# Why are there 12 pentagons and 20 hexagons on a soccer ball?

Edge-attaching many hexagons results in a plane. Edge-attaching pentagons yields a dodecahedron.

Is there some insight into why the alternation of pentagons and hexagons yields an approximated sphere? Is this special, or are there an arbitrary number of assorted n-gons sets that may be joined together to create regular sphere-like surfaces?

#### Solutions Collecting From Web of "Why are there 12 pentagons and 20 hexagons on a soccer ball?"

The possible ways to put polygons together to form a sphere-like object are constrained by Euler’s formula $V – E + F = 2$ (where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number of faces). Equivalently you can think of this as a statement about planar graphs.

Suppose we use $f_3$ triangles, $f_4$ squares, $f_5$ pentagons, etc. Every edge meets exactly two faces, and an edge of type $f_n$ meets $n$ faces, so let’s double-count the number of pairs of an edge and a face next to it: on the one hand, this is $2E$, and on the other hand, this is

$$3f_3 + 4f_4 + 5f_5 + …$$

Plugging this into Euler’s formula gives $V – \frac{f_3 + 2f_4 + 3f_5 + …}{2} = 2$. If in addition the polyhedron is convex and the polygons are regular, there are constraints on the faces that can meet at each vertex coming from the fact that the angles must sum up to less than $360^{\circ}$. (This is one way to prove the classification of Platonic solids.) For example, at most $5$ faces can meet at each vertex if we allow arbitrary faces; this means $3f_3 + 4f_4 + … \le 5V$. (If you really want to, you can allow six triangles to touch at one point, but I would just count this as a hexagon.) If we don’t allow triangles, exactly $3$ faces meet at each vertex; this means $4f_4 + 5f_5 + … = 3V$.

Here is an application in chemistry: a fullerene is a certain type of molecule made from carbon atoms. (One of these, the buckyball, looks just like a soccer ball.) It gives a convex polyhedron in which each face is either a regular pentagon or hexagon. This gives $V – \frac{3f_5 + 4f_6}{2} = 2$ on the one hand, and $3V = 5f_5 + 6f_6$ on the other. Together these equations give $f_5 = 12$ and $V – 2f_6 = 20$; in other words, any fullerene must have exactly twelve pentagons (Twelve Pentagon theorem for fullerene).

(Hexagons are special. One way to interpret this result is that an infinite plane can be tiled with hexagons, so hexagons correspond to zero curvature, whereas since pentagons have a smaller angle at each vertex they correspond to positive curvature. What the above statement says, roughly, is that the total amount of curvature is a constant. This is a simple form of the Gauss-Bonnet theorem, which is closely related to Euler’s formula.)

Here are some other things you can prove, again under the assumptions of convexity and regularity:

• If you only use triangles (as opposed to triangles and hexagons), then $f_3 \le 20$ and $4 | f_3$.
• If you only use squares, then $f_4 = 6$.
• You cannot use only hexagons or higher. (In the Gauss-Bonnet picture, attempting to use heptagons corresponds to negative curvature, and negative curvature does not interact well with Euclidean or spherical geometry; the natural setting for putting heptagons together is instead hyperbolic geometry.)

This is already most of the way to the classification of Platonic solids. If you’re interested in learning more about Euler’s formula, I highly recommend David Richeson’s Euler’s Gem. Extremely well-written and informative. You might also enjoy David Eppstein’s Nineteen ways to prove Euler’s formula.

The soccer ball shape can be formed by truncating an icosahedron—if each of the 12 vertices is cut off, it will leave a pentagonal face and cause each of the 20 triangular faces to become hexagonal.

You’d probably be interested in the Platonic solids, Archimedean solids, Catalan solids, and some of the Johnson solids.

An icosahedron is a platonic solid having 20 congruent equilateral triangular faces ($F=20$), 30 edges ($E=30$) each five meets at each vertex & 12 identical vertices ($V=12$) lying on the spherical surface (with certain radius).

When an icosahedron is (partially) truncated at all its 12 vertices then each of 20 triangular faces become a hexagon & each of 12 vertices produces a new petagonal face. Thus a truncated icosahedron has 12 regular pentagons & 20 regular hexagons.

While the process of truncation produces $12\times 5=60$ new vertices & $12\times 5=60$ new edges in addition to its 30 original edges. Thus it has total $60+30=90$ edges. For a truncated icosahedron, $F=12+20=32$, $E=90$ & $V=60$ which duly satisfies Euler’s formula ($F+V=E+2$). It is also called Archimedean solid.

A soccer ball is analogous to a truncated icosahedron. Hence it has 12 (regular) petagons & 20 (regular) hexagons.