Minimum perimeter of a quadrilateral with an inscribed circle

Question

Let us say a circle is given, and the circle is inscribed to a quadrilateral. How would you prove that the minimum perimeter is achieved when the quadrilateral is a square?

Answer 1

Suppose the inscribed circle has radius 1. In the picture we have a poligon $IHFG$, and we have drawn the perpendicular segments from the center of the circle to the sides of the poligon. The angle $\alpha_1$ is the angle $JOK$, and the rest of the angles are analogous. The segment $OF$ bisects $\alpha_1$, so using the right triangle $FJO$ you have that $\overline{FJ}=\overline{JO}\tan\left(\frac{\alpha_1}{2}\right)=\tan\left(\frac{\alpha_1}{2}\right)$, and also $FK=\tan\left(\frac{\alpha_1}{2}\right)$.

Doing the same for the rest of the segments of the perimeter, we have that the perimeter is $2\left(\tan\frac{\alpha_1}{2}+\tan\frac{\alpha_2}{2}+\tan\frac{\alpha_3}{2}+\tan\frac{\alpha_4}{2}\right)$. So, calling $\beta_i=\frac{\alpha_i}{2}$, you have to minimize $\tan\beta_1+\tan\beta_2+\tan\beta_3+\tan\beta_4$, where $\beta_i\in(0,\frac{\pi}{2})$ and $\sum\beta_i=\pi$.

But $\tan(x)$ is a convex function in $\left(0,\frac{\pi}{2}\right)$, so using Jensen´s inequality, we have that $\tan\beta_1+\tan\beta_2+\tan\beta_3+\tan\beta_4\geq4\tan\left(\frac{\beta_1+\beta_2+\beta_3+\beta_4}{4}\right)=4\tan\left(\frac{\pi}{4}\right)=4$. That is, the minimum value is achieved when $\beta_i=\frac{\pi}{4}$, which is precisely the case of the square.