San José State University
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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 f_{n}, _{m} and f_{p}. The number of edges is denoted as e and the number of vertices of degree k as v_{k}. These quantities have to satisfy certain conditions. First there is the Euler formula:
A facebyface count of the number of edges yields the value nf_{n}+mf_{m}+pf_{n}, but each edge is counted exactly twice so
Similarly a vertexbyvertex count of the number of edges yields a value of kv_{k} but each edge is counted exactly twice so
The Euler formula involves the quantity (v_{k}−e). The above equation implies
The variable e can be eliminated by multiplying the Euler formula equation by 2 and substituting in the above expression to obtain
The two equations coming from counts of edges can be combined as
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 g_{n}, g_{m} and g_{p}. It must hold that
Furthermore a vertexbyvertex count of the number of qgons yields a value of g_{q}v_{k}, but each qgon is counted exactly q times so
Therefore in the equation derived from the Euler formula, 2f_{n}+2_{m}+2f_{p}−(k2)v_{k} = 4, the values of f_{q}'s can be replaced by expressions involving the g_{}'s; i.e.,
The g_{q}'s can take on only a limited number of nonnegative values. If θ_{q} is the interior angle of an qgon expressed as a fraction of a full circle then
The value of the θ_{q}'s are given by
q  θ_{q}  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 g_{q}'s are not independent. Therefore g_{m} and g_{p} can be allowed to take on their limited set of values and g_{n} computed as k−(g_{m}+g_{p}). From these values of the g_{q}'s the number of vertices can be computed through
if this yields an integral value of 4 or greater. From v_{k} the number of edges are computed as kv_{k}/2. The numbers of faces, the f_{q}'s, are computed as g_{q}v_{k}/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.
g_{3}  g_{4}  g_{5}  v_{4}  
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 g_{q} for k=4 the values are negative and/or nonintegral. The numbers of faces and edges for the above combinations are shown below.
f_{3}  f_{4}  f_{5}  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.
g_{3}  g_{4}  g_{6}  v_{4}  
3  0  1  12  
2  1  1  24  
f_{3}  f_{4}  f_{5}  e  Name 
12  0  2  24  twisted hexagonal prism 
16  6  4  48  no known polyhedron 
(To be continued.)
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