# Soccer Ball and Regular Polyhedron

On the article Calculate Pentagons and Hexagons in a Soccer Ball, I calculated pentagons and hexagons in a soccer ball. Here, think about it from another aspect, without using Euler’s formula.

## Connect Centers of Hexagons

Imagine a polyhedron, of which vertexes are the center of hexagons in a soccer ball. Its each face is regular pentagon, and it is composed only from pentagons. It is regular dodecahedron.

The number of faces in a regular dodecahedron is the same as one of regular pentagons contained in a soccer ball. Therefore, the number of pentagons in a soccer ball is 12.

The number of vertexes in a regular dodecahedron is the same as one of regular hexagons contained in a soccer ball. Therefore, the number of regular hexagons in a soccer ball is $$5 \times 12 \div 3 = 20$$ .

Then, the number of faces in a soccer ball is

$12 + 20 = 32 .$

Now, calculate the number of edges in a soccer ball. Add edges of pentagons and hexagons and remove duplication, then

$( 5 \times 12 + 6 \times 20 ) \div 2 = 90 .$

Lastly, calculate the number of vertexes in a soccer ball. A vertex in a soccer ball belongs to 3 faces, so add the vertexes of pentagons and hexagons and divide it by 3, then

$( 5 \times 12 + 6 \times 20 ) \div 3 = 60 .$

Above, we thought polyhedron whose vertexes are the centers of hexagons in a soccer ball. We can calculate if we think about a polyhedron whose vertexes are the centers of pentagons in a soccer ball, too.

## Connect Centers of Pentagons

Imagine a polyhedron, whose vertexes are the center of pentagons in a soccer ball. It is regular icosahedron. It is obvious that it’s the same as a polyhedron made by connecting the centers of the faces in a regular dodecahedron.

## Truncated Polyhedron

Now truncate the vertexes of regular icosahedron.

Truncating the vertexes of regular icosahedron, then soccer ball appears. This soccer ball shape is called truncated regular icosahedron.

If you deeply truncate the vertexes of regular dodecahedron, soccer ball appears, too. On the following regular dodecahedron image, if you truncate along the line, sort of soccer ball will appear. (The line is too deep, so the hexagon is not regular hexagon but hexagons and pentagons appear.)

# Think about Euler’s Formula on polyhedron

Let’s think about Euler’s formula in polyhedron.

## Euler’s Formula

On polyhedron, the following equation is valid.

(vertex count) – (edge count) + (face count) = 2

This formula is also true to plane figure.

Now, let’s think about this formula on plane figure. Polyhedron can be projected to plane figure, so it is true to polyhedron when it’s true to plane figure.

men

## Project Polyhedron to Plane Figure

Polyhedron can be projected to plane figure.

Left is a tetrahedron, and it can be projected to plane figure on the right side. 4 faces of tetrahedron becomes 3 triangle and other, rounding area. Plane is divided into 4 area by vertex and edge.

## Proof

Now, think about plain figure composed vertex, edge and face. Vertex is edge’s start and end point. All edge is connected only at vertex, don’t cross.

Think in mathematical induction. Define $$V$$ as vertex count , $$E$$ as edge count , $$F$$ as face count. The goal is $$V – E + F = 2$$ .

### Case $$E = 1$$

Edge $$E$$ is 1, face $$F$$ is 1, Veftex $$V$$ is 2, then

$V – E + F = 2 – 1 + 1 = 2 .$

The formula is true.

### Case when $$E = k \in \mathbb N$$

Suppose $$E_k$$ , $$V_k$$ , $$F_k$$ as $$E$$ , $$V$$ , $$F$$ when $$E = k$$ and the formula is true.

$V_k – E_k + F_k = 2$

$$V_k$$ and $$F_k$$ is not unique for $$k$$, they are values which meet the formula.

Think of adding a new edge. Edges are connected on vertexes, so new edge should be connected to already existing edges on start or end point

#### When the new edge connect on one point

Vertex count increases by one then $$V = V_k + 1$$. Edge count increases by one then $$E = E_k + 1$$ . Face count doesn’t change so $$F = F_k$$ .

\begin{array}{cl} & V – E + F \\ = & ( V_k + 1 ) – ( E_k + 1 ) + F_k \\ = & V_k – E_k + F_k \\ = & 2 \end{array}

#### When the new edge connect on two point

Without the new edge, edges are connected so one side or the other is rounded by edges. Then, draw the new edge, and the face count increases by one.

Now, vertex count doesn’t change, $$V = V_k$$, edge count increases by one, $$E = E_k + 1$$ , face count increases by one, $$F = F_k + 1$$ . Of course, in the case that start and end point is the same.

\begin{array}{cl} & V – E + F \\ = & V_k – ( E_k + 1 ) + ( F_k + 1 ) \\ = & V_k – E_k + F_k \\ = & 2 \end{array}

From the above, when the formula is true to $$E = k \in \mathbb N$$, it is also true to $$E = k + 1$$ .

Above all, the formula is always true.

With this Euler’s Formula, we can prove Descartes theorem. I wrote the detail on Prove Descartes Theorem on Polyhedron.

# Calculate Pentagons and Hexagons in a Soccer Ball

There are many pentagons and hexagons in a soccer ball. Now, let’s calculate them.

## Euler’s formula

On polyhedron, the following equation is valid.

(vertex count) – (edge count) + (face count) = 2

The proof of Euler’s formula is written on Think about Euler’s Formula on Polyhedron.

Now, using this formula, calculate the number of pentagons and hexagons. ( I calculated faces, etc. without Euler’s Equation on Soccer Ball and Regular Polyhedron . )

## Calculate

Define $$m$$ as the number of pentagons, and $$n$$ as one of hexagons.

On soccer ball, one vertex is common in 3 planes. So, the number of vertexes on a soccer ball is expressed as $$\frac{5m + 6n}{3}$$, the number of edges is $$\frac{5m + 6n}{2}$$, the number of faces is $$m + n$$ .

According to Euler’s formula, the following equation is valid.

$$\frac{5m + 6n}{3} – \frac{5m + 6n}{2} + m + n = 2$$

Now, calculate the above equation and get the value of $$m$$ .

$$m = 12 .$$

On a soccer ball, one pentagon neighbors 5 hexagons, and one hexagons neighbors 3 pentagons. Then, $$n$$ is calculated as the following.

$$n = \frac{5 m}{3} = \frac{5 \times 12}{3} = 20 .$$

From the above, the number of hexagons is 10, and the number of pentagons is 20.

You can calculate them with Descartes theorem. I wrote it on Proof of Descartes Theorem.

## Note

The shape of Fullerene, C60, is the same as soccer ball. Let’s check whether the number of vertexes is 60.

$$\frac{5m + 6n}{3} = \frac{5 \times 12 + 6 \times 20}{3} = 60 .$$

## Reference

Euler’s formula is explained in the following book.