Category Archives: Mathematics

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.)

Mathematics

Proof of Descartes Theorem on Polyhedron

Monday August 10th, 2015 Kenji

On polyhedron, there’s Descartes Theorem besides Euler’s Formula.

Thinking of Polyhedron

Polyhedron is rounded by faces, whatever faces’ figure is. It means there’s a theorem. Now, let’s investigate regular polyhedrons.

About regular polyhedron, the summation of internal angle that gathers one vertex must be less than 360 degrees. If it equals to 360 degrees, vertex doesn’t exist.

Example

Calculate about regular tetrahedron.

A face of it is regular triangle and one of its angles is 60 degrees. At 1 vertex of tetrahedron, 3 faces gather, so summation of angle degrees gathering at one vertex is

\[ 60 times 3 = 180 . \]

Difference from space, 360 degree, is

\[ 360 – 180 = 180 . \]

The number of tetrahedron vertexes is 4, so sum of angle deficit for all faces are

\[ 180 times 4 = 720 . \]

I calculated such angle deficit for all regular polyhedron. The result is below.

table of regular polyhedron angle deficit
	Sum of internal angle on one vertex	Difference from 360 degrees (angle deficit)	Sum of angle deficit for all faces
regular tetrahedron	180 degrees	180 degrees	720 degrees
regular hexahedron	270 degrees	90 degrees	720 degrees
regular octahedron	240 degrees	120 degrees	720 degrees
regular dodecahedron	324 degrees	36 degrees	720 degrees
regular icosahedron	300 degrees	60 degrees	720 degrees

See, the summation of angle deficit over the polyhedron is 720 degree.

Actually, it is true to every polyhedron. This theorem called Descartes theorem.

Now, let’s prove it.

Proof

Write vertex number, edge number and face number of a polyhedron as $ V $ , $ E $ , $ F $ . As I wrote on Think about Euler’s Formula, the following equation is true (Euler’s Formula).

\[ V – E + F = 2 \]

Now, calculate summation of angle deficit. It is explained as ( 360 degree ) × ( vertex count ) – ( sum of internal angle for all faces ).

\[ (360 \textrm{degree} ) \times ( \textrm{vertex count} ) = 360 \times V \]

is clear from definition of $ V $ .

Calculate summation of internal angle for each face.

The summation of internal angle for a face

Write edge number of a face as $ e $ . The summation of internal angle on this face is

\[ (e – 2) \times 180 . \]

Calculate summation of the above value for all faces.

The summation of internal angle for all faces

With $ e $, it is written as the following.

\begin{eqnarray} & & \sum \left\{ (e – 2) times 180 \right\} & = & 180 ( \sum e – \sum 2 ) \end{eqnarray}

$ \sum $ means summation.

$ \sum e $ is the summation of edge count for all faces. It means count twice for each edge, so $ \sum e = 2 E $ . Calculate summation for each face, then $ \sum 2 = 2 F $ .

Now, the value we want to know is

\begin{eqnarray} & & 180 ( sum e – sum 2 ) & = & 180 ( 2 E – 2 times F ) \\ & = & 360 ( E – F ) . \end{eqnarray}

Then, the sum of angle deficit over polyhedron is

\begin{eqnarray} & & 360 V – \left\{ 360 ( E – F ) \right\} & = & 360 ( V – E + F) \\ & = & 360 \times 2 \\ & = & 720 . \end{eqnarray}

Mathematics

Calculate Pentagons and Hexagons in a Soccer Ball

Monday June 2nd, 2014 Kenji

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.

Math Handbook

Code, Mathematics, Diary

(日本語) 数秘術に基づくあの年の運勢

Sunday September 30th, 2012 Kenji

Sorry, this entry is only available in 日本語.

Mathematics

(日本語) 偏差値の計算方法

Monday August 17th, 2009 Kenji

Sorry, this entry is only available in 日本語.

The Life

Category Archives: Mathematics

Soccer Ball and Regular Polyhedron

Connect Centers of Hexagons

Connect Centers of Pentagons

Truncated Polyhedron

Proof of Descartes Theorem on Polyhedron

Thinking of Polyhedron

Example

Proof

The summation of internal angle for a face

The summation of internal angle for all faces

Calculate Pentagons and Hexagons in a Soccer Ball

Euler’s formula

Calculate

Note

Reference

(日本語) 数秘術に基づくあの年の運勢

(日本語) 偏差値の計算方法

A Life Summary of an Gypsy