A prism with pentagons A1A2A3A4A5 and B1B2B3B4B5 as the top and bottom faces is given. Each side of the two pentagons and each of the 25 segments AiBj is colored red or green. Every triangle whose vertices are vertices of the prism and whose sides have all been colored has two sides of a different color. Prove that all 10 sides of the top and bottom faces have the same color.
Solution
We show first that the Ai are all the same color. If not then, there is a vertex, call it A1, with edges A1A2, A1A5 of opposite color. Now consider the five edges A1Bi. At least three of them must be the same color. Suppose it is green and that A1A2 is also green. Take the three edges to be A1Bi, A1Bj, A1Bk. Then considering the triangles A1A2Bi, A1A2Bj, A1A2Bk, the three edges A2Bi, A2Bj, A2Bk must all be red. Two of Bi, Bj, Bk must be adjacent, but if the resulting edge is red then we have an all red triangle with A2, whilst if it is green we have an all green triangle with A1. Contradiction. So the Ai are all the same color. Similarly, the Bi are all the same color. It remains to show that they are the same color. Suppose otherwise, so that the Ai are green and the Bi are red.
Now we argue as before that 3 of the 5 edges A1Bi must be the same color. If it is red, then as before 2 of the 3 Bi must be adjacent and that gives an all red triangle with A1. So 3 of the 5 edges A1Bi must be green. Similarly, 3 of the 5 edges A2Bi must be green. But there must be a Bi featuring in both sets and it forms an all green triangle with A1 and A2. Contradiction. So the Ai and the Bi are all the same color.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
12 Oct 1998