Tag Archives: Polyhedron

Proof of Descartes Theorem on Polyhedron


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}

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.