Twelve pentagons, whatever the hexagons

A football is sewn from twelve black pentagons and twenty white hexagons. The pattern is so familiar that the numbers look like choices — a designer’s decision about how many panels make a ball round enough. They are not choices, or at least only one of them is. The twenty could have been any number: a ball sewn with more, smaller hexagons would be rounder, and a ball with none at all is a dodecahedron. The twelve could not have been anything else.

The same twelve turns up wherever a surface is closed up out of mostly six-sided pieces. The molecule of sixty carbon atoms discovered in 1985 has the football’s pattern exactly. Its cousin with seventy atoms has twenty-five hexagons and, again, twelve pentagons. A geodesic dome built of triangles has twelve points where five triangles meet instead of six, however large the dome is made, and the protein shells of many viruses are arranged the same way. Euler’s formula — corners minus edges plus faces is two — is the reason for all of them, once it is rearranged into a statement about faces alone.

The formula rearranged into a budget #

Take any solid in which exactly three faces meet at every corner. That condition is common rather than special: it holds for the tetrahedron, the cube and the dodecahedron, for every prism, and for most solids built by cutting corners off others. Write $p_k$ for the number of faces with $k$ sides, so that $F = p_3 + p_4 + p_5 + \dots$ counts the faces.

Two counts of the edges follow from the shape of the solid alone. Every edge has two ends and every corner has three edges, so $2E = 3V$. Every edge borders two faces and a face with $k$ sides has $k$ edges, so $2E = \sum_k k\,p_k$. Now take Euler’s formula, $V - E + F = 2$, and multiply it by six:

$$6V - 6E + 6F = 12.$$

Replace $6V$ by $4E$, which is what $2E = 3V$ says, and the corners are gone: $6F - 2E = 12$. Replace $F$ and $2E$ by their sums over the faces, and the edges are gone too:

$$\sum_k (6 - k)\,p_k = 12.$$

Every face has a shortfall of six minus its number of sides, and the shortfalls of all the faces add to exactly twelve. A triangle spends three units of the budget, a square two, a pentagon one. A hexagon spends nothing, which is why the number of hexagons never appears. A face with more than six sides spends a negative amount — an octagon gives two units back — and has to be paid for by extra small faces elsewhere.

The bars above are that sum drawn out. The tetrahedron spends its twelve on four triangles, the cube on six squares, the dodecahedron on twelve pentagons. The triangular prism spends six on its two ends and six on its three square sides. The hexagonal prism’s two hexagonal ends are free, so its six squares carry the whole budget. On every row the dashed line at twelve is met exactly, and the check was made on each solid’s actual faces, not on the formula.

Why the hexagons drop out #

The disappearance of the hexagons is the heart of the matter and deserves a second explanation that does not go through algebra.

A flat sheet can be tiled by hexagons meeting three at a corner — the honeycomb — and the tiling goes on for ever without closing up. Three hexagons fit round a point exactly, their angles of $120°$ adding to a full turn with nothing to spare. A surface made only of hexagons is therefore flat in the sense that matters: nowhere does it bend round on itself. To close a sheet into a ball, something has to bend it, and a smaller face does. Three faces meeting at a corner where one of them is a pentagon leave a gap: $108° + 120° + 120° = 348°$, twelve degrees short of a full turn. That gap is the angle defect of the corner, and a pentagon’s five corners carry sixty degrees of it between them.

The defects of any closed convex solid add to $720°$, two full turns, and that is Descartes’ version of the same theorem. Sixty degrees a pentagon into seven hundred and twenty degrees goes twelve times. A triangle surrounded by hexagons leaves a gap of $60°$ at each of its three corners, $180°$ in all, which is three pentagons’ worth, and a square leaves $30°$ at each of four, two pentagons’ worth. The shortfall of the budget is the curvature a face introduces, measured in units of one pentagon, and a hexagon introduces none.

The two explanations are one theorem seen twice: the algebra counts, the angles measure, and the agreement between them is the content of the statement that Euler’s number and Descartes’ total are the same invariant. What the angles add is intuition about where the curvature sits. On a football it sits in twelve places, at the pentagons, and the hexagonal panels between them are nearly flat — the discrete form of the curvature that a triangle drawn on a globe measures by its excess of angle.

Cutting corners and keeping the total #

The football is not an independent design. It is the icosahedron with its corners cut off — each of the twenty triangles trimmed into a hexagon, and each of the twelve corners, where five triangles met, replaced by a small pentagon. The pentagons of a football are the corners of an icosahedron, and there are twelve of them because the icosahedron has twelve corners.

The picture was made by that operation, carried out on the icosahedron’s coordinates: two new points on each edge a third of the way from either end, a hexagon from the six points on each triangle’s edges, a pentagon from the five points round each old corner. The census of the result is $60 - 90 + 32 = 2$, and its twelve pentagons spend the twelve.

Truncation is a clean way to see the budget working, because it is easy to say what it does to the shortfalls. A face with $k$ sides becomes a face with $2k$ sides, and each old corner where $q$ faces met becomes a new face with $q$ sides. For the icosahedron, $k = 3$ and $q = 5$: triangles become hexagons, which are free, and corners become pentagons, which cost one each. So the budget of the truncated icosahedron is spent entirely at its old corners, twelve times one.

The octahedron has six corners where four triangles meet, so cutting them gives six squares and eight hexagons. Six squares times two is twelve, and the eight hexagons are free again. The truncated tetrahedron works the same way with four triangles; and truncating a cube instead, whose faces are squares, turns each face into an octagon, which gives back two units, while the eight new triangular corners cost three each: $24 - 12 = 12$. Every truncation of every one of the five regular solids balances, and each one is one of the thirteen solids that relax regularity by one word.

Every cage of carbon #

In 1985 Harold Kroto, Robert Curl and Richard Smalley found that vaporised graphite produced a molecule with exactly sixty carbon atoms, and proposed that it was a closed cage in the shape of a football. The structure was confirmed, the molecules were named fullerenes after the architect of geodesic domes, and the three shared the chemistry Nobel prize in 1996. Carbon in graphite forms flat sheets of hexagons, three bonds at every atom. A sheet cannot close into a ball, so a closed cage has to include some smaller rings, and in practice they are always pentagons.

The budget then fixes the pentagons without anybody knowing the geometry. Three bonds at each atom means three faces at each corner; faces of five and six sides only means the budget is spent on pentagons alone, at one unit each. Every fullerene has twelve pentagons, and the atoms decide only the hexagons.

The hexagon count follows by the same algebra. A cage with $n$ atoms has $3n/2$ bonds and, by Euler, $3n/2 - n + 2 = n/2 + 2$ rings. Twelve are pentagons, so $n/2 - 10$ are hexagons: none for the smallest cage, the twenty-atom dodecahedron; twenty for sixty atoms; twenty-five for seventy; forty for a hundred. The count also shows why the number of atoms must be even — $3n/2$ bonds has to be a whole number — and why no cage has fewer than twenty.

The gap at twenty-two atoms is the one fact on this figure that the budget does not supply. The arithmetic permits twelve pentagons and one hexagon; the geometry refuses. Branko Grünbaum and Theodore Motzkin proved in 1963 that a cage of pentagons and hexagons exists for every number of hexagons except exactly one. A lone hexagon cannot be surrounded by pentagons in a way that closes, and the proof is a case analysis of what can sit next to it. So the formula is a necessary condition, never a sufficient one, and the difference between the two is where the real chemistry starts.

Twelve places where the pattern of six breaks #

The same budget holds for the dual arrangement, where the roles of corners and faces swap. A geodesic dome, or any surface built from triangles, has a number of triangles meeting at each corner — its degree. Swapping corners and faces in the argument gives $\sum (6 - d) = 12$ over the corners, so a triangulated sphere in which every corner has degree five or six has exactly twelve corners of degree five.

That is why Buckminster Fuller’s domes, however finely they are subdivided, always have twelve points where the triangles meet five at a time, sitting where the corners of an icosahedron would be. Donald Caspar and Aaron Klug used the same observation in 1962 to explain the protein shells of viruses, which are assembled from identical units and turn out to have twelve five-fold points and a variable number of six-fold ones. Golf balls, climbing frames and the panels of many spherical buildings are all made the same way, and all carry the twelve. The designer chooses how finely to subdivide; the topology of the sphere chooses the twelve. It is also why nobody can wrap a ball in hexagonal chicken wire without the wire buckling: the twelve units of curvature have to go somewhere, and hexagons have none to give.

A budget of six times the Euler characteristic #

Nothing in the rearrangement used that the solid was a sphere, except at one step — the two on the right of Euler’s formula. On a surface with holes, the alternating sum is not two: it is the Euler characteristic $\chi$, which is $0$ on a torus, $-2$ on a surface with two holes, and in general $2 - 2g$ for a surface with $g$ holes. The same multiplication by six then gives

$$\sum_k (6 - k)\,p_k = 6\chi.$$

On a torus the budget is zero, so a torus can be tiled entirely by hexagons — take a honeycomb and glue its opposite edges, as the $4 \times 4$ example does, and every corner still has three hexagons round it. This is a carbon nanotube rolled up and closed on itself, and it is why a nanotube’s walls can be pure hexagons for any length. A tube with open ends capped off is a sphere again, and its caps hold the twelve pentagons between them, six at each end. The pentagons are what close the tube, and they are the chemically reactive points of it — the places where the flatness of the sheet has been spent.

On a surface with two holes the budget is negative. Pentagons would make it worse, and the surface has to be paid for with faces of seven or more sides: twelve heptagons, or six octagons, among any number of hexagons. On the three-holed surface it is twenty-four heptagons, and there is a famous example with exactly that count — Felix Klein’s quartic curve of 1879 can be tiled by twenty-four regular heptagons meeting three at a corner, with $56 - 84 + 24 = -4$. The classification of surfaces says the number of holes is the only thing distinguishing them, and here it decides the one quantity a tiling cannot choose.

What the count cannot place #

The budget says how many small faces there must be. It says nothing about where they go, and the pictures above cannot either, because each shows one solid among the many that share its counts.

The sixty-atom cage has twelve pentagons and twenty hexagons, and there are 1,812 ways of arranging exactly those rings into a closed cage, up to symmetry. Only one of them — the football — keeps every pentagon surrounded by hexagons, with no two pentagons sharing an edge; and it is that one that forms in the laboratory, because two pentagons side by side strain the bonds between them. The isolated pentagon rule, as chemists call it, is a statement about placement, and nothing in Euler’s formula touches it. The same count of faces is compatible with a round ball and with a long lumpy capsule, and the budget cannot tell the two apart.

Nor does the count say that a given list of faces can be assembled. The twenty-two-atom gap is the smallest case. The general statement is Eberhard’s theorem of 1891: any list of small and large faces satisfying the budget can be completed to a real solid by adding some number of hexagons, but the number needed is not given by the formula and can be large. The budget is necessary, and hexagons are the free material that makes it nearly sufficient.

And the figures are drawn with flat faces meeting at edges, but none of the argument needs flatness, regularity or straightness. It counts corners, edges and faces, so it holds just as well for a leather ball whose panels bulge, for a molecule whose rings are slightly puckered, or for a map drawn on a globe with three countries at every meeting point. The equality is topological; the pictures are geometric, and they suggest a precision the theorem never asked for.

Still open: how many cages, and which one forms? #

The number of distinct fullerene cages grows quickly with size. For sixty atoms there are 1,812; for a hundred, 285,914. William Thurston showed in 1998 that the cages with $n$ atoms correspond to points of a lattice in a nine-dimensional space, so their number grows roughly like $n^9$, and the counts have been computed by programs that generate every cage exactly once. What no counting argument settles is which of the many permitted cages a given chemistry actually produces, and why the ones with isolated pentagons dominate so completely at sixty and seventy atoms and so much less completely above.

The four-colour problem was once attacked by exactly this kind of budget. On a map with three countries at every meeting point, the shortfalls of the countries add to twelve, so some country must have five or fewer neighbours — which is how the proof of the four-colour theorem begins, by discharging the twelve units of charge around the map until an unavoidable small configuration appears. On a torus the budget is zero and the number of colours a map can need is seven, not four — Heawood’s bound, computed from the same count with $\chi = 0$ in place of $2$. Which placement questions of this kind can be answered by redistributing twelve units, and which need the whole geometry, is still decided case by case.

A theorem about counts that behaves like a law of nature #

Twelve pentagons is one of the few pieces of pure mathematics that chemists, architects and ball manufacturers all rediscovered independently. It deserves its reputation because it is a law in the strong sense: nothing made of pentagons and hexagons meeting three at a corner can close into a ball any other way, and the proof is four lines of arithmetic on Euler’s formula.

The formula was about corners, edges and faces. Rearranged, it is about curvature — how much bending a face contributes — and says that closing a sphere costs exactly twelve units of it, paid by faces smaller than hexagons, with hexagons free and larger faces giving change. That the same twelve is $720°$ of angle defect, and that it becomes zero on a torus and negative on anything with more holes, is the whole of Euler’s number in a single budget. What a surface is determines what closing it costs; how the cost is spread — where the pentagons sit, how many hexagons fill the rest — is left to the maker, whether that is a chemist, a virus or a person sewing a ball.