IMO 1997 SL 5

Origin: ROM

Problem

Let ABCD be a regular tetrahedron and M, N distinct points in the planes ABC and ADC respectively. Show that the segments MN, BN, MD are the sides of a triangle.

Solution

We shall prove first the 2-dimensional analogue: Lemma. Given an equilateral triangle ABC and two points M, N on the sides AB and AC respectively, there exists a triangle with sides CM, BN, MN. Proof. Consider a regular tetrahedron ABCD. Since CM = DM and BN = DN, one such triangle is DMN.

Now, to solve the problem for a regular tetrahedron ABCD, we consider a 4-dimensional polytope ABCDE whose faces ABCD, ABCE, ABDE, ACDE, BCDE are regular tetrahedra. We don’t know what it looks like, but it yields a desired triangle: for M \inABC and N \inADC, we have DM = EM and BN = EN; hence the desired triangle is EMN. Remark. A solution that avoids embedding in R4 is possible, but no longer so short.