Solution by D. BIDDLE, Member of the Aëronautical Society. F E (Scale, 400 yards to the inch.) Let Q, M, be where plumb-lines let down from the balloons would at the moment of observation strike the earth; then With these data, proceeding to find the position of Q1, draw perpendiculars therefrom to the sides of the triangle cutting AB in D, BC in E, and CA in F; then BQ-BE2 CQ2-(BC-BE)2, CQ,2-CF2 = AQ12 - (CA-CF)2; = = cos A, Again, cos BAQ, cos CAQ1-[(1 − cos2 BAQ1) (1 — cos2 CAQ1)] = whence AQ1 =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. Moreover, 1021151 : 1· = 191 444 1874.8 = The position of M, and height of M can be found by the same formula in any case. But, in this, height of Q. which are the same proportions as in the case of Q. Consequently AM, must be the alternative or second value of AQ1, viz., 348.205; also BM1 = 835.692, and CM, 626.769. Moreover, 1042710 1 = 348·205: 3339·4 = height of M. 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 M,Q, we first find Then AC2+ CM,2-AM,2 2AC.CM1 cos 30° 32′ 11′′, = cos 28° 35′ 14′′. CM1+CQ1 : CM1—CQ1 = tan 60°26′ 17′′. tan & (CQ,M,—CM,Q1), 971-368 282-170=1·7627623 : tan 27° 6′ 54′′, that is, therefore Next, that is, therefore log M,Q1 = 2·7311325, and M1Q1=538·434. Now the difference in the heights of M and Q = 3339-4-1874.8 = 1464·6; therefore (538-4342 + 1464·62) = MQ = 1560.44 distance between the balloons. == = 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. Solution by W. J. GREENSTREET, B.A.; B. H. RAU, M.A.; and others. If A throws 10, this throw is bigger than one of B's possible throws, and if A throws 13, 16, 20, 21, 25, this throw is bigger than 2, 3, 3, 4, 4 respectively; making a total of throws bigger than 17 of B's. Similarly, in B's case there is a total of throws bigger than 16 of A's. Thus the odds are 17: 16 in A's favour. 7584. (By Rev. T. R. TERRY, M.A.)-Prove that the sum of the series whose (p+1)th term is (n + p−1)! (n-p) 2n-p is zero. p! Solution by W. J. C. SHARP, M.A.; W. J. BARTON, B.A.; and others. Denoting the series {2′′-1 = n ! { 2" + n. 2′′-1 + − n! { 2n−1 + (n + 1) 2n − 2 + (n + 1) (2 + 2) . 2n −8 + ... } 1.2 ... =n!.2′′ (1 − } ) - " — n !. 2n − 1 (1 − 1) − n − 1 = n ! . 2′′ . 2n−n!. 2n−1.2n+1 = 0. = 0.] (n 2), 2n−3+ 7738. (By D. EDWARDES.) Prove that if any three lines be drawn from the centre of a triangle ABC to meet the circum-circle in P, Q, R, and the circle through the ex-centres in P', Q, R', (1) the triangle P'Q'R' is similar to PQR and of four times its area; (2) if the lines joining the centroid of ABC with the feet of the perpendiculars be produced through the centroid to meet the circum-circle in L, M, N, the triangle LMN is similar to the pedal triangle of ABC and of four times its area. Solution by B. HANUMANTA RAU, M.A.; SARAH MARKS; and others. 1. The circum-circle of ABC is the nine-point circle of the triangle formed by joining the ex-centres. If O, G, O' are the circum-centre, centroid, and nine-points centre of a triangle, then OG : GO' :: 2 : P::R: R, therefore G is the centre of similitude of the two circles. GP', GQ', GR' are respectively doubles of GP, GQ, GR, therefore triangle P'Q'R' is similar to triangle PQR and is of four times the area. 2. This follows directly from the fact that the feet of the perpendiculars of the triangle ABC are on the nine-point circle. 7737. A2xn-2 ... ... ... (By N'IMPORTE.)-If all the roots of the equation x”—a1x2-1+ an= 0 are whole and positive numbers, prove that (1) an (1+a + + a1) (1 + 2a, +4α2 + +2" an) / 6" is a whole number decomposable into the sum of an squares; (2) a2 (1+ α2+ ...+an)2/4" is a whole number decomposable into the sum of a, cubes; (3) a2 (1 + 2α, + ... + 2”an is decomposable into the algebraic sum of 2"a, squares. Solution by R. KNOWLES, B.A.; E. RUTTER; and others. Let the n roots be p, q, r, ... ± an = (x−p) (x − q) (x − 1') (x — 8) ... x2 — α ̧x2−1+ɑ2x2 − 2 ̧ Putting x = -1 and -, we have 1 + a1 + a + ... + an = (p + 1) (q + 1) (~ + 1) ..... and 1+241 +4α2+ +2" an ... = (2p+ 1) (2g + 1) (2r + 1) ... an = p. q.r.s. 1. First expression = p(p+1) (2p + 1) . & q (q + 1) (2g + 1) ... (12+22+ +p2) (12+22+ +q2) (12 + 22 + • +2·2) = ... ... squares, or an squares. 2. Second expression = p2 (p+1)2. q2 (+1)2... = à (q (13+23+. +p3) (13 + 23 + +93) = sum of a, cubes. 3. Third expression = p2 (2p+ 1). q2 (2q + 1) . r2 (2r + 1) ... ; p2 (2p+ 1) = } [2p (2p + 1) (4p + 1 ) ] − 2 [† p (p + 1) (2p+ 1)] = = Hence the expression = (2p squares) (24 squares) = sum of 2"an squares. 1966. (By the late SAMUEL BILLS.)—Find values of x, y that will make S (p+q2)* + 64 p2q2 ( p2-q) a perfect square. Solution by AsÛTOSH MUKHOPÂDHYÂY. The expression S can be at once seen to be a perfect square, when (1) p = q ; (2) p = 0, S = (q1)2; (3) q = 0, S = (p)2. Developing S, and arranging it in ascending powers of q, we have S = p2 + 68 pбq2 — 122 p1q* + 68 p2q6 + q3, which can easily be put into the form (p++34 p2q2 -639 q1)2 + 68 × 640 p2q6 — [(639)2 — 1]98 ; hence, when S is a perfect square, 68 × 640 p2q6 = (6392-1)98; whence (4) Verification.-Substituting for p2 this value, we get Similarly, if we arrange S in descending powers of q, from the symmetry of the expression we at once see the condition in this case to be q2 = 319 p2, and the value of S is easily inferred from (4) to be 34 NOTE ON BIOT'S FORMULA. By AsÛTOSH MUKHOPADHYÂY. 1. A magnetic needle suspended on its centre of gravity is constrained to move in a vertical plane, making an angle with the magnetic meridian; e is the dip in the magnetic meridian, and o the dip in the vertical plane in which we suppose the motion to be. = Refer the system to rectangular axes, x being at right angles to the plane of the magnetic meridian, y horizontal and in that plane, and z vertical. Then the vertical and horizontal components are FM sin 0, H = M cos 0, and the statical equation of virtual velocities is Foz + Hdy 0. Now, a cos o cos y, z= a sin p, where a is the distance of any particle of the needle from its centre; therefore dy - a sin o cos 484, dz = а соѕ ф бф, therefore F cos H sino cos, or sine cos = cos e sin cosy, there У fore which determine = cos ф = - = cote cos.... and thence the position of equilibrium. = .(1), The equation F cos p H sin cos gives F = M cos tan cos, which gives F, when M, &c. are known. |