IMO 1967 LL POL36

Prove that the center of the sphere circumscribed around a

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IMO 1967 LL POL36

Origin: POL

Problem

Prove that the center of the sphere circumscribed around a tetrahedron ABCD coincides with the center of a sphere inscribed in that tetrahedron if and only if AB = CD, AC = BD, and AD = BC.

Solution

Suppose that the skew edges of the tetrahedron ABCD are equal. Let K, L, M, P, Q, R be the midpoints of edges AB, AC, AD, CD, DB, BC respectively. Segments KP, LQ, MR have the common midpoint T . We claim that the lines KP, LQ and MR are axes of symmetry of the tetrahedron ABCD. From LM \parallelCD \parallelRQ and similarly LR \parallelMQ and LM = CD/2 = AB/2 = LR it follows that LMQR is a rhombus and therefore LQ \perp MR. We similarly show that KP is perpendicular to LQ and MR, and A B D C K M R P L Q T thus it is perpendicular to the plane LMQR. Since the lines AB and CD are parallel to the plane LMQR, they are perpendicular to KP. Hence the points A and C are symmetric to B and D with respect to the line KP, which means that KP is an axis of symmetry of the tetrahedron ABCD. Similarly, so are the lines LQ and MR. The centers of circumscribed and inscribed spheres of tetrahedron ABCD must lie on every axis of symmetry of the tetrahedron, and hence both coincide with T . Conversely, suppose that the centers of circumscribed and inscribed spheres of the tetrahedron ABCD coincide with some point T . Then the orthogonal projections of T onto the faces ABC and ABD are the cir- cumcenters O1 and O2 of these two triangles, and moreover, TO1 = TO2.

Pythagoras’s theorem gives AO1 = AO2, which by the law of sines im- plies \angleACB = \angleADB. Now it easily follows that the sum of the an- gles at one vertex of the tetrahedron is equal to 180◦. Let D′, D′′, and D′′′ be the points in the plane ABC lying outside \triangleABC such that \triangleD′BC ∼= \triangleDBC, \triangleD′′CA ∼= \triangleDCA, and \triangleD′′′AB ∼= \triangleDAB. The angle D′′AD′′′ is then straight, and hence A, B, C are midpoints of the seg- ments D′′D′′′, D′′′D′, D′D′′ respectively. Hence AD = D′′D′′′/2 = BC, and analogously AB = CD and AC = BD.