IMO 1992 LL SPA67

Origin: SPA

Problem

In a triangle, a symmedian is a line through a vertex that is symmetric to the median with the respect to the internal bisector (all relative to the same vertex). In the triangle ABC, the median ma meets BC at A′ and the circumcircle again at A1. The symmedian sa meets BC at M and the circumcircle again at A2. Given that the line A1A2 contains the circumcenter O of the triangle, prove that: (a) AA′ AM = b2 + c2 2bc ; (b) 1 + 4b2c2 = a2(b2 + c2).