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Euler-type Theorems Concerning
the Number of Polygons of Polyhedra

Leonhard Euler discovered in the 18th century that for polyhedra in which any face is a hexagon or a pentagon and every vertex is of degree three there must be exactly twelve pentagonal faces. This is referred to as Euler's Twelve Pentagon Theorem. For the proof of this theorem see Euler. There is a dual to this theorem to the effect that for a polyhedron with only triangular faces and all vertices are of degree five or six there must be exactly twelve vertices of degree five. For the proof of this theorem see Euler2.

There are other theorems of these sorts. First let us consider a few specific cases. Consider polyhedra with only triangular and square faces. Let the number of these faces be denoted as f3 and f4, respectively. Let the number edges be e. All vertices are assumed to be of degree 4 and this number is denoted as v4.

Euler's formula for polyhedra requires

f3 + f4 − e + v4 = 2

If the edges are counted face-by-face the total will be 3f3+4f4. But each edge is counted exactly twice so the total is equal to 2e. Therefore

3f3 + 4f4 = 2e

If the edges are counted vertex-by-vertex the total will be 4v4. But that also is equal to 2e. Thus

4v4 = 2e
or, equivalently
2v4 = e
and hence
v4 - e = −v4

These can be converted into

2(v5-e) = −3v5
2f3 + 2f4 − 3v5 = 4
3f3 + 4f4 − 5v5 = 0

And finally to:

4f3 + 4f4 − 6v5 = 8
3f3 + 4f4 − 5v5 = 0

Subtraction of the second equation above from the first does not lead to a definite value for one of the variables. Instead

f3 − v5 = 8

So, only for certain polyhedra can a conclusion analogous to Euler's Twelve Pentagon Theorem be drawn.

A Generalization of Euler's Twelve Pentagon Theorem

Consider a polyhedron made up of n-gons and m-gons with all vertices of degree k. The equations to be satisfied are then

fn + fm − e + vk = 2
nfn + mfm = 2e
kvk = 2e


2(vk-e) = −(k-2)vk
2fn + 2fm −(k-2)vk = 4
nfn + mfm − kvk = 0

To eliminate fm the last equation can be multiplied by 2 and the preceding equation by m to get

2mfn + 2mfm −m(k-2)vk = 4m
2nfn + 2mfm − 2kvk = 0

The number of vertices and the number of one type of face will be simultaneously eliminated only if

m(k-2) = 2k
or, equivalently
k = 2m/(m-2)

For Euler's Twelve Pentagon Theorem m=6 and k=3. As can be seen these values satisfy the equation. For the case of the triangles and squares considered previously m=4 and k=4. If m=3 then k has to be 6, but the interior angle of any polygons is too large to bring six polygons together at one vertex. Therefore m cannot be 3.

The value of m cannot be 5 because 10/3 is not an integer. Likewise m cannot be 8 because 16/6 is not an integer and the same holds true for any value of m greater than 8. Thus m has to be 4 or 6.

When m is 4 then the subtraction of the equations yields

2(4-n)fn = 16
or, equivalently
fn = 8/(4-n)

There is an integral value of 8 for n=3 but for no other value of n. This is the case which was considered first above; i.e., polyhedra of triangles and squares with vertices of degree four.

When m is 6 then k must be 3 and the subtraction of the equations yields

2(6-n)fn = 24
or, equivalently
fn = 12/(6-n)

There is an integral solution for n=5; i.e, fn=12. This is the Euler twelve pentagon case. Examples of this case are the dodecahedron with 12 pentagons and 0 hexagons and the truncated icosahedron with 12 pentagons and 20 hexagons.


truncated icosahedron

The cases of a polyhedra made up of triangles and pentagons with vertices of degree 4 or 5, the icosideodecahedron and snub icosidodecahedron, fit in with the Euler case. In both cases these polyhedra have 12 pentagonal faces.


snub dodecahedron

There are also solutions to the equation fn = 12/(6-n) for n=4 and n=3. (There are no polygons for n<3 so the solutions for n=2 and n=1 are not relevant.)

The case for n=4, m=6 and k=3 is that of the truncated octahedron with 8 hexagons and 6 squares. It is also the case of the cube with 0 hexagons and 6 squares. The value of f4=12/(6-4) is 6.


truncated octahedron

For the case polyhedra made up of triangles and hexagons, the case of n=3, m=6 and k=3, there is the truncated tetrahedron with 4 hexagons and 4 triangles. This is also the case of the tetrahedron with 0 hexagons and 4 triangles. The value of 12/(6-n) for n=3 is 4.


truncated tetrahedron

So polyhedra made up of triangles and hexagons and vertices of degree 4 must have exactly 4 triangles.

For the polyhedra involving triangles and squares (the case of n=3, m=4 and k=4) there are the examples of the cuboctahedron with 8 triangles and 6 squares, the small rhombicuboctahedron with 8 triangles and 18 squares. It is also the case of the octahedron with 8 triangles and 0 squares.



small rhombicuboctahedron

Thus polyhedra of triangles and squares with vertices of degree 4 must have exactly 8 triangles. There is another polyhedron made up of triangles and squares but with the degree of the vertices being 5 instead of 4. This polyhedron, the snub cube, has 32 triangles and 6 squares.

There are also Archimedean polyhedra involving three types of faces. These are not considered in this analysis. There are also cases of the truncated dodecahedron which has triangular and decagonal faces and the snub dodecahedron with triangular and pentagonal faces. These do not have analogs of Euler's Twelve Pentagon Theorem.

The dual theorems concerning polyhedra with one type of face but vertices of possibly two different degrees is dealt with elsewhere.


The Euler-type theorems concerning polyhedra are of a conditional sort.

These are the cases in which the analysis indicates that if polyhedra exist, no matter how complicated, there will be exactly a certain number of faces of a particular sort. There are other cases of polyhedra that seem to fit into this scheme, such as the dodecahedron and the icosidodecahedron which both have 12 pentagonal but the vertices are of different degrees.

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