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!”

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