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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!

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    $\begingroup$ In your first example solution (triangle circumscribed around the square), what are the side lengths/ angles in the triangle? $\endgroup$ – Zubin Mukerjee Feb 21 '18 at 14:05
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    $\begingroup$ 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" $\endgroup$ – Zubin Mukerjee Feb 21 '18 at 14:08
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    $\begingroup$ @EthanBolker Sorry for the confusion, I mean smallest by the number of vertices of the circumscribed polygon. $\endgroup$ – J. Bubar Feb 21 '18 at 14:13
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    $\begingroup$ @J.Bubar yes I think that is the same thing as the smallest number of sides, so it looks to be correct? $\endgroup$ – Zubin Mukerjee Feb 21 '18 at 14:14
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    $\begingroup$ 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. $\endgroup$ – Zubin Mukerjee Feb 21 '18 at 14:22
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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.

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    $\begingroup$ I have edited your answer to include the justification for it you gave in the comments, I hope you don't mind. (Feel free to edit again if you don't like my wording.) $\endgroup$ – Misha Lavrov Feb 21 '18 at 14:46
  • $\begingroup$ @MishaLavrov Thank you :) $\endgroup$ – Zubin Mukerjee Feb 21 '18 at 16:04

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