How many edges could a cross-section of a polyhedron have? We know that the cross-section of a cube could have 3, 4, 5, or 6 edges. But there could be no more. This can be explained in many ways: (1) The number of edges of a cross-section can't exceed the number of faces of the polyhedron. (2) Consider the cube as a planar graph, if we divide the vertex set into two disjoint partitions, the maximal number of edges spanning between those sets is 6. But both methods only offer a very rough estimate of the upper limit of the number of edges of the cross-section. Is there a more accurate way to calculate how many edges could a cross-section of a polyhedron have?

Consider two cones which are joined along their bases (which are $2n$-sided polygons). Now, adjust the vertices along the base so that they zig-zag when you look at the cones from the side (move the odd vertices towards one cone's vertex and the even ones in the other direction). Now, a cut between the zig-zags will have as many sides as the original polyhedron.