IMO 1967 LL CZS9

Origin: CZS

Problem

The circle k and its diameter AB are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on AB and two other vertices on k.

Solution

The incenter of any such triangle lies inside the circle k. We shall show that every point S interior to the circle S is the incenter of one such triangle. If S lies on the segment AB, then it is obviously the incenter of an isosceles triangle inscribed in k that has AB as an axis of symmetry. Let us now suppose S does not lie on AB. Let X and Y be the intersection points of lines AS and BS with k, and let Z be the foot of the perpendicular from S to AB. Since the quadrilateral BZSX is cyclic, we have \angleZXS = \angleABS = \angleSXY and analogously \angleZY S = \angleSY X, which implies that S is the incenter of \triangleXY Z.