IMO 1999 SL G3

Origin: EST | Category: Geometry

Problem

A set S of points in space will be called completely sym- metric if it has at least three elements and satisfies the following condition: For every two distinct points A, B from S the perpendicular bisector of the segment AB is an axis of symmetry for S. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, a regular tetrahedron, or a regular octahedron.

Solution

Let rPQ denote a reflection about the planar bisector of PQ with P, Q \inS. Let G be the centroid of S. From rPQ(S) = S it follows that rPQ(G) = G. Hence G belongs to the perpendicular bisector of PQ and thus GP = GQ. Consequently the whole of S lies on a sphere \Sigma centered at G. We note the following two cases: (a) S is a subset of a plane \pi. In this case S is included in a circle k, G be- ing its center. Hence its n points form a convex polygon A1A2 . . . An. When applying rAiAi+2 for some 0 < i < n −1 the point Ai+1 trans- forms into some point of S lying on the same side of AiAi+1, which has to be Ai+1 itself. It thus follows that AiAi+1 = Ai+1Ai+2 for all 0 < i < n −1 and hence A1A2 . . . An is a regular n-gon. (b) The points in S are not coplanar. It follows that S is a polyhedron P inscribed in a sphere \Sigma centered at G. By applying the previous case to the faces of the polyhedron, it follows that all faces are regular n-gons. Let us take an arbitrary vertex V and let V V1, V V2 and V V3 be three consecutive edges stemming from V (V , V1, V2, and V3 defining two adjacent faces of P). We now look at rV1V3. Since this transformation leaves the half-planes [V1V3, V2 and [V1V3, V invariant and since V2 and V are the only points of P on the respective half-planes, it fol- lows that rV1V3 leaves V and V2 invariant. This transform also swaps V1 and V3. Hence, the face determined by V V1V2 is transformed by rV1V3 into the face V V3V2, and thus the two faces sharing V V2 are congruent. We conclude that all faces are congruent and similarly that vertices are endpoints of the same number of edges; hence P is a regular polyhedron. Finally, we have to rule out S being vertices of a cube, a dodecahedron, or an icosahedron. In all of these cases if we select two diametrically

opposite points P and Q, then S{P, Q} is not symmetric with respect to the bisector of PQ, which prevents rPQ from being an invariant transformation of S. It thus follows that the only viable finite completely symmetric sets are vertices of regular n-gons, the tetrahedron, and the octahedron. It is not explicitly asked for, but it is easy to verify that all of these are indeed completely symmetric. Remark. On the IMO, a simpler version of this problem was adopted, adding the condition that S belongs to a plane and thus eliminating the need for the second case altogether.