IMO 1985 SL 21

Origin: IRE

Problem

The tangents at B and C to the circumcircle of the acute-angled triangle ABC meet at X. Let M be the midpoint of BC. Prove that (a) \angleBAM = \angleCAX, and (b) AM AX = cos \angleBAC.

Solution

Let B1 and C1 be the points on the rays AC and AB respectively such that XB1 = XC = XB = XC1. Then \angleXB1C = \angleXCB1 = \angleABC and \angleXC1B = \angleXBC1 = \angleACB, which imply that B1, X, C1 are collinear and \triangleAB1C1 ∼\triangleABC. Moreover, X is the midpoint of B1C1 because XB1 = XC = XB = XC1, from which we conclude that \triangleAXC1 ∼ \triangleAMC. Therefore \angleBAX = \angleCAM and AM AX = CM XC1 = CM XC = cos \alpha.