Evdokimov M.A.

A convex polyhedron with vertices at the midpoints of the edges of a certain cube is called a cuboctahedron. When a cuboctahedron is intersected by a plane, a regular polygon is obtained. What is the maximum number of sides it can have?
Let the edge of the original cube, from which the cuboctahedron is obtained, be 1. Consider the sections of the cuboctahedron by a plane parallel to the base of the cube at a distance of $08$. Then the vertices of this $n$-gon must lie on the edges of the cuboctahedron, and no edge can have more than two vertices of the $n$-gon.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f7c9453eb6f3f37ea7g-47.jpg?height=743&width=681&top_left_y=-3&top_left_x=694)

Consider the section of the original cube, which is a regular hexagon (shaded gray in the figure), as well as the sections obtained from this by rotating $90^\circ$, $180^\circ$, and $270^\circ$ about the vertical axis of the cube. Note that the union of the sides of these four regular hexagons is the union of all the edges of the cuboctahedron. We will show that on the sides of one of these four selected regular hexagons, there lie at least 3 vertices of the $n$-gon. Indeed, if on the sides of each such hexagon there are no more than two vertices, then the total number of vertices will not exceed eight. Therefore, the plane of the $n$-gon section coincides with the plane of this hexagon, and a hexagon is obtained in the section of the cuboctahedron. This leads to a contradiction.

## Omвem

8.