Paint each face of a regular octahedron with one of four colors so that the pairs of parallel faces are the same color. How many different colorings are possible if you have four colors of paint and want to use each one?
I. solution. We take the description and symmetries of the regular octahedron as known and do not prove them. We consider the models to be painted as movable.

The 8 faces form 4 pairs of parallel faces, so we paint 1 pair, i.e., 2 faces, with each color. To specify a color arrangement, it is sufficient to give the mutual position of the 4 faces meeting at a single vertex, since there are no parallel faces among them, and thus we have placed one occurrence of each color. We describe the color order at the chosen vertex by directing the axis running into the vertex towards our eyes and, starting from, say, the red face, we give the three colors seen in succession in a counterclockwise direction (the last one is followed by red again).

The first color after red can be chosen in 3 ways, the next in 2 ways, and this determines the color of the last face; thus, the coloring of the 4 faces (converging at a vertex) can be done in $3 \cdot 2=6$ ways. However, the colorings of the entire octahedron resulting from these do not differ from each other. We will show that in the desired coloring of the octahedron, the color order at all 6 vertices is different; from this it follows that the 6 different starting points lead to the same result.

Indeed, the coloring at two opposite vertices is different, because, for example, by painting the upper right face at the front vertex $E$ red, and the others white, green, and blue in sequence (Figure 1), the inner sides of the faces converging at the rear vertex $H$ are seen in the same order of $p, f, z, k$ - with $p$ on the left below - so from the back, i.e., from the outside, the order for $H$ is $p, k, z, f$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3715295c709011cddad6g-1.jpg?height=599&width=558&top_left_y=839&top_left_x=751)

The color order at two adjacent vertices is different because the colors of the two faces converging at their connecting edge appear next to each other in the two descriptions, but in reverse order. Continuing the example, the descriptions of the upper, lower, left, and right vertices are respectively

$$
F: p, z, k, f ; \quad A: p, f, k, z ; \quad B: p, z, f, k ; \quad J: p, k, f, z
$$

This proves our statement.

László Horváth (Hódmezővásárhely, Bethlen G. Gymnasium, 1st year, t.)

II. solution. We can reach the above result more simply if we turn the $F E J$ face of the octahedron perpendicular to our line of sight, paint it red, and write down the color order of the three faces adjacent to it: $f, k, z$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_3715295c709011cddad6g-1.jpg?height=555&width=504&top_left_y=1994&top_left_x=797)

Figure 2

The inner sides of the parallel faces at the back are also seen in the order of $f, k, z$ around the back $p$ face (rotated by $180^{\circ}$), so from the back, looking at the octahedron, the order of the faces adjacent to the $p$ face is $f, z, k$, different from the front.

However, the three remaining colors around $p$ can be placed cyclically in only 2 different orders, and the two starting points lead to the same color order for the (entire) octahedron. (By realizing the two color orders on one model each, these can be rotated so that the faces connecting their corresponding vertices are the same color.)

Márta Schügerl (Budapest, István I. Gymnasium, 3rd year, t.)