[Oblarg]

Originally Posted by DampRobot

Challenge accepted!

In a regular n-gon, the central angle for each side is 180/n degrees, and if the ngon has an inradius r, the side length of each side is S=2r*tan(180/n). I proceeded to construct a isosceles triangle where the center of the n gon is connected with equal length legs to one side of the n gon. The inradius of the n gon is the altitude of that triangle. Thus, r=S/(2tan(180/n)). This means the area of that triangle is A=rS/2=S^2/(4tan(180/n)). Therefore, the area of an N gon with side length S and side number n is A=nS^2/(4tan(180/n)).

The perimeter of a n gon is p=nS. Therefore, p/n=S. Thus, A=p^2/(4n*tan(180/n)).

With p held constant, the limit of A as n approaches infinity converges upwards to the Area of a circle with radius r.

Therefore, a circle is the optimum shape for an n gon of a given perimeter, and the more sides a regular n gon has, the more closely the it approaches the greatest possible area it can have for a given perimeter.

This is one portion of the correct proof, but a crucial (and rather more difficult) piece is missing:

This proves that a circle has a greater area than any regular n-gon, and that the area of a regular n-gon approaches that of a circle as n increases. It does not show that a circle has greater area than any n-gon, regular or irregular. So, to complete the proof, you must show that a regular n-gon has the greatest area of any n-gon with a fixed perimeter.

Actually, if we're going to be really precise, there's an additional step in showing that if a circle has greater area than any polygon with fixed perimeter, then there can be no shape with fixed perimeter with greater area than a circle, but this follows rather trivially from the fact that we can easily approximate any (convex, piecewise-smooth) shape to arbitrary precision with an inscribed polygon.