A plane can intersect at most how many edges of a regular 10-sided prism?
Solution. First, we show that no plane can intersect more than 12 edges of the prism. Each face of the prism is a convex polygon, which any line, and thus the intersection line of the plane of the face and any non-parallel plane, either avoids, intersects at one point, or intersects along a segment. Therefore, a plane different from the planes of the prism's faces intersects at most two edges of any side of the prism. The prism has 12 faces, and each intersection point on an edge belongs to the two faces meeting at that edge, so the maximum number of intersection points is $\frac{12 \cdot 2}{2}=12$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ccd2a535692c3f09b5a5g-1.jpg?height=624&width=542&top_left_y=409&top_left_x=781)

We show that there is a plane that intersects exactly 12 edges of the prism. Let the base be the $A_{1} A_{2} \ldots A_{10}$ polygon, and the top be the $B_{1} B_{2} \ldots B_{10}$ polygon, where the side edges of the prism are the segments $A_{1} B_{1}, \ldots, A_{10} B_{10}$. Let the midpoints of the $A_{1} A_{2}$ and $A_{2} A_{3}$ edges be $C$ and $D$, and the midpoints of the $B_{6} B_{7}$ and $B_{7} B_{8}$ edges be $E$ and $F$. Then $C D$ is parallel to $A_{1} A_{3}$, which, due to the properties of the regular decagon, is parallel to $B_{6} B_{8}$, and the latter is parallel to $E F$. Therefore, the points $C, D, E, F$ lie on a plane $\mathcal{S}$. This plane intersects all side edges of the prism except for $A_{2} B_{2}$ and $A_{7} B_{7}$, because all vertices of the base except $A_{2}$ lie below $\mathcal{S}$, and all vertices of the top except $B_{7}$ lie above $\mathcal{S}$. Thus, $\mathcal{S}$ intersects a total of $4+8=12$ edges of the prism.

Remark. If we are looking for a plane that has the most "common points" with the "edges" instead of "intersecting edges," then the planes of the prism's bases provide the most common points, 20 in total.