San José State University

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Thayer Watkins
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 Euler's Theorem Concerning Polyhedra Composed of Pentagons and Hexagons: There Must Be Exactly 12 Pentagons

Let M be a closed convex polyhedron with no holes which is composed of no polygons other than pentagons and hexagons. Let f, e, v be the number of faces, edges and vertices of M, respectively. Let the number of hexagons be denoted as h and the number of pentagons as p. Then f is equal to h+p.

The Euler-Poincare (oiler-pwan-kar-ray) characteristic of the polyhedron, f-e+v, is equal to 2. This is one equation constraining the values of f, e and v; i.e.,

#### f - e + v = 2 or, equivalently h + p + v - e = 2

If we traverse the polyhedron face-by-face counting the number of edges we will get 6h+5p. We will also count every edge twice. Therefore

#### 6h + 5p = 2e

The number of edges coming together at a vertex has to be three. This will be demonstrated in the Appendix. If we traverse the vertices of the polyhedron counting edges we will get 3v. On the other hand we will have counted each edge twice since each edge terminates in exactly two vertices. Thus

#### 3v = 2e

From this it follows that

#### v + 2v - 2e = 0 or, equivalently 2(v-e) = -v

Now if we multiply the condition that h+p+v-e=2 by two and substitute -v for 2(v-e) we get

#### 2h +2p - v = 4

Now multiply this equation by 3 to get

#### 6h + 6p - 3v = 12

From the conditions 6h+5p=2e and 2e=3v we have

#### 6h+5p-3v = 0

We now have the two equations

#### 6h + 6p - 3v = 12 6h + 5p - 3v = 0

Subtracting the second equation from the first gives

#### p = 12

Thus any closed polyhedron composed no polygons other than hexagons and pentagons and having no holes must have exactly 12 pentagons.

Replacing p by the value 12 gives

#### 3v - 6h = 60 or, equivalently v - 2h = 20 or v = 2h + 20 or, alternatively h = ½v − 10

When h=0, the polyhedron is the dodecahedron having twelve pentagons with 20 vertices and 30 edges. This means that there can be no hexagon-pentagon polyhedron with less than 20 vertices.

Although it is not proven here, no such polyhedron can be constructed with h=1. But for any other number of hexagons a polyhedron made up of only hexagons and 12 pentagons exists.

If h=1 is forbidden then from the above equation v=22 is forbidden. Thus there is no polyhedron composed only of hexagons and pentagons with 22 vertices. This has an application in the matter of buckminsterfullerenes. These are polyhedral configurations of carbon atoms. Such molecules have been created with 60 atoms and designated as C60. Likewise C70 and others have been created, but there is no C22.

## Generalizations

The Euler-Poincare characteristic of a polyhedron with g holes is equal to 2(1-g). (This is a result discovered by the French-Swiss mathematician Lhuilier (L'huilier). This name means he oiler) When g=1 then the characteristic is 0. When the previous analysis is reconstructed with a characteristic zero the resulting two equations are

#### 6h + 6p - 3v = 0 6h + 5p - 3v = 0

Subtraction of the two equations reveal that p must be zero. Thus no hexagon-pentagon polyhedron with one hole can be constructed. Furthermore for polyhedra with two or more holes the Euler Poincare characteristic is negative. For example, for two holes it is -2. The two equations which would arise are

#### 6h + 6p - 3v = −12 6h + 5p - 3v = 0

These equations imply that p=-12. Since a negative number of pentagons have no meaning there cannot be a hexagon-pentagon polyhedron constructed with holes.

While hexagon-pentagon polyhedra with holes cannot exist there can be such polyhedra with cavities. The Euler-Poincare characteristic of the surface of polyhedral solid with a cavity is just the sum of the characteristics for the outer and the inner surfaces. Thus it would have an Euler-Poincare characteristic of 4. This polyhedral arrangement then would have to have 12 pentagons in the outer surface and 12 for the inner surface for a total of 24.

If the Lhuilier formula Χ=2(1-g) is solved for g when Χ=4 the result is g=−1. Thus cavities can be considered be equivalent to negative holes.

## Three Edges Terminate at Each Vertex

First consider the determination of the interior angle of a polygon. The interior angle is the complement of what could be called the turning angle although it is usually called the exterior angle.

As one traverses the polygon the direction of movement must turn from 0 to 2π. Thus the turning angle at a vertex of a regular n-gon must be 2π/n. The interior angle then must be π−2π/n, which reduces to (1−2/n)π. Thus for n=6 the interior angle is (1-2/6)π or 2π/3. For a pentagon the interior angle is (1−2/5)π or 3π/5.

Thus interior angle at a vertex of a hexagon is 2π/3 (120°) and for a pentagon 3π/5 (108°). The sum of the interior angle of the polygons impinging upon a vertex must be less than 2π (360°). There cannot be just two polygons coming together at a vertex. Therefore the only possibilities are: 1. Three pentagons 2. Two pentagons and one hexagon 3. One pentagon and two hexagons. In all three cases there are three edges terminating a vertex.