Mathematical Questions and Solutions in Continuation of the Mathematical Columns of "The Educational Times" Volume 61 (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. 1894 Excerpt: ...point are (en, be, lab cos C). Now the equation to the first circle in the question is 2 cos C 2 agy) = yi (aa), whence, &c. The points in the question are, as is well known, the angular points of Brocard's second triangle. 12205. (W. J. Dokbs, B.A. Suggested hy Qupst. 12162.)--OA, OB are common tangents to two conies, I and II, which touch one another at C. From any point P on I tangents are drawn to II, meeting OA and OB in A and B respectively. From A and B are drawn two more tangents to I intersecting in Q. Prove that (1) P, C, Q are collinear; (2) the conic passing through Q, and touching PA and PB at A and B respectively, touches II at D; and (3) ), D, P are collinear. Solution by R. F. Davis, M.A. Projecting the line AB to infinity, so that A, B become the imaginary circular points at infinity, we have the following simple theorem, which scarcely needs demonstration: --Two conies have a common focus 0 and touch each other at C. Then, if Q, P be the other foci, and P lie on I, (1) C, Q, P are collinear; (2) the circle passing through Q, whose centre is at P, touches II, (3) at the extremity of the focal axis OP. 12198. (Professor Leinekugel.)--Par un point quelconque M d'une tangente a une parabole P, en un point B on eleve une perpendiculaire cette droite, qui rencontre la directrice en A; puis Ton trace AB. Demontrer que la perpendiculaire a AB, issue de M, est tangente i- la parabole. Solution by R. Chartres; II. W. Curjel, B.A.; and others. Draw BN perpendicular to the X directrix. Let perpendicular from M on AB cut AB in R. Produce KA to X and EM to Q. Then angles QMS, SMB = BMQ = MBR, BEM = AEM, BMA = MAX, BAN = NBM, BMN (from A, M, B, N are concyclic) = MBS, BMS; therefore QMS = MBS; therefore EMQ touches the parabola. 12212. (K M. Lang...