Quote:
Originally Posted by Oblarg
It can be, though it's easy to overlook its subtleties and a often pain to iron out the details.
(If you want to be convinced of this, try actually proving the seemingly-obvious thing I posted above).
|
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.