In this section, we will discuss Euler’s theorem on convex polyhedra. Euler is pictured above. A *polyhedron* is a convex connected solid whose boundary consists of a finite number of convex polygons such that each edge is shared by precisely two faces. A *regular polyhedron* is a convex polyhedron whose faces are congruent regular polygons (edges are all the same length) and such that the same number of edges meet at the same vertex.

Here are some examples of *regular polyhedra*

**Cube**

vertices = 8, edges = 12, faces = 6

**Tetrahedron**

vertices = 4, edges = 6, faces = 4

**Octahedron**

vertices = 6, edges = 12, faces = 8

**Dodecahedron**

vertices = 20, edges = 30, faces = 12

**Icosahedron**

vertices = 12, edges = 30, faces = 20

These 5 solids are often called the *Platonic solids* and appear in the writings of Plato.

**Theorem:** The five Platonic solids are the ** only** regular polyhedra.

**Proof:**

To create a vertex, at least three faces must meet at a point and the total of their angles must be less than 360°, i.e the corners of the face must be less than 360°/3=120°. The only polygons meeting these requirements are the triangle (60), square (90), and pentagon (108).

- Triangular faces: each vertex of a regular triangle is 60°, so a shape should be possible with 3, 4, or 5 triangles meeting at a vertex; these are the tetrahedron, octahedron, and icosahedron respectively.
- Square faces: each vertex of a square is 90°, so there is only one arrangement possible with three faces at a vertex, the cube.
- Pentagonal faces: each vertex is 108°; again, only one arrangement, of three faces at a vertex is possible, the dodecahedron, and that exhausts the list of regular 3-dimensional solids.
**QED.**

It will turn out that one can also prove the previous theorem using Euler’s theorem — see below.

Here are some examples of polyhedra which are *not regular *

**Elongated triangular gyrobicupola**

vertices = 18, edges = 36, faces = 20

**Gyroelongated triangular cupola**

vertices = 15, edges = 33, faces = 20

**Metabiaugmented dodecahedron**

vertices = 22, edges = 40, faces = 20

**Nonagonal antiprism**

vertices = 18, edges = 36, faces = 20

If you do a little counting with the surfaces listed above, you will notice the following:* If v is the number of vertices, e is the number of edges, and f is the number of faces, then v – e + f is always equal to two.*

**Theorem (Euler)**: *For any convex polyhedron we have v – e + f = 2.*

The original proof of Euler’s theorem (by Euler) is not a rigorous as it should be (according to modern standards) – as he did not take certain things into account. The first correct proof of Euler’s theorem was by Legendre in 1794. Complicating matters further is the fact there is a bit of a controversy here as to whether or not Euler’s theorem should indeed be called Euler’s theorem after all. Euler’s proof was around 1751 but there are some errors and misunderstandings in the proof. Descartes in 1649 proved v – e + f = 2 but died soon after he wrote down his proof – which some argued was flawed anyway. Unfortunately some of Descartes papers were not discovered until many years later. Some mathematicians refer to v – e + f = 2 as the Euler-Descartes formula. A nice history of this can be found in E. Sandifer’s paper How Euler did it. I’ll also comment that a more in-depth discussion of Euler’s theorem is found in a very readable book Euler’s Gem by D. Richeson — which is required reading for this course. As it turns out there is a more advanced version (with of course a much more complicated proof) of Euler’s theorem for more complicated solids. A great book on this is I. Lakatos, Proofs and Refutations, Cambridge — also required reading.

The key to proving Euler’s theorem is to prove a result (just as interesting) which is true in the plane.

A *graph* if a finite set of points (vertices) which are connected by line segments (edges) such that the graph is connected. Here are a few examples of graphs:

**Example 1.**

vertices = 9, edges = 11, faces = 3

**Example 2:**

vertices = 9, edges = 14, faces = 6

**Example 3:**

vertices = 10, edges = 10, faces = 1

Perhaps you noticed by looking at these graphs that there is an Euler’s theorem here.

**Theorem (Euler):** *In any finite graph v – e + f = 1.*

**Proof:** Let’s start by making the observation that if the graph is just a triangle, then v – e + f = 1 (since v = 3, e = 3, f = 1). The first thing we do with our graph is “triangulate” it. By this we mean that if there is a face that is not a triangle, we add some edges to it so that every face is a triangle. Here is an example of what we mean. Here is our original graph

Notice how there are several faces which are not triangles. Now add in some edges in red to make all the faces triangles.

Notice that every time you add in an edge to make a triangle the number of vertices remains unchanged, the number of edges obviously increases by one, and the number of faces also increases by one. However when you compute v – e + f for the new graph, the extra ones (from the new e and new f) *cancel each other out*. This mean that v – e + f is the same for this new graph. Pretty cool trick!

Now for the next cool part of the proof. Start removing the exterior edges and notice that v – e + f does not change. Just think about it. When you remove an edge, you don’t change the number of vertices but you do also remove a face so v – e + f is unchanged.

Keep on removing exterior edges (never changing v – e + f) until you get a triangle

Notice that for the remaining triangle we have v – e + f = 1. **QED.**

Let us recall Euler’s theorem for polyhedra: v – e + f = 2. Here is the proof of this theorem based on the Euler’s theorem in the plane. Start with your polyhedron. Remove one of the faces. This does not change v or e (important). Now imagine that your polyhedron is flexible, perhaps made of rubber, and that you can flatten it while still keeping its main features, namely the same faces, edges, and vertices. The idea of flattening was due to A. Cauchy. Here is an example of the flattened cube.

Here is an example of a flattened tetrahedron.

Now apply the two dimensional version of Euler’s theorem (v – e + f = 1) to the graph. This means that when we add in the face we removed at the beginning, we get v – e + f = 2. **QED**.

A beautiful consequence of Euler’s theorem is the following:

**Theorem**: *There are only five regular polyhedra: the cube, tetrahedron, octahedron, dodecahedron, and icosahedron.*

Recall that a polyhedron is regular if the faces are congruent regular polygons (each side has equal length and the face is convex) and the same number of edges meet at each vertex.

The proof of this theorem will be an exercise. I’ll give you an outline (posted on Blackboard) but you will have to add in the details as a video presentation.

Final note: Euler’s theorem does not work for non-convex polyhedron. Here are some examples:

**Tetrahemihexahedron**

vertices = 6, edges = 12, faces = 7, v – e + f = 1

**Octahemioctahedron**

vertices = 12, edges = 24, faces = 12, v – e + f = 0.

**Great icosahedron**

*vertices* = 12, *edges* = 30*,* *faces* = 20, * *v – e + f = 2

We see that Euler’s theorem is not quite true if the polyhedron is not convex. In fact things get even worse if the solid has holes in it. Think of taking a cube a cutting out a section of if from the middle.