IMO 1986 SL 14

Origin: IRE

Problem

The circle inscribed in a triangle ABC touches the sides BC, CA, AB in D, E, F, respectively, and X, Y, Z are the midpoints of EF, FD, DE, respectively. Prove that the centers of the inscribed circle and of the circles around XY Z and ABC are collinear.

Solution

We shall use the following simple fact. Lemma. If <k is the image of a circle k under an inversion centered at a point Z, and O1, O2 are centers of k and <k, then O1, O2, and Z are collinear. Proof. The result follows immediately from the symmetry with respect to the line ZO1. Let I be the center of the inscribed circle i. Since IX \cdot IA = IE2, the inversion with respect to i takes points A into X, and analogously B, C into Y, Z respectively. It follows from the lemma that the center of circle ABC, the center of circle XY Z, and point I are collinear.