# Find the maximum area of the regular pentagon

Find the maximum area of the regular pentagon that inscribed a unit square.

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Why should we? Please tell us what you have tried, where the problem showed up etc instead of just giving us orders! –  mrf Feb 19 '13 at 9:47

The largest sidelength of the inscribed pentagon in the unit square is $$\frac{1}{2\cos(\pi/20)\cos(\pi/5)}=0.62573786...$$ This agrees with what I found experimentally using sketchpad, and so I searched the web for verification, finding this website giving details, even a construction of the optimal pentagon. Quite complicated, and the center of the pentagon is not that of the square.

http://mathafou.free.fr/pbg_en/sol118.html

[Look under the "largest pentagon" section. There's also largest hexagon, largest (equilateral) triangle.]

ADDED: description of sketchpad experiment: I drew a pentagon by starting with one side and successively rotating through 108 degrees. Then I drew a random line through one of the vertices, and the parallel line through the furthest vertex from the random line. From there I drew perpendiculars through the pentagon vertices which were furthest apart. In this way a rectangle was obtained, and by measuring the sides of the rectangle and moving the random line around, it became clear there is only one position of the line for which the rectangle ends up as a square. Then for comparison I divided the sidelength of the hexagon by the sidelength of the bounding square, coming up with the 0.625... number. The optimal orientation seems to be the only one for which four of the five vertices of the pentagon lie on sides of the square.

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Is this pentagon really "inscribed" in the square? It seems that one of its vertices doesn't quite make it. –  Ron Gordon Feb 19 '13 at 12:38
It's impossible for all five vertices to lie on the sides of the square. True, a better word would be "contained", so the question would be: Find the side length of the largest regular pentagon which is contained in the (closed) unit square. –  coffeemath Feb 19 '13 at 12:56
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.
This type of problem is often difficult because it is hard to prove you have found the best configuration. In this case there are two obvious candidates: put one side of the pentagon centered on one side of the square and expand the pentagon to maximum height or put one side at a $45^\circ$ angle in a corner and two corners on the other two sides. So calculate how big each is and you have your answer. But maybe there is a better answer?