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If I'm given a circle with radius $r$ and I want to create a polygon with side $n$ (say $n=5$) which can cover the circle fully, then how to prove that a regular polygon is the solution with minimum area?

Martin Sleziak
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Snake Eyes
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    Hint: Find the area of a triangle which has one vertex at the origin and angle $\theta$ at the origin and is tangent to the circle (use a little trig). Then, try to minimize the sum of $5$ such angles which add to $2\pi$. – Michael Burr Jul 04 '15 at 21:22
  • You may be familiar with the proof (first part of this previous Answer) that the polygon of given sides and maximum area is the cyclic one (the one which can be inscribed in circle). I suspect a similar approach would work here, varying the placement of a single side to minimize the area outside the circle and equalize the corresponding interior angles of the circumscribed polygon. – hardmath Jul 04 '15 at 21:36

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If you believe without proof that a minimal area circumscribed $n$-gon exists you can argue as follows:

Let $A$, $B$, $C$ be three successive points of tangency with $\angle(AOB)=\alpha$, $\angle(BOC)=\beta$. Then $\alpha+\beta<\pi$, and the polygonal area contained in the sector $AOC$ of central angle $\alpha+\beta$ is given by $$\tan{\alpha\over2}+\tan{\beta\over2}={\sin{\alpha+\beta\over2}\over\cos{\alpha\over2}\cos{\beta\over2}}={2\sin{\alpha+\beta\over2}\over\cos{\alpha+\beta\over2}+\cos{\alpha-\beta\over2}}\ .$$ Keeping $A$ and $C$ (and with them $\alpha+\beta$) fixed this can be increased if $\alpha\ne\beta$. It follows that the smallest area $n$-gon has to be regular.

Christian Blatter
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    +1. This method can be tweaked to prove the existence also. By moving $B$, we can swap $\alpha$ and $\beta$; so wlog, $\alpha$ is the max and $\beta$ the min of such central angles. If irregular, $\alpha > \frac{2\pi}{n} > \beta$; now, rather than replace $\alpha$ and $\beta$ with two copies of $\frac{\alpha+\beta}{2}$, replace with $\frac{2\pi}{n}$ and $\alpha+\beta-\frac{2\pi}{n}$. Repeat; this terminates. (I learned this trick from Dijkstra's EWD1140.) –  Jul 06 '15 at 12:14