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This question has a recreational flavor, but may not be entirely uninteresting.

Let $P_k$ be a unit-radius regular polygon of $k$ sides, and $P_n$ a unit-radius regular polygon of $n \ge k$ sides. Both are origin-centered. Fix $P_k$, and let $P_n$ rotate about its center by angle $\theta$. Define $s(k,n)$ to be the sequence of the number of sides of $P_k \cap P_n(\theta)$ as $\theta$ varies from $0$ to $2\pi$. For example, I believe that $s(4,8)=(\overline{4,8})$, with the overbar indicating repetition, while $s(5,8)=(\overline{10,9})$. Both are illustrated below. (Reload the page to repeat the [limited] animations.)

          RotReg48
          RotReg58
It is an elementary puzzle to

Q. Determine $s(k,n)$ precisely for all $n \ge k$.

I was interested in a higher-dimensional version of this question, but already, although elementary, it seems to need some care to nail down $s(k,n)$ explicitly.

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There are two situations that can happen around a vertex of $P_k$, either the two nearest vertices of $P_n$ are going to crop off a small corner (thus adding one edge to the intersection $P_k\cap P_n$), or a vertex of $P_n$ coincides with the vertex of $P_k$. In the second situation, once two vertices coincide, $\gcd(n,k)$ of them will coincide. Therefore $$s(k,n)=(\overline{2k,2k-\gcd(n,k)}).$$

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  • $\begingroup$ Neat! Always just two integers repeated. $\endgroup$
    – Joseph O'Rourke
    Commented Oct 17, 2015 at 1:37
  • $\begingroup$ Now I wonder if, permitting translation of $P_n$ as well as rotation, one can achieve intersections with a number of sides equal to every integer from $3$ to $2k$. But that is for a (possible) future question... $\endgroup$
    – Joseph O'Rourke
    Commented Oct 17, 2015 at 13:33

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