Hence the locus is a circle which intersects the ellipse in two coincident points, i.e., touches the ellipse; but, since + = 0 can only be true, ney b 7689. (By N'IMPORTE.)-If the two roots of the equation x2- a1x + a2 = 0 are whole and positive numbers, prove that (1) 3α2 (1 + α1 + α2) (1 + 2α1 +4α2) is a whole number decomposable into the sum of a squares; (2) a2 (1+a1 + a)2 is a whole number decomposable into the sum of a cubes; (3) a22 (1 + 2a ̧ + 4a2) is decomposable into the algebraic sum of 4a, squares. Solution by B. HANUMANTA RAU, M.A.; R. KNOWLES, B.A.; and others. Let m, n be the roots of the equation x2 - a1x+a2 = 0, then a1 = m +n and ag = mn. 1.32 (1+a1 + a2) (1 + 2α1 + 4α2) = = m (m + 1) (2m + 1) . ¡n (n + = 3mn (1 +m)(1 + n)(1 + 2m)(1 + 2n) 1) (2n + 1) [12+22+ ... + m2] [12 + 22+ +n2] = sum of mn squares. 2. 122(1+a1 + a2)2 = √5m2n2(1 +m)2(1 + n)2 = [{m (m + 1) ]2 . [§n (n + 1)]2 +m3] [13 +23+ +23] = sum of mn cubes. = [13+23+ ... ... 3. The expression m°n2 [1+2 (m + n) + 4mn] now = 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 + 2α1 +4α2) = (sum of 2m squares) × (sum of 2n squares) algebraic sum of 4mn or 4a, 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, 4, 4, the differences between the eccentric angles of the points of contact, and is equal to 2 cos 0 cos o cosy. 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. x are 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 ρ 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 λε + μ + v = 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 3. 30 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. : 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. Solution by D. BIDDLE, Member of the Aëronautical Society. Let Q1, M, be where plumb-lines let down from the balloons would at the moment of observation strike the earth; then cot 84° 10′ 10′′ =3 •1021151 rel. val. of QA = 553, BC (cos cos A = = = •1876869 = rel. val. of M,A = 791, and CA = AB2+ CA2-BC2 2AB. CA M1B ,, M,C 399, therefore - 3640771 = cos 111° 21′ 3′′, cos B cos C = 9313687; sin B = 6514951. With these data, proceeding to find the position of Q, draw perpendiculars therefrom to the sides of the triangle cutting AB in D, BC in E, and CA in F; then = AQ,2-AD2 = BQ,2-(AB-AD)2, BQ-BE2 CQ,2-(BC-BE), CQ,2-CF2 = AQ,2 - (CA-CF)2; Again, cos BAQ1 cos CAQ1— [(1 − cos2 BAQ1) (1 — cos2 CAQ1)]1 = cos A, .. 1-cos2 A=cos BAQ1+cos2 CAQ1-2 cos A. cos BAQ,. cos CAQ1· This results in the following equation: = whence AQ, 348.205 or 191.444. The latter only serves, as the balloon Q is over the triangle; hence BQ1 459 466 and CQ 344 599. More = = over, 1021151 : 1 = 191.444 1874.8 = height of Q. The position of M, and height of M can be found by the same formula in any case. But, in this, which are the same proportions as in the case of Q. Consequently AM1 must be the alternative or second value of AQ1, viz., 348.205; also 835.692, and CM1 BM1 = Moreover, = 626.769. The distance between the two balloons is the hypotenuse of a rightangled triangle, of which one side is the difference of their heights and the other side is represented by M1Q1• To find M1Q, we first find ZACM1+ ACQ1 59° 7' 25". 971.368 CQ,M1 Then CM1+CQ, CM1-CQ1=tan 60°26′ 17′′; tan (CQ,M,—CM,Q1), that is, therefore Next, that is, = sin CQ,M1: CM1 = sin (ACM1 + ACQ1): M1Q1 therefore log M1Q1 = 2·7311325, and M1Q1=538·434. Now the difference 7629. (By BELLE EASTON.)-A and B throw for a certain stake, A having a die whose faces are numbered 10, 13, 16, 20, 21, 25; and B a die whose faces are numbered 5, 10, 15, 20, 25, 30. If the highest throw is to win, and equal throws go for nothing; prove that the odds are 17 to 16 in favour of A. |