## 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!”