TITLE: Find the maximum area of the regular pentagon
QUESTION [6 upvotes]: Find the maximum area of the regular pentagon that inscribed a unit square.

REPLY [2 votes]: It is easier to start with a regular pentagon with vertices on the unit circle, one of them at $(1,0)$, and finding the smallest square containing it. Consider supporting lines $x \cos\phi +y \sin\phi =f(\phi)$. Then $f$ periodic with period ${2\pi\over 5}$, and  one has
$$f(\phi)=\cos\phi\qquad\left(-{\pi\over5}\leq\phi\leq{\pi\over5}\right)\ .$$
We  want to know
$$s:=\min_{0\leq\phi\leq 2\pi}\biggl\{\max\bigl\{f(\phi)+f(\phi+\pi), \ f(\phi+{\textstyle{\pi\over2}}) +f(\phi-{\textstyle{\pi\over2}})\bigr\}\biggr\}\ .$$
Because of symmetry reasons it is enough to consider the $\phi$-interval $\bigl[0,{\pi\over10}\bigr]$, in which each of the $f$-expressions on the right has a unique form.
When this $s$ has been determined it is easy to compute the area ratio between the pentagon and the resulting square.