IMO 1981 SL 17

Origin: USS

Problem

Three equal circles touch the sides of a triangle and have one common point O. Show that the center of the circle inscribed in and of the circle circumscribed about the triangle ABC and the point O are collinear.

Solution

Let us denote by SA, SB, SC the centers of the given circles, where SA lies on the bisector of \angleA, etc. Then SASB \parallelAB, SBSC \parallelBC, SCSA \parallelCA, so that the inner bisectors of the angles of triangle ABC are also inner bisectors of the angles of \triangleSASBSC. These two triangles thus have a common incenter S, which is also the center of the homothety \chi mapping \triangleSASBSC onto \triangleABC. The point O is the circumcenter of triangle SASBSC, and so is mapped by \chi onto the circumcenter P of ABC. This means that O, P, and the center S of \chi are collinear.