Mathematical Questions and Solutions Volume 67 (Paperback)

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1897 Excerpt: ...(P-W. Flood.)--Euler says it would be difficult to find x so as to make p-2pV + x + x a square; but show that it is not difficult. 13361. (W. C. Stanham.)--Construct (a) a right-angled triangle having its angular points on the given rectangular hyperbola xi--y2 = d2 (rectangular coordinates), and its centroid at the point (J- cos, a tan ), f being any given angle; and show (b) that, both in this case and in the case in which the centroid is to be at the point ( - sec p, -Ja cot f), the determination of the coordinates of the angular points of one such triangle will not necessitate the solution of an equation of a degree higher than the second. Solution by the Proposer, Professor Radhakrishnan, and others. Let G be the point (iJCOS(f, -tan(f), or (jy y), and let CG = r and L GCA = 29. (a) Draw CP to the curve, bisecting GCA; CO at right angles to CP, meeting PG in O; and QOR parallel to the diameter conjugate to CO. Then PQR is tlie triangle required. For OR2 = OQs = CO2 + CP2 = OP2, since conjugate diameters (sic) are equal, and diameters at right angles Note on Quest. 5517. By H. W. Cuejel, M.A. The Solution to this Question on page 25 of Vol. Lx. is evidently wrong, OP being the semi-diameter down which a particle will fall in the same time as down OA. If OR is the semi-diameter of quickest descent, R is evidently one of the points of contact of the circle (if there is one) passing through the centre and having double contact with the ellipse. Hence at R the normal equals the distance of its point of intersection with the major axis from the centre. Let p = the excentric angle of R. Then Hence, if 242 a2, i.e., if e 1/-/2, the value of p given by this equation is imaginary, and OA is evidently the semi-diameter of quick...