IMO 1985 SL 10

Problem

2b.(VIE 1) Prove that for every point M on the surface of a regular tetrahedron there exists a point M ′ such that there are at least three different curves on the surface joining M to M ′ with the smallest possible length among all curves on the surface joining M to M ′.

Solution

If M is at a vertex of the regular tetrahedron ABCD (AB = 1), then one can take M ′ at the center of the opposite face of the tetrahedron. Let M be on the face (ABC) of the tetrahedron, excluding the vertices. Consider a continuous mapping f of C onto the surface S of ABCD that maps m + neı\pi/3 for m, n \in Z onto A, B, C, D if (m, n) \equiv (1, 1), (1, 0), (0, 1), (0, 0) (mod 2) re- D B C A B C A M6 M1 M2 M3 M4 M5 @ M ′ A C B B A C spectively, and maps each unit equilateral triangle with vertices of the form m+neı\pi/3 isometrically onto the corresponding face of ABCD. The point M then has one preimage Mj, j = 1, 2, . . ., 6, in each of the six preimages of \triangleABC having two vertices on the unit circle. The Mj’s form a convex centrally symmetric (possibly degenerate) hexagon. Of the triangles formed by two adjacent sides of this hexagon consider the one, say M1M2M3, with the smallest radius of circumcircle and denote by @ M ′

its circumcenter. Then we can choose M ′ = f( @ M ′). Indeed, the images of the segments M1 @ M ′, M2 @ M ′, M3 @ M ′ are three different shortest paths on S from M to M ′.