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Section 10.2 Tiling the Sphere

Now imagine trying to tile the sphere with regular (spherical) polygons. Which regular polygons can be used?

We start off similarly to the approach used in Section 10.1. Again, a regular polygon with \(n\) sides can be divided into \(n\) centeral triangles using radial lines, and again the central angle must be \(\frac{2\pi}{n}\text{.}\) And again we assume that \(k\) such polygons meet at a point, so the interior angles are \(\frac{2\pi}{k}\text{.}\) So the angle sum for each such triangle is formally the same as before, namely \(\frac{2\pi}{n}+\frac{2\pi}{k}\text{.}\) However, this angle sum is no longer equal to \(\pi\text{.}\) Rather, the angle sum for any triangle on a sphere will always be greater than \(\pi\text{,}\) that is,

\begin{equation} \pi \lt \pi \left(\frac2n+\frac2k\right)\label{stri}\tag{10.2.1} \end{equation}

which reduces to

\begin{equation} k \lt \frac{2n}{n-2} .\tag{10.2.2} \end{equation}

The case \(k=1\) can at best be interpreted as the \(0\)-sided “polygon” consisting of a single vertex and no sides, and \(k=2\) corresponds to a dihedron, which is a fancy word for a hemisphere, but with an arbitrary number of sides and vertices, all lying on the equator. For \(k\) to be bigger than \(2\text{,}\) we must have \(n\le5\text{.}\) If \(n=2\text{,}\) we get a lune, known in this context as a hosohedron. Any number of lunes can be used to tile the sphere, so long as the angle is chosen appropriately.

There are five remaining cases, each corresponding to one of the Platonic solids, “inflated” to lie on a sphere. As shown in Figure 10.2.1, these cases are \(n=3\text{,}\) \(k=3\) (tetrahedron), \(n=4\text{,}\) \(k=3\) (cube), \(n=3\text{,}\) \(k=4\) (octahedron), \(n=5\text{,}\) \(k=3\) (dodecahedron), and \(n=3\text{,}\) \(k=5\) (icosahedron).

Figure 10.2.1. The five Platonic solids, expanded onto a sphere.

In each of these last five cases, we can determine not only the number of allowable sides (\(n\)) and how many polygons meet (\(k\)), but also how many polygonal tiles we need to cover the sphere. Since the area of a spherical triangle is proportional to its excess, the area of each polygon is \(n\pi r^2\left(\frac2n+\frac2k-1\right)\text{,}\) using (10.2.1), where the constant of proportionality is the radius of the sphere. Since the total area of the sphere is \(4\pi r^2\text{,}\) the number \(p\) of tiles must be given by

\begin{equation} p = \frac{4k}{2n+2k-kn}\tag{10.2.3} \end{equation}

which correctly yields \(4\) “faces” for the tetrahedron, \(8\) for the octahedron, \(20\) for the icosahedron, \(6\) for the cube, and \(12\) for the dodecahedron.

Finally, we can investigate the lengths of the sides of our tiles. Unlike in Euclidean geometry, each tiling requires the tiles to have a specific size. We can use the Supplemental Law of Cosines in spherical geometry to determine these lengths. This relation, which we state without proof, says that the length of side \(c\) of a spherical triangle is given by

\begin{equation} \cos\left(\frac{c}{r}\right) = \frac{\cos(C)+\cos(A)\cos(B)}{\sin(A)\sin(B)}\tag{10.2.4} \end{equation}

where \(A\text{,}\) \(B\text{,}\) \(C\) denote the angles of the triangle, with \(C\) opposite side \(c\text{,}\) and where \(r\) is the radius of the sphere. Here, we have broken our \(n\)-sided polygon into \(n\) central triangles, with central angle \(C=\frac{2\pi}{n}\text{,}\) and remaining angles equal to half the interior angle of the polygon, so \(A=B=\frac12\left(\frac{2\pi}{k}\right)\text{.}\) Thus,

\begin{equation} \frac{c}{r} = \arccos\left( \frac{\cos\frac{2\pi}{n}+\cos^2\frac{\pi}{k}}{\sin^2\frac{\pi}{k}} \right) .\tag{10.2.5} \end{equation}

Similarly, the “radius” \(\rho\) of each tile is

\begin{equation} \frac{\rho}{r} = \arccos\left( \frac{\cos\frac{\pi}{k}+\cos\frac{\pi}{k}\,\cos\frac{2\pi}{n}} {\sin\frac{\pi}{k}\,\sin\frac{2\pi}{n}} \right) .\tag{10.2.6} \end{equation}
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