12 pentagons!

I’ve been reading Sternberg’s “Group Theory and Physics”, Cambridge University, reprinted 1999. On pages 43 to 44 he says, “… every fullerene has exactly 12 pentagons. This is not an accident.”

The stable structure of carbon which has 60 carbon atoms arranged like the vertices of a soccer ball is called a buckyball. It turns out that, in similar structures, we can have any even number, greater than 18, of carbon atoms except for 22. This is equivalent to polyhedra having 12 pentagons and any number of hexagons except 1.

This family of structures consists of polyhedra whose faces are either pentagons or hexagons. In chemistry they are labeled by the number of carbon atoms, so they talk about $C_{20}, C_{22}, ... C_{60}, ... C_{72}, ....\$

I find it unforgettable and marvelous that the number of pentagons is always exactly 12. And I can prove it.

Sternberg’s proof is somewhat different from mine, because I choose to use two general equations which we have seen before. One equation depends on the fact that every edge separates exactly 2 faces; the other depends on the fact that every edge joins exactly 2 vertices.

Imagine that we have a polyhedron. Choose one face, and count the number of edges it has. Do this for all the faces.

We will have counted each edge twice. Let $f_k\$ be the number of faces with k edges; then, for example, $f_5\$ is the number of pentagons. We have the general formula

$2 e = 3 f_3 + 4 f_4 + 5 f_5 + 6 f_6 + ....\$.

(I think I first showed you that here.)

For example, the term $3 f_3\$ says that we counted 3 edges for each triangular face we touched. But the 2e on the LHS says that we will have counted each of those edges as part of another face, too. And yes, an edge can join two different kinds of polygons.

Similarly, if we count the number of edges by counting the number at each vertex, we will have counted each edge twice. Let $v_k\$ be the number of vertices with k edges; we have the formula

$2 e = 3 v_3 + 4 v_4 + 5 v_5 + ....\$

(I believe I have not shown you that equation, except in the very special case 2 e = 3 v, for triangulations.)

We also have Euler’s formula in general

$\chi = v - e + f\$

and

$\chi = 2\$

in particular.

Let’s see what the smallest case looks like: 12 pentagons.

This is the dodecahedron, one of the five platonic solids. (Now is a good time to remark that people used to think of them as 3D solids, but for quite some time mathematicians have treated them as 2D surfaces bounding 3D volumes. Similarly, the sphere is the surface which bounds the ball.) We see that there are three edges at every vertex.

It has $\{v, e, f\} = \{20,30,12\}\$.

I note that it has 20 vertices. It corresponds to $C_{20}\$. That’s why the list starts at 20.

What about a soccer ball (buckyball)? Mathematica® knows this as the TruncatedIcosahedron.

It has $\{v,e,f\} = \{60,90,32\}\$. This corresponds to carbon 60. Or to the most common soccer ball. (Not all soccer balls have 32 faces, apparently.)

Note also that every vertex has 3 edges.

Not every polyhedron — not even every regular polyhedron — has 3 edges at every vertex. Here is the regular 20-sided Platonic solid, the icosahedron:

Incidentally, it has 5 edges at every vertex. We are going to assume 3 edges at every vertex.

Here is the question. If

• every face of a polyhedron is either a pentagon or a hexagon,
• and every vertex has three edges,

what are the possible polyhedra?

• any such polyhedron must have 12 pentagons;
• it can have any number of hexagons other than 1.

What I will actually prove is that no number of pentagons is possible, except 12.

Our general formula for edges and faces

$2 e = 3 f_3 + 4 f_4 + 5 f_5 + 6 f_6 + ....\$

becomes

$2 e = 5 f_5 + 6 f_6\$,

because $f_k = 0\$ for every other k.

Our general formula for edges and vertices becomes

$2 e = 3 v_3\$

because $v_k = 0\$ for every k ≠ 3.

The total number of faces f is

$f = f_5 + f_6\$,

and the total number of vertices v is

$v = v_3\$.

Thus, we wish to investigate solutions of the following 4 equations:

$2 e = 5 f_5 + 6 f_6\$

$f = f_5 + f_6\$

$2 e = 3 v\$

$2 = v - e + f\$.

It turns out that Mathematica® is extraordinarily cooperative if I write separate equations for

$\chi = v - e + f\$

$\chi = 2\$

and consider a set of 5 equations:

$\begin{array}{l} 2 e=5 f_5+6 f_6 \\ 2 e=3 v \\ \chi =-e+f+v \\ f=f_5+f_6 \\ \chi =2\end{array}\$

When I tell it to eliminate $\chi\$, Mathematica® tells me that $f_5 = 12\$:

$2 e=3 v\land f=f_5+f_6\land v=2 (f-2)\land f_5=12\$

It’s not too difficult to work this out by hand; you can do it without Mathematica.

This algebra places no restrictions on the number of hexagons. An it’s easy enough to exhibit solutions for any number of hexagons. Even for just one hexagon! Ruling out that case is a challenge; the proof dates only from 1963.

We know a couple of cases. We know that $f_6 = 0\$ (no hexagons) is a dodecahedron. And $f_6 = 20\$ (32 faces total) is a soccer ball or buckyball.

Whether or not we can actually build a polygon with any value of $f_6\$ is another question. It turns out that there is only one nonnegative value of $f_6\$ which does not work: we can’t have just one hexagon.

We can have a polyhedron with any number except n=1; that is, only $C_{22}\$ cannot exist. But i haven’t proved either that $C_{22}\$ cannot exist, or that anything does exist, other than $C_{20}\$ and $C_{32}\$, which I have drawn.

I am not about to try to prove either that $f_6 = 1\$ cannot be constructed, or that every other value can be. Maybe someday, but not today. (Among other things, I have no idea if all the hexagonal faces can be chosen the same size, nor even if they are all regular hexagons. My geometric intuition is nil — but the algebra is clear. Well, maybe it isn’t: the algebra says that simply by counting vertices, edges, and faces we can rule out anything that doesn’t have 12 pentagons. It says northing about the shape or even about the existence.)

Exactly 12 pentagons. I think that is amazing. And the equations I used are nicely general, and can be used for other things – such as showing that there can be no regular polyhedra other than the 5 Platonic solids.

For more about buckyballs, you can try this Wikipedia article.

Here is a drawing that suggests why we can’t add just one hexagon to a dodecahedron.

Within my personal library, the best reference on polyhedra is Hartshorne’s “Geometry: Euclid and Beyond”. (See my bibliographies page.)

Ah, a book newly acquired since I drafted this is: Richeson, “Euler’s Gem”, Princeton University 2008; most of it should be accessible to a high school student.

My reference for the two “2 e” equations is Firby & Gardiner, also on my bibliographies page.

A proof that $f_6 = 1\$ cannot exist can be found in Branko Grunbaum’s “Convex Polytopes”, Springer, 2nd Edition, 2003. (Also newly acquired. It’s a graduate text.)

9 Responses to “12 pentagons!”

1. Daniel Lichtblau Says:

The Mathematica part can be done in one go, using the function Reduce[]. Just give the appropriate equations and plausible inequalities, and specify domain as the integers.

In[2]:= Reduce[{2*e==3*v,2*e==5*f5+6*f6,v+f5+f6-e==2,
v>=1,e>=1,f5>=1,f6>=0}, {v,e,f5,f6}, Integers]

Out[2]= C[1] \[Element] Integers && C[1] >= 10 && v == 2 C[1] &&
e == 3 C[1] && f5 == 12 && f6 == -10 + C[1]

Daniel Lichtblau
Wolfram Research

• rip Says:

Hi Daniel!

I recognize you from the Mathematica newsgroup. It’s nice to see you here.

Thank you for the Mathematica tip. I’m not an expert; I just get by.

Rip

2. Leona Says:

Hello Rip – your site is sooooo interesting. I could learn a lot from you! I came across it when google sent me to one of your color theory / HSB articles.

Btw: regarding 12 pentagons, you might find it interesting to look up/find out about the ISEA DGGs (Icosahedral Snyder Equal Area projection, for Discrete Global Grids).

(ps: have you somehow hacked wordpress so it serves up drupal pages?!!)

3. rip Says:

Hi Leona,

Thanks for mentioning the Icosahedral Snyder Equal Area projection — it’s news to me.

I have to ask: drupal pages?

(And no, this is ordinary, free WordPress.)

Rip

4. Leona Says:

Oh… neat. Must be a theme designed to look like Drupal then. Nice. 🙂

5. rip Says:

Oh, you’re right.

This theme is called “Garland”: “A flexible, three-column theme with customizable colors. Design based on Themetastic for Drupal by Stefan Nagtegaal and Steven Wittens.”

rip

6. Dan Says:

I don’t know if the definition of a fullerene may exclude the following, but Euler’s formula has a caveat that must be stated and potentially investigated; the solid cannot have a hole.

7. rip Says:

Hi Dan,

You are, of course, correct. I was speaking only of polyhedra which have no holes. Nice catch.

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