IMO 1979 · Problem 2

A prism with pentagons \(A_1A_2A_3A_4A_5\) and \(B_1B_2B_3B_4B_5\) as top and bottom faces is given. Each side of the two pentagons and each of the line-segments \(A_iB_j\) for all \(i,j=1,\ldots,5\), is colored either 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. Show that all 10 sides of the top and bottom faces are the same color.

Official text of the problem. Official problem paper (PDF)

Solution

The hypothesis says that no triangle whose sides have all been colored is monochromatic. All subscripts below are read modulo \(5\).

We first observe that any three vertices of a pentagon contain two consecutive vertices. Indeed, otherwise the successor of each selected vertex would be a distinct unselected vertex, requiring at least six vertices.

Fix a vertex \(A_i\). Among the five segments \(A_iB_j\), at least three have the same color, say \(c\). By the observation, two of their endpoints are consecutive vertices \(B_j,B_{j+1}\). The side \(B_jB_{j+1}\) must therefore have the other color, say \(d\).

Let \(A_k\) be either neighbor of \(A_i\) on the top pentagon. If \(A_iA_k\) had color \(c\), the triangles \[ A_iA_kB_j \qquad\text{and}\qquad A_iA_kB_{j+1} \] would force both \(A_kB_j\) and \(A_kB_{j+1}\) to have color \(d\). But then \(A_kB_jB_{j+1}\) would be a monochromatic triangle of color \(d\), a contradiction. Thus both sides of the top pentagon incident with \(A_i\) have color \(d\).

Since \(A_i\) was arbitrary, the two sides meeting at every top vertex have the same color. Consequently, all five sides of the top pentagon have one color. Interchanging the color names if necessary, suppose this color is red.

For each bottom vertex \(B_j\), at most two of the segments \(B_jA_i\) can be red: three red segments would have two consecutive top endpoints, forming a red triangle with the red side between them. Hence every bottom vertex is joined by green segments to at least three top vertices.

Now consider any bottom side \(B_jB_{j+1}\). Each endpoint has at least three green neighbors among the five top vertices, so the two sets of green neighbors intersect, since \[ 3+3>5. \] Choose a common green neighbor \(A_i\). The triangle \(A_iB_jB_{j+1}\) then forces \(B_jB_{j+1}\) to be red.

Thus all five bottom sides are red as well, proving that all ten sides have the same color.

The solution was checked against John Scholes' IMO solution archive (Kalva).