22nd ASU 1988 Problem 12 In the triangle ABC, the angle C is obtuse and D is a fixed point on the side BC, different from B and C. For any point M on the side BC, different from D, the ray AM intersects the circumcircle S of ABC at N. The circle through M, D and N meets S again at P, different from N. Find the location of the point M which minimises MP.
Take A' on the circle S such that AA' is parallel to BC. Let the ray AD meet S again at P'. Then ∠MNP' = ∠ANP' (same angle) = ∠AA'P' (A'AP'N cyclic) = ∠A'DB (BC parallel to AA') = angle MDP (opposite angles). So MDNP' is cyclic, so P must be P'. Since P is a fixed point, independent of M, we minimise MP by taking M as the foot of the perpendicular from P to BC. 22nd ASU 1988 © John Scholes jscholes@kalva.demon.co.uk 26 June 2002