IMO 1990 Problem A1 Chords AB and CD of a circle intersect at a point E inside the circle. Let M be an interior point of the segment EB. The tangent at E to the circle through D, E and M intersects the lines BC and AC at F and G respectively. Find EF/EG in terms of t = AM/AB.
By Theo Koupelis, University of Wisconsin, Marathon ∠ECF = ∠DCB (same angle) = ∠DAB (ACBD is cyclic) = ∠MAD (same angle). Also ∠CEF = ∠EMD (GE tangent to circle EMD) = ∠AMD (same angle). So triangles CEF and AMD are similar. ∠CEG = 180 o - ∠CEF = 180 o - ∠EMD = ∠BMD. Also ∠ECG = ∠ACD (same angle) = ∠ABD (BCAD is cyclic) = ∠MBD (same angle). So triangles CEG and BMD are similar. Hence EF/CE = MD/AM, EG/CE = MD/BM, and so dividing, EF/EG = BM/AM = (1- t)/t. s are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 31st IMO 1990 © John Scholes jscholes@kalva.demon.co.uk 22 Sep 2002