Minimize area of n-gon circumscribed around unit circle

Solution 1:

Divide the circumscribed $n$-gon into $2n$ right triangles, each with one vertex at the center of the circle, one vertex at a vertex of the polygon, and the third vertex at the point of tangency of a side of the polygon to the circle.

If the central angle in triangle $k$ is $\theta_k$, we have $0<\theta_k<\pi,\ k=1,\dots,2n, \sum_{k=1}^{2n}\theta_k=2\pi,$ and the area of the polygon is $$A:= \frac12\sum_{k=1}^{2n}\tan\theta_k$$

The area of the regular circumscribed $n$-gon is $$\frac{2n}2\tan\frac{2\pi}{2n} =\frac{2n}2\tan\left(\frac1{2n}\sum_{k=1}^{2n}\theta_k\right)\leq \frac{2n}2\sum_{k=1}^{2n}\frac1{2n}\tan\theta_k=A,$$ where the inequality follows from the convexity of the tangent function on $[0,\pi/2).$

NOTE

I wasn't able to do this without appealing to convexity. Without using convexity I was able to show that if a minimum exists, it is the regular $n$-gon, but I couldn't show existence, since the domain isn't compact.