IMO 1973 SL 5

Origin: FRA

Problem

A circle of radius 1 is located in a right-angled trihedron and touches all its faces. Find the locus of centers of such circles.

Solution

Let O be the vertex of the trihedron, Z the center of a circle k inscribed in the trihedron, and A, B, C points in which the plane of the circle meets the edges of the trihedron. We claim that the distance OZ is constant. Set OA = x, OB = y, OC = z, BC = a, CA = b, AB = c, and let S and r = 1 be the area and inradius of \triangleABC. Since Z is the incenter of ABC, we have (a + b + c)−\to OZ = a−\to OA + b−−\to OB + c−−\to OC. Hence (a + b + c)2OZ2 = (a−\to OA + b−−\to OB + c−−\to OC)2 = a2x2 + b2y2 + c2z2. (1) But since y2 + z2 = a2, z2 + x2 = b2 and x2 + y2 = c2, we obtain x2 = −a2+b2+c2 , y2 = a2−b2+c2 , z2 = a2+b2−c2 . Substituting these values in (1) yields (a + b + c)2OZ2 = 2a2b2 + 2b2c2 + 2c2a2 −a4 −b4 −c4 = 8S2 = 2(a + b + c)2r2. Hence OZ = r \sqrt 2 = \sqrt 2, and Z belongs to a sphere \sigma with center O and radius \sqrt 2. Moreover, the distances of Z from the faces of the trihedron do not exceed 1; hence Z belongs to a part of \sigma that lies inside the unit cube with three faces lying on the faces of the trihedron. It is easy to see that this part of \sigma is exactly the required locus.