ad+bc)jinB ^ ^ p ^ ^,^., sm 6 ^ Xac + b d){ ad + be) ab + cd . . ab A' cd . ■„ ^ iJabcd .". sin 6 = , sm B = ac -^ id ac -\- bd •■'°'^ Lac + bdf fac ~ bd-f \acJt-bd' a 1 _ Tf ///■ --> A/7 tan2 cos 5 i(? abed S" 2 1 + cos 6 ac {acf (acf If ffic < bd tan2 ^^ ac ~ bd " abed S^ {bdf " {bdf If 5 = 90°, 1= 45°, ,: 1 = tan - = - 2 ac Jabcd ac sj~'i.:bd = ^ ac 5. Let §(? meet §'(?' in I, and let CD be conjugate to CP. Then §P . Pq = CZ)'' = PG . PO'. :. QP : Pff :: PS' : PQ, .: the triangles PQff, P&'Q' are similar. .-. the angle PQF = PG'F. .: a circle will go round PQG'F. .: the angle G'FQ = G'PQ = a right angle. 6. Since the parabola passes the origin, its equation is of the form (ax + Wf + V + 2^5' = 0- WEEKLY PROBLEM PAPERS. 218 The equation to the directrix, Salm. Con. p. 269, is x^{g^ - fa) - ya{ff^ - fa) = \{J^ + /) . . {A) If y = 0, a; = or - "4 «2 y . g If a; = 0, y = or - l(, .-. = k. 2/ i32' ■ ■ /a2 .-. g^ -fa = a^{Hk - ah) ; /2 + / = aW + /3V<:2, .-. (^) becomes %a^{&k - ah)(fix - ay) = oVi^ + 0%2 . {B). The equation to a straight line through the origin perpendicular to this is ax + fy = 0, .-. ^^ = ^ (C). .: eliminating a and between (5) and (C) we have for the locus of the intersection 2j:tf{kx + hy){x^ + /) -|- }?f + fc* = 0. By Salm. Con. p. 196, the equation to the axis is "■^ + py -^-+j2 - -^ht-^^' ^ :^+i .'rr?h — k , a = mx - - — ; — 3", where m = - -• 1 + m' p 7. Tripos 1878. Tuesday morning. No. 5. Paper LXXXl. 1. Lety = «'' = x . X . X . . . to jo''-i factors. Since X is prime to jo, .•. y is prime to jb .-. ^P-i = 1 + M{p). 214 SOLUTIONS OF 2. Let C be the centre of the circle, and produce CO to meet the circumference in A. Let P^OA = a, P^^OPi = ^ = P^Ol'^, &c., Then n^ = 2^. CPi^ = C02 + 0Pi2 -200. OPi cos COPi. .-. a2 = ^2 _|. 0Pj2 ^ 2b . OPi cos a. ■' '-"^ ~ *^) TT^" = ^-^1 + 24 cos a. OP I So (a2 - i^) -}- ^ OP2 4- 2« cos (a + jS), &c., .: OP, + 0P,+ . . +OPn-= («2 - i2)^_i^ + oF + ... +^j + 2^-^. where (S" = cos a + cos (a + ^) + . . . + cos (a -f » - 1/3) = sm I a -f- -— — j3 j sin ~- coseo - n ■ . n^ . n = 0, since sin ^ = sin w = . 3. Produce AO to meet the circle round ABG in B. Make the angle AFG =■ ALB, and ABE = ^2)a Then circles will go round BBQV, GDHE. Then AGF = ABD = sup. of AGO = sup. of AHE. .: FG is parallel to HE. .: AH = OG. .: BA.AF= DA. AG, and CA.AE = BA.AH = DA. OG. .-. BA.AF+ CA. AE = DA(AG + GO) = DA. AO = AO'^ + AO . OD ^ AG' + BO . OG. 4. If we project the ellipse into a circle, the inscribed triangle is equilateral, and the sum of the squares on the sides is equal to | X num of the squares on two diameters at right angles. In the ellipse the diameters corresponding to these are conjugate, and the sum of the squares on two conjugate diameters is equal to the sum of the squares on the axes ; hence the theorem. 5. Draw PN and RM perpendicular to the axis, and RZ' perpendicular lo AZ, and let the normal meet the axis in G. WEEKLY PROBLEM PAPERS. 215 Then AM : AS :: RZ : ZS :: MP : PG -.-.MN-.NO. Now NG = ^AS, .: MN = 2AM, .: FZ = BZ'. 6. Let the first resultant make an angle with the direction of P. Then in the first case and E^ = P^+ Q' + 2PQ cos a sin (a — ) sm {p + 4>) P + R (1) (2) (3) P . From (2), - sin a — sin (a — (^). R sin d cos ^ — cos a . ^ sin a. .'. cos = P + Q cos a R ' .". cos (a — (jj) = COS a cos (jb + sin a sin (j) _ P cos a + § R From (3), (P + P)(sin 5 cos ^ -j- cos 6 sin 0) = (§ + P) {sin (a - 0) cos d — cos (a - 0) sin 6} •■• ^ {(-P -f- RKP + Q cos «)-(-(§ + RKP cos a + §)} .-. tan 6 ■■ cos 6 ~R~ (Q + R)P Bin a - (P 4 P) § sin a} R(P - Q) sin a Rip +Q+{P+Q) cos a\ + r^+Q^ + iPq cos a (P - g) sin g P + Q + P + (P+5)oosa by (1). 216 SOLUTIONS OF 7. Let A be the highest, the lowest point of the vertical diameter, B any point on the circumference. Draw BD perpendicular to AO, and malte the angle BOP = BOA, producing AB to meet OP in P. Draw PX perpendicular to BD, and BS perpendicular to OP. Then since the velocity at is due to height OJ), BD is the directrix of the path described. Since BO is a tangent, the focus is iii OP. Since OS = OD, S is the focus. The triangles OBP, OB A are equal in all respects, as are also BKP and ABB ; and OS = OD, .: SP = DA = BK, and the angle SPli = KPB. „•. P is a point on the parabola, and BP is a tangent Since OP = OA = const, the locus of P is a circle, centre and radius equal to the diameter of the given circle. Paper LXXXII. 1 2» = (1 + 1)- = 1 + {« + ?'"-^H + K'' - !)(« - 2) _ 1 , (n±^ , {n + \)n{n ~ 1)(» - 2) , - + ^2- + 14 + ■ • • 2. Let A, B, C, D, B be consecutive angular points of the polygon, and let BD meet AG in M and EG in N. Then the angle ACB = BAG = BDG = CBD. .: MB = MG. Now the angle BOD = 2-11^ TT, .-. MBG = MGB = -. Similarly DCN = NDG = -• .'. the .triangles BGM, CDN are similar ; and BG = CD, .'. the triangles are equal in all respects. .'. if a denote the length of AB, the required area = area of polygon — n. X triangle BGM na' . TT = -— cot - 4 n n . « - 2 2 sin TT AVEEKLT PROBLEM PAPERS. 217 sin'' - nn^ . IT nn^ rt —-cot • r- 4 n 'I . li! sin « rt =s ■ — I cot tan - 1 4 V n «/ 4 V n vcfi . lit : cot — • 2 n 3. Draw AI> perpendicular to BG. Then circles will go round AFBC and ABDB. .: BF.BA = BD.BC, and CE . CA = CD . CB .: AB.BF+ AC.CE = BC{BD + DC) = BC^ 4. Since the angle BPG is the supplement oi A, .: the circles round PBC and PAC are each equal to the circle round ABC. .: A'P = B^P = C'P = R. .: P is the centre of the circle round AB'C. Since AlP — ^'(7 and B!P = B'C, we have two isosceles triangles on the same base PC, :. PC is perpendicular to AB'. .: A'B' is parallel to AB. Similarly it may be shewn that B'C is parallel toBC, and'C'A' to CA. .'■ A'O, which is perpendicular to BC, is also at right angles to B'C. .: is the orthocentre of A' B'C Since the circles, centres A' and C, are each equal to the circle round ABC .'■ CA! = CB' = CO. .'. C is the centre of the circle round A'OB' ' Similarly it may be shewn that A and B are the centres of the circles round FOC, C'OA!. ^ ,^ , . . , ,, , , Consider the triangle BPC. Its orthocentre is A, and the centre of its circum-cirole is A'. Since AO = A:P, .-. AA' bisects OP. .: the centre of its nine-point circle, which is at the middle point of AA, is also at the middle point of OP, and the radius - iA'P = iP. Consider the triangle B'OC. Its orthocentre is A , and tlie centre of its circum-circle is A. ■■■ the centre of its nine-point circle is at the middle point of OP, and the radius = iAO = jB. Consider A'B'C. Its orthocentre is 0, and the centre of its circum- circle is P, and radius = A'P = P. _ . ^ . , .-. the 8 triangles have the same mne-point circle. 5 In general five conditions are required to determine a conic. Since the focus is the pole of the directrix, the focus being given is equivalent to two conditions, and only one ellipse can be described having its focus at a given point and touching the sides of a triangle. 218 SOLUTIONS OF Let ABC ho a triangle, P the orthocentre, and JV the centres of the circum- and nine-point circles. Then the angle OJB = PAC, each being the complement of AGB. .: since AB and AC are tangents, we see that if either or P is a focus, the other is also, and the centre is at N, which is the middle point of OP. Again, since the centre is the pole of the line at infinity, the centre being given is equivalent to two conditions, and /. only one ellipse can be described touching the sides of a triangle, and having its centre at the centre of the nine-point circle. .'. by the converse of the first case, the foci are at and P. 6. Let 0, ff be the centres of the small and large circles, C and D their points of contact with the beam, A and B the points where they touch the plane. Produce BA and DC to meet in H. Then AHG — a. Since the circle is in equilibrium, and is acted on by forces at A and G, .: the resultant force on the cylinder acts at G along CA. .: the resultant force on the beam at C acts along AC. Similarly the resultant force on the beam at D acts along BB, which is parallel to CA. .: if ^be the middle point of the beam, lEF perpendicular to AB, and EG parallel to AC or BB, the resultant of P and W acts along EG. P ein PEG am PEG Bin PEG . ™_ , a •'• r = s-taGlP = ^Kegb = sliTJ^ = ^""^-^^ = '"" 2 for 2EGF = TT - a. 7. Let the tangent at P meet the tangent at A in T. Join SF cutting the curve in Q. Then ST bisects the angle ASP. .: ASF = 60°. Draw the ordinate PM, and join AP. Then PM = 2AY. The parabolic area AQP = f the triangle ATP. (Bes. Gon. p. 162.) .'. parabolic area ASP = parabolic area AQP + triangle ASP = I ATP + ASP = lAY.rM+lAS.PM. Now PM = 2Ar - 2AS tan 60°, PJf2 AM - V^„ = AS . tan2 60'-', iAS .: parabolic area ASP = AS^ (tan 60° + J tan^ 60°) = AS^ 2 VsT So „ ASp = AS^ (tan 30° -f J tan^ 30°) = AS'- . ~ ■ WEEKLY PROBLEM PAPERS. 219 And lime in AP : time in ^^ :: area ASP : ASp :: 27 : 5. Paper LXXXIII. 1. Let X denote the rateable value of the University. Then - is the amount it pays. .•. the amount paid by the town = )-, and the rateable value of the town = 8r, .•. the rateable values are as 1 : 8. The total amount of rates paid = '-• -|- _ = — • The total Value of rateable property = x + %x =^ Vx, .: the average rate is £^ in the £. .-. in the £ the University would save £ (^ ^ ii) = ^ • 9" = a °^ their present payment. 2. (1) = ^(6»6 -f 15n^ + lOn.^ - n}. By trial we find that this is integral when « = 1, 2, 3. Suppose it is integral when n = p. Then writing jo + 1 for ii in the expression and subtracting its value when n = ^, we have ^ |30/ + 120/)3 + 180p2 4- 120;) + 30J, which is integral. , .•. if the expression is integral for any value of n, it is also integral for the next consecutive value of n. Now we know it is an integer when « = 3, .'. it is an integer when n = 4, .'. by induction it is always integral. (2) = ^{2n« + 6»6 -'r 5«* - w^). By trial this is integral when «j = 1, 2, 3. Suppose it is integral when » = p, then as in (1), the difierenoe of its values when n. = p and « = ^ + 1 is J^(12/ + 60p5 + 120^3 + 120/)2 _|. 60;j - 12), which is integral. .-. as before, the proof follows by induction. 220 SOLUTIONS OF 3. Let be the point of intersection, Q, M, K the feet of the perpendiculars on BC, CA, AB. Then GK = OK ^!° ^^-f = OB smB = iBEsmB. sin O&K Similarly fflT = | CF sin C ; FH = % AD sin A. Also 0X= OBsmOBE= 'iEB.sinEBK = ^bsmA = Jffsinif. Similarly OH = ^ c sin ^ = J a sin (7 ; Off = ^ 5 sin C = ^ c sin A .•. area of triangle GHK = '■^^{a' -{■ b^ + (?) sin A sin 5 sin C. . „ F&.QE.HF 8 AD.BE.CF . . . T, . „ .: E = — = — — - sin A sin B sm C 4.AFGS 27 4./\FGII ^ 4 AD.BE.CF ^ a^ + 6^ + c2" 4. Let the chord through S cut the inner curve m B, p and the outer in P', p'. Let the tangents at JP, p meet in JR, those at B', p' in ii*. Produce BR to meet jo'iJ' in T, and produce ^iJ to meet PR' in 2". Then since tangents at the extremities of a focal chord intersect at right angles on the directrix, .•. RTJST' is a rectangle. Since SB and SS are perpendicular to the chord, .•. BB' passes through S. Let TT' and BB' meet in F, and draw PK' perpendicular to the directrix of the outer parabola. Then the angle FT'B' = T'R'F = T'B'E'. :. TT' is parallel to the directrix. 5. Let the tangents at A and B meet in T. Draw PB, BE, FF perpendiculars on TA, TB^ AB. Similarly draw §2X, Q^, QF' per- pendiculars on the same lines. Then by Casey, p. 33, PF^ = PB.PE; QF'^ = Qfil . Qfi'. And FB = QV. PE FF^ BB? Similarly, if PQ, QQ' be perpendiculars on the tangent at C, we BG ^ BS^ QG' QS^' shall have — — = / PB.PS -f ^ BE.PG ^ PM^ \QR.qs) QE'.QG' qm' WEEKLY PROBLEM PAPERS. 221 *6. If two heavy particles be projected in the same vertical plane at the same instant frotti two given points A, B, with the same velocity u in directions AQ,, HQ and meet at the point P after a time t, then if PQ be drawn vertically upwards and equal to ^ fffi, it follows that if gravity did not act the particles would pass through Q at the same instant. , This requires AQ = ut = BQ. .: the directions of projection, viz. AQ, BQ are equally inclined to AB, and .•. the sum of their inclinations to the horizon is const. Again, let A, B be the two points of projection, and let AB = 2«. Bisect AB in S, and let Q be any point in the line through E at right angles to AB. Then if tt be the velocity of projection, AQ = tit = BQ. Draw QP vertically down- wards and = ifffl. Then we require to find the locus of P as Q moves along the fixed line BQ. Through E draw Elf parallel to QP and FN parallel to QE. Then QPNE is a parallelogram. Then AQ = lit, .: AQ oc i since k is constant. And QP^ifft^ .-.QPc^.f-, .: AQ" o= QP, .'. AQ" = 'KQP where X is some constant. On EN take a point such that X.OE = AEK Then is a fixed point. ,-. AQ^ - AE" = X{QP - OE), .: QE" = PN" = \.0N. .: the locus of P is a parabola having ON for a diameter, i e. a vertical line through the middle point of AB. The ordinates to that diameter are perpendicular to AB. 7. Draw AD at right angles to BC. Since the resultant passes through the circum-centre and the orthocentre, the sum of the moments of the forces about each of these points will vanish. If p be the radius of the circum-circle, the perpendicular from the chcum-oentre on BG = poos A. The perpendicular from the orthocentre on BO = BD cot C = c . cos B cot C, = — — cos B cos C. sin /i .■.PcosA+ QcosB + RcobC = ^ "° {2 " ^^ ~ ^'^1 ^ coK^ - g) _ ...„ = ^^cos(5-0- AC Bin AE'C sinB sinB sin A From these we can write down the values of g, r, v, w. .: f (w-v) = -,— cos^B - C) {cos{A - B) - cos(C- A)} sin^ A ^ fl3 WEEKLY PROBLEM PAPERS. 223' {l + cos'(5 - C} {cos {A- B) - cos (C - A)) where Cj = cos (5 - C), Cj = cos (^ - B), C^ = cos (C - A). Similarly j2(a - «,) = -^ (1 + t?^) (C3 - Ci), sin" ji ''^'^ ~ ''^ ^ blS^ ^^ + ^3) ^^^ ~ ^^'' .*. p^(w — ») + ^^(w — a») J- '■^(» — k) = 0. 4. Since ad is parallel to BC, .'. arc CW = arc Ba = arc ^i, for AOa, BOb are diameters. .•. the angle dbO = ACb. .: Eb = BC = JEf, since bCf is a right angle, heing in the semi-circle bCB. .'. £ is the middle point of if; and is the middle point of bB. .'. OB is parallel to BC, 5. Let S be the common focus, CL one of the asymptotes. Draw SD perpendicular to CL and BX perpendicular to the axis. Then since CL is a tangent to P, .'. B is on the tangent at the vertex of P. But since CL is an asymptote, .". Bis on the directrix of B". Let R be one of the points of intersection, and let the tangent to P at iB meet the axis in A'. Draw the ordinate KN, and let A be the vertex of B nearer S. Then since R is on P, .: SB = SA', and A'X = WX. Since B is on if, .-. SB : NX :: SA : AX .: SA' : AX :: SA : AX. .; A' is the further vertex of S. 6. If C be the centre, TC bisects PQ and is .'. the direction of the resultant of the forces TP, TQ. .: if C be fixed the ellipse will remain at rest. 7. The equation to the normal at P is as; hy 2 70 , ,, :; r^ = a^ - I- (^) cos 9 sm0 If 6 be the eccentric angle of the point § where {A) meets the ellipse, the point (a cos 6, b sin S) will lie on {A). 224 SOLUTIONS OF a^cosB 42 sin 5 ., ,, • :r ^^ = a' - b\ cos <^ sin

+ i^ cot 4 (5 + 0) =0. Let (1^, r)) be the coordinates of the middle point of PQ. Then f = ^ (cos 6 + cos ib) = a cos — 1_5 cos ^~ S I? = - (sin 6 + am (h) = b sin — ^ — - cos — ~^» 2 'e. 2, ■■■- = J cot — ^ = - -5 tan 0, cos sin 1 .•. substituting in (yi), we have .,. .3.,2 .. (,2_y.)2 Paper LXXXV. 1. Let 2:, :r + 1, y, y + 1 denote the digits. Then lOa: + w+l=y{y + \), .■.lU+l=f + ij . . (!) and 1002:+ % + ;!;+ I=9(y + l)2,.-.l01« + 10y+l = 9(y2 + 2y + l) (2) By (1) 99a; + 9 = 9/ + 9y, By (2) 101;!; + 1 = 9/ + 8y + 9, /. 2a; + y = 17 ; .'. y = 17 - 2.r, WEEKLY PROBLEM PAPERS. 225 .-. from (1) 11a; -j- 1 = (17 - 2,t) (18 - 2x) = 306 - 70;i; + ^x\ .: 4a;2 - 81.1; + 305 = 0, .-. a; = 5 or i«. Taking the integral value, y = 17 — 10 = 7, .'. the numbers are 56, 78. 2. Let u, 0, y, 8 he the vertices of the triangles described on the bases AJi, BC, CD, DA respectively. Produce aA and yD to meet in X. Then XAh = ~ - A; SAD = '^ ; XAD = ^J ~ A. 2 4 4 Similarly XDA = — - Z); .-. AXD = A + D - Z- J 4 2 = (i + ^^^T + (7-2 + ^^^'y - K7-2 + ^'^)(r2 + ^^0=°^^ = ^' 4. £.^ - ac fin (^ + J) + Xr3 + DX2 _ 2^X. JTi) cos X + ^2a{AX - DXcobX) + V'^r(-OA' - ^-X'cosJT) = ^% £.^ _ ac sm{A + D)-\-d^+ J2ad cos Z)^X + ^-Icd cos ^i).T =.'f j^t- acsm{A+D)+d''-\- ^iadcoa (^ -\-A\ - V2cicos (^ +D\ = - + - - «csin(^+Z')+«/2 + a(?(sin^- cos^ + C(^(sini) - cosD), Similarly by producing aB, yCio meet in Y, we get ay2 = ^Vl--«csin(54-C) + ^^ + *«(sinC-cosC)+«5(sini?-cos5), .-. 2ay2 = ^2 + ,^2 + c2 + (^2 + &(sin ^ - cos A) + f/i(sin 5 - cos B) -\- bc{&in C - cos C) + cd{s'm D - cos 2)), and from symmetry, 2/38^ is equal to the same quantity. 3. On BG, CA, A^ describe the squares BCLK, CANM, ABHG, and join GN, ML, KII. Q 226 SOLUTIONS OF Then GN"^ = GJ^ + AW'' - IGA JNooa GAN = ^2 -I- c2 + 2bc cos A = 2(^2 + (,2) - a\ Now if S denote the area of a triangle whose sities are a. b, c, tV 4. First let 1) and ^ be on the same side of BC. Produce AD and CB through D and B to meet itt -ff. Then the angle ADB = sup. ot ACB = sop. of ^i?C = ABE\ and Z^y^^ is couinion to the two triangles DAB, ABU; .: they are similar. .-. DB : BA :: BE : AB. . . . (1) Also the angle ADC = ABC = ACB, and DAC is common to the two triangles ADC, ACE; .'. they are similar. .: DC : DA :: CE : AC (or AB) (2) .•. from (1) and (2) by subtraction we have DC — DB : DA :: BC : AB, i.e. in a constant ratio. Similaily it can be shewn that when D and A are on opposite sides of BC, DC+ DB -.DA :: BC: AB. WEEKLY PROBLEM PAPERS. 229 5. Let the tangents at Q and Q meet in T, and let CT, which is one of the eqiiiconiiigate diameters, meet QQ' in K Then since TQO, TQO are right angles, .-. 0, Q, Q, T are concyclic. And qV^ = Cr. FT, .: Cis on the same circle, 6. Let CD be the diameter conjugate to CA. Produce CA to meet the circle in B, and draw CB, a tangent to the circle. Then CE^ = CA.CB = CA" + CA . AB. CTP Now AB = \ chord of curvature in direction AC = ■ CA .•. CE'^ = CA^ + CD^ = «2 ^ ^2 = square of radius of director circle. .■. the two circles cut orthogonally. 7. Let a denote ^'s velocity before impact; a', «' the velocities of A and B estimated along AB after impact, u\ the velocit}' of A measured at right angles to AB after impact. Let e be the elasticity, m, m' the masses of A and B. Then mu cos a = mv! + m'v' ; eu cos a = »' — u', mucosa + OTffK cos a = {m + w') »', ., ,'=(!+,) !!^i^ (1) m — -era' , , .•. K' = a cos a J-, and Mj = asm a, «« + »i .•. if F be the "velocity of A after impact, F^ = a'2 + i!) u' a + b' a + 6 a " a - 1/ ' 230 SOLUTIONS OF ,(._3.)^=(^{. + 2.-M^f «(» - bf { ^ I A,2_, ^ 4a(a + b) f (ci + b)* , a{a + ^.) (« + 2/y) ; 4^(^ + ^^1 a - b I «2 i- a- b ) ., y(^ _ 3.)^ + 4.(/ + ^.) = ^-|^±|-' . 3«2(« + i/ = 12 ('^ + ^)' . "'(« + ^)' = 12.A Again I (fl + (« + 5) + . . . + 4' = - {4a2 + 4ai(» - 1) + (a - 1)2 i^} = «a3 + «i?2(« - 1) + ?^^_zJ) i2. And «^ + (» + 5) (/ - i) + • . • :^ a[a + {u - 1) b} + {a + b) {a - b + (n - 1) i} + . . . = a^ + a^ - b^ + a^ - W + . . .ton terms + (« - 1) b{a + ia + b) + {a + 2b) + .. .ton terms} = K«2 -52 (12 + 22+... to a- 1 terms) + "J^-IiJ} b{2a + {n -l)b] = ««2 + «i^(, _ 1) + ^2 /("-l) (, _ 1 _ !^) .-. given expression = ?0«jli) i2 {3(„ - 1) - 6(« - 1) -I- 2(2» - 1)J ^ «(a2 - 1) ^2 12 WEEKLY PEOBLEII PAPERS. 231 2. 2^/6■+ 1 = V2-r|:W^= 5 (l + ^^J = 5^14-1 ?i^_i 2^3 1.3 23.6^ _ 1.3.5 26.3^ \ So 2^6 - 1 _ f. 1 2^6^ 1 23.3 1.3 23.6^ 1.3.5 2°.3g \ V 2 ■ 6^ 2 1 ■ 6* 3 1 ■ 66 4 1 ■ 68 ' ' 7' , ,- „ ,/, 1 23.3 1.3.5 26.. B2 \ .•.4N/b = 2.5(^1- -.— ___._^-...^. 3. 2 {sin (a + 2^) sin (a + 30) + sin (a + 3/3) sin a + sin a sin (a + /3)} = COS0 - cos (2a + 50) + cos 3/3 - cos (2a + 30) + cos0 - cos (2a + 0) = COS0 + 2 cos 20 COS0 - cos (2a + 30) (1 + 2 cos 20) =■ (1 + 2 cos 20) {oos0 - cos (2a + 30)} = 2(1 + 2 cos 20) sin (a + 0) sin (a + 20), .•. the given expression _ sin (1 + 2 cos 20) ^ sin + sin 30 - sin ^ sin 30 " sin (a + 30) sin (a + 30) "~ sin (a + 30)' 4. Let n. be the number of sides in tlie one whose angles are measured in degrees, m the number in the other. Then if D be the number of degrees in an angle of the 1st, 0- the number of grades in an angle of the 2nd, ^ = 2(«-^_90. g^.2(«'^-2)_ioo. ,„i^^ff. « m, .: mn = 20» — 18to, 20« ^ 20 »+ ^8 - 18 ^ 20 ^^^ !z + 18 » -t- i« n+m .: « + 18 is a divisor of 3G0. Now 3G0 = 1 X 360 = 2 X 180 "= 3 X 120 = 4 X 90 = 5 X 72 = 6 X 60 = 8 X 45 = 9 X 40 = 10 X 36 = 12 X 30 = 15 X 24 = 18 X 20, . n -\- 18 may have any one of the eleven values 360, 180, 120, 90, 72, 60, 45, 40, 36, 30, 21 232 SOLUTIONS OF Tlie 12th value 20 is inadmissible, since n > 2. .: n = 342, 162, 102, 72, 54, 42, 27, 22, 18, 12, 6, m = 19, 18, 17, 16, 16, 14, 12, 11, 10, 8, 5. 5. Lei CBHF be any quadrilateral. Produce BE and CF tliroiigli E and F to meet in A, and CB, FF through B and E to meet in 1). We will first prove that the circum-circles of the four triangles thus formed pass through a common point. Let the circles round AFF and BFL intersect in K. Join FK, KB, KB. Then the angle FKD = FEE + EEI) = FAB + BBC = it - ACB. .-. the circle round CDii' passes through X. Join AK, KB. Then AKB = BKE+EKJ = BBE+ CFE= tt - ACB. .-. the circle round ABC also passes through K. Let 0,P, Q,Jihe the centres of the circles round BED, AEF,ABC, CDF. Then since the line joining the centres of two intersecting circles is perpendicular to their common chord .'. ItO and OP are perpendicular respectively to KB and KB. .: the ans:le FOB. = EKD = ABC. Similarly RQ and FQ are perpendicular to Cffand AK, .: PQR = sup. of AKC = n- - ABC. .: FOB + PQR = 2 right angles. .-. a circle will go round OPQR. It can be shewn that the point K also lies on this circle. For OKE = 5 - POK =1 - B/)K = 5 - EBK, Ji £1 £1 and VKE = | - OPK = | - EAK, .: OKP = T - {EBK + EAK) = AKB = tt - ACB = FKV = TT - ORP, since RO and RP are perpendicular to KD and KF. .: OKP + ORP = 2 right angles. .'. Xlies on the circle round FOR. Note. — By the aid of XXVIL No. 5 we can shew that the ortho- centres of the triangles fornied by four straight lines in a plane lie on a straight line, viz. the directrix of th? parabola which can be described touching the four given straight lines. G. Let {x'/) be the point from which the tangents are drawn. Let px -{■ qy = 1 (1) be the equation to the tangent to '- -[■ ^ = I. WEEKLY PROBLEM PAPERS. 233 Then a^f + W'q^ = 1 (2) The equation to a straight line through (.r'/) at right angles to (1) is q{!C - x') - p{y - y) = 0, or qx — py = qx' — pi/'. If this touches -„ + —„ = I, we have a2 ^ fl2 ' a\^ + ^Y = {qx- - py'f (3) and since {x'y') is on (1) /. px' -\- qy' = \ (4) We have to eliminate^ and g between (2), (3) and (4). From (2) and (4) «y + b\^ = 1 = p^x"^ + aV^ + ipqx'y'. From (3) ^Y + ('Y = t^^ -\-fn"^ - ^Pqx'y'. ... ^2(^2 + 0^) + ^2.„2 + i2) = (p2 + g2) (^.-2 + y2)^ .-. j92(a2 + 02 - a;'2 - y2) - g2(:!;'2 +y2 - a^ - i^). . (5) Prom (3) ^pYx'Y^ = [p^lfi^ - y'^ + q%a^ - x'')Y. ... (6) .'. substituting in (6) the value of the ratio p'' : g^ from (5), the equation of the locus becomes 4xy(_a^ + /32 -.r2 - /) {x^ + y^ ~ a' - U^) = {[X^ - «2) {X^ - a^ - (y2 - 42) (/ _ 02)|2. . (7) If J^ be any point on the curve, the origin, and a^ ^ ^32 > „2 ^. ^2 then if 01^ (= a:^ + y^) be '> a^ -|. ^2^ the expression on the left is negative, and .'. the expression on the right, which is a perfect square, is negative, which is impossible. .\ no part of the curve lies outside the larger circle. Similarly it can be shewn that no part lies within the smaller circle. If we put x^ +y^ = a^ + 02, (7) becomes = {{X^ - a2) {x^ +y2 _ „2 _ 42)j2. Now since a^ + ^^ > a? + i% we cannot have a;2 + / - a^ - 42 = 0, .-. ;r2 _ a2 = = / - 02. .-. the curve meets the larger circle in the real points given by ^2 = ^2 1/^ = 02^ and it has been proved that no part of the curve lies outside this circle, .". the curve touches the larger circle at these points. Similarly it can be shewn that the curve touches the smaller circle in the points given by x^ = a', / = ^2. 234 SOLUTIONS OF 7. Let be the point of projection, A the position of any one of the particles at the time t. Eefer the position of A to three rectangular axes through as oiigin, the plane of xy being horizontal, and the axis of z vertical. Let «i be the initial velocity of A, a, the angle of projection, and let OM, the intersection of the plane of ^'s projection with the plane of xy make an angle A with Ox. Draw MN perpendicular to Ox. Then if Xy, yi, ^l be the coordinates of A, xi = ON = OM cos A = ti^ cos % cos A . t, 1/1 = MN = OM sin A = iti cos a^ sin A . t, i'l = AM = »isinai. t - ^fffi, with similar expressions for the coordinates of Ji and C. Now MC' = (.r^ - x,y + {y, - y,Y + {z, - z^\ and since the quantities n-i, nj, A, &c. are constant .•. BC"^ oc f^ = cfifi suppose. .-. the area of the triangle ABC = <2 sj-ih^c' + Ic'^d^ + -la'U^ - a'^ - 0* - c^ oc A Next suppose that A and B are projected in the same vertical plane from the point 0. Then if the plane of the triangle passes through 0, it is obvious that JB passes through 0. Draw AM and BN perpen- diculars on the horizontal line in tlie plane through 0. Let AOM=e. Then OM = w cos a. t; AM = msma .t - Igfi. . ,. AM .: tan.5 = — - ■- OM .". i = (tan a Similarly i = (tan;3 - .•. tanfl ('ycos(3 — ncosa) = 'ysin(3 — 2«8ina, 2u cos a f , u sin a - v sin „•. t = < tan a — ff I u cos a — vc.osp _ 2 nv sin - a) g ' a cos a - c cos (3 w sin a . < - iffi" u cos a . J .tane)2«'=°^« , .. 2» cos tan^) /3 WEEKLY PEOBLEM PAPERS. 235 Papek LXXXVIII. 1. (1) \/.r"+ffl + \/j + 2a = iJa; + Ua - iJx + 8», .-. x+a+a;+2a+-2'Jx^+iia.v+'ia^ = x+6a-\-x+3a-2'J^+^ax+18i^, .: Va;2 + 3aa; + 2^2 = 3a - ' Va-^ + 9a.T + ISa^, .-. a^ + 3fla- + 2a2 = 9b2 + a;2 + Qaa: + ISa^ - 6a Va;2 4. 9«« + l^o^, .-. 36(a;2 + 9«ar + IPa^) = (26a + 6a;)2 = 625a2 + SOOai •+ SCa^, .-. 23a2 = _ 24aa;, .-.«•= - f|«. (2) '-s + >-3 - '•l) 2 4r2r3 ^ (^e + ca + ab ~ s^s"^ - a^ - be) 4(s - af{s - ~b){s - c) ' • a(s - a) cos2 " = (^g + ca + ab - s^jas^ - a^ - abc) ' 2 4(« - a){s - b){s - c) .: the given expression on the left ^ (bn + ca + ab - .t^){/(g + 6 + c) - a^ - b^ - c' - 3abr} 4{s - a){s — b){s — c) WEEKLY PEOBLEM PAPEUS. 237 Now if we substitute " '^ " for s, and multiply out, we find s^a + i + c) - a^ - Ifi - c^ - lahc = l{a^ + al? + b^c + be^ + c^a + ca^ - a? - b^ - (? - iabc), and 4(s — a){s — b)(s - c) m a. similar manner can be shewn equal to ^ the same expression, .". the given expression = § (Ac + ca + «5 - s^). 4. Let F be the middle point of JBD. Since JEF cuts the sides of the triangle SOD, .: BF . BA . CE = FB . CA . EU, and Bf = FB\ .: BA.CE = CA. ED. .: CE : ED :: AC : AB :: BC : CA. 5. Let AL be the fixed line, S the fixed point. Draw 8A perpen- dicular to AL, and produce ^ to Xso that SA'= AX. ThroiighXdraw XX parallel to ^i. Then XX is a fixed straight line. Let I' be any position of the angular point of the right anffle, so that SY, TP are tho positions of the arms. Produce ST to meet XX in X, and draw through X a straight line at right angles to XX, meeting YP in P. Join SP. Then SY : YK :: 8A : AX. .: SY = YK. And YP is common to the two triangles SPY, EPY, and at right angles to SK. .: the angle SPY = SPY, and SP = PIC .'■ the straight line YP meets the parabola at P. Suppose that it also meets it at P'. Join SP' and draw P'lC perpendicular to XX, and produce PY to meet XX in T. Join ST. Then in the triangles SPT, EPT, SP, PT = KP, FT, and the angle SPT = KPT, .: the angle PST = PET = a right angle. Now SP = PXand SP' = P'K', .: SP : SP' :: PK : P'K' :: TP : TP'. .'. TS biaecia the angle between PS and P'S produced, and is .•. at right angles to the line bisecting the angle PSF'. But TS has been shewn to be perpendicular to SP. .'. P' must be indefinitely near to P. .'. the straight line TP passes through two consecutive points on the curve, and is .'. a tangent 6. Let x^ — y^ = a^ be the equation of an equilateral hyperbola. If »y be any point on the curve, the equation of the normal at x'y' is .r/ + i'xf = 2.ry. This expresses the relation between (a'/) and the coordinntes of any point (ay) on the normal. If the point on the normal be given, {h, k) suppose, to find a-y we have 2x'i/ = /y -V kx' {A) 238 SOL0TIOKS OF This represents a curve passing through the points where the normals from {h, k) meet the given hyperbola. Now {J) represents a hyperbola whose asymptotes are parallel to the axes. .'. it is rectangular. Now by Bes. Con. Art. 138, the four points in which two rectangular hyper- bolas intersect are such that any one of them is the orthooentre of the triangle formed by the other three. 7. Tripos 1875. Tuesday morning. No. 5. Pafek LXXXIX. 1. (1) Sr[uare and transpose. ^ a - X ' ^ a-\- X . a^ + x^ ^ ^ - 4 Vf T + 2 ■ • a2 _ 3.2 - ■ — 2" _ .r_2 _ 5-4 Vr ,2 b - iJb +.4 (2) For a: + y write z, and add 1 to both sides of the first equation. ;V + (0+ 1)2 = 136) 2xz{z + 1) = 120 I X'. .-. zx + z + I = ± l^ \ zx - {z +1) = ± 4 j .-. « -M = ± 10, ± 6 ; . . ^ = 9, - 11, 6, - 7, and «.r = 2 -f 1 ± 4. .-. (1) 9a: = 10 ± 4 = 14 or 6 (2) - ll.r = - 10 ± 4 = - 6 or - 14 (3) 5a: = 6 ± 4 = 10 or 2 (4) - Ix = - 6 i 4 = - 2 or - 10 WEEKLY PROJBLEM PAPERS. 239 The corresponding values ofy are given byy = ^ — ;r. (3) Subtract the second equation from the third. .: ^ - z + x(s - s) = 0, .: {x + l)(y - z) = Q. ;!; + 1 = 0, though true, gives an indeterminate equation to determine y and z. If y — z = 0, for z write y in (1), .-. / + 2y = 3, .-. y = 1 or - 3 ; .-. z = 1 or - 3. If we substitute these values in the given equations, we get the single value X = — 1. 2. A + B-\- C=Tr, .-. 2^ + 25 + 2<7 = 27r, .-. (tt - 2A) + (tt - 25) + (tt - 2C) = jr, and sin (tt — iJ) = sin iA ; cos (tt — 2A) = — cos iA. .: all formulae which hold for A, B, C hold for 1A, 2B, 2G by changing the signs of the cosines. Again 5^ + 55 + 5C = Stt, .-. (2n- - 54)H-(25r-55) + (2^-5(7)=^-, and sin (27r — 5A) = — sin hA ; cos (27r - 5A) = cos 5^. .•. all formulae which hold for A, B, C hold for 6A, bB, 6C by changing the signs of the sines. Now by Todh. Trig. Art 114, sin '2d + sin 25 + sin 2C = 4 sin A sin 5 sin C, .-. sin 10^ + sin 105 + sin 10(7 = 4 sin hA sin 65 sin 5(7, and by the same Article, when A ■{■ B -\- C = -■ 1 = tan B tan C + tan C tan ^ + tan A tan 5, .". cot A + cot 5 + cot <7 = cot ., i = JCsin 6 = AB sin (00° - ^) ._ a? + b'^ + ab = -=^ f23in2<^ + 2sin2(60°-^)-f-2sin<^sin(C0'>-(;^)( = 4?! {l _ cos20 + 1 - cos (120 - 20) + cos (20 - 60°) - cos 60°) ^d^ {a _ cos20 + 2sin3O°sin(8O° - 20)} = I AB\ Now area oi ABC = ^ AB . AB sin 60° = ^ ^£3 = -yj («' + i' + '■'^J = „-4 («^ + 4^ + ^=)- v3 2 V3 Note. — To construct tljis triangle see Catalan, Geom. JEl. p. 66. 4. Let D be the middle point of RQ. JiCQ = 2RPq, .-. DGq = RPQ. .-. a circle will go round QGNP. .: RNG = MNP = GQM; and RCN = QGif, .: the triangles J2(7i\^, QGM ave equiangular, and .-. similar. 5. Let the parabolas have a oomirion tangent at G. Through Cdraw any two chords CpP, GqQ. Join PQ, pq. Then since all parabolas are shuilar cui-ves .•. GP : Up :: GQ : Gq. .: PQ and pq are paralleL If one of the chords be turned about C until it becomes indefinitely near to the other, PQ and pq become the directions of the tangents at P and p. .". the tangents at the extremities of any chord through G are parallel. By the converse of this we obtain the required theorem. 6. By symmetry the circle touches the axis minor at the centre of the ellipse. .". taking this point as origin, its equation is of the form a:^+f + Ax = . . . . (I) We can write the equation of the ellipse in the form y2 + L^ ^2 _ ^2 = . . . (2) WEEKLY PROBLEM PAPERS. 241 Since the circle and ellipse have double contact, the equation of the circle is y^ + - .T^ — W' = {nix + tif, or f + a:2 (i^ - «A - "imnx - {V- + n") = . . . (3) Since (1) and (3) represent the same circle, .-. ^^ - ^2 = 1 ; «2 = - i2, .-. mV = -Ja^ - b\ a' e2 2A ■ .-. (1) becomes x^ -\- f ± - ^ifi _ yi^ = 0. and the radius of this circle is evidently - ^Ja? — L'. 7^ If V denote the velocity in feet per second, v = 88. If r be the radius of the curve,^ the vertical acceleration, the horizontal acceleration If iV denote the number of oscillations in any time, I the length of' the pendulum, N cc ,^ " . = '^j and the resultant acceleration = a/ a- + ^ I /new number of oscillationsy _ new resultant acceleration \forraer „ „ / former ~ ~ ~ "- ^1+' rY' \i20j ^ gh'-'-> ••• i = (^ + liff)* - 1 = s'ij ne^rfy. »2 88 X 88 ,_ ^ , ., r = - V3U = — -^i — v 30 feet = \ mile nearly. g 61 ■ ' It 242 SOLUTIONS OF Paper XC. 1. /S = 3 4- 2 -(- 29 + 36 + 137 + 122 + 429 + 200 -\- ... + ... qx^S = 3qx'^+ Sq3fi + 20qx'^+ 50qx^ + l^Sqx^ + ... r^fiS = 3r.rfi + 8rx* + 20^5 ^ sg^^s ^ ,., Add, and assume that the coeflBcients of x^, x*, :c^ vanish. .-. 3?-+ 8q+ 20p+ 50 = 0^ = 0. 8>-+202+ 50/; + 128 = }- 20r + 502 + 128/) + 338 = - From these equations we get^ = - 5, q = 11, r= — 6. WEEKLY PROBLEM PAPERS. 243 By trial we find that these values of p, g, r make the coefficient of x^ vanish. „ ^ 3 + .rg + 3p) + x^ (20 + 8p + 3g) _ ___3- W5 +_5.r2 — ^^ " ■■ ■■ (i_^)C] ---'^ by partial fractions. 1 - 6a; + IW - 6«3 (1 - x) (l - 1x) (1 - 3x) 1 1— bj; 1 — i!.r \ — a: .: the «"• term is 3"-i - 3 . 2''-i — 1. .•. the «*'' term of the given series _ d'^-T- - 3.2"-i - 1 1^ 3 1_ ,, „ . 2'» - 1 3» - 1 6™ - 1 .'. the sum 01 n terms = — ^ — = — - — r- 2"-i 2.3™-2 5.6«-i (3) This can be treated like No. I. p. 19. The »** term is — -—= — - — ; — ^ ■ This can be written 1.3.0 . . . (2» + 1) 2;^ f (2« + l)-l ) * 1.3.6...(2« + I) 1 2» S ■" ~ * I 1.3.5. .. (2» + l) ~ 1 . 3 . 5 . . . (2» - 1} j ' and the first term is — i -I t-^ — t f ... the Slim of » terms = i \l - 1 ^^\ g . . . (2n + l) } ' (4) Employ the method given in Arts. 7 and 8, pp. 5 and 6. The ra** term of the given series is n{>i + l)Vn = («3 + 2»2 + »)ffl7i = «?an + 2K2are + nan, .: S» = n\n + 3)2™-3 + «(h + l)2»-i -f m^™^! = «2»-3 (b^ + 7» + 8). 2. In the formula cos "■ — ^ ~ a^-^" Ty • • • for a write 2r, 4x, 6a:, &c., and neglect powers of x higher than the e2 2t4 solutions of fourth. The numerator on reduction becomes — -r- x*. and the de- 4 ! 192 nominator becomes — - a*. .•. the limiting value of the given fraction = ?51^ = 105. 192 3. The angle J?,^iCi =~- ABJ = B, jBiCi = BiC + CC, — i cot ^ + . ' = -. — . ■ ^ , ^ sin C sin^sinO .•- if ^1 be the radius of the circle round A-^B^Ci, -r, B^Ci a fsin B sin cos A . _. .1 i^i = — - =, \ + sm.4} zsin^i 2 sm^ sm 5 sm C I sin^ J = ^r~- — !—■ — ^:— : — ^ (2 sin^ ^ + 2 sin 5 sin (7 cos yf]-. 2 sm ^ sin ^ sin C ^ ' Now 2 sin^ sin C cos ^ = sin^ {sin (.4 + C) - sin (^ — C?)} = sin^iS - %va.{A-\- C) sin (^ - C) = sin^ 5 — sin^ A + sin^ C. . ^ _ ^ sing A + sin'-g + sin^ G ^ ^ 2 sin ^ sin ^ sin G Similarly we can shew that ^^ _ ^ sing^ + sin" J + sin^ ^ ^ r %v^A + sin^ .g + sin^ O ^ ^ 2 sin ^ sin ^ sin (7 V 2 sin ^ sin 5 sin C ) . „ „ l^\T? A + sin2 B + sin2 Cn" . . li/n = it 1 : rr : ;= 1 • \ 2 Sin ji sin B eiu C J 4. Join A'B, and produce BIf to meet A'B in B. Produce AlBl to meet BB in (?. Then since BUB is a transversal of A'BO, .-. BB.GB . EA' = GB . A'B! . BB, or AC.B'B.EA: = AA' . CB' . BB, .: ABB' is a transversal of JBC. WEEKLY PEOBLEM PAPEES. 245 5. Let the equations to (J) and {£) be ^2 = iax (1); ^2 _ _ ^gj. ^2) The pole of {h, Ic) with respect to B is ylc = - 2a(.r + h) (3), .■.;;:=- (g + h)- Substitute for x in (1), .-. / - iyh + 4ai = (4), .•. if (3) is a tangent to {J), (4) must have equal roots, .•. Ic^ = iah, . •. (J) is the locus of {h, k). G. The equation to PQ is ^ ~f = ~, , where (a^V") are tne co- ay ^V ordinates of P. If (I, J/) be the coordinates of Q, j; - / I - .r ay ^v Since (^, ij), (af, y') are on the ellipse, . >; V ^' = 1 = ^' 4- •'■". • ■ ^2 + o2 i2 "^ a^ »'-/' = _ ^ - ^'' ('>^ Now (2) is satisfied by the coordinates of P, F and Q. .: dividing (2) by (1), the equation of Q^ is 7. Let e be the elasticity, » the velocity, a the angle of projection, a the distance of the point of projection from the wall. 246 SOLUTIONS OF Then the time to the wall = and the time back again Also the whole time V cos a a ev cos a 2» sin a g .: v^ . sin 2a = aff{\. -]- e). Now the right-hand side is constant; .". c is a min. when sin 2a is a max., i.e. when a = - • 4 Paper XCI. 1. Let X, V denote the two quantities. Their H.M. is , • The G.M. between x and ^^•^- is x /J -^ ■ The G.M. between X + y ^ X + 1/ -^ and;/ isy f./-2f- — .-. by question 2 V2 'Jxy = Va- + y ( V^ + Vy), •■• (« + 5')^ + 2 'Jxy {x + y) = 8:!:^, .•. a: + y = — vay ± v 9j!;y = — 4 sxy or 2 vay. The value a; + y = 2 v.ry gives us .r = y, which is inadmissible. .-. ^ +y = - 4 V^ ,, ^ + 4 y^ + 1 = 0, ■•■ /v/- = - 2 ± Vs, WEEKLY PROBLEM PAPERS. 247 2. (m + 1)2 = «i2 -f 3m + I > (ra -f- l)(m + 2) {m + 1)2 = »j2 + 7ra + ^ > (ra + 3)(»! + 4) fm + ^^-^J =m^+ (4n - l)m *+ (^^^^T >(» + 2»)(» + 2»-l) ... (^ + 1)2 (^ + 1)2 ... (^ + ^^J J > (». + 2«) ... {m + 1), ••• (»» + I) (»» + i) ■■• [«i + —2~) > ( ^ I / S. Let y — 1, y, y + 1 be the lengths of the sides. Then s = —, s - a = , «-5 = '^,«-(; = -:- > 2 2 2' 2 .-. ^2 = ,(, _«)(,_ 4) (.-«)= t' . 3(y2 _ 4). .". 3(y2 — 4) must be a perfect square, = .r2 suppose. Then :;: is a multiple of 3. One solution is easily seen to be ^r = 6, y = 4. Consider the equation x^ - 3^^ = I. J_ 1 + 2 + ^^ = i + ri2i .-. Todh. A Iff. Art. 639, one solution is .i; = 2, y = 1. By Art. 643 we obtain the other solutions thus. (2 - V3)» (2 + -^3)" = 1 = (x- V3y) (.y + s'3j/). Let .r - \/3y = (2 - VS)" ; a: + ^S.?- = (2 + a/3)« , . •. ^ = i {(2 + V3)'' + (2 - v^3)" } ; y = _1^ {(2+ ^'3)« - (2 - ^3)-}. By giving to n hi succession the values 1, 2, 3, 4 we get a; = 2,7, 26, 99, y .: Todh. Jiff. Art. 645, we have ^ = pn ± qm, where ^ = 6, j = 4, and m and tt are corresponding values of a; and y in (A), .-. y = 4, 14, 52, 194, 724, &c. 2,7,26,99, ■ ■ l^ 1, 4, 15, 56, ... 3 248 SOLUTIONS OF 4. Let be the circum-centre, P the ortho-centre. Bisect OP in N. Then N is a fixed point. Now the radius of the nine-points' circle = ^ the radius of the circum-circle, and is . '. given, and its centre is at N. .: the nine-points' circle, which passes through the middle points of the sides, is fixed. 5. If She the other focus, ITq = SQ, since Qq is a diameter. Since St bisects the angle PSq .'. the angle qSl = FSl = Slq, .-. qt = Sq. Similarly TQ = SQ. .'. TQ -'r tq = Sq + SQ = Sq + m/ = AA'. 6. Let PSQ be the focal chord, Q being nearer the vertex than P. Let the normals at P and Q meet the axis in G, 0' ; and let the tangents intersect in the directrix at Z. Let T denote the tension of the string, which acts along the tangents at P and Q. Then Wi:T::X : sin SQP :: 1 : cos SPZ, Ws-. T::l: sinSG^'Q :: 1 : cos SQZ, and SQZ ='^ - SPZ, .-. T == W, cos SPZ = Fj sin SPZ, . ?> = tan SPZ=^-? ■ ^2 = tan SQZ = ^, W.^ ^ SP W-, ^ SQ . ^1 , ^^ - 7 = \ cot a. .: c[> = tan-i (J cot a). WEEKLY PKOBLEM PAPERS. 249 Paper XOIL 1. (o- - af + (o- - by + (o- - cf - 3(0- - a) { a riglit angle, FB^ > PQ2 + QB^ by Euc. II. 12. „ QBB „ §52 > QR^ + RB^, &c. .-. AJ? > AF^ + Pq^ + . . . . -{- KJ^. 5. The circle on PQ as diameter touches the parabola atP and passes through K. Bes. Parab. Art. 50. .'. PA'§ is a right angle. 6. After the pie.ce has been cut out let x be the distance of the centre of gravity from the hollow end, and let A be the area of a transverse section. Then X (as. - "^ . l] == Aa .^ - A . ~ . - \ n. J 2 n 'i I I — 3 2 na^ — /2 .•. X 1{m - I) WEEKLY PROBLEM PAPERS. 251 .•. the distance moved = Put i . "-^ na^ — l^ ti. 2(m-l) 2 2m 2 na- I -" .: P - al+ %my - 2/y = 0, .-. ^2 _ ;(a + 2,/) + Inay = (1). When ^ is a max. (1} has equal roots. .•. the point of support must be under the extremity of the bore. 7. Tripos 1878. Tuesday morning. No. 9. Paper XCIII. Vl - a.'= 1 + z2 1 + a-a: ' 1 - W (i - a.4)(i _ ^) .: since a-, y, z are all unequal, the- given equation reduces to = XZ{X^ - 02)(1 _/) + 2y(22 _ y2)(l _ 3.4) + yx{f - X^){1 - sfi) = (A — xz^ -t- y^x — y^ + 2^y — zy^y (J) — xyz{x'^y^ — a?y'^ ■\- y'^z^ — y^z"^ -\- ^x^ — zV) {B). Consider the expression A. It vanishes when we put x =y, ot x = z, OT y = z, and is a symmetrical expression of the 4tli degree. .-. A = Eipn — y){y - z)(z — x){x -{■ y + z), where E is some numerical constant. Put a; = 2, y = 1, « = 0, and we get E = \. Now consider B. It vanishes when we put x = y, or x = z, or y = z, and is a symmetrical expression of the 5th degree. .-. B^{x-y) {y -z){^- x) [C{x^ +y^ + z^) + D{yz + zx + xy)] Equating coefBcieuts of like terms, we get C = 0, Z) = 1. 252 SOLITTIONS OF .•. the given equation becomes = (.r - y)0 - 2)(« - .v) {x -ty + z - 3:yz{xy -\-yz-\- zx^ = a;+y + z - aryz(xy+yz + zx). 2. sin-i A + cos-i I = 2 sin-i | = 2 tan-i | = tan-i -— ^ -r = tan-l (- =^), ,-. - 5^ = tan jtan-i (x+1) -[- tan-i -^-\ a- - 1 „ sin BJR _SS.^ '^CR ^ g sin CAR sin i< ~ AB,^ AR sin C ' Bin C'-^g _ CQ ^25§^ 2 sin BAQ sin ~ ^§ ~ ^§ sin ^ ' .-. sin BAB . sin C4§ = 4 sin (7^72 sin BAQ. 4. Draw Ji? perpendicular to CZ>. Then since tlie circles are equal, .•. the angle Bt)C = BCD. .: the triangle BCD is isosceles ; .'. E is the middle point of the base, and B lies on the circle described on AB as diameter. 5. Tripos 1878. Monday morning. No- 12. 6. Tripos 1875. 2nd Tuesday afternoon. No. 1. 7. Take the point of contact as origin, the common tangent, and the line joining the centres as axes. Let the coordinates of P at any time be (x^. y-^. Then those of P' at the same time are {— Xi, — y^. Similarly if «, v be the component velocities of P parallel to the axes, — u, — V will be those of P'. .: if (,t, y) be the relative coordinates of P' with respect to P, x = 2xi, y = 2yi ; and the component relative velocities of P' with respect to P are 2m, 2». .'. P' will appear to P to be moving in a circle whose radius is equal to twice the radius of either circle. WEEKLY PROBLEM PAPEES. 253 Paper XCIV. 1. Tripos 1878. "Wednesday morning. No. 5. 2. (1). Square and transpose. . 2 - (o3 -j- 42) (tan2 6 + cot^ 6) + 2aH^ = 2 V{l - «^(tan2 6 + cot^ ff) + ««} {l - ^^(tan^ 6 + cot^ 6) + 4«}. On squaring both sides and reducing, we obtain («2 - b'Y (tan^fl - cot2fl)2 = 0, .■. tan 6 = ± cot^, .'. 6 = ktt ± - ■ f2). a cos 5 + 4 sin 5 = — j^; « + -7^ i = a cos - + 4 sin -, 4 4 .•. « fcos 6 -,cos I ) = - 4 f sin5 - sin ^Y •■• ^'*'^^'' 5 - 8 = """' '"■ *^° I2 + 8 j = ^' .-. ^ + IT = ^ + tan-l t 2^8 « 1 ^ (3). 4 sin3 5 - sin 35 = — = 4 sin^ | - sin ~, . 371 4 ( sinS fl - sinS f j = gin 3fl - sin - = 3 sin 6 - 4 sin^ .-. 8 (sinSfl - sinS |) = 3 (sin 6 - sin |), = 3sinfi - 4sin35 - 3sin| + 4gin3^, 254 SOLUTIONS OF .". either sinfl - sin - = 0, .•. sin 6 = sin -, .•. 6 = iitt 4- (— 1)" _, or sin^ 6 -\- sind sin - -|- sin^ _ = -, 4 4 8 ■■■ (^^"' + 24/^T = '' ••■ ^ = - + (- 1)" {^--^ (- 4)} ■ 3. Express all the terms as sines and cosines, transpose all to the left- hand side, and bring to a common denominator. Then sec^5 + sec^C"-!- 2sec5secCcoB^ — sec^eeoCsin^ (tan^-f- tanC) = . — ,— Icos^ C + cos^S -I- 2 cos 5 cos CcosA - sin^^] cos^Jcos^C^ > = — 3-^ — — {cos^ B -\- icosB cosCcosA + cos{C-^J)cos(C- J)\ = ■ — s-^^r- — s-Tv °os B {2 cos AcosC - (cos C + A ->r cos C - A)] cos'B cos' (7 '■ ' = 0. 4. Consider the point C as a point circle. Then BD is the radical axis. .-. by Casey, p. 113. Cor. 1, CF^ = 2C0 . MN, where is the centre of the given circle, M is the intersection of 1)B and CO, and iV is the foot of the perpendicular from B on CO. 5. Tripos 1875. Wednesday morning. No. 6. 0. Describe an ellipse passing through 0, and confooal to the given ellipse, and let C be a point on the confocal near to 0. The tangents from and 0' to the given ellipse intersect in A and B. The normals at and C to the confocal bisect the angles AOB, AO'B. Now, with foci A and B, describe a conic passing through 0. Since its normal bisects the angle AOB, the two conies touch, and .'. ultimately they both pass through 0', and .'. their normals at coincide. .'. the two curves have the same curvature at 0, and their central chords of curvature are equal. ■vT t T \ J c 1. n product of focal distances Now central chord of curvature = 2 , distance from centre . OA.OB ^ OS. OH " ON 00 ' WEEKLY PROBLEM PAPERS. 255 7. The ball is reflected each time at the same angle as it struck tlie circumference. .". during the time of falling only the vertical velocity is altered, and .•. the intervals between the reflections are equal. Let u be the velocity of projection. Since the number of reflections depends only on the horizontal motion of the ball, we may suppose the ball to move in an equilateral polygon inscribed in a circle ou a smooth horizontal table. The length of each side of the polygon is 2r sin ''L, .: if t be the time of reaching the water, ui = 2>ir sin -, But if A is the depth, 9 Now if s be the space to which M is due, k^ = Igs. .". s r2 / . ffV Paper XCV. 1. (1). The n^^ term = n{ji + 2) = n{n + 1) + a. .-. the sum of n terms ^ n{n ■VV){n-\- 2) ^ r^J) ^ j„(^ + j) ^^^ + 7). 3 2 (2). The »tii term = n{n + l)^ = «(« + 1) (» + 2) - r,{n ■\- 1). .•. the sum oin terms _ ^(^ + l)(;. + 2)(;. + 3) .(;, + l)fa^2) ^ ^,,,(;, + l)(^ + 2)(3« + 5). 4 3 ' (3). The «th term = ^^—^^^2^,^-^ 256 SOLUTIONS OF .•. the sum of n terms = =■ i 1 - — r, \ • See Errata. 8 1 {Zii + 1/ J 2. Let X, xr, tcr^ denote the numher of degrees in the angles of the 1st triangle, 3.r, ZxU, ^ixW' the number in the angles of the 2nd. Then we have X + xr + xr^ = 180 \ 3x + 3xB + 3rf2 = 180 ( xr^ + 3.riE2 = 240 ) ^ -\- r -\- r^ _ 1 + r + c^ _ „ ■ ■ 1 + li + B?' ~ ' ~W^fiW ~ '*'' .-. ^R\ + 3^2 = 4^2 + 4r + 4, .-. 9i22 = ^2 + 4r + 4 = (?■ + 2)2, .-. 3i2 = r + 2, .-. 3(1 + r + ?-2) = 9(1 + iJ + i22) = 9 + 3r + 6 + ?-2 + 4!- + 4, ... ^2 - 2»- - 8 = 0, .-. (r - 4) (?• + 2) = C, .■.r = i, i2 = 2, a' = 1^, , .^, , ,. ^ 180 4 X 180 16 X 180 •. angles of the 1st m degrees are ■— -, • — , — , < . . . , TT Al! 1677 ,, „ 1st in circular measure are — , — , -^— , ill. ^1 ^1 11 >> ''^^ II 11 1) 7 ' "7 ' T ' 3. (1). The expression on the right = _ {log. 10 -h -,+ - . -^ + -._ + .. .} 3 , ,. , 1 (3 , 1 32 , 1 33 , = -log. 10 + _|- +-._+-._+ . =^|,iogao-i,iog.(i-|) 3 , T„ 1 1 125 = j-0log.l0-_loge^ WEEKLY PROBLEM PAPERS. 257 3 , ,. 1 , 103 = i loge 10 - i log* 10 + logs 2 = log, 2. In finding the coefficient of afl no term higher than x"-'^ will be required. If k is even, the sigh of each term will evidently be positive, and if it is odd, the sign will be negative. .-. coef. oin^ = (- 1)"/ — ^ 1 i 1 i h . . .1 ^ ' \l(n - 1) ^ ^ti - 2) ^ 3(» - 3) / _ 2(- 1)" (1,1,1, . 1 ) - ^i~ It + 2 + 3 + • ■ • + ^rrr]' since each fraction will occur twice. 4. Produce J/B to meet CN in 5'. Then MD : AH :: ND : NE :: DE' : EC. But AE = EC, .: MD = H^' ; and BD = DC, .: BM and CN are parallel. 5. Let the tangent at F meet the conjugate diameters CD, CD' in if and N. Draw the ordinates Pm, Pn, m being on CD. Then Ci)2 = CM.Cm = CM . Pn, cm = CW". Cfe = CN .Pm, .: CD'^ . Ciy^ = CM. Pm X CN. Pn, •:OM.Pm^^^,.:ACPM^^-^ 6. Let the equations to the tangents be y = miX + Ci; y = m^x -^ c.^ The coordinates of the foci are {x, y), {- x, - y). 258 SOLUTIONS OF Then since SF . ST = £C^ = SZ . S'Z', . (y - Wia; - g;) {- j/ + mix - gj) ^ (y -m,x- C2)(- s/4-m^ - g,J 1 + V 1 + V ' .: &' = y - ^2 + 2xy . L^^l^a »Ji + OTg = ^2 — «2 + 2;ry cot (a + /3). 7. Tripos 1878. Tuesday morning. No. 7. Paper XCVI. 1. Tripos 1875. Monday morning. No. 2. 2. Tripos 1875. Wednesday morning. No. 7. 3. Let a be the least side, B the radius of the circum-circlo. Then .g_.2ie, .•.| = i^ = g. .-. ^ = fl 4. Let ^', 5', C" be the feet of the perpendiculars from D on BC, GJ, AB. Then DA' . BC = 2 ^ BBC = DC.DBsm A, 1 _ BC 1 _ IB.BA ^ DA ■'■ DA' ~ Bin A ' DB . DC ~ DA . DB . DC ~ yi suppose, .-. i)J . i)^' = n = DB. DB' = i)C. DC by symmetry. Now ^'0'2 = i),^'2 + Z)C'2 - 2DA'.DC' . cos A' DC = 2 /J_ , J_ ■ 2cos ^ 1 (5^2 + DC^ + 2DA. DC cos B) = -J^-4£! .A'C DAKDC^^ ' DAKBC^ liAG DA. DC WEEKLY PROBLEM PAPERS. 259 Now d.A'C =^ DA'. DC sin B = — >^ sin B, DA: DC ' ■ d = <"' sin B y -P-^--Pg _ f^-BinB _ DA.DB.DO 1 iJ^.iJC "" fi.AC ~ AC 21 ' 2S' .-.idB^ = DA.BB.DC. 5. Produce PO to meet the curve in P', and join P§, Pff. Then since PC = CP', and QC = C§', . ■. PqPq is a parallelogram. .-. the angle TPQ = PP'Q = P'P^-. To each add CFT. .: the angle CPQ = IPQ'. 6. Draw ST, SZ, ST', S'Z' perpendiculars from the foci on the tangents. Then the angle SPY = S'PY', .: sm^ SPT = sin SPY. sin S'PF' ^^ _ ^ ^^ 6P S'P CD'' sin^SQT = sin^OT-.sin^'C^ = I " ff' = S .-. sin2 SPT + sin^ SQT = BC^ (^^ + -^) = const, since CD and CV are at right angles. Salm. Con. Art. 159, Ex. 1. 7. If /be the acceleration of W, the acceleration of P will be tif. The kinetic energy of the system at the end of time t is \Pr?ffi WWp^. (^) The work done by gravity during the time ^is 4 (P - ^) nfgi^ {S) .: equating (^) and (P) we have fSizJh _nP-W •' n^P -\-W '^ fi^P + W'^' £ 2 2G0 SOLUTIONS OF Papee XCVII. 1. Tripos 1875. Wednesday morning. No. 5. 2. 2 sinS e = sin (A - 6) {cos {B - 0) - coa(£ + C - 20)}, .-. 4siaS5 = sm(A + £ - C - 6) + sin (J + G - £ - 6) -Bin{£+0- A-e)-Bin(A + £+C-3e) = sin {2C+d) + sin (2£ + ff) + sin (2 A +6)- sin 3(9, .-. 3sini9 = 4sin3fl + siu35 = sin e (cos 2C + cos 25 + cos 2^ + cos 6 (sin 2(7 + sin 2B + sin 2A), .a 3 - co3 2^ - cos2B - cos2C , . cot 6 = — sin 2^ + sin 25 + sin 2C 4(1 + cos ^ cos 5 cos C) m ji ,7, ■ ri — I- ■,(. = ^ . . — 7—: — =r-^ — yi — ', Todh. Trzfl'. Cap. vm. Ex. 18. 4sin^sin5smC ' ^ ^ = cot ^ + cot if + cote. „ „ „ Ex.28. 3. Consider T as a point circle. Then iZ is a point on the radical axis. 4. Tripos 1878. Monday morning. No. 10. 5. The equation to the normal to the parabola y^ — i^ ^t the point {x', y) is y - / = - -J (H! - x)- If {?i, k) be a point through which this normal passes, k - y = - Ji (h - x'). Tlie equation which determines the ordinates of the points in which normals through (A, k) meet the parabola is y' + (I - h) ii/ - '^ = 0, .'. if yi, ^2' ^3 ^^ til® roots of this equation, ^1+^2 + ^3 = 0, ^2^3 + nyi + yxVi = ' V2 ~ ^V' ^'■''2^3 = 2 • WEEKLY PROBLEM PAPERS. 261 Now twice the area of the triangle (^i, yi), (JTj, yj), {x^, y^, 2 A = y\{H - ^s) + y4.H - '^i) + .'/3<'^i - ^2) = f {y\{yi - yi) + y^^yi - y^^) + y-J^yi - yi')} = J- (yi - yi)(.y2 - ysXya - yi)- Now ^3 = - (yi + ^2). ■■• (yi - ^2)^ = - %iy2 - ^ (2 ~ V Similarly (^2 - yi7 = - Syz^s " ^ (2 ~ '')' and (yg - yO'' = - ^s^i " ^ (2 " '^)' ... _ 4/2A^ = - (yi - y^)^ (y2 - Vzf {y^ - yj" - {3yiy2 + ^ (r- ^)} {%2y3 + ^ (^ - '^)) {^3^1 + ^ (^ - '')} = 27 yi Vya' + 9^ (^ - ^) yi y2y3(yi + y2 + ys) Now since (h, k) ]ies on the curve y + ^ (/ - 2a.-)' = - «^ we see that Z iv i6A^ = «2,,.^=!^^= const. If the point (;i, h) he taken so that two of the normals coincide, the area of the triangle will vanish, and .■. a = 0. 262 SOLUTIONS OF .-. the locus of the point from which only two separate uormals can be drawn to the parabola y^ = h, is the curve / + 4 (^ - 2^^)' = 0- 6. If {h, Jc) be the point of intersection of the normals, the points where they meet curve y^ = Ix are given by the intersection of the parabola with the curve Jc - y = - I {h - x), or 2.ry = y(2A - I) + Ik. This represents a rectangular hyperbola whose asymptotes are parallel to the axes, and whose centre is given by the equations 2x = h — I, 1y — 0. The latter shews that the centre always lies on the axis of the parabola. 7. Tripos 1875. Tuesday morning. No. 12. Paper XCVIII. 1. Let / be ^s income for any given year. First suppose I to increase in A.P. common difference h. Then /+ 3, J + 25, 1+ 3«, J + 4«, i"+ 55 . . . represents his income for the 2nd, 3rd, 4th, 5th, 6th .. . years. .-. if c denote the percentage, the income tax on the 3rd, 4th, 5th, 6th .. . years is i±h^ i±n^ i±u^ i±u _ _ ^ ^.^.^^ .^ ^^ ^_p_ ,,^^ „„„„„„ C C C B difference being -. c Next let the income increase in G.P., the common ratio being r. Then his income for the different years beginning with the first is /, Ir, Ir^ . . . .: if d be the percentage, the tax for the 3rd, 4th, 6th ... years is Bd ' Sd ' 3cl . . . ^ which is a G.P. common ra,tio r. WEEKLY PEOBLEM PAPERS. 263 2. «2 = 2%^, Kg = 3[(1 + ]c)y^ _ 22«j^| = 2,njc = 3.2. ?/i/i;2, «4 = 4l(l + 4)^3 - 3?«2^} = #3^ = 4 . 3 . 2 . «i/£^ &c, •■ oil! ^ l!2! ■'"213'! '^ "' = «i + «i^ + «i I?, + «i I?, -f ... 3. Let^= sine. !H^ - Bin 25 . !il^ + sin 35 .i^^ 1 2 ' 3 ■" C7 = cos e . !!L.^ - oos 26 . ?i5l^ + cos 35 . fi5i^ 1 2 ^ 3 ••• 1 2 ^ 3 ■" = log (1 + fi9« sin 5), ^ /. 1 + «" sin 5 = eO + Si = eC, gSi^ .: 1 + sin (9 (cos 5 + J sin 6) = eO . (cos /S + s sin a?), .-. eO cos /S = 1 -\- sin fl cos 5 ) (1) fiC sin S = sin 25 J (2) .-. cot S = cosec 25 + cot 5 = 1 + cot 5 + cot^ 5. Also squaring and adding (1) and (2) e2C = 1 + sin 25 + sin^e .-. 2C = log {1 + sin 25 + sin^ 5}. ' 4. Let J be the centre of the inscribed circle. Draw ID, HE perprn- dicular to AB. Then D and B are points of contact and AE = »■ AI) = s — a, where 2s = a -{- -^ c. Then AE : AD :: OA : I A, .: s : s - a :: OA^ : OA . lA :: OA^ : be, from similar triangles OAB, I AC. 264 SOLUTIONS OF And from the same triangles, OB : IC :: OA : AC, OB' IC^ OA^ s - c s s ~ c bo ab bo a s — a s — a or.'' s - b ab s — a OA^ OB^ 00^ _ J "be ca ab Similarly 5. Let A, A!, bethe vertices of the ellipse and cone, and let the focal spheres touch OA in H and ^'. Then A'O - AO = OF' - A'F' - {OE + EA) = OF" - A'F - OF - AS = A'F - AS = A'S - AS= SS'. .'. the locus of is a hyperbola, whose foci are A and A'. 6. Tripos 1878. Tuesday morning. No. 6.. 7. On AB as diameter describe a circle, centre 0. Draw FN perpen- dicular to AB meeting the circle in R. Let AOB = 6. Then the a tangent at P makes an angle — with the horizon. it The length Pq, = arc AP = 2 . chord AS, and PQ is parallel to AE. If i and f be the times down PQ and AR, ,, 2PQ 4AR . ^ ^ .2AR gsm- gsm- ffsm- .: t = 'J'l.V. Now t is constant, being the time of descent to the lowest point of a vertical circle. .'. ^ is constant. Paper XCIX. 1. (1). Let X = (/r^ + 3a;2 + 34r + 37)4 ; r = (a:3 - 3a;2 -f 34x - 37)i. WEEKLY PEOBLEM PAPERS. 265 Cube both sides of the given equation. .-. 8 = J? - r3 - 3xr(x - r), and x - r = 2, .-. a;2 + 11 = xr = {(a;3 + 34ir)2 - (3a:2 + 37)2U, .-. «« + 33:2:4 + 363:r2 + 1331 = x^ + 59.r* + 934^2 - 1369, .-. 26a;'' + 571:1:2 _ 270O = 0, .-. :r2 = 4 or - .^/ ;.-.:!;= ± 2, ± 15 V^^ (2). Multiply (1) bj'y, and (2) by s, and then subtract. x*^ - %fx + 48y(y2 - 1) = A - 8/a; - 6a:(a;' - 4) = 0, .-. if-x^-%i/ + ix = Q, .-. (2y-a;)(4/ + 2,ry + a:2-4) = 0. Put a: = 2y in the given equations. .•. ^ = ± 1, a: = ± 2. Again, add the given equations. .-. a»{x + 2^) - %f{x + 2y) - 12(a:2 - 4/) = 0, .-. (a: + 2y) (a; - 2^) (a;^ + 2a:y + 4/ - 12) = 0. Put:.= -2y. .-. y = Ili±_^ :. = ^1^^. ^ s 4 ' 2 (3). Put 1 - 16/ = a, 1 - 16a:2 = jr. .-. n/k - V» = 2(a; + y) ] . , , , ai . >• and K - p = 16(a:2 - /). • • n/k + V» = 8(a: - y) J Square and add. .-. - 2(« + ») = 4(a; +y)2 + 64(a: - yf, or 1 - 8a:2 - 8/ = (a: + ^)2 + 16fa: - y)^. Let a: +y = a, a; - y = 3. .-. 1 = 5a2 + 20/32 = 3a2 - /32 3 ^ 1 Also from the 2nd given equation, f = 3o2 — j3' 7 From these equations we can obtain the four values of x and y. 2. Let the inscribed circle touch .^fCin E. Then O^is perpendicular to AC. .: AE = AO cos ^, EC = OE cot ~ = 0^ sin ~ cot ?, 266 SOLUTIONS OF C B ^ COS - _, cos - r1 .-. b = JS+EC = ^0 ("cos - + isin ~\ = AO sin - ' sin - 2 2 .'. AO = ^ sec - sin - . See Errata. 2 2 Now ax = ^ = Wr cos -, &c (1) Sin — 2 .-. {A) = 2(iVyz2 + c2aV.r2 + aWxhf) - oV - ^V^ - «'.^* = {ax -\- by ■\- cz) {by -\- cz — ax) {ez -\- ax - by) {ax + iy — cz) ,,_,./ A , B , C\ [ B , C A\ = (4^»-)*( cos - + COS - + COS - j ( cos - + cos - — cos- j multiplied by 2 other factors. „ A , B , . IT - A It - B IT - n Now cos - + cos — \- cos -i = 4 cos cos cos 2 2 2 4 4 4 Todh. Trig. Cap. viii. Ex. 20. , 1 / A , . A\ \( B , . B\ \( G , . C\ = 4 . — — cos - -f sin -- — ;^ I cos - + sin - — — cos — f- sin - , V2 V 4^ ^J ^/^\ 4^ 4/^2^ * 4/' and COS-+ C08--L - cos - = 4 cos ; — cos — '- — cos — ■ — .MEx.21. 2 2 2 4 4 4 , 1 / A , . A\ 1 / B . B\ 1 / a . C\ .■.{A)= 4(4ie?-)<(cos2 2- sin2 -^jYcos^^ - An^-^ /cos^^ -sin^^Y = 4(4i2)V cos2^ cos2-? cbs^^ Ji Z /i = 4(42J)M ^ . X- ■ ^, from (1) = {a -{- i + tf)^ ary^', since 4B = —--, and r = - • o s WEEKLY PROBLEM PAPEES. 267 3. Let ABCD be the quadrilateral. Let E, F, 6 be the middle points of BD, AC, EF, and let H, Z, K, M be the middle points of EG, CD, DA, AS. Then since AK = KD, and AM = MB, .: KM is parallel to BD. .: SZKM is a parallelogram having its sides parallel to AC and BE. Similarly EHFK is a parallelogram having its sides parallel to AB and CD. Since the diagonals of a parallelogram bisect each other, .•. the middle points of HK and LM are at G. Then by Todh. Euc. Ap. Prop. 1, GA? + GS^ =r 2(A3P + GM^), &c. ,-. 2{GA^ + 6B^ + 6C^ + GE^) = EK^ + LM^ + 1{AB^ + BC^ + CD^ + EA^). Now AB^ 4- EC^ = •KJ.F^ + BF^), CIP + i)^2 = liAF"^ -f- i^lD^), . •. AW' + 5(72 _|_ crz)2 + EA^ = 4^J^ + 1(,BF'^ + i?'2)3) = ^C2 + 4(JJ'2 + ^i^a) = AC^ + 5i)2 + 4S'J'2, .-. ■il^GA^+GB^'-^GC^^-\-GE^)=EK'^+LM■^-{■1EF'^-\-\{AC^^BE'^). (1) Again, (5^2 + (?(72 = 2^^^ ^ 2CS'P = 2a = iSP + EP, .: S'P = £P. Similarly S'Q = EQ. . .: S'PE, S'QE are isosceles triangles. .•. PQ bisects S'E at right angles in Z. Now Iflies on a fixed circle, centre S, and S' is a fixed point, .". by Todh. Euc, Ap. p. 332, the locus of ^ is a circle, centre C ; and since PQ is at right angles to S'Z at the point Z, .: the envelope of PQ is a conic of which the locus of Z is the auxiliary circle. Also, C is the centre and S' one of the foci. .•. the other focus is S. It will be seen from a figure that if SE < SS' PQ will cut SS' between S and S', and the locus is a hyperbola. If ^Sff > SS', PQ will cut SS' produced, and the locus is an ellipse. If SE — SS', PQ will always pass through the focus S, which may be considered as a point ellipse. See Errata. ^GS SOLUTIONS OF 6. On AC take a point E such that EC = OA, and on BD take i?" so that FD = BO. TheC. ofG. of ^at^and~at Cis E, sines AE. ^— = EC .^ AC AG AC AC and the weight which we may suppose placed at E will be 0A+ PC ^ J AG Similarly the C. of G. of — - at B, and — - at B is F, and the weight JjB BD = 1. .•. instead of 5 particles at A, B, C, B, 0, we have now 3 equal particles at 0, E, F, and their C. of G. will be the C. of G. of the triangle OEF. We Vill now find the C. of G. of the quadrilateral. Bisect OE in G, so that G is the middle point of AC. Join BG, GT), and take points K, L on BG and DC so that KG = iBG, GL = \BG, and join KL., which is parallel to BB. Then Xis the 0. of G. of the triangle ABC, and L is the C. of G. of ABC. .: Mis the C. of G. of the quadrilateral ABCB, so that LM ^ AABC ^ BO _ BF me: ~ AABC ~ OB ~ fb' . . Gilf produced passes through F, and the C. of G. of ABCB lies on GF. Similarly it can he shewn to lie on EH. .: it is the point of intersection of G'^and HE, and /. coincides with the C. of G. of the triangle OEF, i.e. it coincides with the 0. of G. of the 5 particles at A, B, C, B, 0, by what has been shewn. 7. Vr+l = »r + Vr-l, .*. Vr = »r+I - Vr-i; Vr = Vr-l + Sr-2, .'. Vr — Vr-i = Vr-\ ; .: Vr (Vr — Vr-i) = Vr-l (Or+1 - Vr-l) ; . Sr ^ Vr-l Vr+1 = Vr-V ~ Vr-2 Vr = Vr-i^ ~Vr-SVr-l = Pjj' ~ Si»3, and pj = Pj5 -j- pj, = X^Bi^ — (X + l)vi^, for »2 = Xvi, = V (X2 - X - 1). WEEKLY PROBLEM PAPERS. 269 Papee C. 1. From (1) and (2) we have y + « z + x' .: z{x - y){y + z){is + a;) = x\z -\- x) - f{y H- z) = z(^ - f) + a;* - /, •■• ^(y + ^)(« + «) = «(«^ + isy + /) + (^ + y)(-!^^ +/), since x # y, .•. reducing, we have xz^ — A + 9/z^ — y% = afi -\- x^t/ -\- xy^ -^ y^ — i?. . {A) Similarly from (2) and (3) we obtain on reduction x^z — xz^ + x'^y — xy^ — y^ -\- y^z + yz^ -\- z^ — x^. . {£) Adding (A) and (B) we have ^^2 _j. y3 _^ ^2^ = Q^ and y ^ Oj .: X -{■ y -{- z = 0. a? Again yz -] = yz — x^, since y ->[■ z = — x = yz + x[y + z) „ „ = yz -\- xy -\- xz. , 2. Tripos 1878. 2nd Monday afternoon. No. 2. 3. Let 1, 2, 3, 4, 5, 6 be the feet of the perpendiculars from on DC, DB, BA, CA, GB, BA. Then 134 is the pedal line of the triangle ADC, 125 of BCD, 236 of ABD, 456 of ABC. The circle on OD as diameter evidently passes through 1, 2, 3, i.e. the circle through 1, 2, 3 passes through 0. Similarly we can see that lies on each circle described round the triangles formed by taking any three of the sis points 1, 2, 3, 4, 5, 6. That is the point is the intersection of the circles round the four triangles formed by the four pedal lines. .'. if we describe a parabola touching these four pedal lines, the focus will be at 0, and the feet of the perpendiculars from on the four pedal lines will lie on a straight line, viz. the tangent at the vertex of the parabola. See LXXXVII. No. 5. 270 SOLUTIONS OF This question is solved analytically in Smith's Conies, p. 83. From the form of the solution there given we see that if be fixed, and any number of points be taken on the circumference, and triangles formed by joining these points by threes, and the pedal lines of all these triangles be drawn with regard to 0, the feet of the perpendiculars from on these pedal lines are coUinear. 4. Let S and ff be the foci of the fixed ellipse, S' the other focus of the moving ellipse. Let IF'P, TQ^Q be two common tangents, P and Q being on the fixed ellipse. Then the angle ETQ = 8TP = S'TQ. .: B, S', T are collinear. Draw SY, HZ, S'Z' perpendicular to TP, and let b, V be the semi- minor axes. Then BZ.ST = b\ S'Z' . ST = V^, .: ET : S'T in a constant ratio. .•. HS' : HT in a constant ratio. Now the locus of S' is a circle, .•. the locus of ^ is a circle. Todh. Sue. Ap. p. 332. 5. Let (^1, yi), {x^, y^), (s:^, j/z) be the coordinates of P, §, P, and let the equation to the parabola be y^ = iax. Draw the ordinates PN,pit. Then PN.pn = Aa\ .■. the equation to the diameter Ap'isy = — — The equation to QP is S/3 = -> = '^ - y^ yi {X- y2 + y3 ■ i A ax 4a2 yi + y. _ y^y^ I 4«2 H+^s y2 + y3 yi the equation to AS 'is ^t 4« + «(y2 + yz) ^ yx •! _ 4a2 yi {x-a) 16a3 {x - I ■>) = y{yxyiy% + «2 ^(yi + y2+y3)}- The symmetry of this equation shews that we should obtain the same equation for the lines BS, and CS. .'. A, B, and Clie on a straight line passing through the focus. 6. Let A denote the area of the first lamina, a the distance of its centre of gravity from the vertical line through the centres. Then since the thickness of each is the same, the masses are pro- portional to the areas, and .■. to the squares of the radii. WEEKLY PROBLEM PAPERS. 271 .•. the distance of the centre of gravity of the pile from the vertical line , , _ Ja\l^ + 8^ + 5^+ ... +(2/i - 1)4 ~ AlA." + 32 + 52 + . . . + C2« - l;^j ■ Now (2» - 1)5 = (2ffl - l)|(2;z - 1)3 - 4} + 4(2» - 1) = (2» - 3)(2« - 1)(2« + 1) + 4(2;z - 1), ,■. the sum of « terms of the series in the numerator _ (2n - 3) {2H - 1) (2n. + 1) (in. + 3) 3.3 , 4.2 4.2"*" __ {i«^ - 9) (in^ - 1) - 9 j_ 4^3 ■~ 8 ' = «2(2»2 - 1). And (2» - 1)2 = 4»(» - 1) + 1. .'. the sum of li terms of the series in the denominator 3 = I (4«2 - 1) .•. the required ratio is 3ffl(2»2 - 1) : 4»2 — 1. 7. Let CA, CB be sections of the given plane made by a vertical plane passing through the pulley, and at right angles to the common section, and let A and B be the positions of the particles at any time. Then if O be the 0. of Q-. of A and B, m and n the masses of A and B, AQ- : OtB :: n : m. Take CA and CB as axes of x and y, and draw QM, GN parallel to them meeting CA in N and CB in M. Then from similar triangles CA : CN :: AB : BG :: m + n : m, • CA = (m + n)- ■ Similarly CB = (m + n) ^■ m tt .: I = CA -[■ CB ^ {m + n)- + (m + 9i) ^ 272 SOLUTIONS OF x V ••• - + r = : + n which is the equation to a straight line, and is the loous of (?. Let T be the tension of the string, and suppose A to be moving up the plane. . T Then the acceleration of A upwards is g sin a, m, T and „ B downwards is ^ sm ^ — - > T . . T and these are equal, /. - —a sin a = a sin fi — - • .•. T = a . ■ (sin a + sin j3) • m -]- n ,: if/be the acceleration of ^, / = — ^ — (a sin a -!- « sin 3 — (m 4- n) sin aX m -{- li '■ ' = — ■" — (n sin — m sin a). If SJ'be the loous of (?, we may for simplicity suppose w^ to be at H and Wj *t -^• Then the acceleration of w-^ along TiW = - 'Jw{' + w^ — "iw-iW^ cos EOF = — i^w-^ + w.? + 2%2«2 "OS (" + /^) = -^1 suppose, with a similar expression for F^. And, by Todh. Mech. p. 295, the acceleration of the 0. of. G. w^F-x + wj^i W1 + W2 _ „ Wi sin a ~ OT2 ^i" q-^^-^2 '^2»i^ + M'a^ + 2wi«)2 COS (a + 0). APPENDIX. •s APPENDIX. ADDITIONAL PEOBLEMS. Paper VI. 7. An ellipse is described having for axes the tangent and normal at any point P of a fixed ellipse, and touching one of the axes of the fixed ellipse at its centre. Prove that the locus of the focus of the moving ellipse is two circles, of radii a ± 5. Paper VII. 5. liAA'BB', BB'GC, CC'Ad' be three circles, and the straight lines A/i', BF, GO' cut the circle A'B'G' again in u, /3, y, respectively, the triangle ajSy will be similar to ABG. Paper XI. 6. Prove that the asymptotes of the curve lU-2 + 24.-!^ + 4/ - %x + 16^ +11=0, are given by the equation {lU + 2y + 9) (.T + 2y - 1) = 0. Trace the curve, find the lengths of its axes, and prove that the equation of its director circle is se"^ +~/ + 2.r - 2y = 1. Paper XII. 7. AG, GB are chords at right angles'in a circle, F is any point on the circumference. PA, PB, PG represent forces. Shew that the locus of the extremity of the straight line which represents theif resultant is a circle. T 2 276 SOLUTIONS OF WEEKLY PROBLEM PAPERS. Papek XVII. 7. Along the sides of a regular hexagon taken in order act 6 forces represented by 1, 2, 3, 4, 6, 6 respectively. Prove that their resultant will be represented by 6, and that its direction will be parallel to one of the sides, and at a distance from the centre of the hexagon equal lo 3^ times the radius of the inscribed circle. , Papee XVIII. 1. A road runs from A to meet another at right angles. Shew that there are two points on the second road which may be reached in the same time from A whether we travel by road or across country, the rates of travelling by road and across country being as 7 : 6. Also shew that for places between these the quickest route is across country, and the quickest for all other places is by road. 2. If 16 be added to the product of four consecutive odd or even numbers, the result is always a pquare number. For odd numbers its last digit in four cases out of five is 1, in the remaining case 6. For even numbers the last digit in four cases out of five is 6, in the remaining case 0. 3. If I, m, n be the distances of any point in the plane of a triangle ABC icom its angular points, and d its distance from the circum-cenlre, shew that P sin 2 A + ffl2 sin 2JB + «^ sin 2C = 4(^2 + d^) sin A sin S sin C, R being the radius of the oiroum-oircle. 7. A conic passes through the centres of the four circles which touch the sides of a triangle. Prove that the locus of its centre is the circumscribing circle. Paper XIX. 2. Solve the equations (1) x* + a.i = iax{x^ + «2), a y ' z X ' b z X y c 4. If R be the radius of the ciroum-cirole, shew that the area of the triangle «= f ^ {sinS A cos (J? - C) + sin^ 5 cos (C - ^) + sin^ Coos {A - Ii% APPENDIX. 277 5. If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by two lines equally inclined to the bisector, one terminated by the base and the other by the rircum-circle. Paper XX. 3. If the area of a qnndrilateral he V(« — a) {s - b) (s — c) [s — d), shew that it can be inscribed in a circle. 4. AB, CD are chords of a circle intersecting in 0, and JC,BB meet at P. If circJes be described about the triangles AOC, BUD, the angle between their tangents at O will be equal to ^FB, and their other common point will lie on OF. .5. A uniform rod AB rests with its ends on a rough circular wire in a vertical plane, and the equilibrium is limiting. Shew that the vertical through the centre of the rod meets the circle through A, B and the centre of the wire jn two points, in one of which the directions of the resultant actions at A and B meet. 6. 2ff and 2i are the major and minor axes of an ellipse. V/ith centre as centre, and radii a, b, a -\- b circles are described, and a radius vector OPQR is drawn meeting them respectively in P, Q, R. If a parallel to the minor axis drawn through P meet a parallel to the major axis drawn through § in S, then S is & point on the ellipse, and SB is the normal at 8. 7. Defining the angle at v/hioh two circles cut to be that in which no part of either circle lies, prove that if the circles {x - b) {x - 5') +/ = 0, (.r - a){x - a)+f = 0, cut at an angle 6, {a - a'f (b - b'Y sin^ 6 + 4(i' - a) (b - a) (i' - a') (b - a') = 0. Paper XXI. 1. The population of a town at the end of any year can be found by subtracting eleven times the population at the end of the previous year from ten times the population at the end of the succeeding year. Nine years ago the population was 1210, eleven years ago it was 1000. Prove that it increases in G.P. I n „ 2. Prove that 5 sin-l — ^^ -'r 2 sm- Vso 25ViO 278 SOLUTIONS OF WEEKLY PROBLEM PAPERS. 3. Sum to infinity the series (1) sin (9 + i sin 20 + ^, sin 3d + V, sin 46 + ... 2 2' 2^ (2). sin 6 — - sin 35 + -^ sin 56 — - sin 76 -j- ... 3 5 7 (3) sin fl + i sin Zd + sin 56 + -^ sin 76 + ... 4. The three perpendioulars from the angles A, B, C oi & triangle on tlie opposite sides meet the sides in D, E, F. If D, E, F be given, shew how to construct the triangle ABC. 5. P is the orthooentre of a triangle, Q any point on the circum- circle. Shew that FQ is bisected by the pedal line of the triangle with respect to the point Q. 6. If \ be a variable parameter, the locus of the vertices of the hyperbolas represented by is the curvo (x'' + t/'^' = »^(a:' — y^). Paper XXII. 2. If the impossible root o! x^ -\- qx -\- r = Ohe a -{■ ^ 'v — 1, shew that 02 = 3ai _f. 2. 4. Through the angular points of a triangle JBC draw straight lines perpendicular to the lines bisecting the angles. If A, P be the area and perimeter of the original triangle. A', P' those of the new triangle, prove that (1) 4AA' = Paic ; (2) PP' = 4a' foos:^ + cos ^ + cos ^) . 6. If ABCD be a quadrilateral inscribed in a circle, and the sides be produced to meet in i^and G, prove that the bisectors of the angles at F and meet at right angles. 7. Chords of a hyperbola are drawn through a fixed point. Shew that the locus of their middle points is a hyperbola similar to the original hyperbola or its conjugate. APPENDIX. 279 Paper XXIII. and X, y, z be unequal, prove that each member of these equations = a;+y + ^ — «»• 2. A besieged garrison is provisioned for a certain number of days ; after 10 days one-sixteenth of the men are killed in a sortie, when it is calculated that by diminishing the daily rations by one-fifth it will be able to hold out for 30 days longer than was first supposed. Subse- quently 150 men with a quantity of provisions equal to half what is still left come in ; by which it will be enabled to increase the time it can still hold out by one-fourth. How many men were there originally? and for how long was it provisioned ? Paper XXV. 6. Out of a wooden cylinder is cut a cone of the same base, and the hole is filled up with lead. If lead be nine times as heavy as wood, and if the centre of gravity of the whole be at the vertex of the cone, shew that the height of the cone : the height of the cylinder :: sin 18° : 1. Paper XXX. 1. Prove that (1) the coefficient of x^+'^y'^+^ in the expansion of (1 -■^^)(1 -y) .g {m-\-n )\ \ — X — y rtiXnX (2) the coefficient of a^—'^ in the expansion of {(1 - ^) (1 - ex) (1 - A) (1 _ c%))~\ in ascending powers of x is (1 - e")(l - ct+i)(l - C +g) (1 - c) (1 - c2) (1 - c3) 5. A circle is described about a triangle ABG, and from any point 2) lines TiB, DC are drawn cutting the circle in two points 2" and § whose pedal lines intersect in 8. Prove that the angle S is equal to the difference between the angles A and D. 280 SOLUTIONS OF WEEKLY PROBLEM PAPERS. Paper XXXI. 1. In a bag there is a number of tickets marked with the natural numbers from 1 to n^ -{- 1. Every number is marked on each of r tickets, and every square number rf confers a prize of m shillings. A person can draw one ticket from the bag. Shew that the value of his expectation is ■ o", -,,, J. ^: - 2. If (1 + a; + m^T = Po + Pi^i! + P^^"- + . . , + Pr./ + . . . prove (1) Pn = P„2 - Pi2 + Pa' - . . • mp _ 1 {{in-\)\ _^ (2^-4)1^^-1) (2»-7)!_ 1 ^' («-l)l\ n\ '{ii--6)\'^ 21 "(k-G)! '■■/■ 5. A, B, C, D are four points not in one plane. If AB is perpendicular to GD, and AC is perpendicular to BB, then will AD be perpendicular to BC. 6. TP, TQ are tangents to a parabola whose focus is S. LM, a third tangent cuts them in L and M. Prove that the triangles SPL, STM are similar. Hence shew that TL : LP :: QM : MT. Paper XXXII. 1. Of three events it is 2 to 1 against the first and second happening, 3 to 2 against the second and third, and 9 to 1 against the first and third. Shew that the odds against all three happening are 5«y3 - 1 to 1. 2. is the centre of gravity of a triangle. AO, BO. CO are produced to points B, E, F such that AB = l.AO, BB = m.BO, CF = n. CO. Find the values of I, m, n so that the sides of the triangle DBF may pass through the points A, B, C. 5. A, B, C, B are four points in space. AB, AC are divided in F, F so that AB: FB :: AF : FC. BB, BC are divided in ff, R so that BQ: OB:: BR : RC. Shew that the lines OF and EF will intersect. APPENDIX. 281 Paper XXXTII. 1. A river flows from P to Q, a distance of 12 miles, at a uniform rate. £ starts at 12 o'clock from Q to row to P, and A starts at 5 minutes past 12 to row from Q to P and back again. A overtakes £ a mile from Q; he rows on to P, and at once turning back meets 5 two miles below A. A reaches Q 35 minutes after if reaches P. Find the times at which A passed P, and the rate of the stream. 5. Any point P is taken on a given segment of a circle described on a line AB, and perpendiculars AGr and BS are let fall on BP and AP respectively. Prove that Gil touches a fixed circle. Pafer XXXIV. 1. A policeman walks round his beat uniformly during his hours of duty. Shew that the chance of my meeting him, if I walk in the opposite direction down a street, which is -th of his beat, at a rate m times his, is 1 + -, where m > 1. 2. lii.'^+c z Also solve the problem when the condition in italics is removed. ) then will ( aKv^ + i^' + cV + abcxj/z = 0, fl' + i^ + c' = 5alic. C. - + «.-= I. - + by- = e y ^ 4. Prove that the distance between the centre of the inscribed circle and the orthocentre of a triangle is 2i2 (vera^ vers 5 vers C - cos^ cos 5 cos C)", where B, is the radius of the ciroum-circle. Papek XXXV. 2. Solve the equations : — (1) ^ + V.r2 = i'x^ - 1 { A^ + ^ - 'Jx^ - 4 282 SOLUTIONS OF WEEKLY PEOBLEM PAPERS. = ^. TiJ' (2) a;(^ + 2)2 = 1 + a3 ; a; +y = ^ + a: i yz (3) a{y - r) + b{z - x) + e{i; - y) = ) {^^ - y)(y - 2)(2 - -r) = c?3 . X +y + z = e ) 3. Prove that in a triangle where a < e, cos «^ \ L , a T, , nin + 1) ^^ .(. + !)(. + 2) .3^^^^^ 3 1 c^ Paper XLIII. 6. An ellipse and hyperbola are described so that the foci of each are at the extremities of the transverse axis of the other. Prove that the tangents at their points of intersection meet the conjugate axis in points equidistant from the centre. Paper XLIV. 1. li X, y, z are in G.P. when ^j is subtracted from each ; and z, y, x are in G.P. when 6^ is subtracted from each; and a, z, y are in ti.P. when ^3 is subtracted from each ; prove that ^— + ^- + -J- = 0. e^- x^ 6^- y^ 6^- z 4. If TA, TB be tangents meeting a circle in A and B, and TCQD be any chord meeting the circle in G and 2), whilst Q is the middle point of the chord CJ), shew that TQ bisects the angle AQB, and the length of TQ varies as the sum of the lengths of ^§ and^§. 5. T is any point on the tangent to a parabola at Q. Prove that the tangent at T to the circle round TQS touches the parabola. Paper LI. 1. Prove that the value of the expression \'a' cos^ (p ^ ifi sin2 <^ -[. fJifi sin^ ^ + i^ cos^

-\- i^ sill''' (^ V «2 sin^ cp -|- ^2 cos^ lies between — |- - and 5. ABC is a triangle inscribed in a conic whose centre is 0, and Oa, Ob. Oo are drawn to the middle points of the chords. From any point P on the conic, Pa, P/3, Fy are drawn parallel to On, Ob, Oc to meet the sides in a, /3, y. Prove that the points a, ;S, y are collinear. Paper LV. 3. The sides of a triangle are in A.P., and its area is to that of an equilateral triangle of the same perimeter :: 3 : 6. Shew that the greatest angle is 120°. 5. PAQ, PBC are two semi-circles which touch internally at P, PQO being the common diameter. Through P draw a secant PAB such that the area of the triangle ABC may be a max., and shew that for this position of the secant the area of the triangle QA3 is also a max. Paper LVII. 2. In the continued fraction 1 x x^ (1 - «:) + (1 - ^2) + (1 - x') + (I - a;7) + " • • shew that the «tii convergent is • a-a V where a-n = a— ^^ -- ^"^^ + ^"^^ - a;-*^ +...+(- ])"-i . aj-n", 7. Prove geometrically that if a line be drawn through a focus of a central conic making a constant angle with a tangent, the locus of the point of intersection is a circle. Paper LIX. 5. ABC is a triangle, any point, in tiie same plane or not ; P, Q, R points in OA, OB, OG. Bit, CQ intersect in L ; CP, AR in M; AQ, BP in N. OL, OM, ON cut BC, CA, AB in D, E, F. Prove that AD, BE, CF are concurrent. 284 SOLUTIONS OF WEEKLY PROBLEM PAPERS. fi. A semicircular piece of paper is folded over so that a particular point P on the bounding diameter lies on the circular boundary. Shew- that the crease-line always touches a fixed conic. 7. A straight line of given length moves so that its extremities always lie (1) on a fixed ellipse, (2) on a fixed parabola. Find the locus of its middle point in the two cases. Paper LX. 2. Shew how to find n if the sura of n terms of the series 1 -f- 5 + 9 + 13 + . . . be a perfect square ; and find the first two values of k greater than unity. 4. From a point A on the outer of two concentric circles tangents AP,AQ are drawn to the inner. AP, QP meet the outer again in T, K Prove that PP : PQ :: PT^ : RA^. Paper LXF. 4. The sum of the reciprocals of the distances of a fixed point from tangents to a circle at the extremities of any ohord through the point is constant. Paper LXVII. 7. In the system of pulleys in which each string is attached to a Taar supporting the weight, find at what point of the bar the weight must be attached if there are two movable pulleys. Also shew that if the weight be then doubled, it will descend with acceleration = ■^. 16 Paper LXX. 2. If .r2,r3 + j/^i/i = Tjri + y^i/j, = x^t^ + y^ij^ = 1, and f^i = x^ij^ - 3-3^2, ("^2 = •■^s.'/i - '^t.l/s' ^s = ^iV-i - ''iVv shew that (fi + (?2 + APPENDIX. 285 Paper LXXV. 6. P is any point on a conic circumscribing the ^triangle ABG, and the diameters which bisect the chords parallel to PA, PB, PC meet the tangents at A, B, C in the points D, E, F respectively. Shew that D, E, F lie on the polar of P. Papek LXXVI. 7. A particle of elasticity e is projected from a point in the wall of a square room in a direction whose projection on the floor makes an angle 6 with the wall. Shew that if the particle after striking each wall in succession returns to the point of projection, then e(/i + 1) cot 5 = «/j. + 1, fj. : 1 being the ratio in which a horizontal line in the side of the wall is divided by the point of projection. Papee LXXVII. 5. If in a rough inclined plane the ratio of the greatest force to the least force which, acting parallel to the plane, will just support a given weight on the plane be equal to the ratio of the weigiit to the pressure on the plane, prove that the coefficient of friction is tan a . tan^-, where a is the inclination of the plane. Paper LXXVIII. 7. A projectile is discharged with velocity v at an elevation u, and n seconds afterwards a second one is discharged after it so as to strike it. If v', a' be its velocity and elevation, prove that 2 vv' sin (a - a) = {v cos a + »' cos a) gn. Paper LXXIX. 3. Prove that (l)l = tan^^{tan^+2tan|^-F... + 2«-itan|^ + 2-i} 286 SOLUTIONS OF WEEKLY PROBLEM PAPERS. e (2)2 cos 5 cos - 1 1 cos 5 1 2 cos^- cos^- cos^-cos^- 2 2 2 2^ a a cos 6 cos - cos — C0S2- C0S3- C0S2- and sum to n terms the series Bec2 5 + 2^sec^i6 + 24sec2 22 5 + . . • + 2^"-^ 8602 2""^ 5. 4. On the sides of a triangle as bases are described externally three similar isosceles triangles. Prove geometrically that the lines joinin;,' the vertices of these triangles with the opposite vertices of the given triangle are concurrent. 5. Shew that the equation of the envelope of a circle described upon a chord of the circle {x — of -^ y^ — c^ passing through the origin as diameter is (x^ + y^ + o^ — c'^){x'^ — lax + y^ + «" — '-'^) = '^y^- Prove also that the maximum distance of a point on the envelope from the centre of the given circle is c n 2. Paper LXXX. 7. A tennis ball is served from a height of 8 feet. It just touches the net at a point where the net is 3 ft. 3 in. high, and hits the service line, 21 feet from the net. The horizontal distance of the server from the foot of the net is 39 feet. Prove that the angle which the direction of projection makes with the horizontal is tan~^ Hi's i ^"'^ 'Ca^i- the horizontal velocity of the ball is about IGO feet per second, the plane of projection being perpendicular to the plane of the net. Paper LXXXV. 4. S and H are the foci of a hyperbola, and FT, tlie tangent at T, cuts an asymptote in T. Prove that the angle STP = PUT. APPENDIX. 287 Paper LXXXVIII. 7. One end of a string is fixed to a beam, from wliioh it passes downwards and under a movable pulley of weight P, then over a fixed pulley, then under a second movable pulley of the same weight, and then the other end is attached to the first movable pulley. A weight W is attached to the second movable pulley, and all the straight , portions of the string are vertical. Prove that there will be equilibrium if ^ = P. Also, if W > P, the downward acceleration of W will be W - P W+bP^' Paper XCII. 7. If on a rectangular billiard table whose sides are a, h, a ball describe a rectangle whose sides are c, d, prove that the coefficient of elasticity between the ball and the sides of the table is fad — bc^ fbd — ac \bd — ac) \ad — be Paper XOIII. 5. The envelope of a perpendicular drawn to a normal to a parabola at the point where the normal cuts the axis is a parabola. Prove also that the focal vector of the point of the parabola at which the normal is drawn meets the envelope at the point where the perpendicular touches it. 6. Shew th at if w-^ + yi^ = s-^ + j/^ = x^ + ^^ = «', and xiifi - n) + ^aiys - ^i) + ^3(^1 - ^2) = 0, then a-1.T2.r3 + yi^s^a = a?. A straight line cuts in 3 real points the curve a;' + ^^ = «'• Shew that their centroid, if it lie on either axis of coordinates, will be at the origin. . Paper XCIV. 1. Prove that 3 . 81"+i + (16» - 64) 9''+i - 320»2 - 144» + 243 is a multiple of 212. 5. The diameter d oi a circle is divided into 2» equal parts, and straight lines are drawn from any point in the circumference to each 288 SOLUTIONS OF WEEKLY PROBLEM PAPERS. point of division. Ji Oi, a^ ■ ■ ■ ain-\ be the lengths of the lines so drawn, prove that in the limit, when the number of parts is increased indefinitely, a^ - a} -\- a.} - a^ ■\- . . . ■\- uHn-\ — - ■ Papeb XCV. 7. From a point at a distance d from a plane whose inclination is j3, two particles are projected simultaneously with velocities u and v in two diflFerent directions parallel to the plane and at right angles to each other. Prove that they will strike the plane simultaneously at points A and B such that JB^ = ^ (a2 -[- ^2) sec ^. Paper XCVI. ,,2 ^2 J.2 1. If » = a + '-, 2 = 4 + V' r = c + -, prove that a b e I /l - '/ ~ '" ^ = 1 /l - '•-■?' ] = 1 /l - ^ ~ g ) ; and eliminate x, y, z from the equations iiz zx (ry a , b , c „ X ^ ■' y z X y z 2. If a, /5, c, (? be the sides of a quadrilateral taken in order, and <^ the angle between the diagonals, shew that the area of the quad- rilateral is J(a2 _ j2 + ^ _ (f2)tan<|). Paper XCVII. 1. Prove that the max. and min. values of :fi + ?ipx^ + 3ji + r are 2j)' — S/ig + /• ± 2(^'' — g) . 4. If tangents be drawn to a fixed circle from any point on another circle, the envelope of the chord of contact is a conic. APPENDIX. 289 7. If a particle of mass m fall down a cycloid under the action of gravity starting from the cusp, prove that the pressure of the particle upon the cycloid at any point is 2mff cosi|r, where yjr is the inclination to the horizon pf the tangent to the cycloid at the point ; also shew that the resultant acceleration = ff. Papek XCYIir. 6. Two equal uniform ladders, each of length / and weight w, are freely jointed at A and are connected by a rope FQ. A man whose weight is W goes b feet up one of the ladders. If the ground be smooth, prove that the tension of the rope W6 + wl B 2a Ja'^ _ (,2 where 2c is the length of the rope in feet, and a = AP = AQ. Paper XCIX. 4. Two triangles SAC, BA'C are inscribed in a circle on the common base BG, and the pedal lines of the triangles BAG, BA'C are formed with regard to the points A! and A respectively. Shew that these two lines and the nine points' circles of the two triangles intersect in the same point. Papek C. 2. Eliminate 6, having given X cos {6 — a) + 1/ cos 6 = '2a sin {d + y) cos 6 cos {6 — a), a; sin (5 — a) +y sinfl = 2a {sin (5 + 7) sin 5 cos(d — a) — cos/S cosfll, o + /3 + y = T. IT SOLUTIONS OF ADDITIONAL PEOBLEMS. Paper VI. 7. Let CD be the diameter of the fixed ellipse conjugate to CP, and on the normal at P take PH = PH' = CD. Then by Salm. Con. Art. 181 (a), the angle HCH' is bisected by one of the axes of the fixed ellipse. ,". the other axis touches an ellipse whose foci are H and ti' , and which passes through C. Also CH = a — b, CR' = a -\- b. Paper VII. 5. The mgle Pay = ^aJ'+A'ay=SB'J'+J'C'C=BJJ'+CAA'=SAC. The angle aPy = B'^y - B'^a = jr - B'C'C - B'A'a = TT - B'BC - (n- - B'A'J) = B'BA - B'BC = ABC. .: the angle ^ya = BCA. .: a/3y is similar to ABC. Paper XI. 6. The condition that the equation ax^ + iixy + bf + 2i/a; + 2fif + c = should represent two straight lines is aba - af - bff' - clP' -\-ygh = Q (1) The asymptotes of the given equation are l\.-fi + 24.ry + 4/ - 2a? + 16y + 11 + X = . . (2) where \ is such a quantity as will make (2) represent two straight lines. APPENDIX. 291 .-. by substitution in (1) we find X = - 20. .-. the required equation is 11.t2 + li^xy + 4/ - 2.r + 16y - 9 = 0, or (lU + iy + 9)(.r + 2y - 1) = 0. The centre is the intersection of the asymptotes, and is given by ll.r + 2y + 9 = ; ;;; + 2y - 1 = 0. .-.w = -\,y =\. Transferring to parallel axes through the centre, the equation becomes ll.i;2 + 1\ay + 4/ = 2, The equation of a concentric circle is x'' -\- y"^ — 'fi. The equation of the common chords is ft' - y ^' + ^2.^ + (2 - y ^^ = 0- These coincide and are the axes when .-. the semi-axes are v jj, v — \. .'. the curve is a hyperbola, of which one branch lies entirely between the positive axis of x and the negative axis of y, the centre and the other branch being in the opposite quadrant. The equation of a pair of tangents from slif to the hyperbola is (n.r2 + 24.ry + 4/ - 2ar 4- 1% + ll)(lla-'H 24.iy + 4/^ - 2.i''-|- 16y+ll) = {l].r/ + Vl{xy' + x'y) + 4yy' - (.r + x') - 8(y + /) + n}^. These two tangents will be at right angles if the sum of the coefBcients of x^ and y^ is zero. .-. 15(ll.r'3 + 24;i-y + if- - 2^ -t- \JfT73 > 5(a + y), i.e. if 24(y2 _|- ^2^ — 50ay is positive, i.e. if 24 ^y - Jj (^y - 'Jj is positive, «'.«. if V does not lie between — and — . •^ 3 4 2. Lot « bo any number. Then by elementary algebra we find n{n + 2){n + 4)(» + 6) + 16 = («« + 6« 4- 4)2. APPENDIX. 293 First let n be odd, = 2;; + 1, .-. iV" = »2 + 6» + 4 = 4/ -1- 16^ + 11. Let p = 6q + r, .: N = IOO92 + 120g/- + 4^2 + 16^ ^ ^ = M+ 4r2 + IBr + 11, where if is a multiple of 10. If r = 0, JV= Jlf+11; r = \, iV=J/+31; ?• = 2, JV=J!f+59; r = 3, iV"=ilf+95; r = 4, iV"=Jlf+139. ,•. when r = 0, 1, 2, 4, the last figure of N^ is 1. when r = 3, the last figure of N^ is 5. Next let w be even, and = 2p. Then as before we get iV = Jf + 4!-^ + 12r + 4. If ?- = 0, JV" = ilf+4;r = l, i\^= i¥+ 20 ; »• = 2, iV^ = Jf + 44 ; r = 3, iV= Jf+76; »• = 4, i\r = j!f+116. .'. when r = 0, 2, 3, 4, the last figure of N^ is 6. when r = 1, the last figure of IP is 0. 3. Let P be the point, the circumeentre. Join and P to the points A, B, C. Let GOP = e. Then /2 = iJ2 + i^ - 2i2c? cos (25 - 6), n? = S? Jr d^ - 2M cos (2^ - 6), n^ = P? + d^ - 2Sd cos 0. .: 1? sin 2^ + m^ sin 25 + a^sin 2C = (52 + (f2)(Bin 2^ + sin 25 + sin 26') - 25)te2 - aa; + 2a2) = 0, APPENDIX. 295 .-. ar = «, or I (1 + \'^. .-. y = i » = «, or i (1 db xf^), £ a 2 4. ZsmA = I, Bin 2^ = 2sin^co8^ = ^ . ^' +/ - "', .: B? so? Acos{S - C) = £? sin2 J.sm{B+C) cos (5 - C) = ^'(sin2B + siii2C) 8 ^ tf 2 / e2 -|. a2 _ ^2 g2 -I- ja - gi v • 4 \ cV ■•■ a2ii2 j , the given expression on the right ^ 2 6&W ~ 3 ■ 4ffl242c2 ,R 5. Let ^^C be a triangle. Bipect the^^ angle at ji by j4J), a;id in BD take any point F. Join ^^, and at the point A, make the angle DAGr = DAF, & being on the circum-circle. Then the angle £AF = CAO, and ABI' = ACQ in the ?ame segment, .■. the triangles ABF, AGO are similar, .-. JBA : AF :: AG : AG. .: AB.AC = AF. AG, Paper XX. 3. Let AB = ai BC= b, CB = c, DA =^ d. Then DI? = a^ + d^ - lad cos A = i? + ^2 _ 2fo cos 0, ... (fl ^ d^ - ]^ - e^ = lad cos A - Ibc cos (7, ... (a + fff _ (6 _ cf = iad cos2^ + He sin^^ and (« - dy - (i + c)2 = - iadsafi^ - Uccos^^, .: 16(* - «)(« - i)(s - c){s - d) = 16 (be sin^- + ad cos^^j ( be coa^- + ai sin2-^ 296 SOLUTIONS OF WEEKLY PROBLEM PAPERS. = 16^( ocsm-cos-+«asm— cos- 1 +ao(;a( cos-cos; — sin- Bin- ) \ ■= 4(4c sin C -\- ad sin A)^ + 16 abed cos^ — Z_ = (area of quadrilateral)^ + 16 abed cos^ — T — . .*. if area = v (s - a){s — 6){s — c){s — d), we must have cos — 31 — =0, .'. A-\- C = 180°. .*. the quadrilateral can be inscribed in a circle. 4. Let OM, OV be the tangents at to the circles round AOG, BOD respectively. Then the angle COE = OAC, and COF = OBD, because they are in alternate segments. .-. FOE = OAC ~ OBD = APB. Now let OP produced meet the circle round AOO in H. Then OP . PH = AP.PO == BP . PD, .: H is also a point on the circle round BOD. 5. Let W be the weight of the rod, B, B' the total resistances at A and B. Let C be the centre of the wire. Join CA, CB. Since the rod is in equilibrium, the directions of W, R, B' meet in a point. Let this be 0. Since the equilibrium is limiting, the angle OAG = OBC. .: 0, A, B, C are points on a circle. This gives the required result. 6. Produce PSto meet the major axis in N, and draw the ordinate QM. Then SN ^ QM, and PN^ = AN.NA', .: SN^ : AN". NA' :: QIP : PN^ :: 0<^ : OP^ :: P : a\ .: S is on the ellipse. Produce BS to meet the major axis in G. Then 09 : QS (= M2f) :: OB : OP. And MN:ON :: PQ : OP, .-.OG-.ON ■.-.OB.PQ-.OP^ :: (a + b){a - b) : a^ :: c^ — }? : a\ ,". SO is the normal at S. APPENDIX. 297 7. The abscissa of a point of intersection {x', /) is given by {x' - a)[x' - a') ={x^ ~ b){a/ - b'), .: a/{a + a' -b - b') ^ aa! - bV, (a - b){a -b') , , {c/ - b)(r'' - *') a + a — b — b a + a' - b - 6' (« + a — - )'' Tlie equations of tlie tangents at (x', y') are 2(a;a:' + yy') - (a + a')(a: + y) + 2(za' = 0, 1{xx + yyO - (5 + b'){x + «;')+ 2ii' = 0. a \ d - U b ■\-ti - 1x' .'. tan e = -^ ~ / , , (a + a' - 2a:')(i + ^' - 2ar') 2/(g -f a' - a - 5') (a - b'){b - a') +_ (a - b) {b' - a') i f {a + a ' - b - b'Y ^ {(a - b')(b - a') 4- (a - b)(b' - a')? •'• ~ sin2 6 cos2 e = iy'-(a + a'- b-b')^ + {(a - b'){b - a') + (a - b){b' - a')f == _ 4(a - b) (a - b') (a- - b) (a' - b') + (a - i')' (»' - b)^ + 2(a - i) (a - *') (a' - 5) (a' - b') + (a - 4)2 (a - by = {(a - 4') (a' - 4) - (a - 4) (a' - 4')}' = {-(a-aO(S-4')}^ /. (a - a^ib- b'f so? 6 + 4(5' - a) {b - a) {V - of) {b - a') = 0. Papee XXI. 1. The population 10 years ago = 12100 - 11000 = 1100, .*. the numbers in three consecutive years were 1000, 1100, 1210, which are terms of a G.P. common ratio being ^J. 2. We have to prove that 6 sin-i -j^= = j = ? - 2 sin-i VSO 4 25ViO 298 SOLUTIONS OF WEEKLY PROBLEM PAPERS. 1 3 Let a = sin-i —;^, /3 = sin-i rr^ — — ^i .•. sin 5a = 5 sin a — 20 sin' a + 16 sin" a 1_ /. _ 20 16 \ _ 2879 ~ ^5^0^ 60 "^502/ ~ 6V2 . „ 3 o 79 . „„ 237 "'^^ = 2-5;/ro''=°^^ = 26;yi^' •••-°2^ = ^ .-. cos2 2/3 = ,4 (5i» - 2372) = 4 X 9709456, ,-. cos 2/3 = ^^ 51U 510 56 /. sin (j - 2)3) = —1 (cog 2/3 - sin 2/3) = sin 5a. v/2 _ 3116 - 237 ^ 2879 ~ 6V2 6V2 *Tliis may also be solved by noticing tbat 1 3 V sin-i — — : = tan-i i, sin-i — — - — = tan-i t^, VSO 25v/ro .-. 5 tan-i } + 2 tan-i A = 3 tan-i ?- + 2 tan-i j^- = tan-i f + 2 tan-i ^ = tan -1 i + tan-i i = j, Euler's Series. Note.— From this q^uestion can be obtained the expansion of jr given in XXI. No. 1. 3. (1). Let S=sm6-^l sin 16 + 1 sin 3(9 + ... C = cos e +i cos 2fl + ^2 cos 35 + ... .-. 0^-81 = ^+ h-^'i + 2j e3« + ... = -{' + T + (f)'+- APPENDIX. 299 .-. (2 - e'i){€+Si) = 26"', .-. (2 - cos 5 - i sin e)(C+Si) = 2(cos 6 + i sin 6), .'. equating real and unreal parts, (2-co8e)C + S.Biue = 2 COS 6 "j (2 - cos 5) /S - C . sin fl = 2 sin fl J ■ Eliminating C, we find S =- * ^™ ^ 5 — cos (2). Let (S = sin fl - i sin 3fl + J sin 5^.., C = cos^- J cos 35+ I cos 55... .-. C+Si -- e« - JeS'i +ie5« _ 1 ''0 + Hw"? + i<»«'0* + -} = i log ^+'^" . 2j ^ 1 - je« ...2«-2^=logL±iJ, 1 - «■«« •. (l + sin5- Jcos5)(cos2C+«sin2C)-e-2S= 1 -sin5 + icosfl, .•. equating real and unreal parts, (1 + sin 5) cos 2C+ cos 5 sin 2(7= (1 - sin ( (1 + sin 6) sin 2C - cos 5 cos 2<7 = cos 5 . e^s If we square and add these equations, we shall eliminate C. .'. 26*3 (1 _ sin 6) = 2(1 + sin 6), .: S = i log ]'^^'^y ^ ' 1 - sin 5 (3). L,et S=Bmd-\-- sin 35 + t sin 55 + . .. C = cos 5 + - cos 35 + y, cos 55 + ... 4 i^ }• 300 SOLUTIONS OF WEEKLY PROBLEM PAPEKiS. .•. proceeding as in (1) we find S = a— ,—,„-.- „-^» 9 + 16 sin^ 4. Since JD bisects the angle EDF, we must bisect the angles at 2), ^, F, and draw lines through these points at right angles to the bisectors of the angles. 5. Let Q be any point on the circumscribing circle, QD perpendicular on £C, and let F be the orthocentre. Bisect FQ in M. Then since /' is a centre of similitude of the nine points' and circum-ciroles, .v M is on the nine points' circle. Join MI) cutting A/i in M. Join J.F and bisect it in U, and let AF cut £C in X Then M, U, X are on the N.P.C. And MTJ — \Q,A, .: the angle in the segment of the cireum-circle cut ofi by QA = the angle in the segment of the nine points' circle cut off by MU. .-. the angle QCA = MXlf. Since M bisects FQ, .: MX = MD. .-. the angle MXU = QDM. .: QCA = QDM. .: a circle will go round QODB. .". QEC is a right angle. .•. E is the foot of the perpendicular from Q on AC, and is .". a point on the pedal line. .'. the pedal line bisects FQ,. For other solutions, see Casey, p. 36, and Catalan, p. 37. 6. The axes bisect the angle between the asymptotes. .•. their X —~ 1/ 2^y • equation is — = — - ■ .'. elimmating A, we have for the locus of 2 A the vertices, (x^ — y^ - «^) {^^ — y) = — 4a;2y-, i.e. (x^ — y'^Y + 4a.y = (fiijc'' — y^), or (a-^ + if)''' = a^ix^ - y^). Paper XXII. 2. If a + /3 V — 1 be one impossible root, the other will be a — 3 v — 1. Let y be the third root. Thenx^ + gx-i-r = {ie -a + ^'J-~l){x -a- ^ ^^l){x- y) = {x' - 2ax + o2 + (32) {x - y) = afi-(2a+ y)x^ + (a2 + ^^ + 2ay)x - (a? + j32)v, .: y= - 2a, g = a' + /32 + 2ay = a2 + 32-4a2 = 32-3a'''. APPENDIX. 301 4. The angular points of the new triangle are evidently the centres of the escrihed circles of the original triangle. .". by Todh. Trig. Cap. XVI. Ex. 34, A = — = _ (1) Again, if O^O^O^ be the angular points of the new triangle, ri = BOi cos - , r3 = BO^ cos -> B G B A „ cos - cos — COS - COS - .-. O1O3 cos - = n + rj = fl h c .— — .-. O1O3 = a 4 / cos % c cos C 2 a \ cos - -I C . C\ cos — Bin - A^ . A. cos - sin - 2 2/ cos ■ 2 2« B — — cos - sin^ 2 .-r, B aU B M B , ,,> = 4iE cos -^ = — cos - = 4— cos -, by (1) .'. PP' = 4A' (cos - + cos - + cos - j ■ 6. Let the bisectors of the angles F and Q meet in 0. Produce TO to meet AD in E. Then the angle FOO = UGD + OKD = OGD + A + AFO='U^-A-B)+iin-A-I>)+A = n-?-±^ = ^- 7. Let B be the middle point of any chord through the fixed point 0. Take the asvmptotes as axes, and let the chord meet the axis of .« in Q', and the axis of y in P'. Then since PF = QQ', B is the middle point of P'Qf. Let (h, k) be the coordinates of 0. The equation to the chord is t^ + —^ = 1 ; also 7^' + 7^ = h Cq ' CP' cq^ CP cq; ' CP' . „ , OT CP' Now, if g, jj be the coordinates otB,i = -^,v= -^' .: the locus of B is ^ ~J + "LJzl = 0. ^5 2ij 302 SOLUTIONS OF WEEKLY PROBLEM PAPEES. .*. transferring to ( -> 5), we have hh xy =--, which represents a similar hyperbola. This question may also be solved geometrically as follows : Using the same letters as in the analytical proof, bisect CO in C, and through G' draw EOF parallel to P'^, meeting CP' in E and CQ' in F. Join ie^, RF. Through C draw lines parallel to OF and Cqi,. Let the first meet T^R in Y and C(l in K. Let the second meet UP in X, and CP in if. Then M is the middle point of PQ, and .•. also of P'§'. Also ^ and P are the middle points of CF, C^. .: ER, PR are parallel to the asymptotes. .•. in the parallelogram CERP the complement YC'RX = complement EK = const. .-. RX. RT = const. .". locus of i2 is a similar hyperbola, centre C. *This may also be considered as a particular case of the following general proposition. In any conic the locus of the middle point of chords which pass through a fixed point is a similar conic. Describe a similar and similarly situated conic passing through 0. Let POO'P' be any chord intersecting this conic in 0, 0' and the original conic in P, F- Then since the same diameter bisects both PP' and 00', .-. OP = O'F ; and if R be the middle point of 00' or PP', OR = \00'. .: since is a fixed point, the locus of i2 is a conic similar and similarly situated to either of the other two conies. Paper XXIIL 1, Put each expression = R. .*. x^ — myz = Ri? (1) ^ — mzx = Ry^ (2) ^ — mxy = R^ (3) From (1) and (2), tfi - y'^ -\- mz{x - y) = R(x^ - y*), .". x^ -{- xy -{- y^ -{- mz = R(x + y). APPENDIX. 803 Similarly ^2 + ^^ 4. ^z + ^^ = ^(y ^ 2)^ .'. subtracting x^ - z' + y{x - z) -\- m{z - x) = B{x - z), . .'. R = X -y- y -\- z — m. 2. Let X denote the original number of men, y the required number of days. Then xy = lOx + 4 . i§r(y + 20) ; .-. y= ]0 + 3(y + 20); .-. y = 100. ^'^° JT15-0 = i = •■• ^' = ^(-^ + ^^°) = •■■ ^ = ^^°- Papeb XXV, 6. Let It, X be the height of the cylinder and cone. Then the vol. of cylinder : vol. of cone : : A : - > o ,; weight of cylinder : weight of cone :: A :'-■ Also, the heights above the base of the C. of G-. of the cylinder and cone respectively are ^> j • Now the cylinder with the lead cone inside it is equivalent to the wooden cylinder + a cone of 8 times the weight of the wooden cone. .•. since the C. of G. of this coincides with the vertex of the cone, its height above the base is x. 7. * 4. 8 ■'^ • ■■" _ "'2 ^ '4 3 _ 3^_+4r2 •'• ^ - 8x ~ eA + l&x' .: 4x^ + 2x/i - W = 0. .: X = h . — — = = h sin 18°, since x is positive. 304 SOLUTIONS OF WEEKLY PROBLEM PAPERS. Paper XXX. 1. (1) (L-J^)J1^A = l-.r-y + .r, ^ , ^ f, 1 (,^ )j_. 1 — X — y \ — X - y > = 1 +a:y{l +(.r +y) + (^+y)^+ + (^+y)°'+"'+ •■••}, .•. coefficient of arm+l y^+l = coefficient of ar»> y" in (a; + y)""^-™ _ (m + ;?) ! «7i ! » I (2) Let ((1 - z){\ - cz\\ - ^ + l). ..(;? + >?- 1) _ ^ »^g- + l)---('g-!-»- -4) »! ■ (« - 3) ! (2^^_1)^! _ ^ (2 ;? - 4) ! ;z(« - 1) (2^ - 7) ! \ (;? - 1) M « 1 ■ (/^ - 3) ! ' 2 ! {n-fj)\ X 306 SOLUTIONS OF WEEKLY PROBLEM PAPERS. 5. Let ABG be the plane of the paper. Let Q be the foot of the perpendicular from I) on this plane. Join QA, GB, QG. Produce AC to meet BO in F, and produce 00 to meet AB in B. Then the plane EDO is .perpendicular to the plane ABC, and also to the line AB, and BBO is perpendicular to ABO, and also to AF. .: in the triangle ABO, AF is perpendicular to BO, and OB to AB. .'. BO is perpendicular to AO. .". AB is perpendicular to BO, since the plane AUG is perpendicular to the plane ABC. 6. Let ZM touch the parabola in B. Join SR. Then the angle SLP = SBL = tt - SRM = ir - SMQ = SMT, and (SPi = STM, .: the triangles /SPi and STM axe similar. Similarly /SIT and SMQ are similar triangles. .-. TL : QM :: ZS : MSirom similar triangles SZT, SMQ, ■.:ZP:MT „ „ SPZ, STM. Paper XXXIL 1. Let ^1, /ig. P3 be the respective probabilities. Then p^p^ = -, ^2;»3 = gi i'si'i = Jq. •"• iPiPiPif' = 3—53' •■• PiPiPs = g-j^> .: the odds against all three happening are 5 v3 — 1 to 1. 2. If the points F, A, E are collinear, then AOEF = AOAE + AOAF. Now AOEF = iOE . OP sin EOF = i(m - 1)(« - 1) 05 . OCsin BOG = (m - 1)(« - 1) AOBC. AOAE = iOE. OA sin AOE = (m - 1) AAOB. AOAF = iOA . OP sin AOF = {n - \) AAOC. : {m - 1)(» - 1) ABOC = (m - 1) AAOB + («-!) AAOC. But ABOC = A^OjS = AviOa since is the C. of G. .•. («M — 1) (» — 1) = »J — 1 -|- « — I, .-. (ra - 1) (» - 1 - 1) = K - 1 == a - 2 H- 1, APPENDIX. 307 /. (ot - 2) (« - 2) = 1~1 To solve these equations, multiply Similarly (n — '2) (I — 2) = 1 v*^^™ together, extract the square root ' Coi the product, and divide by each (/ - 2) (m - 2) = 1 J equation. .-. / - 2 = OT - 2 = » - 2 = 1. .: I == m = „ = -6. 5. Join CB. Then EF and GIT are each of them parallel to BO, and .". also to one another, and .-. a plane can be drawn through them. .•. F0 and SH, being in this plane, intersect. Paper XXXIII. 1. Let X, y, z be the rates per hour of A, B, and the river respectively. 12 12 A takes hours to go from Q to P. and hours from P to 0. X - Z ^ X + z 1_ 1 ^ ■*■ 12"^ X - z" y - z— ^^^ 12^ u; - z^ x + z y - z-^' J:^2 _12_ ^ _6_ _ 1 X - z w -\- z y — z 2 " ^ '' These are three simultaneous equations of the first degree, the unknowns being , , . From them we find X -\- z X — z y — z x + z = 8, X— z = i, y — z = 3. .: x = 6,y = 5, z = 2. .: A overtakes 5 at (^ + i) hours after 12 o'clock, i.e. at 12.20, and A meets B again at ^ hours after 12, i.e. at 3.20. The rate of the stream is 2 miles per hour. 5. The points G and IT are on the seimicircle described on AB as diameter. Bisect AB at C. Then C is the centre of the semicircle. .-. the angle GCH = 2GAH = 2 X comp. oiAPG = constant, since the angle AFB is constant. .*. Gllis a chord of the semicircle of constant length, and .•. always touches a circle, centre C. £2 o08. SOLUTIONS OF WEEKLY PROBLEM PAPERS. Paper XXXIV. 1. The necessary condition that we should meet is, that when I start, the policeman must be either in the street or within such a distance of the end of it that he will reach it before I do, i.e. he must be within -til of the length of the street from it. .". he must be in a particular m - ( 1 + - jth of his beat, and the chance of this is -^ — n \ ml inn n If I walk in the same direction as the policeman, when I start, the policeman must be within M — _ yh of the length of the street from the starting point. The chance of this is — .•. the chance of our mn, meeting (the condition in italics being removed) is ^ fm ■{■ 1 m — 1\ _ Im 1 " V mn mn j imn n 2. ff? _ 4 + cf. = ) z X \ a - + bl ~ c = { y ^ I a. b c x^ y z z y X ifl + i^A + i,/.?! +■!')= 1 --51-. Again, in the first equation for x writing — , &c., we obtain ax , cz . hv iV \ ^ b — -f- V — = a, .'. b- ■\- c - = a. by rz z y .-. the equation is unaltered. .-. from (J), aV + 4y + cV + abc xyz = 0. APPENDIX. 309 Now multiply the three equations together. .-. .-. abc = U^^ +c^ (ca i + cb^-,+ a^- + ah -i^ \ z y' ^ y * y^ ''i = lahc + a(a2 - ilc) + b{JJ^ ~ 2ca) + e{c^ - lah) = cfi ■\- }fi -\- (? — ^ahc. 4. Let P be the orthocentre, /the centre of the inscribed circle. Then Li = — ?1- , AP = '2R cos J, sin - 2 the angle I^P =^ -i? - ^ = ^^, r = - B, = -—-, .: r = AR sm — sin - sin • s 4*S' z A z .:■ IF^ = lA^ + AP^ - 1IA . AP cos lAP r'' . ,-nt ■> J .Pr COB A B - C iR^ cos^ A - A — cos . , A ' -A 2 sm'' -• sm - = 47J2/4sin2-?sin2r^+cos2J , . B . . B - G — 4 sm - sm - cos A cos — - — 2 2 2 = 4iJ2| (1 - cos B) (1 - cos 0) + cos^ A - cos^ sin^ sin C — 4 cos A %\v? - sin^ - I £i ^ J = 4-ffi2 {(1 - cos 5) (1 - cos C) - cos A cos 5 cos C - C0S^(1 - C03j)(l - cosC)} = 4iJ2 {(1 - COS i?) (1 - cos C) (1 - cos A) — cos .1^ cos B cos (7 j . ... jp = iR {vers A vers 5 vers C - cob A cos 5 cos (7 } L 310 SOLUTIONS OF WEEKLY PROBLEM PAPERS. Papsk XXXV. i , (■ / / X — iJx' — 1 2. (1). v'(a;2 - 1) iV:!;2 + a; - >Jx^ - A = -j = !x - >Jx^ - 115, •■• n/^2 - 1 {2x^ - Ix'Jx'' - l} = {x - •Jx' - 1)3, . 2a-2 V^^TTl _ 2^(.t2 - 1) = a:3 - 3j2 >J'^^^^\ + 3a:(^2 - 1) - {x^ - 1) Vi2~rT^ .-. f,ifi - hx = (6^2 - 1) \/a-2 - 1, .-. 36;r6 - 60;i'3 + 25:!;2 = {x"^ - 1)(36»* - 12.r2 + 1), .-. 12.1-4 - \%x^ -1=0, . a:^ = -^-- — ^ — ^j .". a; = ± f j-i±-2 "^'Jd " ^ 2^/3 (2). From (1) and (3), 1 + a? = x{{y - zf + i} From (2), = xli - a;)^ + a}, .'. x^ - Sa'^ + 3a: — 1 = a^^ .: x — 1 = a, .•. a; = as + 1, ■ •■ (y + 2)= = J _|_— == 1 _ « + a2, ... ^ + ^ = + s/l _a + a' (y - zf = l-a + a'-l:^{a- If, .: ff - z = ± {a - i), . ■. y and z are known. (3). a{y - z) + biz - a:) + c(a; - y) = .^ - y; = " ) > - 7/) = ) and (y _ r) + (2 _ a:) 4- (a: _ y) = . . £ — Xt, — e c — a a — b .-. d^ = {h - c) (c - a) (a - 6) ijs. .-. B is known. X - If = {a- b)R, 2y - 2z = (25 - 2(;)ffi, .■. a: + y - 2z = (« + 4 - 2c)R, and ar + y + 2: = e, .-. z = ^ /« - (a + 4 - 2c)iE|. Similarly x = i{e - [b + c - 2a)B)j, y = i {«-(<; + «- 2i)i?}: APPENDIX. 311 3. Let C = 1 + It" cosB + "^^ + ^^ - cos 2JS + ... c 2 I c'^ (S = «- sin ^ -^ — ^—-i- — '- - sm 25 + ... c 2 1 e^ .-. C+ &■ = 1 + «%Bi + <«+ ^) I.SBZ + ... c 2 ! c^ = C (c — a cos 3 — i . a sin .B)-" = c" (4 cos ^ - i . b Bin A)-'^ in = , (cos nA + J sin »y^). cos ra^ - - C = — < 1 + n- cosB +-^ . — ' -cos 25 +...(, Papek XLIII. 6. Let S, S' be the foci of the ellipse, H, S' those of the hyperbola. Let B be the extremity of the conjugate axis of the ellipse, b that of the hyperbola. Let 5 be a point of intersection, and let the tangents to the ellipse and hyperbola at F meet the conjugate axis in T, t, and draw Pn perpendicular to CB. Then CH^ - CB^ = CS\ .: CB^ = Cm - CS^. And CS"" + W = CR\ .: Cb^ = Cff^ _ CS^ .: CB^ =. Cil .: Cfi.CT = CB^ = Cb^ = Cn . Ct, .: CT = Ct. Paper XLIV. 1. (^ _ e^f =.f^y- 6,){z - e,) ; (y - 0,Y = (^ - 5.)(^ - e,) -, {z - ^3)2 = {x- 6i){y - 8,) ; .-. s^ - 2x6^ + ei' =yz- 6^{y + ^) + 6^, ^ 2x — y — z . ^ _ Q ^ 2.r^ - -ry- xz-x^+yz ^ {x — y) (x - g) ^ ^ 2x — y — z "x - y — z 312 SOLUTIONS OF WEEKLY PEOBLEM PAPERS. .-. Similarly y - £., = ^ — '—- ; z - 0.,= ^—- — '-^ ^ ; 2j/ — z - X Zz — X ~ y , 1 1 1 Ix — y — z 111 — z - .v^ ■' e^-x'^ e.2-1/^ e^- z {x-y){z - x) ' (y - z){x - <,) 22 — X — ?/ (« - a;) (y - z) If we reduce the right-hand side to a common denominator, the numerator = 2^(y -z)-{y -;- ^)(y - z) + 2y(z - x) - (z + x){z - x) + 1z{x - y) - {x +2/){^ - V) = 0. 4 Take the centre of the circle. Then 0<^ is perpendicular to CI). .: circles will go round TJQO and TQOB. .-. the angle JQT = AOT = TOB = TQB, i.e. TQ bisects the angle AQB. Again, since TJQB is a quadrilateral in a circle, Tq.AB^-TB.AQ + TA. BQ = TA{AQ -I- BQ). .: Tq = ^-^ {AQ + BQ). .: TQ ^ AQ + BQ. 5. Let TP, TQ be two tangents to a parabola, GD a third tangent. Then the circle round TOD always passes through S. Let CD change its position, always touching the parabola, until it is indefinitely near to the position TQ. Then in its limiting position the circle will pass through two consecutive points on TP, and will .-. touch TP at T. Paper LI. 3. Lety = Va^ eos^ 4> + F- sin^ ^ + sJa"^ sin^ cos^4,) = a2 ^ J2 4- \/(a^_ i2j2sin2 2<;, + .ia2i2 y2 has its max. value when sin^ 2i^ = 1. APPENDIX. 313 ■. y2 = a2 + i2 + V(a2-i2)2 + 4g2^2 = 2(«2 + h% .: y = 'J-Ad^+2b\ \ y^ has its min. value when sin^ 2<^ = 0, >■ .-. y2 = ^2 ^. ^2 _^ 2«6, .-. y.= a-Yb. 3 Next let y = -^ — ^ ^. -^ V a^ cos2 1^ + 1^2 gin2 ^ Va2 sin2 ip -\- d^ cos^4> sJcfi Bin2 ^ + &2 C032 ^ ^/^a oog2 (^ -j. ^2 3in2 (^ . 2 ^ g2 ^ ^2 _[_ ^(^2 _ y ) si„2 2^ _(_ 4a3^3 (a2 _ J2) ?Hli^ + ^2^2 _ ' 4(a2 + ^2) ^^ (fl2 _ A2) sin2 2<^ + 4«2i2 V(a2 - 52) sin2 2^'4^2i2' .•. y2 lias its min. value when sin2 2(^ = 1. • • -/ (a2 _ ^,2)2 + 4^.i3 I V(<^2 _ i2j2 + 4^262 «2 + ^2 "t" ^. «' + A' V2«2 + 2i2 ^2 has its max, value when sin2 20 = 0, . 2 ^ a2 + ^2 2 ^ ('2+iY • ?/ = ^ + ^ = ^ -L I ■ ■ ^ «2j2 + a5 W4 / ' ■ ■ -^ «A a'^ b' 5. In a circle if Pa be drawn from a point P on the circumference perpendicular to the side BC of an inscribed triangle, it is parallel to Oa the line joining the centre to the middle point of BC. .: the required theorem is obtained by projecting orthogonally the proposition ' If perpendiculars be drawn from a point on a circle to the sides of an inscribed triangle, their feet are collinear.' Papek LV. 3. Let a, « + 5, « + 25 be the lengths of the sides of the 1st triangle. Then « + 4 is the length of each side of the equilateral triangle. 314 SOLUTIONS OF WEEKLY PEOBLEM PAPEES. The area of the 1st = / 3(a + ^) a_+_3b a + b a_-_b V 9 ■ 9 ■ O ■ Q 2 ^ (a + *) "^{a + 5ii)(a - 4). The area of the equilateral triangle = -^ (« + Vf, .: «/(« + 34)(a - A) : a + « :: 3 : 5, .-. 25(a2 + 2»5 - 3i2) = %c? + 2a5 + !?■), .: 16a2 + 32ai5 - 84*2 = 0, .-. A = |-«, or - fa. If A be the angle opposite the side a + 24, then taking 4 = f ff, 2«(ffl + 4) 2.^.f.a2 ^' If we take 4 = — fs, the greatest side is a, ••'°'^ 2(«+24)(«+4) - 2".lTf:^^ .,..^-120. 5. Let P^3 he the secant required. Join QA, QB. Then the angles FAQ and PBC are right angles, .-. AQ is parallel to SG. .: A ^5(7 = QBC. Now the triangle QBC is a max. when B is the middle point of the aru PBO. .'. PBC is an isosceles right-angled triangle, as is also PAQ. .•. A is the middle point of the arc PA(^. .: the triangle QAG is a max. and QAC = QAB. Paper LVII. 2. ^-' J2 ~ (l-:!;)(l-:!.'3)+^ ~ 1 - a:3-j-a:4 "" ^-i_.t-4_1 = ^rZ"l 33 "" (1 - iEj(l - x^){l - afi) + x\l - x) + x{i -Ifi) _ \-afi-{-^ _ x-'^ - X-* -{■ x-^ _ 0-3 ~ i~- afi + ^"~x^ ~ x-'^ - x-i + x-^ - 1 ~ 0-3- i' Assume that the law holds for ^""^ and .^!^. 5»-2 Sn-l APPENDIX. 315 Then ^^-2 = C-l)"-l.«(«-2)V™-2, ffn-2 = (-l)''-l.a<'»-2)''((rn_2-l), Pn-1 = (- 1)" . aK"-!)^™-!, qn~i = (-!)» .d^l)\ be the eccentric angles of the extremities of a chord of length 21. Then iP = a^coa - cos (fif + ^^(sin 6 - sin (j))", ,. p = («2,i,2^ + i2eos^l+_^) sin^i^. 316 SOLUTIONS OF WEEKLY PROBLEM PAPERS. If (,c, y) be the coordinates of the middle point of the chord, X = ^ a(cos 6 + cos 9) = « cos — - — - cos -, y = J i(sin 6 + sin 0) = i sin — ^— cos — — -^, .■. i^J.-2 + a2y2 ^ g2J2 (jos^ — — i, and i%2 + «4^2 = a2i2/'a2 gin^ 1±-^ + ^2 cos^ ^-^) cos^ .'^ ~ '^ 2 ' ■■ Ui.v^ + aY b*^^ + a*f ,,2^,2 ,„,2 ^ - '^ 1_ (2). Let y^ = px be the equation to the parabola. Since the coordinates of any point can be put in the form {pm^, pm), let the coordinates of the extremities of a chord of length 21 be {pnii^, pm-^ and {pm^, pm^. Then iP = p^{m^ - w.ff + p'^im-^ — m^y = p%m-y - OTj)^ {(swi + Wg)^ + 1). If (,T, y) be the coordinates of the middle point of the chord, ix = ^{m-? + ml'), 2y = p[mi + m^, .'. 4(^2 _ ^,^^ = i^'tyi^ -\- im-im^ + OTa^ — 2»Zi^ — im^) = - fk"h - mij\ and 4/ + i)^ = i''{(»'i + ""'H^ + 1}, .-. 16(y2 _ ;,.r:(^y2 +^i) = - i'Vi - «2)' {('»i + '^2)' + 1 } APPENDIX. 817 Paper LX. 2. The»ti> term of the given series = i{>i - 1) + 1. .•. the sum of « terms = 2(» — l)n. -\- n = 2n^ - n = y^ suppose. .-. 16(k - i)2 - 8/ = 1, or x'^ -?y^ = 1, where x = in — 1. ar = 3, y = 1 is obviously one sohition, .-. (:!^ - >/8y)(.r + J8y) = 1 = (3 - V8)'"(3 + 78)™- Let X - Vay = (3 - Vs)'", .-. x + 78y = (3 + v'S)"', .-. 2^ = (3 - Ji)"' + (3 + V8)™. By giving m different values, we can obtain as many values as we p'ease of ,r, and .•. also of n. 1£ m = 1, X = S, in = 4, .: n = 1. m = 1, X = VI, in = 18, inadmissible. ?» = 3, a; = 27 + 72, 4^ = 100, .-. n = 25. m = 4, ar = 81 + 432 + 64, in = blB, inadmissible. m = b,x = 243 + 2160 + 960, in = 3364, .-. n = 811. .". the first two values of n greater than unity are 25, 841. 4. Draw TK a tangent to the inner circle, and produce RPQ to meat the outer circle in S. Join AS. Then MP : BQ :: SP^ : HP . HQ :: RF^ : TK^ ;: RP^ : ACl\ since TK = AQ, being tangents from points on the outer circle. .-. AR = AS. .: the angle JJTi' = ARQ, and RPT = APQ = AQP. .-, RT^ ■.RA'^:: RP^ : AQ^ :: RP : RQ. Papbb LXI. 4. Let P be the given point, RE the polar of P, C'PG any chord through P. Produce G'C to meet RH in R. Draw CD, CD, PE per- pendiculars on RE. Then EP passes through the centre 0. Let the tangents at C, C meet in B on RE. Join 00. Draw PQ, PQ, PO, Cf perpendiculars on EC, EG', GO, J' respectively. 318 SOLUTIONS or weekly problem papers. Then, OP . OE = (iad.)2 = 00. 00, OP. OF =00.00, .: Bubt. OP.EF= OC. CO. .: OP . CD = 00 . PQ. {J) Similarly we can prove OP . O'V = OO . Pq. {B) Now since iJCPC is a harmonic range OR ^ C'B Pii . -L+J- =1-. CO ^ G'jy PE ,: from {A) and {B), 1 , 1 WO ■'■-pq + rpq = oFTTE=^ ""''''■ Note. — Eeciprocating with respect to P we obtain the theorem : ' The sum of the focal distances of a point on an ellipse is constant.' PArER LXVn. 7. Let A, B, be the points on the bar at which the strings from the pulleys are fastened, and let D be the point at which the weight is suspended. Let CI) = x. Then if a, b, c be the radii of the pulleys whose strings are fastened at A, B, C, OB = 2c - h, BA = n - a, A corresponding to the lowest pulley. Then we have forces P, 2P, 4P acting upwards at A, B, C, respectively, and IP downwards at D. .: taking moments about 0, iP.CD = 2P.BB + P. AD, .: ix = ic-2b -2x + 2c + 6 -a- a; = 6c - b ~ a-3x, ^c — b — u When there is equilibrium, W = IP, .: if the power be moving at any time with acceleration /, and if f be the acceleration of the APPENDIX. 319 weight,/' = }f .f. Suppose the power to be 2P instead of P. Then after any time t the kinetic energy which the system possesses will be equivalent to the work done by the system against gravity in that time. .-. \f-f>' . 2P + \p^W = \ffg . 2P - if'PgW. .: ^P . 2P + i/2 . 7P = ifff . 2P - \fg . 7P. ... 2 . 49/2 + 7/2 = g{Uf - If), .: 16/ = ff. Paper LXX. 2. x^p:^ + ^1^2 = 1, XiXs + yjya = 1 . _^_ = _Ji_ =;^^iiLM2= 1. Similarly -^^ = -^^ = I£i±Ma = I and ■•^3 ^ ^3 ^ ■•''3'^i + ^3^1 ^ 1 . .'. we have to prove that X. yi "^ yi. n yiyi3/% »■«■ y2y3(^2 - ^2) + •■• = ('^a - «'3)(^3 - '^OC*! - ^a)- The left-hand side = ^2^3(^2 - ^3) + ^3^1(^3 - '^i) - y^y^k^i. - s's + ^3 - «i) = {y-az - yiy2)(^2 - '^ii + (ys^i - yiy2)(^3 - ^^i) = y2(y3 - yiX^2 - -^s) + yifys - y2)(^3 - ^i)- Now ^1(^3 - y^ = x-SsCi - ir3) ; yaCya - yi) = '^2(«i - ^•3), /. the left-hand side = xix-^ - XgXsi - X3) + Xi{Xi - X3){Xg - x^) = (X2 - X3){X3 - Si){Xi - ^^2). 320 SOLUTIONS OF WEEKLY PROBLEM PAPERS. This may also be proved as follows. Let i) = (^ + (?2 + ^i, D = ^i>yi> 1 = H,yx^ 1 - r,2- -y^' = ^11^11 1 - o-^ 2^2, ^2, 1 ^2,^2.1 -^2-^1-^2^1 ^2-2, ^2, ^3> ys. 1 ^3,^3, 1 - ■■'■3^1 -^3^1 'J^3,y3>0 = (1 - ^1^ - yi')^- Now d^di = X = ^3-'!^i+y3yiJ ^2% + y2y3 = 1, 1 yi' + yi^ 1 = 1 - ^i' - i'l^ : f^i i - «'2 + <^3 = C k 6?2(?3 Paper LXXV. 6. Lot" 1- ■^ = 1 be the equation of the conic round ABC. Let ih, k) be the coordioates of F, {a cos a, b sin a) the coordinates of A. The equation of AF is y — i sin a = (.r - a cos a), h — a C03 a .•. the equation of diameter conjugate to AF is _ _ V^ h — a cos a cfi' h — b sin a ' i.e. b\h — a cos djx + (^Uc — b sin a)y = 0. The equation of the tangent at A is cos a + ^ sin a = 1, a b . : the line whose equation is b^{/i-acosa)x-i-a\k — bsma)y-\-ab{bxooBa + aysina-ab) = passes through 3. This reduces to b^Ax 4- aHv — a^b^ = 0, or"— j. -^ = 1, which is the equation of the polar of P. Similarly it may be shewn that the polar of F passes through E and F. APPENDIX. 321 Paper LXXVI. 7. Let AB be tbe horizontal line in the side of the wall, S the point of projection. Let AB, BG, CD be the sections of the walls made by a horizontal plane passing through AD, and let F, G, II be the points of impact in AB, BC, CD respectively. Let BE = a, EA = /la, AEF =• 6. Then AF = fxa tan 6, .: FB = a + //«(! - tan 6), and tan BF(r = e cot 6. BQ = BF\,!m BFQ = ae cot 6 + ^ae (cot (9 - 1), .-. C& = (ij. -\- 1)« (1 - « cot ff) + fiae. tan CGH = e tan BQF = e . - tan 5 = tan 6, e .: CJI = (u + 1)« (tan — e) + /lae tan 6, .: BE = «(/x + 1) (1 + e - tan 6) - /x«e tan 0. tan BEE = e tan CIIG = e cot 0, .: BE = ae{fi. + 1} |(1 + e) cot 6 — \\ - fj-oe' = ^a, .: e(fi + 1) cot 6 = e/i + 1. Paper LXXVII. 5. Let i'l, -P2 be the greatest and least forces. Then Pi : ^ :: sin a -+ fi cos a : 1, and Pi '• Jf^ '■'■ sin a — 11 cos a : 1, and ^ : iJ ; : 1 : cos a. ,■. (sin a -\- fi cos a) cos a = sin a — /li cos a, .-. ;. = sin''(l - c°«") = tan a tan^ ^ • cos a(l + cos a) 2 Paper LXXVIIL 7. Let the projectiles come into collision t seconds after the second is discharged. At the moment of collision their horizontal and vertical distances from the point of projection must be respectively eqnal. Y 322 SOLUTIONS OF WEEKLY PROBLEM PAPERS. .•. Fcos a . (n + () = V' cos a' . t, and Tsin a . (« -\- t) - \g{n + tf = F' . sin a' . ( - igfl, .'. i{Vs\n a — V sin a' — gri) = n,{\gn — ^sin a)^ and Vii cos a = t{V' cos a' — V cos a), .-. r?" sin (a - aO = igri{r cos a + V cos a'). Paper LXXIX. 3. (1). Since tan .r = cot x — 2 cot 2.r, the given expression on tlie right = tan "^ i cot - — , — 2 cot — + 2oot - - 22cot^ 4- 2'«-2 cot'L - 2''-i cot -„ + 2"-i I = tan ^';— \ cot '^ - 21-1 cot ^ + 2"-i \ = I. *'2). This question should he stated as follows. Prove that cos fl cos - o = _^_+ 1. ^^li + I •l^_ , „ 5^2 „d) .,t) ^2' .,$ „e ,d "^ ■■ ■ COS'' - COS'' - COS- -, COS'' - cos-* - cos- - 2 2 2^ -i -I- -l^ a Since 2 cos- - = 1 + cos 6, . o ^ !._ _!. '^""'^ ^ \_ I 1 COS 6 ^ COS''- COS''- cos^- cos^- 2 2 2 2 a a Similarly, by changing 6 into -, -, . . . cos - 1 , 1. _A 2 22 APPENDIX. 82S cos^ - 93 .2. 23 ,". substituting, we have 1 2 = cos"- 2 ,1 cnsfl (9 cos - 1 + ■ 23 .r 2 - -^^ 22 = given series. (3) Bec2^ + cosec2^ = 2^00860^2^; .*. sec'''ji = 22cosec23.^-co8ec2.4. 22(sec22^ + cosec22^) = 2«cosec222.i; .-. 22seo22^ = 2*cosec222^ 22(rt-i) (sec^ 2"-i 2^ + cosec^ S"-! A) = 22" cosec^ 2" ^ ; .-. 22«-i sec22'»-i^= 22» cosec^ 2" ^ - 22'>-2cosec2 2"-i^. .". by addition, the given series = 22ncos3c2 2»^ — cosec^.4. 4. Let ABC be the given triangle, A' EC the vertices of the isosceles triangles on BC, CA, AB. Let AA', BB', CG' meet BC, CA, AB in a, ;3, -/. Then since the isosceles triangles are similar, AB : AC ■.-.AC: AB", and the angle CAB' = BAC, . : the angle CAC = BAB: I'^uc. vi. 15, the triangles ABB', ACC are equal. Similarly the triangle BCC = BAA', and CAA' = CBB'. Now A BAA' : A CAA' :: Ba : Ca, and A CBB' : A ABB" :: C^ : A^, and A ACC : A BCC ■.-.Ay: By, .: Ba.C^. Ay = Ca. By. Afi, .'. Aa, B^, Cy are concurrent. 5. (^-af+f (!)• The equation of a chord through the origin \s y — mx (2). From (1) and (2), (1 + m^x^ - laic + fl2 - c2 = 0. by 324 SOLUTIONS OF WEEKLY PROBLEM PAPEES. If (.rj,^]), (.r^i/,) be the points where (1) iiitevseots (2), the equation of the circle described on the intercepted chord as diameter is = (x - x^) {x - x^) + (y - yi) (y - ya) = x^ + f - x(xi + x^) - y(yi + yj) + XyV^ + yiyj = («^ + /) 0- + m^) - 2« - itnciij + (. 330 SOLUTIONS OF WEEKLY PROBLEM PAPERS. Then A = area of ABC = ial sin X, A' = area otADG = ^i^sin/x, A _ ff sin X _ sin (f^ + ^i) sinX _ cot jj + cot (^ ' ' A' d sin fi sin {

= Rl sin ( - 6) (3). .-. Wb cos ^ = ij;{sin (5 + ) - sin (^ - (9)} , = iRl sin 5 cos 0. . •. -77 = -ffi san fl (4). .-. from (1), (3), and (4) T(j — a) sin (f> -{■ w- cos (j) = Tl aia ) and (4) that the locus of Q is still a circle. It will often be found that the question of finding the envelope of a straight line can be put in the following form. If two straight lines include a constant angle, and one of them always passes through a fixed point, it is required to find the envelope of the other line when the vertex moves alo:)g a, fixed straight line or circle. 336 SOLUTIONS OF WEEKLY PROBLEM PAPERS. 6. Now since the tangent at the vertex of a parabola is the locus of the foot of the perpendicular from the focus on a tangent, .•. if we have two straight lines at right angles, one of which always passes through a fixed point whilst the vertex moves along a fixed straight line, the other line will envelop a parabola having the fixed point for focus, and the fixed straight line for tangent at the vertex. V. If in 6 the two linos be inclined at ««y constant angle, the envelope will still be a parabola. Besant, Parab. Art. 36. 8. Since the auxiliarj' circle is the locus of the foot of the perpendicular from the focus of a central conic on a tangent, we see that if we have two straight lines at right angles, one of which always passes through a fixed point whilst the vertex moves along a fixed circle, the other line will envelop a conic having the fixed point for one of its foci, and the fixed circle for its auxiliary circle. From this it follows that if one of the lines passes through the focus of a given conic, and the angular point moves along a concentric circle, the other line will envelop a oonfocal conic. 9. Suppose that the lines, instead of being at right angles, include a constant angle. Let S be the fixed point, IPS, PQ any position of the two lines. Draw SQ perpendicular to FQ. Then since the angle SPQ is constant, and SQP is a right angle, .•. SP : SQ in a constant ratio. But the locus of P is a circle, .■. by 5, tlie locus of § is a circle, and .•. by 8, the envelope of PQ is a conic having S for a focus and the locus of Q for the auxiliary circle. Hence we see that if we can find the locus of the vertex, we can at once determine the envelope required. If the locus is a straight line, the corresponding envelope is a parabola. If the locus is a circle the envelope is an ellipse, hyperbola, or point, according as the fixed point is within, without, or on the circumference of the circle which is the locus of the foot of the perpendicular from the fixed point on the line whose envelope is required. From 7 and 9 we at once deduce the following theorem. If a triangle of given species has one angular point fixed, and if a second angular point moves along (1) a given line, (2) a given circle, the third will also move along (1) a given line, (2) a given circle; for the side opposite the fixed angle envelopes (1) a parabola, (2) a central conic. See Casey, p. 71 ; Catalan, Giom. p. 81. APPENDIX. 337 NOTE ir. Geometbical Maxima and Minima. It is usual to treat questions of this class as isolated problems, each requiring its own' special mode of solution. It will be seen, however, from the examples which are solved in different parts of this work, that problems relating to this subject can in general be treated by the following simple method. If the variation of a geometrical magnitude be continuous, when the magnitude has its max. or min. value its rate of change is zero ; in other words, when a varying quantity is a max. or min. its values in two consecutive positions are equal. .'. to find when it is a max. or min. we have only to assume two consecutive values to be equal, and we at once deduce the required result. As illustrations we will work out a few examples. 1. A and B are two points on the same side of a fiied line. It is required to find the point on the line at which -AB subtends a max. angle. Let P and Q be two points on the line which are indefinitely near to each other and such that the angle APB = J.QB. Then a . circle will go round JPQB, and the given line, which passes through the two consecutive points P and §, is a tangent. Hence the con- struction ; through A and B describe a circle touching the given line. The point of contact is the point required. 2. On a given chord as base inscribe the max. triangle in a circle. Assume the inscribed triangle to have the same magnitude for two consecutive points on the curve. The line joining these points is parallel to the base by Euc. i. 37. But the line joining two consecutive points on a curve is a tangent. .'. the tangent at the vertex is parallel to the base, and the max. triangle is isosceles. 3. Shew that the max. triangle which can be inscribed in a circle is equilateral. Suppose each of the sides in turn to remain fixed, whilst the opposite vertex moves along the curve. By 2, the triangle is in each case a max. when the tangent at the vertex is parallel to the base. .'. when all the sides vary, the tangents at the angular points are parallel to the opposite sides, and the triangle is equilateral. Questions which appear to depend upon the variation of two mag- nitudes can sometimes be made to depend upon only' one. 6oS SOLUTIONS OF WEEKLY PROBLEM PAPERS. 4. Given two points A and £, find the point P on a given line or circle Buoh that AP^ + BP^ may be a min. Let C be the middle point of AB. Then AP'^ + BP^ = 2AC^ + 2GF'. Thus we have only to find when CP^ is a min. Let P and P* be two consecutive points on the line- such that CP = CP'- Then since the angle PGP" is very small, CPP' and CP'P are ultimately right angles, and P is the foot of the perpendicular from C. In the case of the circle CP is at right angles to the line joining P and P', i.e. to the tangent at P. .: P is the point of intersection of the circle with the line joining to the centre. If CP be produced to meet the circle again in Q, AQ^ + B^ is a max. 5. ABC is a semicircle, centre and diameter AG. . ADO is another Kemicirole, centre 0', on the opposite side of AO. It is required to draw a chord at right angles to AO such that the portion intercepted between the curves shall be a max. Let P§, P'Q be two equal consecutive values of the chord. Then FP and QQ, which are the tangents at P and Q, are parallel by Euc. i. r.Z. .". OP, O'Q are parallel, and if PQ meets Off in iV, the triangles OPN, O'QN are similar. .■. O'N : ON :: 1 : 2. .-.the point N is determined. 6. In LXVII. 3, is given a geometrical proof of the theorem that the max. quadrilateral whijh can be formed from four given straight lines is inscriptible in a circle. From this it is easy to shew that of all plane rectilineal figures which can be formed from straight lines which are given both in number and length, the max. is inscriptible. For if A, B, C, D be four angular points taken in order, then ABCD is inscriptible. Similarly if B be the next angular point to B, the points BCBE will lie on a circle, and these two circles are the same, for they are both described about the triangle BGI). By proceeding in this manner we may shew that the theorem holds for any number of straight lines. If we suppose the number of sides to be increased and their lengths diminished indefinitely, we see that the area of a circle is greater than the area of any other closed figure having the same perimeter. 7. Again, since {ax + byf + {bx - ayf = {a^ + b^) {x^ + f), .-. if ax + by be given, and a and b be constants, x^ +^2 jg ^ min. when ~ = ■j- • The geometrical meaning of this gives a solution of the question ' In the base of a triangle find a point such that the sum of the squares on the perpendiculars drawn from it to the sides shall be a min.' APPENLiIX. 339 If the point be within the triangle, x, y, « its perpendicular distances from the sides, we at once deduce that is^ + ^^ + «^ is a min. when - = - = -, and the value of the min. sum is evidently a c u^ + i'^ + c^ It may be noticed that the point thus determined is the 'Symmedian'' point. See Paper LXVIIL, 4. 8. By considering {ax + byf — {ax - 6yy = iabxy, -ve see that the middle point of the base of a triangle is such that the product of the perpendiculars from it to the sides is a min. ; and we deduce that the point within a triangle such that the product of the perpendicular distances from it to the sides is a min. is the centroid of A3 the triangle, and that the value of the min. product is —r- : For further treatment of this subject, see an elementary treatise by the present author on Geometrical and Algebraical Max. and Min. without the aid of Differential Calculus, where it is shewn that the problem of finding the max. or min. values of geometrical magnitudes depending only on the point line and circle is reduced to the much simpler one of finding its positions of symmetry. THE END. Richard Clay & Sons, dread street hill, london, e.c and Bufig'ay, Sti_ffolk.