Given a regular $n$-gon $Q$, there are many polygons $P$ that entirely contain $Q$, and such that all $n$ vertices of $Q$ lie on edges of $P$. These circumscribing polygons $P$ have different numbers of edges. What is the smallest number of edges possible for a circumscribing polygon?
In the following pictures, I present the solutions for $n=4,5,6$ for which the smallest circumscribed polygon are triangles, and for $n=7$ for which the smallest circumscribed polygon is a quadrilateral.
Is the solution of this problem known for general $n$?
Thank you for your help!
Given a regular $n$-gon $P$, the smallest number of vertices for a circumscribing polygon $Q$ is $$\max \left(\, 3,\, \left\lceil \frac{n}{2}\right\rceil \,\right).$$ Each side of $Q$ can contain at most $2$ vertices of $P$; since all $n$ vertices of $P$ need to lie on the sides of $Q$, $Q$ needs to have at least $\frac n2$ sides.