IMO 1978 SL 12

Origin: USA

Problem

In a triangle ABC we have AB = AC. A circle is tangent internally to the circumcircle of ABC and also to the sides AB, AC, at P, Q respectively. Prove that the midpoint of PQ is the center of the incircle of ABC.

Solution

Let O be the center of the smaller circle, T its contact point with the circumcircle of ABC, and J the midpoint of segment BC. The figure is symmetric with respect to the line through A, O, J, T. A homothety centered at A taking T into J will take the smaller circle into the incircle of ABC, hence will take O into the incenter I. On the other hand, \angleABT = \angleACT = 90◦implies that the quadrilaterals ABTC and APOQ are similar. Hence the above homothety also maps O to the midpoint of PQ. This finishes the proof. Remark. The assertion is true for a nonisosceles triangle ABC as well, and this (more difficult) case is a matter of SL93-3.