Tiling the Euclidean Plane

Section 10.1 Tiling the Euclidean Plane

Imagine trying to tile the Euclidean plane with regular polygons. Which regular polygons can be used?

It's easy to see by trial and error that the only regular polygons that tile the plane are equilateral triangles, squares, and regular hexagons. These tilings are shown in Figure 10.1.1. But why is this true?

Figure 10.1.1. The three regular Euclidean tilings of the plane, using triangles, squares, and hexagons.

A regular polygon with \(n\) sides cna be divided into \(n\) central triangles, using radial lines connecting the vertices of the polygon to its center. Each central angle is clearly \(\frac{2\pi}{n}\text{.}\) Each of the other angles of these triangles is precisely half of the interior angle \(\alpha\) of the polygon. Since the angle sum of a Euclidean triangle is \(\pi\text{,}\) we therefore have

\begin{equation} \frac{2\pi}{n}+2\frac{\alpha}{2} = \pi\tag{10.1.1} \end{equation}

or equivalently

\begin{equation} \alpha = \left(\frac{n-2}{n}\right)\pi .\tag{10.1.2} \end{equation}

Among other things, this result shows that the interior angles of a polygon with \(n\) sides add up to \((n-2)\pi\text{.}\)

Furthermore, if \(k\) polygons come together at each vertex, those angles must also add up to \(2\pi\text{.}\) Thus, each interior angle \(\alpha\) must be \(\frac{2\pi}{k}\text{.}\) We have therefore obtained a condition for our polygons to tile the plane, namely that

\begin{equation} k = \frac{2\pi}{\alpha} = \frac{2n}{n-2}\tag{10.1.3} \end{equation}

must be an integer. Since \(n\) is clearly greater than \(2\text{,}\) \(k\) is a decreasing function of \(n\text{,}\) and it is now easy to see that the only integer solutions are \(n=3\text{,}\) \(k=6\) (six triangles), \(n=4\text{,}\) \(k=4\) (four squares), and \(n=6\text{,}\) \(k=3\) (three hexagons).