San José State University

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Thayer Watkins
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 On the Algebraic Determination of the Structure of Polyhedra

This is an effort to recast the problem of determining the polyhedra having certain qualitative characteristics from a geometric question into a numerical question. If this can be done it would make geometric analysis much easier because ist is much easier to do an exhaustive search of the numerical possibilies than trying geometrically to find all the possibilities. It is at least plausible that such can be done.

## The Platonic Polyhedra

Consider first the case of polyhedra composed of one type of polygon and all vertices being of the same degree. (The degree of a vertex is the number of edges terminating at that vertex.) This is the case of the Platonic polyhedra.

Let n be the number of sides of the polygon and k the degree of the vertices. Both n and k must be greater than of equal to 3. Let the number of faces, edges and vertices be denoted by f, e and v, respectively.

If the number of edges are counted face-by-face the total count will be nf. But each edge is counted twice so that total is equal to 2e; i.e.,

#### 2e = nf or e = nf/2

A polygon that has n edges also has n vertices. A vertex that has k edges coming together also has k polygons meeting there. A face-by-face counting of the vertices gives a total of nv, but each vertex is counted k times. Therefore

#### kv = nf or v = nf/k

If the polyhedron has no holes then the numbers of faces, edges and vertices must satisfy Euler's formula for polyhedra:

#### f -e + v = 2

Substituting the expressions for e and v in terms of f gives

#### f - nf/2 + nf/k = 2 or, equivalently (1 - n/2 + n/k)f = 2 which reduces to f = 4k/(2k - nk + 2n)

For f to be positive (2k-nk + 2n ) must be positive. Since both n and k must be greater than or equal to 3, the conditions to be satified are:

#### 2k -nk + 2n > 0 n ≥ 3 k ≥ 3

Consider the following display of the values of (2k-nk+2n) for various values of n and k.

The Values of (2m-nm+2n)
n\k3456
33210
420-2-4
51-2-5-8
60-4-8-12

The combinations of n and k for which (2k-nk+2n) is positive are shown in red. As can be seen there are only five combinations where this is so. From the relationships previously derived

#### f = 4k/(2k-nk+2n) e = nf/2 v = nf/k

the characteristics of the Platonic polyhedra can be computed and identified.

 (n,k) faces edges vertices name (3,3) 4 6 4 tetrahedron (3,4) 8 12 6 octahedron (3,5) 20 18 12 icosahedron (4,3) 6 12 8 cube (5,3) 12 30 20 dodecahedron

## The Archimedean Polyhedra

The Greek mathematician Archimedes delineated thirteen polyhedra having a high degree of regularity but less than that of the Platonic polyhedra. These polyhedra have faces of two or more types of polygons rather than a single type as in the case of the Platonic polyhedra. First consider the Archimedean polyhedra having exactly two types of faces and vertices of a single degree. Let n and m stand for the number of edges of the polygonal faces and k for the degree of the vertices. Let fn and fm denote the number of n-gon and m-gon, respectively, faces of the polyhedron. The number of degree k vertices is denoted as vk. Again the number of edges is denoted as e.

Euler's formula must be satisfied; i.e.,

#### fn + fm − e + vk = 2

A face-by-face count of the number of edges gives nfn + mfm but each edge is counted exactly two times so

#### nfn + mfm = 2e

A vertex-by-vertex count of the edges gives kvk but since each edge is counted twice, once at each end point,

#### kvn = 2e

There are thus three equations to be satisfied by the four variables; i.e.,

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

The variable e can be easily eliminated. Euler's formula involves the term vk−e. An expression for this quantity can be derived from the third equation; i.e.,

#### 2(vk-e) = −(k-2)vk

When this expression is substituted into the Euler's equation and the second and third equations are combined the result is the following set of two equations in three unknowns.

#### 2fn + 2fm = (k-2)vk + 4 nfn + mfm = kvk

These equation can be solved for fn and fm in terms of vk. To eliminate fm the second equation can be multiplied by 2 and the first equation by m to get

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

Subtraction of the second equation from the first yields

Likewise

#### fm = [(n(k-2)-2k)vk + 4n]/(2(n-m))

Suppose n=3, m=5 and k=4. Then the solutions reduce to

#### f3 = [5*2-8]v4 + 20]/4 = v4/2 + 5 and f5 = [3*2-8]v4 + 12]/(-4) = v4/2 - 3

These equations indicate that v4 must be divisible by 2. For the icosidodecahedron v4=30 and thus f5=15-3=12 and f3=15+5=20.

icosidodecahedron

Now consider other values for v4. If v4=6 then by the above equations f5=0 and f3=8. Then 2e=4v4=24 and thus e=12. These are the specifications for the octahedron. So v4 can be 6 or 30.

Suppose v4=8. This would mean that f5 would have to be 1 and f3 would have to be 9. The number of edges would have to be 16, arrived at both as one half of 4v4=32 and as one half of (5f5+3f3)=(5+3*9)=32. A polyhedron having these specifications is not easily envisioned. A pentagonal pyramid would use one pentagon, five triangles, ten edges and six vertices. There would be left over four triangles, six edges and two vertices. This is not enough to form a triangular pyramid (a tetrahedron). However even if there were enough parts to form another polyhedron this would be invalid beause the Euler-Poincare characteristic of two disjoint polyhedra would be four. Furthermore the degrees of the vertices of a pentagonal pyramid are 3 and 5 and not 4. It is not possible to say at this point that there is no geometric configuration corresponding to the solution to the constraining equations in which there are 8 vertices all of degree 4 with 16 edges having 1 pentagonal face and 9 triangular face. Yet these quantities satisfy the constraining equations. If such a configuration were to exist it is unlikely to be anything like the intuitive notion of a polyhedron.

This is a very important point. There may be configurations that satisfy the equations which are not those of a simple polyhedron. Thus there is not a one-to-one correspondence between the solutions to the equations and geometric polyhedra. It is possible that there is some extension of the concept of a polyhedron which would establish such a correspondence.

Before leaving this case suppose v4=10. This would mean f5=2 and f3=10. The number of edges would be 20. The numbers of faces and edges are the same as two pentagonal pyramids, but not the number and degrees of the vertices.

For v4=12, f5=3, f3=11 and e=24. Again it is difficult to conceive a polyhedral configuration that would satisfy these conditions.

For v4=14, f5=4, f3=12 and e=28.

## Analogs of the Snub Dodecahedron

snub dodecahedron

The snub dodecahedron is composed of pentagons and triangles with vertices all of degree 5. For this case the analysis would end with the pair of equations

#### 2f3 + 2f5 - 3v5 = 4 3f3 + 5f5 - 5v5 = 0

The solutions for f3 and f5 in terms of v5 are

#### f3 = 5*v5/4 +5 and f5 = v5/4 − 3

The equations thus indicate that v5 must be a multiple of 4.

For the snub dodecahedron the 60 vertices imply that the number of pentagons is 12, but a polyhedron of this type with more vertices of degree 5 would not have just 12 pentagonal faces. The number of triangular faces for the snub dodecahedron is 80 and the number of edges is 150.

Consider the case of v5/4=12. Then f5=0 and f3=20. The number of edges is 30. This is the specification for the icosahedron.

If v5=16 then f5=1, f3=20 and e=40. This is a solution to the equations but it is difficult to conceive of a corresponding geometric configuration.

## Divisibility Requirements for the Number of Vertices

For polyhedra having n-gon and m-gon faces with vertices of degree k the solutions found previously were

#### fn = [(m(k-2)-2k)vk + 4m]/(2(m-n)) and fm = [(n(k-2)-2k)vk + 4n]/(2(n-m))

These can be expressed as

#### fn = p*vk/q + 2m/(m-n) and fm = vk/q − 2n/(m-n)

where it is presumed that m>n. The divisor q is equal to 2(m-n)/[2(n+k)−nk]. The factor p is equal to [mk-2(m+k)]/[2(n+k)-nk]. Thus for m>n, vk must be divisible by q.

For example, if m=5, n=3 and k=5 then q=4/(2*8-3*5)=4 and p=(5*5-2*10)/(2*8-3*5)=5.

For n=5, m=6 and k=3, q=2/(2*8-5*3)=2 and p=(6*3-2*9)/(2*8-5*3)=−1.

For n=3, m=6 and k=3, q=2*3/(2*6-3*3)=2 and p=(6*3-2*9)/(2*6-3*3)=0.

Almost all of the relevant cases are tabulated below.

 n m k q p Example vk 3 6 3 2 0 TruncatedTetrahedron 12 4 6 3 2 0 TruncatedOctahedron 24 5 6 3 2 −1 TruncatedIcosahedron 60 3 4 4 1 0 Cuboctahedron 12 3 5 4 2 1 Icosidodecahedron 30 3 5 5 4 5 SnubIcosidodecahedron 60 3 4 4 1 0 Rhombicuboctahedron 24

## Conclusions

The equations which constrain the number of faces, edges and vertices of different types having solutions which do not appear to correspond to any concievable geometric configuration. Thus there is not a one-to-one correspondence between the solutions to the equations and geometric polyhedra. The equations do indicate in most cases a limitation on the number of vertices; i.e., that the number must be divisible by a certain number which is a function of n, m and k.

(To be continued.)