San José State University
Department of Economics

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 Polyhedra with Three Types of Faces and One Type of Vertex

The Archimedean polyhedra are those with vertices all of the same kind in terms of their degree, i.e., the number of edges terminating at them. If a vertex has k edges terminating at it, it also has k polygons impinging upon it. Most of the Archimedean polyhedra have only two types of polygonal faces, the Platonic polyhedra have only one type, but a few of the Archimedean polyhedra have three types of polygonal faces. The purposes of this material is to create a system for generate the combinations of polygonal faces in type and number which correspond to polyhedra. A system of this sort was created for polyhdedra of no more than two types of faces but that method does not extend to three types.

Let n, m and p be the numbesr of sides of the polygonal faces with n<m<p. The number of such faces are denoted as fn, m and fp. The number of edges is denoted as e and the number of vertices of degree k as vk. These quantities have to satisfy certain conditions. First there is the Euler formula:

#### fn + fm + fp − e + vk = 2

A face-by-face count of the number of edges yields the value nfn+mfm+pfn, but each edge is counted exactly twice so

#### nfn + mfm + pfn = 2e

Similarly a vertex-by-vertex count of the number of edges yields a value of kvk but each edge is counted exactly twice so

#### kvk = 2e

The Euler formula involves the quantity (vk−e). The above equation implies

#### 2(vk−e) = −(k-2)vk

The variable e can be eliminated by multiplying the Euler formula equation by 2 and substituting in the above expression to obtain

#### 2fn + 2m + 2fp − (k-2)vk = 4

The two equations coming from counts of edges can be combined as

#### nfn + mfm + pfp − vk = 0

The key to generating the characteristics of polyhedra are the numbers of the three types of polygons coming together at a vertex. These are denoted as gn, gm and gp. It must hold that

#### gn + gm + gp = k

Furthermore a vertex-by-vertex count of the number of q-gons yields a value of gqvk, but each q-gon is counted exactly q times so

#### qfq = gqvk for q=m, n, p. and hence fq = gqvk/q

Therefore in the equation derived from the Euler formula, 2fn+2m+2fp−(k-2)vk = 4, the values of fq's can be replaced by expressions involving the g's; i.e.,

#### 2gnvk/n + 2gmvk/m + 2gpvk/p − (k-2)vk = 4 and thus [2gn/n + 2gm/m + gp/p − (k-2)]vk = 4 and hence vk = 4/[2gn/n + 2gm/m + gp/p − (k-2)]

The gq's can take on only a limited number of nonnegative values. If θq is the interior angle of an q-gon expressed as a fraction of a full circle then

#### gq < 1/θq

The value of the θq's are given by

#### θq = (1/2−1/q)

 q θq M 3 1/6 5 4 1/4 3 5 3/10 3 6 1/3 2 7 5/14 2 8 3/8 2 10 2/5 2

Furthermore the gq's are not independent. Therefore gm and gp can be allowed to take on their limited set of values and gn computed as k−(gm+gp). From these values of the gq's the number of vertices can be computed through

#### vk = 4/[2gn/n + 2gm/m + 2gp/p − (k-2)]

if this yields an integral value of 4 or greater. From vk the number of edges are computed as kvk/2. The numbers of faces, the fq's, are computed as gqvk/q. These all have to be nonnegative integral values for the solution to be valid.

The types of faces in the Archimedean polyhedra go up to ten sided polygons, decagons.

Here are the results of the computation for n=3, m=4, p=5 and k=4.

 g3 g4 g5 v4 4 0 0 6 3 1 0 8 2 2 0 12 1 3 0 24 3 0 1 10 1 2 1 60 2 0 2 30

For all other combinations of the gq for k=4 the values are negative and/or nonintegral. The numbers of faces and edges for the above combinations are shown below.

 f3 f4 f5 e Name 8 0 0 12 octahedron 8 2 0 16 twisted square prism 8 6 0 24 cuboctahedron 8 18 0 48 rhombicuboctahedron 10 0 2 20 twisted pentagonal prism 20 30 12 120 rhombicosidodecahedron

As can be seen above, there is only one case in which there are actually three types of faces present in the polyhedron. Any actual polyhedron will correspond to a numerical solution so there cannot be any polyhedra having only triangular, square or pentagonal faces with vertices all of degree 4 other than those shown.

Although the method may capture all of the actual polyhedra there may be numerical solutions that do not correspond to any realizable polyhedron. Consider the case of n=3, m=4, p=6 and k=4. In addition to the cases of no hexagons, which are the same as the cases of no pentagons in the previous tables, the method generates the following.

 g3 g4 g6 v4 3 0 1 12 2 1 1 24 f3 f4 f5 e Name 12 0 2 24 twisted hexagonal prism 16 6 4 48 no known polyhedron

(To be continued.)