IMO 1986 SL 20

Origin: CAN

Problem

Prove that the sum of the face angles at each vertex of a tetra- hedron is a straight angle if and only if the faces are congruent triangles.

Solution

If the faces of the tetrahedron ABCD are congruent triangles, we must have AB = CD, AC = BD, and AD = BC. Then the sum of angles at A is \angleBAC + \angleCAD + \angleDAB = \angleBDC + \angleCBD + \angleDCB = 180◦. We now assume that the sum of angles at each vertex is 180◦. Let us construct triangles BCD′, CAD′′, ABD′′′ in the plane ABC, exte- rior to \triangleABC, such that \triangleBCD′ ∼= \triangleBCD, \triangleCAD′′ ∼= \triangleCAD, and \triangleABD′′′ ∼= \triangleABD. Then by the assumption, A \inD′′D′′′, B \inD′′′D′, and C \inD′D′′. Since also D′′A = D′′′A = DA, etc., A, B, C are the mid-

points of segments D′′D′′′, D′′′D′, D′D′′ respectively. Thus the triangles ABC, BCD′, CAD′′, ABD′′′ are congruent, and the statement follows.