5. A polyhedron with vertices at the midpoints of the edges of a certain cube is called a cuboctahedron. When the cuboctahedron is intersected by a plane, a regular polygon is obtained. What is the maximum number of sides this polygon can have?

(M. A. Evdokimov)
Solution.
![](https://cdn.mathpix.com/cropped/2024_05_06_b324deed733a73d5f064g-2.jpg?height=370&width=892&top_left_y=1484&top_left_x=584)

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 more than two vertices of the $n$-gon can belong to one edge. Consider the section of the original cube, which is a regular hexagon (see fig.), as well as sections that are obtained from this by rotating by $90^{\circ}, 180^{\circ}$, 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 the four chosen 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 section of the $n$-gon coincides with the plane of this hexagon, and a hexagon is obtained in the section of the cuboctahedron. This leads to a contradiction.