A Neat Result About Regular Polygons and Circles

In geometry class, we learn about this magical constant called $\pi$. After briefly being wowed by the fact that its decimal expansion goes on forever without repeating, we are introduced to the formulas for the circumference and area of a circle:

$$C = 2\pi r \quad \text{and} \quad A = \pi r^2.$$

As we put these simple formulas to use in our boring homework assignments, we might be left with the feeling that $\pi$’s irrationality was the most interesting thing about the lesson.

But there’s something fascinating about these formulas that I didn’t notice until much later: They both involve the same constant $\pi$. This means that the circle’s area and circumference are closely related to each other by

$$Cr = 2A$$

Sure, it’s not too surprising that the circumference happens to be related to the radius by some irrational constant, and the same goes for the area. But why should the circumference constant and the perimeter constant be the exact same number? Especially since, for most shapes, perimeter and area have no apparent relationship.

Well, there are some simple explanations for this, but they all have a qualitatively similar “calculus” flavor. In this quick post I’ll share another explanation with a slightly more “geometry/trig” feeling. In fact, we’ll see that circles aren’t the only shapes with this special perimeter-area relationship. It’s also true for all regular polygons! (Well, sort of…)

The Polygon Relationship

First, let’s make things simpler by just sticking to a radius of $1$. The remarkable fact about the circle is then that

$$C = 2A.$$

Since it’s intuitive that circumference scales linearly with respect to the radius while area scales with respect to its square, this special case is all we really need.

Similarly, we’ll focus on regular polygons with radius $1$. That is, polygons inscribed in circles of radius $1$. Thus the distance from a polygon’s center to any of its vertices is 1.

Let’s use $P(n)$ and $A(n)$ to denote the perimeter and area of a regular $n$ sided polygon of radius 1. We might hope that $P(n) = 2A(n)$, mirroring the circle’s formula. But sadly that’s not true (although it’s pretty close). To see what the true relationship is, let’s create formulas for $P(n)$ and $A(n)$. We’ll need to use some trig, but you don’t need to pull out your calculator.

The Perimeter

Let’s start with the perimeter, $P(n)$. To do this, we can split up the $n$-gon into $2n$ congruent right triangles. I’m using $n = 5$ for the figure below.

Since the radius of the polygon is $1$, the length of the vertical leg of the triangle is $\sin(\theta)$. And since each of the polygon’s $n$ sides is composed of two of these segments, the perimeter is given by

$$P(n) = 2n \sin(\theta).$$

Of course, we would need to know what $\theta$ is to actually compute this (it’s $360\degree / 2n$), but the exact value won’t be important.

The Area

We can find the area with a similar construction. But this time I’ll use a $2n$ sided polygon. I’ll represent this angle with $\phi$ to make it clear that it is a different angle than the $\theta$ we had for the $n$ sided polygon. To keep consistent with the previous figure, the shape below is a $10$ sided polygon.

As before, the vertical leg of each triangle has length $\sin(\phi)$. But this time we also need the other leg, which is $\cos(\phi)$. Thus the area of each triangle is $\frac12\sin(\phi)\cos(\phi)$. Since there are $4n$ triangles, the total area is

$$A(2n) = 2n \sin(\phi)\cos(\phi).$$

If you’ve been keeping up on your trig identities, you should be seeing the double angle sine formula jumping out at you:

$$\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha).$$

Therefore,

$$A(2n) = n\sin(2\phi).$$

The Equality

Finally, since the polygon we used for the perimeter has half as many sides as the one we used for the area, its central angle is twice as big. That is,

$$\theta = 2\phi.$$

This means that

$$P(n) = 2n\sin(\theta) = 2n\sin(2\phi) = 2A(2n).$$

And with that we have our final result. The perimeter of a regular polygon is equal to double the area of the polygon with twice as many sides!

$$P(n) = 2A(2n).$$

Wrapping it Up

How does this show that $C = 2A$ for the unit circle? Well, if we think of a circle as an “infinity sided regular polygon”, then “doubling” the number of sides still gives us a circle. Therefore

$$C = P(\infty) = 2A(2 \cdot \infty) = 2A(\infty) = 2A.$$

Although this use of infinity is not mathematically proper, (though I would argue that it’s no worse than other intuitive proofs), it is easily converted into a more precise argument involving limits.

I discovered this when I completely misinterpreted a homework problem in a calculus class, and I went off working in an orthogonal direction. It’s nothing earth-shattering, but it was a great surprise to stumble upon something more pretty than a good mark.