Illustrating isoperimetic problem with regular polygons

Given a curve of a fixed length, how do you maximize the area inside? This is known as the isoperimetric problem.

The answer is to use a circle. The solution was known long before it was possible to prove; proving that the circle is optimal is surprisingly difficult. I won’t give a proof here, but I’ll give an illustration.

Consider a regular polygons inscribed in a circle. What happens to the ratio of area to perimeter as the number of sides increases? You might suspect that the ratio increases with the number of sides, because the polygon is becoming more like a circle. This turns out to be correct, and it’s not that hard to be precise about what the ratio is as a function of the number of sides.

For a regular polygon inscribed in a circle of radius r,

$\text{Area } = \frac{n}{2} r^2 \sin\left(\frac{2\pi}{n} \right )$

and

$\text{Perimeter} = 2n r \sin\left(\frac{\pi}{n} \right )$

For a regular n-gon inscribed in a unit circle, we have

$\begin{align*} \frac{\text{Area}}{\text{Perimeter}} &= \frac{(n/2) \sin(2\pi/n)}{2n \sin(\pi/n)} \\ &= \frac{n \sin(\pi/n)\cos(\pi/n)}{2n \sin(\pi/n)} \\ &= \cos(\pi/n)/2 \end{align*}$

We used the double-angle identity for sine in the second line above.

As n increases, the ratio increases toward 1/2, the ratio of the area of a unit circle to its circumference. In a little more detail, the difference between the ratios for a circle and for a regular n-gon goes to zero like O(1/n²), based on a Taylor expansion for cosine.

Here’s a plot of the ratios as a function of the number of sides.

Ratio of area to perimeter

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