﻿ Regularities of the Platonic and Archimedean Polyhedra
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 Regularities of the Platonic and Archimedean Polyhedra

The Platonic polyhedra are those with a single type of polygonal face and with the same number of edges meeting at each vertex. This latter number is called the degree of a vertex. The Archimedean polyhedra have more than one type of polygonal faces but the vertices are all of a single degree. They are sometimes referred to as convex, by which is meant that they are the surface of convex polyhedral solids. The data for the Platonic and Archimedean polyhedra are given in the table below ordered first by the number of vertices and then by the degree of the vertices.

 v k n fn m fm p fp e Name 4 3 3 4 6 tetrahedron 6 4 3 8 12 octahedron 8 3 4 6 12 cube 12 3 3 4 6 4 18 truncated tetrahedron 12 4 3 8 4 6 24 cuboctahedron 12 5 3 20 30 icosahedron 20 3 5 12 30 dodecahedron 24 3 3 8 8 6 36 truncated cube 24 3 4 6 6 8 36 truncated octahedron 24 4 3 8 4 18 48 rhombicuboctahedron 24 5 3 32 4 6 60 snub cuboctahedron 30 4 3 20 5 12 60 icosidodecahedron 48 3 4 12 6 8 8 6 72 truncated cuboctahedron 60 3 3 20 10 12 90 truncated dodecahedron 60 3 5 12 6 20 90 truncated icosahedron 60 4 3 20 4 30 5 12 120 rhombicosidodecahedron 60 5 3 80 5 12 150 snub icosidodecahedron 120 3 4 30 6 20 10 12 180 truncated icosidodecahedron

where:

• v = Number of vertices
• k = Degree of vertices
• n = Number of sides of first type of polygonal face
• fn = Number of first type of polygonal face
• m = Number of sides of second type of polygonal face
• fm = Number of second type of polygonal face
• p = Number of sides of third type of polygonal face
• fp = Number of third type of polygonal face
• Number of edges

There are alternative names for some of the polyhedra. The truncated octahedron is sometimes called the greater rhombicuboctahedron and the truncated icosidodecahedron is also called the greater rhombicosidodecahedron. The cube, to be consistent with the naming of the other polyhedra, could be called the hexahedron.

The most notable regularities in the above table are that the numbers of edges are multiples of six and the numbers of faces and vertices are all even. However there is the triangular prism which has two triangles and three squares which has nine edges, a number not divisible by six or two. The polygonal prisms are polyhedra that satisfy the conditions defining the Archimedean polyhedra but are generally not included simply because they are so simple and constitute an infinite set. The cube happens to be a member of that set. For more on such polyhedra see the Prosaic Polyhedra.

## General Conditions Which are Satisfied by Polyhedra, Regular and Irregular

The Euler-Poincare characteristic Χ for a geometric structure is f-e+v, where f, e and v are the number of its faces, edges and vertices, respectively. The faces are two dimensional, the edges one dimensional and the vertices zero dimensional. Letting fq denote the number of faces with q edges and vj the number of vertices of degree j, then

#### Χ = Σqfq − e + Σjvj

The value of q must be greater than or equal to 3.

The regular convex polyhedra all have an Euler-Poincare characteristic of 2. Thus one condition satisfied by them is

#### Σqfq − e + Σjvj = 2

A face-by-face count of the edges yields a value of Σqqfq, but this is equal to 2e because each edge is counted exactly twice since it is the edge of two faces. Thus another condition satified by the polyhedra is

#### Σqqfq = 2e

A vertex-by-vertex count of the edges yields a value of Σjjvj, but this is equal to 2e because each edge is counted exactly twice since it impinges upon exactly two vertices. Thus another condition satified by the polyhedra is

#### Σjjvj = 2e

The first condition involves the term Σjvj−e. This term can be constructed from the third condition as

#### 2(Σjvj−e) = −Σj(j-2)vj

If the first equation is multiplied by two and the above expression substituted in, the result is

#### Σq2fq − Σj(j-2)vj = 4

The second and third condition can be combined to yield the equation

#### Σqqfq − Σjjvj = 0

Thus the quantity e has be eliminated from the equations. Subtracting the first of the above two equations from the second yields

#### Σq(q-2)fq − 2Σjvj = −4

But Σjvj is just v so the above equation can be put into the form

#### Σq(q-2)fq = 2(v −2)

This means that if the number of vertices is changed by an amount Δv then the changes in the number of faces Δfq have to satisfy the condition

#### Σq(q-2)Δfq = 2Δv

For two polyhedra with the same number of vertices but a different degree for the vertices, the changes in the number of faces have to satisfy the condition

#### Σq(q-2)Δfq = 0

For example, consider the change from an icosahedron which has 20 triangles and 0 hexagons to a cuboctahedron which has 8 triangles and 6 squares. The changes are then Δf3=−12 and Δf4=+6. Then

#### (3-2)*(-12) + (4-2)*6 = -12 + 12 = 0

For polyhedra made up of n-gons and m-gons

#### Δfn/Δfm = −(m-2)/(n-2)

Thus the tradeoff, so to speak, between n-gons and m-gons is (m-2)/(n-2). Thus for m=4 and n=3 that tradeoff is 2/1 and hence to get 6 more squares the number of triangles has to be reduced by 2 times 6 or 12.

## Divisibility of the numbers of faces, edges and vertices

For polyhedra having vertices of a single degree k it holds that

#### kvk = 2e

This means that for polyhedra whose vertices are of an odd degree (3 or 5) it must be that the number of vertices is an even number. It also happens that even for the Platonic and Archimedean polyhedra with vertices of an even degree the number of vertices are even, but it would be desirable to have a mathematical proof rather than the results of examining all of the cases.

As of yet there is no proof that the number of edges of the Platonic and Archimedean polyhedra must be divisible by six; they just are. As was noted previously the triangular prism illustrates that the divisibility of the number edges by six does not hold for all polyhedra, even those with just two types of faces and vertices all of a single degree.

(To be continued.)

## The Number of Polygons Meeting at a Vertex

If k edges meet at a vertex then k polygon faces meet at that same vertex. Let gq be the number of q-gons impinging upon a vertex. Then

#### Σqgq = k

The interior angle θq of a q-gon, measured as fraction of a full circle, is (1/2−1/q). The constraint on the gq's is that the sum of the interior angles of the polygons meeting at a vertex must be less than a full circle; i.e.,

#### Σqθqgq < 1

A vertex-by-vertex count of the q-gons will give a figure of gqvk. But each q-gon has q vertices so it is counted q times. Thus

#### gqvk = qfqor, equivalently gq = qfq/vkfor all q

Since gq must be an integer these equations impose a constraint upon the values of the fq's.

The values of the gq for the Platonic and Archimedean polyhedra are given in the table below. The values are determined from the above formula, but can be verified by a simple count in the image which can be accessed by clicking on the name of the polyhedron.

 v k n gn m gm p gp e Name 4 3 3 3 6 tetrahedron 6 4 3 4 12 octahedron 8 3 4 1 12 cube 12 3 3 1 6 2 18 truncated tetrahedron 12 4 3 2 4 2 24 cuboctahedron 12 5 3 5 30 icosahedron 20 3 5 3 30 dodecahedron 24 3 3 1 8 2 36 truncated cube 24 3 4 1 6 2 36 truncated octahedron 24 4 3 1 4 3 48 rhombicuboctahedron 24 5 3 4 4 1 60 snub cuboctahedron 30 4 3 2 5 2 60 icosidodecahedron 48 3 4 1 6 1 8 1 72 truncated cuboctahedron 60 3 3 1 10 2 90 truncated dodecahedron 60 3 5 1 6 2 90 truncated icosahedron 60 4 3 1 4 2 5 1 120 rhombicosidodecahedron 60 5 3 4 5 1 150 snub icosidodecahedron 120 3 4 1 6 1 10 1 180 truncated icosidodecahedron

For polyhedra of no more than two types of faces, n-gons and m-gons, and vertices all of the same degree k the constraining equations can be reduced to a set of two equations in three unknowns, fn, fm and v.

#### 2fn + 2fm − (k-2)v = 4 nfn + mfm − kv = 0

These two equations can be solved for fn and fm as linear functions of v. These take the form of

#### fn = (rv + 4m)/s fm = (tv − 4n)/s with m>n

where r, s and t are integers which are a function of m, n and k. Specifically s=2(m-n), r=mk-2(m+k) and t=−nk+2(n+k). Expressions of the form 2(m+k)−mk occur often enough to define a function φ(x,y)=2(x+y)−xy.

Thus the solutions take the form

#### fn = (−φ(m,k)v+4m)/s fm = (φ(n,k)v−4n)/s

It is presumed that m is the larger of m and n. Since the number of m-gons at a vertex, gm=mfm/v, must be greater than or equal to zero

#### gm = m(φ(n,k)v − 4n)/sv = (m/s)(φ(n,k) − 4n/v) ≥ 0 and hence φ(n,k) − 4n/v ≤ 0 and thus φ(n,k) ≤ 4n/v so, if φ(n,k)<0 v ≥ 4n/φ(n,k)

It is notable that this lower limit for v depends only upon n and k and is independent of m.

Let M be the maximum number of m-gons that can meet at a vertex. Then

#### gm ≤ M so m(φ(n,k) − 4n/v) ≤ M and hence φ(n,k) − 4n/v ≤ M/m and consequently φ(m,k)−M/m ≤ 4n/v and thus if φ(m,k)−M/m<0 v ≤ 4n/[(M/m)−φ(n,k)]

So, for any combination of m, n and k the allowable number of vertices is limited to a particular range from 4n/φ(n,k) to 4n[(M/m)−φ(n,k)].

Within that range the allowable values of v are those such that

#### gm = m((φ(n,k)−4n/v)/s) is a nonnegative integer.

This is equivalent to

This reduces to

#### φ(n,k)−sj/m = 4n/v and thus v = 4n/[φ(n,k)−sj/m] for some nonnegative integer j

It is also requiered that fn and fm be nonnegative.

Consider the case of n=3, m=5 and k=4. For j=0, 1, 2 the values of v are 6, 10 and 30. These correspond to the octahedron, the skewed pentagonal prism and the icosidodecahedron, respectively. For all higher values of j the computed values of v are negative. Thus the only allowable values of v are 6, 10 and 30.

For the case of n=3, m=5 and k=3 the only nonnegative values of v generated for integral values of j are v=4 and v=20. These are the case of the tetrahedron and the dodecahedron.

For n=3, m=6 and k=3 the allowable values of v are 4, 6 and 12. The value of v=6 corresponds to the cases of the tetrahedron and v=12 to the truncated tetrahedron, but the case of v=6 is a puzzle. It would have to have 9 edges, which is not an even number. It should have 4 triangles and one hexagon. The one hexagon would require all of the vertices and 6 of the edges. A figure with those specificatitions can be constructed but it is simply a flat hexagon with three pairs of vertices connected and not a polyhedron. The table below gives the allowable number of vertices for the various possible combinations of n, m and k. The symbol {. . .} denotes a set and {_} denotes the empty set.

 n m k Allowablev 3 4 3 {4, 6, 8} 3 4 4 {6, 8, 12, 24} 3 4 5 {12, 24} 3 5 3 {4, 20} 3 5 4 {6, 10, 30} 3 5 5 {12, 60} 3 6 3 {4, 6, 12} 3 6 4 {6, 12} 3 6 5 {12} 3 8 3 {4, 24} 3 10 3 {4, 60} 4 5 3 {8, 10, 20} 4 5 4 {_} 4 6 3 {8, 12, 24} 4 6 4 {_} 5 6 3 {20, 30, 60} 5 6 4 {_}

As illustration of the analysis behind the above table take the case of n=3, m=4 and k=4.

 Integerj Vertices v4 f3 f4 e polyhedron 0 6 8 0 12 octahedron 1 8 8 2 16 twisted cube 2 12 8 6 24 cuboctahedron 3 24 8 18 48 rhombicuboctahedron

The twisted cube has a square at the bottom and a square at the top which is turned 45° with respect to the bottom square and with the top and bottom squares connected with triangles. The fact that there are exactly 8 triangles for each polyhedron illustrates an analog of Euler's Five Pentagon Theorem . In this case, any polyhedron whose faces are either triangles or squares and whose vertices are all of degree four must have exactly eight triangular faces.

Another interesting case of this sort is the polyhedra whose faces are either squares or hexagons and with degree 3 vertices.

 Integerj Vertices v3 f4 f6 e polyhedron 0 8 6 0 12 cube 1 12 6 2 18 hexagonal prism 2 24 6 8 36 truncated octahedron

Such polyhedra must have exactly 6 square faces.

To show that the integers generating the number of vertices are not necessarily sequential consider the case of n=3, m=5 and k=3.

 Integerj Vertices v3 f3 f5 e polyhedron 0 4 4 0 6 tetrahedron 3 20 0 12 30 dodecahedron

The other values of the integer j generate nonintegral values for all the variables.

Here is one last illustration, that of polyhedra having faces that are either triangles or pentagons and vertices all of degree 4.

 Integerj Vertices v4 f3 f5 e polyhedron 0 6 8 0 12 octahedron 1 10 10 2 20 twistedpentagonalprism 2 30 20 12 60 icosidodecahedron

## Conclusions

The numbers and types of faces of the Platonic and Archimedean polyhedra are the nonnegative integer solutions of a set of equations. Some solutions of this sort correspond to the polyhdera which are polygonal prisms. Such prismatic figures have the same regularity as the Archimedean polyhedra. There is only one case in which such a solution did not correspond to a polyhedron. Thus the geometric problem of the existence of particular arrangements of polygonal faces, edges and vertices can be reduced to a numerical one.