Poly-π

Clem Padin

8 November 2024

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While reading William Dunham’s The Mathematical Universe, I came across this:

A novice, introduced to circles for the first time, would rather quickly recognise an essential fact: All circles have the same shape… We note by contrast that not all triangles have the same shape, nor all rectangles. Behind this unexciting observation lies one of the profound theorems in mathematics: that the ratio of the circumference to the diameter is the same for one circle as for any other.

What does Dunham mean by “all circles have the same shape”? We would all agree that this is true, regardless of a circle’s size, but more formally, we can say all circles are geometrically similar. And while this is obviously true for circles, it is equally true for equilateral triangles or squares. And it occurred to me that this “profound theorem” Dunham refers to is as true for these polygons as it is for the circle, provided we can define their diameters. This is easy for a certain subset of polygons: regular polygons! Just as $\pi$ represents the constant ratio of circumference to diameter for all circles of all sizes, regular polygons have an analogous constant, which I’ve named ‘polygon-$\pi$’, or ‘poly-$\pi$’ for short.

The triangle

Let’s start with the equilateral triangle. For consistency, I’ll define the triangle’s radius $r$ as the distance from the centre of the triangle to any vertex. Its diameter is then twice its radius.

For an equilateral triangle with sides of length $L_3$, the counterpart to the circle’s circumference is, of course, the triangle’s perimeter, $3L_3$. So, we can define the $\pi_3$ value for all equilateral triangles as \begin{equation*} \pi_3 = \frac{3L_3}{2r}, \end{equation*} where the subscript ‘3’ indicates the number of sides.

To find $L_3$ in terms of $r$, we can draw a line from the centre of the triangle to each vertex. This divides the equilateral triangle into three equal isosceles triangles with sides of length $r$ and three central angles of 120°.

Dropping a perpendicular from the centre to any side divides $L_3$ and the central angle in half, creating a right-angled triangle and allowing us to observe the following relation:

\begin{equation*} \sin(60°) = \frac{L_3}{2r}. \end{equation*}

Replacing $L_3 / 2r$ with $\sin60°$ in the expression for $\pi_3$ gives us
\[\pi_3 = 3\sin(60°) \approx 2.598.\]
This value is the equilateral triangle’s poly-$\pi$ value! So, for any equilateral triangle, the ratio of the triangle’s perimeter to its ‘diameter’ equals $\pi_3$ or about $2.6$.

The square

Let’s now examine the square. Again, we define a radius $r$ as the length from the square’s centre to a vertex and the length of one side as $L_4$.

Now the ratio of the square’s ‘circumference’ to its ‘diameter’ can be written as
\begin{equation*} \pi_4 = \frac{4L_4}{2r}. \end{equation*} Note that the central angles in the square are 90° and the four radii divide the square into four isosceles triangles with base angles of 45°.

Again, dropping a perpendicular divides $L$ and a central angle so we can write
\begin{equation*} \sin(45°) = \frac{L_4}{2r}. \end{equation*} Substituting $\sin( 45° )$ for $L_4 / 2r$ in our expression for $\pi_4$, we get \[\pi_4 = 4\sin( 45° ) \approx 2.828.\] This is the square’s poly-$\pi$ value.

The hexagon

The poly-$\pi$ value for a hexagon turns out to be quite simple.

Note that the inner angles are $360° / 6 = 60°$. The other angles in the six triangles of the hexagon are then also $60°$ and so we have six equilateral triangles. This then means that $r = L_6$. As a result,
\begin{equation*} \pi_6 = \frac{6L_6}{2r} = \frac{6L_6}{2L_6}, \end{equation*} and so \[\pi_6 = 3.\]

$n$-sided polygons

In each subsequent regular polygon we employ the same technique. That is, the $n$th poly-$\pi$ is given by
\begin{equation*} \pi_n = \frac{nL_n}{2r} = \frac{P_n}{2r}, \end{equation*} where $P_n$ is the perimeter.

A perpendicular drawn from the central angle $360°/n$ divides $L$ and the central angle such that
\begin{equation*} \sin\left(\frac{1}{2}\times \frac{360°}{n}\right) = \frac{L_n}{2r}, \end{equation*}
so that
\begin{equation} \pi_n = n\sin\left(\frac{180°}{n}\right). \tag{1} \label{eq:pin} \end{equation} This now is the poly-$\pi$ value for a regular polygon with $n$ sides. Note how the result does not depend on any physical characteristics of the polygon except the number of sides!

The table below displays the poly-$\pi$ for several polygons. The values have been rounded.

And here’s a plot of the poly-$\pi$s up to 10:

We can see (as we would expect) that as $n\to\infty$, $\pi_n → \pi = 3.14159\dots$

Using poly-$\pi$

When given the radius of a circle, we can calculate the circumference, and vice versa, because of the constant of proportionality between them expressed as $\pi$. This is also the case for a regular polygon because of the existence of its poly-$\pi$. So for a pentagon whose perimeter is, say, 15 we can find its radius (the distance from the centre of the pentagon to a vertex) by rearranging \eqref{eq:pin} to get \[r = \frac{P_5}{2\pi_5}.\] Using the table above we find that $\pi_5 = 2.94$, so \[r = \frac{15}{2(2.94)} \approx 2.55\] Unlike a circle, however, in addition to using poly-$\pi$s to calculate the perimeter or radius, we can also calculate the length of a polygon’s side. Using \eqref{eq:pin} and the value derived above for the radius, we get
\begin{align*} L_5 = \frac{2\pi_5r}{5} \approx \frac{2\times2.94\times2.55}{5} \approx 3.0, \end{align*} which is, of course, what we should get as the perimeter is 15 and the figure is a pentagon, so each side is $15/5=3$.

Poly-$\pi$ and area

The area of each $n$-sided polygon is given by the sum of the areas of the $n$ inscribed triangles. A perpendicular drawn from the central angle to a side ($L_n$, the triangle’s base) gives us the height of one of the inscribed triangles and can be written as \[r \cos\left( \frac{1}{2} \times \frac{360°}{n} \right) = r \cos\left(\frac{180°}{n}\right).\] So, for a regular $n$-sided polygon, its area is given by
\begin{equation*} A_n = n\left(\frac{1}{2}L_n\right)\left[r\cos\left(\frac{180°}{n}\right)\right] = \frac{nrL_n}{2}\cos\left(\frac{180°}{n}\right). \end{equation*} Solving \eqref{eq:pin} for $L_n$ we get
\begin{equation*} L_n = \frac{2\pi_n r}{n}. \end{equation*}

Combining these last two equations, we can write the area of an $n$-sided polygon in terms of $r$, its radius, as

\begin{equation} A_n = \pi_n r^2 \cos\left(\frac{180°}{n}\right). \tag{2} \label{eq:An2} \end{equation} Note how similar this equation is to the equation for the area of a circle! And, in fact, note that as $n\to\infty$, $180°/n \to 0$, and $\cos( 180°/ n ) \to 1$. Thus, we can state that for $n$-sided polygons, as the number of sides increases or as $n \to \infty$, $A_n → \pi r^2$.

We can also express the area in terms of the length of a side. Using \eqref{eq:pin} but this time solving for $r$, we get \[ r = \frac{n L_n}{2\pi_n}.\] Substituting this into \eqref{eq:An2}, we get
\begin{equation} A_n = \frac{1}{\pi_n}\left(\frac{nL_n}{2}\right)^2 \cos\left(\frac{180°}{n}\right). \tag{3} \label{eq:An3} \end{equation} Though not as elegant as \eqref{eq:An2}, it does allow us to determine the area of a regular polygon given the length of one side. Again we can consider what happens as $n\to\infty$. If we have a fixed side length $L$, our area formula becomes \[A_n=\frac{1}{\pi_n}n^2 \left(\frac{L}{2}\right)^2 \cos\left(\frac{180°}{n}\right).\] Now, we know $1/\pi_n →1/\pi$, and $\cos(180°/n) \to 1$ as before. Since we’re holding the side length constant, $(L/2)^2$ is just a constant. But $n^2$ certainly goes to infinity! In hindsight, this is obvious—if we keep the same side length, but add more and more sides then clearly the area must get larger and larger.

Using poly-$\pi$ to calculate the area of a regular polygon

Using \eqref{eq:An2} we can easily find the area of any regular polygon if we know its radius. So for a hexagon with a radius of 12, we can find its area as follows: \[ A_6 = \pi_6 r^2 \cos\left( \frac{180°}{6} \right).\] We know $\pi_6$ is exactly 3, so \[ A_6 = (3)(12)^2\cos\left(\frac{180°}{6}\right) = 432 \cos(30°).\] What is the length of a hexagon’s side given such an area? Solving \eqref{eq:An3} for $L_n$, we find

\begin{equation*} L_n = \frac{2}{n}\left(\sqrt{\frac{A_n \pi_n}{\cos(180°/n)}}\right). \end{equation*}

For the above hexagon, that is

\begin{align*} L_6 &= \frac{2}{6}\sqrt{432\times 3 \times \cos(30°)/\cos(30°)} \\ &= \frac{1}{3}\sqrt{1296} = 12. \end{align*} Although, we knew that from the start—given this is a hexagon, we already know that the radius and the length of a side are equal. Since we proposed a radius of length 12, the side ($L_6$) should also be of length 12.

Conclusion

A $\pi$-like value exists for regular polygons. It emerges naturally once we noticed that regular polygons at different scales have `the same shape’, just like circles. Given a measurement for a polygon, poly-$\pi$s make it easy to find others. For example, from the side length we can find the distance from its centre to any vertex (its radius); from the area we can find the perimeter.

Additionally, this concept might help people get a better idea of what $\pi$ really means. The special symbol and the `funny’ name of $\pi$ may create a mystique (not entirely undeserved) that can distract from its true meaning. Introducing poly-$\pi$s helps to clarify that $\pi$ is not a mysterious quantity exclusive to circles but rather a simple ratio—and such ratios appear in many different geometrical objects.

Clem Padin

Clem graduated from the University of Arizona with a physics major and a mathematics minor and has continued studying mathematics ever since. He is pictured next to the little library that he built for his neighbourhood.

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