## Platonic Solids – A Prelude to Euler’s Polyhedra Formula

This post was motivated by the book “Euler’s Gem: The Polyhedra Formula and the Birth of Topology” by David R. Richeson.

Plato was a greek philosopher and founder of the Academy in Athens, the first institution of higher learning in the western world. A lecturer of the Academia, and also a friend of Plato, Theaetetus is credited to be the first to give a complete proof of the existence of five (and no more than five) regular convex polyhedra.  Plato believed that it must be some cosmological reason for this fact and he proposed an atomic model in which all the matter was composed by four elements and that
these four elements must be shaped as regular convex polyhedra. The fifth element, the dodecahedron, would be the material from what the gods created the universe itself.

In the following, we are going to present three different arguments proving that only five convex regular polyhedra can exist. We start saying few words on polyhedra’s cousin in a lower dimension, the polygons.

## Polygons

Grab a pencil and position it on a piece of paper, let’s say over a point $p_0$. Draw a sequence of non-crossing straight lines (edges) such that the last edge has endpoint $p_0$. The figure you created is a polygon. One class of polygons are those that are regular, namely those polygons in which every side has the same length. We can go further and restrict the class of regular polygons to those who are also convex. Convex polygons are those in which the hand turns the same way during its drawing. There are some nice theorems involving regular convex polygons

Theorem [External Angles Sum]: The sum of the external angles of a k-regular convex polygon is equal to $\displaystyle 2\pi$.

Theorem [Internal Angle]: The internal angle of a k-regular convex polygon is equal to $\displaystyle \frac{\pi(k-2)}{k}$.

The first theorem comes from the closeness of a polygon. Since I start and finish its drawing at the same point, my pencil must describe a full turn on the paper. By letting $x$ be the internal angle of the $k$-regular convex polygon, we can write each of its external angles as $\pi - x$ and then apply the first theorem to derive the second.

The class of regular convex polygons is infinite. For any $k$, one can always construct a $k$-regular convex polygon. Despite its symmetric properties, so much attention on regular convex polygons may be explained by the result below:

Theorem [Best ratio area/perimeter]: Among all the polygons with $k$ sides and a given perimeter $L$, the $k$-regular convex polygon is the one with the largest area.

The last result shown here talks about coverings of the plane by polygons. A set of adjacent polygons that divides the plane in a set of regions is called a tiling of the plane. A tiling in which the same polygon is used all over again is called an uniform tiling, and a uniform tiling that uses regular convex polygons is called of regular tiling.

Theorem [Few regular tilings of the plane]: There exist only three regular tilings of the plane.

Choose any $k$-regular convex polygon and let’s try to cover the plane with it. In order to do that I must be able to arrange my polygons in such a way that I fill all the gaps. In other words, if I center a circle of any radius on a vertex of my tiling, then the circle must completely lie in the tiling. Mathematically, there must exist a positive integer $q$ such that:

$\displaystyle 2\pi = q \cdot \frac{\pi(k-2)}{k}.$

From which one derives

$\displaystyle k = \frac{2q}{q-2}.$

The last equation defines the hyperbole in the figure. We do not care for negative values of $q$ or $k$, and we consider only the part of the graph in the first quadrant. We know that $k\geq 3$ and then we try this value:

$3 = \frac{2q}{q-2} \Rightarrow q = 6.$

The graph tell us that for values of $q$ larger than $6$, $k \in [2,3)$ and we can limit our analysis to three cases.

$\begin{array}{lr} q=5 & k = \frac{2 \cdot 5}{5-2} = \frac{10}{3} \\[0.1in] q=4 & k = \frac{2 \cdot 4}{4-2} = 4 \\[0.1in] q=3 & k = \frac{2 \cdot 3}{3-2} = 6 \end{array}$

Therefore, a regular tiling of the plane must use equilateral triangles, squares or regular hexagons.

## Polyhedra

A polyhedron can be described in the same way as a polygon, except that now you are allowed to draw your figure in a three-dimensional space. The class of regular convex polyhedra is defined analogously as done for polygons, but while in the latter the class contains infinity elements, regular convex polyhedra is a very selected group with only five members. Those are exactly the platonic solids.

## Theaetetus Proof

In a polyhedron, pick any vertex $v$ and it must have at least three adjacent vertices, otherwise would not be possible to have a closed region in the space. Moreover, consider the angle formed by each pair of adjacent faces around $v$ and denote by $\sigma$ its sum. Observe that $\sigma$ must be less than $2\pi$ and we denote the difference $2\pi - \sigma$ the angle defect of $v$.

Recall that each face of a regular convex polyhedron is of the same polygon type. Let’s enumerate the possibilities:

1. Equilateral triangle face:
1. Three faces at each vertex $v$: Angle defect = $\displaystyle 2\pi - 3 \cdot \pi/6 = \pi/2$.
2. Four faces at each vertex $v$: Angle defect = $\displaystyle 2\pi/3$.
3. Five faces: Angle defect = $\displaystyle \pi/3$.
4. Six faces: Angle defect = 0. Not possible.
2. Square:
1. Three faces: Angle defect = $\displaystyle \pi/2$.
2. Four faces: Angle defect = 0. Not possible.
3.  Regular Pentagon:
1. Three faces: Angle defect = $\displaystyle \pi/5$.
2. Four faces: Angle defect = $\displaystyle -2\pi/5$. Not possible.
4. Regular Hexagon:
1. Three faces: Angle defect =0. Not possible.

It is useless to keep trying, and then we identified the only five possible types of convex regular polyhedron.

## Using Euler’s formula for polyhedra

The celebrated formula of Euler that relates the number of vertices (V), edges (E) and faces (F) of a polyhedron it also gives us a proof for the five platonic solids.

$V - E + F = 2$.

In a regular convex polyhedron, each face has the same number $m$ of edges and each edge is part of exactly two faces, which let us to write

$mF = 2E$.

Similarly, each vertex of the polyhedron has $n$ edges connected to it. Therefore,

$nV = 2E$.

Solving for $F$ in Euler’s formula:

$\begin{array}{c} mF/n - mF/2 + F = 2 \\[0.1in] F = 4n /(2m - mn +2n). \end{array}$

We know that $m,n \geq 3$ and  $F > 0$. Hence,

$2m - mn + 2n > 0$.

$\begin{array}{lcccc} m/n & 3 & 4 & 5 & 6\\[0.1in] 3 & 3 (OK) & 2 (OK) & 1 (OK) & -2 \\[0.1in] 4 & 2 (OK) & 0 & & \\[0.1in] 5 & 1 (OK) & -2 & & \\[0.1in] 6 & 0 & & & \end{array}$

## Legendre’s proof

Let S be an sphere of radius one. A great circle of S is any circle of radius one that lies on the surface of S. A spherical lune is the region of the sphere delimited by two great circles, as shown in the figure.

By Pbroks13Image:Regular digon in spherical geometry.png, Public Domain, Link

The great circles of a lune meet at two antipodal points with the same angle, let’s say $a$. We can easily compute the area of the lune $L$.

$\begin{array}{c}\displaystyle \frac{a}{2\pi} = \frac{L}{4\pi} \\[0.1in] L = 2a. \end{array}$

Lunes can be used to compute areas of triangles on the surface of sphere. A triangle is formed by three great circles, and for each pair of them one has a lune. We can cover all the sphere by summing up all the lunes and their symmetric relatives, as shown in the figure. Consider the triangle $ABC$ and denote by $\sigma$ its area. Let $L_x$ the corresponding lune of angle $x$.

By Peter MercatorOwn work, CC BY-SA 3.0, Link

$\begin{array}{c} 2L_a + 2L_b + 2L_c - 2\sigma = 4\pi\\[0.1in] \sigma = a + b + c - \pi. \end{array}$

We can extend this result to any polygon on a sphere. For a polygon with $n$ sides I can partition it in $n-2$ triangles. Therefore, the $n$-side polygon area on a shpere, $\sigma_n$, is given by

$\sigma_n = \text{Sum of internal angles} - (n-2)\pi .$

There is a drawing to help us remember this formula. Each vertex is labelled with its value, each edge of the polygon is labelled with $-\pi$ and the polygon itself is labelled with $\pi$. The sum of all the labelling gives the area of the polygon on a sphere of radius one.

Now we are ready for Legendre’s ingenious proof. Start with any convex polyhedron and surround it by a sphere of radius one (you always can do that by rescaling the polyhedron properly). Then project the polyhedron on the sphere by emanating rays from the center of the sphere. For each face of the polyhedron, do the labeling as described above. Note that the adjacent angles of each vertex of the projected polyhedron sum up to $2\pi$. Moreover, each edge contributes twice, one for each face it belongs. Therefore,

$\begin{array}{c} 2\pi V - 2\pi E + 2\pi F = 4 \pi \\[0.1in] V - E + F = 2. \end{array}$