This post has been in my drafts folder for a while. With the World Cup here, it’s time to post it!

It’s a rule that a 3D assembly of hexagons must have at least twelve pentagons in order to be a closed polyhedral shape. This post takes a look at why this is true.

First, some examples from nature. The stinkhorn fungus *Clathrus ruber*, has a largely hexagonal layout, with pentagons inserted. The core of HIV has to contain twelve pentagons (shown in red, in this image from the Briggs group) amongst many hexagonal units. My personal favourite, the clathrin cage, can assemble into many buckminsterfullerene-like shapes, but all must contain at least twelve pentagons with a variable number of hexagons.

The case of clathrin is particularly interesting because clathrin triskelia can assemble into a flat hexagonal lattice on membranes. If clathrin is going to coat a vesicle, that means 12 pentagons need to be introduced. So there needs to be quite a bit of rearrangement in order to do this.

You can see the same rule in everyday objects. The best example is a football, or soccer ball, if you are reading in the USA.

The classic design of football has precisely twelve pentagons and twenty hexagonal panels. The roadsign for football stadia here in the UK shows a weirdly distorted hexagonal array that has no pentagons. 22,543 people signed a petition to pressurise the authorities to change it, but the Government responded that it was too costly to correct this geometrical error.

**So why do all of these assemblies have 12 pentagons?**

In the classic text “On Growth and Form” by D’Arcy Wentworth Thompson, polyhedral forms in nature are explored in some detail. In the wonderfully titled *On Concretions, Specules etc.* section, the author notes polyhedral forms in natural objects.

One example is *Dorataspis*, shown left. The layout is identical to the D6 hexagonal barrel assembly of a clathrin cage shown above. There is a belt of six hexagons, one at the top, one at the bottom (eight total) and twelve pentagons between the hexagons. In the book, there is an explanation of the maths behind why there must be twelve pentagons in such assemblies, but it’s obfuscated in bizarre footnotes in latin. I’ll attempt to explain it below.

To shed some light on this we need the help of Euler’s formulae. The surface of a polyhedron in 3D is composed of faces, edges and vertices. If we think back to the football the faces are the pentagons and hexgonal panels, the edges are the stitching where two panels meet and the vertices are where three edges come together. We can denote faces, edges and vertices as f, e and v, respectively. These are 2D, 1D and zero-dimensional objects, respectively. Euler’s formula which is true for all polyhedra is:

\(f – e + v = 2\)If you think about a cube, it has six faces. It has 12 edges and 8 vertices. So, 6 – 12 + 8 = 2. We can also check out a the football above. This has 32 faces (twelve pentagons, twenty hexagons), 90 edges and sixty vertices. 32 – 90 + 60 = 2. Feel free to check it with other polyhedra!

Euler found a second formula which is true for polyhedra where three edges come together at a vertex.

\(\sum (6-n)f_{n} = 12\)in this formula, \(f_{n}\) means number of n-gons.

So let’s say we have dodecahedron, which is a polyhedron made of 12 pentagons. So \(n\) = 5 and \(f_{n}\) = 12, and you can see that \((6-5)12 = 12\).

Let’s take a more complicated object, like the football. Now we have:

\(((6-6)20) + ((6-5)12) = 12\)You can now see why the twelve pentagons are needed. Because 6-6 = 0, we can add as many hexagons as we like, this will add nothing to the left hand side. As long as the twelve pentagons are there, we will have a polyhedron. Without them we don’t. *This is the answer to why there must be twelve pentagons in a closed polyhedral assembly.*

So how did Euler get to the second equation? You might have spotted this yourself for the f, e, v values for the football. Did you notice that the ratio of edges to vertices is 3:2? This is because each edge has two vertices at either end (it is a 1D object) and remember we are dealing with polyhedra with three edges at each vertex. so \(v = \frac{2}{3}e\). Also, each edge is at the boundary of two polygons. So \( e = \frac{1}{2}\sum n f_{n}\). You can check that with the values for the cube or football above. We know that \(f = \sum f_{n}\), this just means that the number of faces is the sum of all the faces of all n-gons. This means that:

\(f – e + v = 2\)Can be turned into

\(f – (1/3)e = \sum n f_{n} – \frac{1}{6}\sum n f_{n} = 2\)Let’s multiply by 6 to get, oh yes

\(\sum (6-n)f_{n} = 12\)There are some topics for further exploration here:

- You can add 0, 2 or 10000 hexagons to 12 pentagons to make a polyhedron, but can you add just one?
- What happens when you add a few heptagons into the array?

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Image credits (free-to-use/wiki or):

Clathrus ruber – tineye search didn’t find source.

HIV cores – Briggs Group

Exploded football – Quora

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The post title comes from “Pentagrammarspin” by Steve Hillage from the 2006 remaster of his LP Fish Rising

An interesting topic! I think, the formula at the bottom has to be sum(fn) – 1/6 sum(n*fn). About the polyhedron with 1 hexagon and 12 pentagons, it would have to have 13 faces, 33 edges and 22 verticies to fullfill Euler’s criteria, but seems to not exist, because Euler’s criteria is only necessary. There are further criterias to make it a necessary and sufficient condition, proven bei Ernst Steinitz, but they do not specify the shape of the faces. So yes, a polyhedron with 13 faces, 33 edges and 22 verticies exists, but it doesnt have to be 12 pentagons and 1 hexagon (if i understood it correctly).