# Smallest circumscribed polygon around regular polygons

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 $n$?

• I suspect that for a regular $n$-gon, the smallest circumscribed polygon with the minimum number of sides will be a $k$-gon, where $$k = \max \left(3, \left\lceil \frac{n}{2}\right\rceil \right)$$ The reason for this guess is that each edge can contain $2$ of the regular $n$-gon's edges. The proposed formula does work for your four examples, and for a regular octagon, for which a square is the smallest circumscribing polygon. A pentagon (not regular) should be the smallest circumscribing polygon for a regular $9$-gon. Edit: thanks Wouter, added "with minimum number of sides" – Zubin Mukerjee Feb 21 '18 at 14:08
• Let me rewrite the question $$\,$$ Given a regular $n$-gon $Q$ with all sides length $1$, there is a set $S = \{P\}$ of polygons $P$ that entirely contain $Q$, and such that all $n$ vertices of $Q$ lie on edges of $P$. The polygons $P$ in $S$ have different numbers of edges. Let $k$ be the minimum number of edges over all polygons $P$ in $S$. What is the $k$-gon that is in the set $S$ with the smallest area? $$\,$$ Hopefully that is the question you want to ask. I suspect that $k = \max \left(3, \lceil n/2 \rceil \right)$, but I'm not sure what shape that $k$-gon will have. – Zubin Mukerjee Feb 21 '18 at 14:22
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.