Solution by the PROPOSER. == Let O, the centre of the pulley, be taken as origin, the axis of y being vertical; then, by the principle of virtual velocities, if y, y' be the coordinates of P and Q. Py+ Qy' = const., or y +λy' being the radii vectores of P and Q, r+r' = const. restricted as defined in the question, = = A; also, r and r 2—2Lr+ My = N, or (B-)2- 2L (B − r') + M (A —λy') = N, which establishes a linear relation between y' and r', and therefore represents a Cartesian oval as defined. 7615. (By W. NICOLLS, B.A.)-If u1 + U2+ Uz = c represent a surface of revolution, the origin being the centre of revolution, and u1, uq, uz containing respectively all the terms of the first, second, and third degrees in x, y, z; prove that u, is perpendicular to the axis of revolution and a factor of 3. Solution by J. P. JOHNSTON, B.A.; ELIZABETH BLACKWOOD; and others. Considering the equation of the cubic surface in cylindrical coordinates (z, r, e) as an equation for r, and taking the axis of z as the axis of revolution, it is evident that it can only contain even powers of r since all sections perpendicular to the axis are circles. Therefore the equation is of the form z [z2 + r2 ƒ1 (0)] + az2 + r2 ƒ1⁄2 (0) + bz + c = 0, where ƒ1 (0) and ƒ (0) are quadratic functions of cose and sin 0. Hence it appears that if his be put in the form u3+2+1 = c, u1 is perpendicular to the axis of revolution and a factor of uz. [Mr. NICOLLS' theorem may be slightly generalised thus:-All cubics of revolution can be written in the form LC + C' + L = 0, where C and C' are cones and L a plane perpendicular to the axis of revolution.] μ 7445. (By C. LEUDESDORF, M.A.)—A particle, describing a circular orbit about a centre of attractive force u (distance)-3 tending to a point on the circumference, is disturbed by a small force ƒ tending to the same point; prove that the variations of the diameter (2a) and of the inclination to a fixed straight line in the plane () of that diameter which passes through the centre of force are given by the equations Solution by D. EDWARDES; MARGARET T. MEYER; and others. If the attraction be μD-5 and the velocity of projection that from in finity, the orbit is u = (2a) -1 sec (0−w), where h2 stants vanishes, we have ·(1). which arises from the variation of the con (. da de + a sec (0−) 0......(2). Cosec = de da 2a2 cos (0-w) de' миз f h2 h2u dw 2a cos3 (0-) do' a sec (0-0) this gives by (2) dw 8a1f cost (0-w). de h2 8a3 (2) 1 sec2 (0 — w); de 1 But = hu2 = dt da - a sec (0-w) dt dw dt 7498. (By A. MARTIN, B.A.)-If a straight line be drawn from the focus of an ellipse to make a given angle a with the tangent, show that the locus of its intersection with the tangent will be a circle which touches or falls entirely without the ellipse according as cos a is less or greater than the excentricity of the ellipse. Solution by Rev. J. L. KITCHIN, M.A.; J. O'REGAN; and others. The equations of the tangent and of a line through the positive focus at angle a with tangent are or [kx (x-ae)+y (ky-ae)]2 = a2 (y — kx + ae k)2 + (a2 — a2e2) (ky + x— ae)2, which becomes k2 (x2+ y2)-2ae ky + a2e2 a2 (1 + k2) Hence the locus is a circle which intersects the ellipse in two coincident points, i.e., touches the ellipse; but, since ey + b b = 0 can only be true, k as regards the ellipse, as long as y or <-b; hence the circle touches or = 7689. (By N'IMPORTE.)-If the two roots of the equation x2-a1 x + a2 = 0 are whole and positive numbers, prove that (1) ¿α2 (1 +α1 + α2) (1 + 2α1 + 4α2) is a whole number decomposable into the sum of a squares; (2) a2(1+a+a)2 is a whole number decomposable into the sum of a2 cubes; (3) ag2 (1 + 2α, + 4a) is decomposable into the algebraic sum of 4a2 squares. Solution by B. HANUMANTA RAU, M.A.; R. KNOWLES, B.A.; and others. Let m, n be the roots of the equation x2 - α1x + α 2 0, then a1 = m + n and ag = mn. 1.32 (1+ a1 + a2) (1 + 2a1 + 4a2) = = = 3mn (1 +m)(1 + n)(1 + 2m) (1 + 2n) 1)(2n + 1) [12+22+ +m2] [12 + 22 + +n2] = sum of mn squares. ... ... 2. 1a2(1+a1+a2)2 = ‚¿m2n2(1 +m)2(1 + n)2 = [μm (m + 1) ]2 . [3n (n + 1)]2 [13 +23+ +m3] [13+23+ +n3] = sum of mn cubes. now 3. The expression m^n2 [1+2 (m+n) +4mn] = m2 (2m + 1) . n2 (2n + 1) ; m2 (2m + 1) 2 × } [2m (2m + 1)(4m + 1)] − 4 [μm (m + 1)(2m + 1)] 2.12+22+2.32+ 2 (2m-1)2+(2m)2 = algebraic sum of 2m squares. Thus a22 (1 + 2α, +4α2) (sum of 2m squares) × (sum of 2n squares) algebraic sum of 4mn or 4a2 squares. 7563. (By Rev. T. C. SIMMONS, M.A.)-Show that the ratio of the area of a triangle inscribed in an ellipse to the area of its polar triangle depends only on 0, 9, 4, the differences between the eccentric angles of the points of contact, and is equal to 2 cos e cos cos. Solution by MARGARET T. MEYER; Professor NASH, M.A.; and others. The points of contact are a cos a, b sin a, a cos B, b sin B, a cosy, b sin y, where (a, B, y) are the eccentric angles; and the equations of the tangents COS a + sin α= 1, &c. are a Area of triangle in the ellipse = ab [sin (a-8) + sin (B-) + sin (y-a)]. Area of triangle contained by the three tangents is 4 sin (a-B) sin (6-7) sin (y − a) = 2 cos(a-B) cos (8− y) cos (y− a) = 2 cos 0 cos cos. 7698. (By R. LACHLAN, B.A.)-Show that (1) four circles can be drawn cutting the sides of a triangle in angles a, B, y respectively; (2) if their radii be p, P1, P2, P3, and they cut any other straight line in angles, P1, P2, P3, then Solution by Rev. T. C. SIMMONS, M.A.; B. HANUMANTA RAU, M.A.; and others. Consider first the circle whose centre lies within the triangle; let di, da, da denote the distances of its centre from the sides, and P its radius; then So, if p1 be the radius of that circle whose centre lies beyond the side a, and, since the expressions always give real values for the four radii, four circles can always be drawn. Again, let the equation to the new line referred to the given triangle as triangle of reference be = 0, and let the perpendiculars on it from the four centres be respectively P, P1, P2, P3; then 7695. (By J. O'REGAN.)-Two persons play for a stake, each throwing two dice. They throw in turn, A commencing. A wins if he throws 6, B if he throws 7: the game ceasing as soon as either event happens. Show that A's chance is to B's as 30 to 31. Solution by D. BIDDLE; W. J. GREENSTREET, B.A.; and others. Out of 36 ways of throwing two dice, 6 may be turned up in 5 ways, viz., 1+5, 2+4, 3+3, 4+2, 5+1; and 7 may be turned up in 6 ways, viz., 1+6, 2+5, 3 + 4, 4 + 3, 5 + 2, 6 + 1. There are therefore 31 chances against throwing 6, but only 30 against throwing 7. The probability that B will have a throw after A is accordingly ; but that A will throw again after B, only 38. 1192. (By the EDITOR.)-In order to ascertain the heights of two balloons (Q, M), their angles of elevation as set forth hereunder are observed, at the same instant, from three stations (A, B, C) on the horizontal plane, whose distances apart are AB 553, BC = 791, CA = 399 = yards, (Q, A) denoting the elevation of Q at A, &c. : (Q, A) (Q, B) (Q, C) = 84° 10' 10" | (M, A) = = = 84° 2′ 50′′ 76° 13′ 46.5" | (M, B) = 75° 57′ 1′′ It is also observed that only one of the balloons (Q) is vertically over the triangle ABC. Show that the heights of the balloons Q, M are 1874.8, 3339-4, and that their distance apart is 1560·4. |