IMO 1979 SL 4

Origin: BUL

Problem

A pentagonal prism A1A2 . . . A5B1B2 . . . B5 is given. The edges, the diagonals of the lateral walls and the internal diagonals of the prism are each colored either red or green in such a way that no triangle whose vertices are vertices of the prism has its three edges of the same color. Prove that all edges of the bases are of the same color.

Solution

Let us prove first that the edges A1A2, A2A3, . . . , A5A1 are of the same color. Assume the contrary, and let w.l.o.g. A1A2 be red and A2A3 be

green. Three of the segments A2Bl (l = 1, 2, 3, 4, 5), say A2Bi, A2Bj, A2Bk, have to be of the same color, let it w.l.o.g. be red. Then A1Bi, A1Bj, A1Bk must be green. At least one of the sides of triangle BiBjBk, say BiBj, must be an edge of the prism. Then looking at the triangles A1BiBj and A2BiBj we deduce that BiBj can be neither green nor red, which is a contradiction. Hence all five edges of the pentagon A1A2A3A4A5 have the same color. Similarly, all five edges of B1B2B3B4B5 have the same color. We now show that the two colors are the same. Assume otherwise, i.e., that w.l.o.g. the A edges are painted red and the B edges green. Let us call segments of the form AiBj diagonal (i and j may be equal). We now count the diagonal segments by grouping the red segments based on their A point, and the green segments based on their B point. As above, the assumption that three of AiBj for fixed i are red leads to a contradiction. Hence at most two diagonal segments out of each Ai may be red, which counts up to at most 10 red segments. Similarly, at most 10 diagonal segments can be green. But then we can paint at most 20 diagonal segments out of 25, which is a contradiction. Hence all edges in the pentagons A1A2A3A4A5 and B1B2B3B4B5 have the same color.