Cross-sections of a pyramid

Question

^{I found this problem on the Art of Problem Solving forum, but it's gone unanswered there. Let's see if PSE can do better!}

Let $n$ be a natural number greater than 4, and consider a pyramid whose base is a regular $n$-gon. Is it possible to construct a plane whose intersection with this pyramid is a regular $(n+1)$-gon?

_{(I do know the yes/no answer to this question [spoiler: it's in the link above], but my geometric intuition/visualisation isn't good enough for me to come up with a proof.)}

Answer 1

Answer:

It can't be done.

A pyramid with a regular $n$-gonal base has $n+1$ faces, so the path of intersection must go through every face exactly once. This means that the plane may not pass through any corners of the pyramid, otherwise at least one of the faces incident to that corner would not be intersected.

We now know how the plane must intersect the pyramid: The tip of the pyramid must be on the same side of the plane as exactly one other corner. This can be seen using the fact that the path of intersection corresponds to a Hamilton path of the dual polyhedron, which is conveniently also a regular $n$-pyramid.

Now let's first consider the case $n \geq 6$. The pyramid is inscribed inside a cone, and the plane intersects this cone in a conic section which is not a circle (since the plane is not parallel to the base of the pyramid). The regular $n+1$-gon is inscribed inside a conic section which is a circle. Now this non-circle and the circle coincide in at least $5$ separate points, those points are the intersection of the plane with $n-1$ of the spokes of the pyramid. Since five points are enough to uniquely determine a conic, this is a contradiction.

Now let's tackle the case $n = 5$ separately. Let $ABCDEF$ be the regular hexagon resulting from the intersection, where $A, B$ are the two vertices lying on the base of the pentagonal pyramid. Since $AB$ is parallel to the base, $DE$ is also parallel to the base. But if the pyramid is also regular, this means that $EF$ is longer than $DE$ (any $F'$ on the same spoke as $F$ such that $|EF'| = |DE|$ have at least the same altitude as $D$ and $E$ and therefore cannot be as low as $F$ is). This is a contradiction to the regularity of $ABCDEF$.