Minimum circumsribed square

Question

I want to find the minimum circumsribed square of the regular polygon with 2n edge. n is the odd number.I have known the method by observing some simple condition. But how to prove it strictly.

Answer 1

$n$ is odd; write $n=2p+1$. Let us represent the regular $2n$-gon by the $2n^\text{th}$ roots of unity in $\Bbb C$, i.e. $U:=\{e^{i\frac{2k\pi}{2n}}\mid k\in\{0,\ldots,2n-1\}\}$. We can rotate this polygon by multiplying the roots by $e^{i\alpha}$. Considering various symmetries, we can restrict $\alpha$ to $[0,\frac\pi{2n}]$. The width and height of the rotated polygon (i.e. the width and height of the smallest rectangle which contains the polygon) are then, respectively, $$\max_{x,y\in U}\Re(xe^{i\alpha}-ye^{i\alpha})\quad\text{and}\quad\max_{x,y\in U}\Im(xe^{i\alpha}-ye^{i\alpha}).$$ We can easily see that these two values are $$2\cos(\alpha)\quad\text{and}\quad2\cos(\frac\pi{2n}-\alpha),$$ i.e. $(x,y)=(1,-1)$ and $(x,y)=(e^{i\frac{p\pi}n},e^{-i\frac{p\pi}n})$ respectively.

$2\cos$ is nonincreasing on $[0,\frac{\pi}{2n}]$, so to minimize the height and width, it suffices to find $\alpha\in[0,\frac\pi{2n}]$ such that $2\cos(\alpha)=2\cos(\frac\pi{2n}-\alpha)$. We solve and get $\alpha=\frac\pi{4n}$, so the minimal square circumscribing the regular $2n$-gon of diagonal length $2$ is a square of width $2\cos(\frac\pi{4n})$.