Substituting, we get AQ 3.15.11.0+Y4.18 (21+ 25 g) 1.0 = = = = 3.15.11.0+Y4.13.1*1.0.2 Y3'.15.11.0 +Y4.18.11.2+Y4.13.5*2.0' by mere inspection of the table. We must now apply Rule 5 to each of these three terms thus :— Hence the first termy.15.12.1.0', which = Y3.15.12.0 by inspec tion. This term is now elementary, and we will denote it by E,. NOTE.-A statement of the above form is said to be elementary when the application of Rule 5 to any of the variables will introduce either no fresh limits among the other variables, or only such as are implied in the limits already stated. The second and third terms may be shown in like manner to be already elementary. We will denote them by E, and E, respectively; so that we have [This Solution, though, at the Proposer's request, subsequent in publication, was written before the second article on "Symbolical Language,' which appeared on p. 100 of Vol. XXVIII. of the Reprint. A more regular order of the letters will be obtained by interchanging y and z both in the Question and Solution. The present order was adopted because the Proposer was at first under the erroneous impression that the shortest solution would be obtained by making the coefficient of vary first; and the first solution sent to the Editor was in accordance with this order of integration.] 5515. (By E. P. CULVERWELL, M.A.)-If r be one of the four normal distances of a point P from an ellipse, and p the parallel central perpendicular on a tangent line; prove that, if {(pr)-1} vanishes, then P lies on the director circle of the ellipse; and state the corresponding theorem for an ellipsoid. Solution by E. W. SYMONS; the PROPOSER; and others. Let the coordinates of P be (a, B), and of Q, the foot of the normal, be (x, y); then we have a2a2 (b2+pr)2 + b2ß2 (a2 + pr)2 = (a2 +pr) (b2 + pr). If the coefficient of pr in this equation vanish, the sum of the reciprocals vanishes. That coefficient is [pr] [2 (a2b ̈a2 + b2ß2a2) — 2 (a2ba + b2aa)]. Hence sum of reciprocals vanishes if_a2b2 [a2 + B2 — (a2+b2)] = 0, that is if a, B lies on the director circle. to say, In the ellipsoid the sum of reciprocals evidently vanishes when (a,B) lies on an ellipsoid. 5467. (By CHRISTINE LADD.)-If three conics touch each other and have a common focus, prove that the common tangent of any two will cut the directrix of the third in three points which lie on one straight line. I. Solution by S. JOHNSTON; C. BICKERDIKE; and others. Reciprocating the theorem in Question 5356 with respect to any point, we have the theorem in question. II. Solution by H. W. HARRIS, D. EDWARDES, and others. First projecting and then reciprocating, we reduce the question to the following: "Three conics touch each other at a common point C, and have two points A and B common to all; prove that the three polars of the line AB with respect to each of the conics lie on a right line." Take ABC for triangle of reference, and let the three conics be then, since they have a tangent at C (which we take as a, B) common, we have = L 13 m1 m2 M3 But the three polars of AB, or y, have for their coordinates (11, m1, -n1), (l2, M2, −n2), (13, M3, —N3); and the condition that these should lie on a right line is which is evidently satisfied if the relations above hold good. 5299. (By L. H. ROSENTHAL.)-Solve the simultaneous equations, x3- ax2 + (b−2y) x + ay—c = 0 ........ x2y-axy - (y2- by + d) = 0......... Solution by the PROPOSER; J. RICHARDS, M.A.; and others. To solve these equations, take the general biquadratic, viz., x4-ax3 + bx2- cx + d = 0, and assume it to be the product of two quadratic factors, viz., χα-λα + μ and x2-vx+p; then, equating coefficients, we have the four equations, λ+ y = (, λν+μ +ρ == ..(1), ..(2). b, μv + pλ = c, μp = d......(1, 2, 3, 4), between any three of which we can linearly eliminate either of the pairs λ, μ, or v, p; the result being a relation between the other pair; and therefore, if we get two such relations from the four equations (1, 2, 3, 4), we shall have equations sufficient for the determination of these quantities. Now, eliminating and p between (1), (2), (3), and (1), (2), (4), we have respectively 1 0 入-a = 1 0 λα = 0; 1 μ-6 0 μ -d or, expanding, and V λ 1 μ-6 μ -c = 0, λ3 παλ? + λ (0 – 2μ) + αμ -c= = 0, μλαμλ (με - όμ + d) = 0, respectively. Hence, from the foregoing, we see that the solutions of the two latter equations are λ = a + ß, μ = aß ; where a, B, y, d are the roots of x-ax3 + bx2- cx + d 0. It is evident also from the above that there are six distinct (in one of the two equations) pairs of equations of the form proposed, whose solutions are the same. = Otherwise, to solve these equations directly, assume x2-ax + b = A, when the proposed equations become respectively, = Solve for y from (5) in terms of x and A, substitute for it in (6), and eliminate x between result, and x2-ax÷b-▲ 0; we thus get for ▲ the equation ▲3-b▲2 + (ac− 4d) ▲ +4bd — c2 — a2d = 0, the well-known equation for aß+7d, where a, ß, y, d are the roots of x-ax3+ bx2 - = 0......... .(7) ; therefore x2- ax + b = aß + yd, and hence, substituting for a and b in terms of the roots of (7), we find corresponding to the root aß+yd, λ = a+ ß and y+; similarly for the other two roots ay + ßd and ad+By. Again, from (6), substituting for A, aß+yd we get for y the values aß, yd, and similarly for the other values of A. Thus we see that the equation whose roots are a + B for a biquadratic may be written in the convenient and I believe new form, (x2-ax+b)3 − b (x2 — ax + b)2 + (ac — 4d) (x2 — ax + b) + 4bd — c2 — a2d = 0. It is also evident that the equation in A is the same as the equation which would be obtained for u+p, from the four equations (1, 2, 3, 4), and thus answers an elimination question connected with those four equations. Again, to obtain the equation in y whose roots are aß, &c., we have, eliminating ▲ between (5) and (6) and solving for x, x = y (ay-c) y2-d stituting in (6) and reducing, we get for the required equation, (y2 — d)2 (y2 —by + d) + y2 (ay−c) (cy—ad) = 0. sub The same equations may be formed for an equation of the nth degree by supposing it to be the product of a quadratic and an equation of the (n-2)th degree; and we can get two determinants of the (n-1)th and (n-2)th degrees respectively in a containing both A and μ; between which eliminating either we get required equation in the other. tli Similarly, by supposing an equation to be the product of two equations of the mth and (n-m)th degrees, we can find the equations whose roots are Σma, Zmaß, Σmaßy, &c. &c. This method, however, becomes very troublesome in equations beyond the sixth degree. 5441. (By D. EDWARDES.)-Prove that sin 2a sin (6-7) cos (0− a) + sin 28 sin (y-a) cos (0-8) + sin 2y sin (a-B) cos (0-y) sin 2a sin (87) sin a) + sin 28 sin (y-a) sin (0-8) + sin 2y sin (a- ẞ) sin (0 − y) tan (a+B+y-0). Solution by R. TUCKER, M.A.; D. EDWARDES; J. O. JELLY; and others. Let a+B+ y = x; then we have α 4 (numerator) = 4 {sin 2a sin (B-) cos (B + y −k) + ... + ... } = 2 sin 2a {sin (26— «) — sin (2y−k)} +. +... == cos (2α-2B + k) — cos (2a + 2ẞ−k) — cos (2a — 2y+K) + cos (2a + 2y −k) +cos (28-2y+k) — cos (2ẞ + 2y−k) — cos (2B — 2a + k) + cos (2B + 2α — K) + cos (2y-2a + k) —cos (2y + 2a —K) — cos (2y−2B + k) + cos (2y +26−k) − 2 sin ê {sin (2a — 2ß) + sin (28 − 2y) + sin (2y—2a)} ; +... +... 4 (denominator) = 2 sin 2a {cos (2y-k) — —cos (28−k)} + = 5475. (By C. TAYLOR, M. A.)-Prove the following construction for tangents to a conic :—Take a point T at a distance TN from the directrix, and divide ST in t so that St: ST = AX: TN, where A is the vertex, and X the foot of the directrix. About S draw a circle with radius SA, and from t draw tangents to the circle cutting the tangent at A in V, V'. Then TV, TV' will touch the conic. Solution by R. TUCKER, M.A.; S. TEBAY, B.A.; and others. 5564. (By Professor BENJAMIN PEIRCE, F.R.S.)-Find the probabilities at a game of a given number of points, which is played in such a way that there is only one person who is the actual player, and when the player is successful he counts a point, but when he is unsuccessful he loses all the points he has made and adds one to his opponent's score. Solution by SEPTIMUS TEBAY, B.A. The player may be successful a1 times in succession, and then fail; after this he may be again successful a times, and then fail a second time; and so on for i periods. Now, in order that the player may be finally successful, there must be i-1 failures, enabling the player to score a; points. Therefore, if n be the number of points in the game, we must have The equation a1+as+.... + a¿ = n−i + 1. ... admits of (n−1—ai)! (i-2)! (n-i+1-a;)! solutions, each of which terminates with a; successes. Hence the number of points scored by the player on this (i-2 (i-2)! · [(n − i + 1). x (x + 1) ... (x + i− 3) – (x − 1) x ... (x + i −3)] |