Two regular $2n$-sided pyramids $(n \geq 2)$ (with a regular polygon base and equal side edges), whose bases are congruent, are joined by placing their congruent bases in coincidence, such that a convex polyhedron composed entirely of triangular faces is formed. What is the maximum number of sides that a plane section of the resulting solid can have?
The axis of the polyhedron formed by the combination is the straight line connecting the apexes of the two pyramids. We understand an axis of the polyhedron to be a line that is the axis of symmetry for every cross-sectional domain of a plane intersecting the said line.

The edges of our polyhedron are grouped as follows: 1) the base edges of the pyramids, which, after combination, form a $2n$-sided regular polygon with center $O$ in a single plane, consisting of $2n$ edges; 2) $n$ planes intersect the axis, each containing 4 pyramid edges, forming a kite that is symmetrical to the axis. The equal sides of the kites are the side edges of the combined pyramids (opposite to each other).

Each edge of our polyhedron lies in one of the mentioned $n+1$ planes, but only in one. A cutting plane intersects each of these planes in a straight line, thus intersecting the perimeter of each polygon at most at 2 points. Therefore, the number of intersection points formed by the edges of the polyhedron is at most $2(n+1)$, meaning the resulting plane section can be a polygon with at most $2(n+1)$ vertices and sides.

Such a polygon with this many sides can indeed be obtained, for example, by intersecting the body with a plane that passes through two points on the perimeter of the $2n$-sided polygon, opposite to each other and not coinciding with the vertices, and does not contain the axis. This plane passes through the intersection points of the diagonals of the kites, thus intersecting their perimeters at 2 points each. These points are all distinct, as the plane does not pass through any vertex.

Benkö Bálint (Sárospatak, Rákóczi g. III, o. t.)