## Getting rid of triangles, and other ways to Euler

That proof that any convex polyhedron made of pentagons and hexagons must contain at least 12 pentagons rests on Euler’s polyhedron formula: If a convex polyhedron has $V$ vertices, $E$ edges, and $F$ faces, then

$V-E+F=2$

So how is that proved? Lots of ways, some of which rely on the Jordan curve theorem, so then you need a proof of that. An outline of one that doesn’t use the Jordan curve theorem, Triangle Removal, is as follows.

First, transform the polyhedron into a planar graph. You can think of this as removing one face of the polyhedron, and then stretching the boundaries of that face and deforming the other edges until all the edges lie in a plane. Or you can think of it as putting a light source near one face and a plane on the opposite side: the shadow cast onto the plane is the planar graph. This graph has the same number of vertices and edges as the polyhedron, and apparently one fewer face — until you realize the entire plane exterior to the graph can be thought of as a face too.

Okay, we have our graph. Now do the following: Find any interior face that has more than three edges, and draw a diagonal through it from one vertex to another. Doing that increases the number of edges by 1, and divides one face into two, but leaves the vertices unchanged. That means $V-E+F$ is unchanged. Repeat that until all the interior faces are triangles.

Now do this:

1. Identify a triangle that has two edges on the border of the graph. If you can’t find one, go to step 2. If you can find one, remove those two edges. This eliminates two edges, one face, and one vertex, so again $V-E+F$ is unchanged. Now repeat step 1.
2. Identify a triangle that has one edge on the border of the graph. Remove that edge. This eliminates one edge, one face, and no vertices, so again $V-E+F$ is unchanged. Now repeat step 1.

Keep repeating as long as you can. Here I get hand wavy, because I’m less than clear on the argument, but: You don’t want to do step 2 if you can do step 1, because that leads to dead ends — specifically, a graph broken up into two disconnected graphs. Always do step 1 if you can, step 2 only if you can’t do step 1, and the graph will stay connected.

For instance, starting from a (kinda irregular) square pyramid. Here it’s turned into a planar graph with one quadrilateral and three triangular faces (the fourth triangle mapping to the exterior of the graph). The quadrilateral is divided into two triangles, and then triangles are removed one at a time.

Since the graph stays connected, the process continues until there’s only one triangle left. At this point we stop and take a tally: There are 3 vertices, 3 edges, and 2 faces (the interior of the triangle, and the exterior). So $V-E+F = 2$, and since every step we took preserved the value of $V-E+F,$ this proves $V-E+F=2$ for the original polyhedron.

Something about that proof appeals to me, even though there’s that one part I’m hazy on.

The Sum of Angles proof I find clearer, but less cute. Here again you turn the polyhedron into a planar graph with straight edges. Now add up the interior angles of all the faces (including the exterior face). For face $i$, which has $k_i$ edges, the sum of the interior angles is $(k_i-2)\pi$. Summing over all the faces,

$\Sigma_{i=1}^F(k_i-2)\pi = \left(\Sigma_{i=1}^Fk_i-\Sigma_{i=1}^F2\right) = (2E-2F)\pi$

But we get the same result if we sum the angles meeting at each vertex. For each interior vertex, the sum is $2\pi$, adding up to $2\pi(V-k)$ where $k$ is the number of vertices on the border. For the vertices on the border, it’s $2(\pi-\theta_i)$ where $\theta_i$ is the exterior angle at the $i$th vertex of the exterior polygon, and the 2 is because we include the (equal) contributions of the internal and external polygons. But the sum of the exterior angles of a polygon is $2\pi$, so the exterior vertices contribute a total of $2\pi k-4\pi$. Altogether, then, the sum is $2\pi V-4\pi$.

So $(2E-2F)\pi = 2\pi V-4\pi$, or $E-F = V-2$,

Either way, no Jordan curve theorem required! Yay.

## Hexagonal filth

I was amused by this report of a petition to change the representation of soccer balls on UK signs. At present the image is of a ball divided into brown and white hexagons. At least on one side.

The problem is, if it’s hexagons all the way around, it’s a geometric impossibility. You can’t have a convex polyhedron consisting solely of hexagons, regular or otherwise. In fact: If a convex polyhedron has only hexagons and/or pentagons as faces, then there must be at least 12 pentagons.

The proof is based on Euler’s polyhedron formula: If a convex polyhedron has $V$ vertices, $E$ edges, and $F$ faces, then

$V-E+F=2$.

For our hex-and-pent-agon polyhedron, consisting of $h$ hexagons and $p$ pentagons, $F = p+h$, of course. What’s $E$? Well, each hexagon has 6 edges and each pentagon has 5, but that counts each edge twice since each edge belongs to two faces. So $E = (5p+6h)/2$.

Now, at each vertex, at least 3 edges meet. If you add up the number of edges at each vertex over all vertices you get at least $3V$; but since each edge contributes 2 to that total, $3V \le 2E$.

But from Euler’s formula,

$3V = 6+3E-3F \le 2E$

or

$6-3F \le -E$

and substituting our formulas for $F$ and $E$,

$6-3p-3h \le -(5p+6h)/2$

or

$12-6p-6h \le -5p-6h$

or

$12-p \le 0$

or

$p \ge 12$.

QED.

(The equality holds if every vertex has 3 edges — or equivalently, 3 faces. In particular, if you’re using only regular hexagons and pentagons, then there isn’t room at a vertex for more than 3 faces to meet there. So for a polyhedron made of regular hexagons and pentagons, or any polyhedron with exactly three hexagons and/or pentagons meeting at each vertex, $p = 12$.)

Hence the government must be petitioned. As The Aperiodical says, “Ban this hexagonal filth!”