San José State University
Thayer Watkins
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& Tornado Alley

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
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
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)

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) facesedgesverticesname
(3,3) 464tetrahedron
(3,4) 8126octahedron
(3,5) 201812icosahedron
(4,3) 6128cube
(5,3) 123020dodecahedron

The Interior Angle of Polygons

The interior angle is the complement of what could be called the turning angle although it is usually called the exterior angle.

As one traverses the polygon the direction of movement must turn from 0 to 2π, a full circle. Thus the turning angle at a vertex of a regular n-gon must be 2π/n. The interior angle then must be π−2π/n, which reduces to (1−2/n)π. Thus for n=6 the interior angle is (1-2/6)π or 2π/3. For a pentagon the interior angle is (1−2/5)π or 3π/5.

Another, and more convenient, way of expressing the interior angles of polygons is as a fraction of a full circle. For an n-gon this is (½−1/n). Thus for a triangle it is 1/6; for a square 1/4, for a pentagon 3/10 and for a hexagon 1/3.

Restrictions on the Number of Polygons at a Vertex

The sum of the interior angle of the polygons impinging upon a vertex must be less than one full circle. There cannot be just two polygons coming together at a vertex. Therefore the largest polygon such that three could come together at a vertex is the pentagon. Three hexagons would involve there being a polyhedral angle of one full circle (2π radians or 360°) at a vertex. This would be a flat surface.

If there are k edges coming together at a vertex then there are k polygons impinging upon that vertex. Thus the constraint for the Platonic polyhedra is

k(½−1/n) < 1
or, solving for 1/n
1/n + 1/k > ½

For k=3 this means (1/n)>1/6 and hence n<6. For k=4, (1/n)>(1/4) and hence n<4. For k=5, (1/n)>3/10 and hence n<10/3 so n=3. Altogether then for k=3, n must be in the set {3, 4, 5}, for k=4 or 5 n must be 3. Thus the only combinations (n, k) for the regular polyhdra involving one type of face are five cases found previously. This analysis however introduces the constraint imposed on the number of polygons at a vertex.

The Archimedean Polyhedra

The great 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

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


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
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.


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.

Now suppose v4=10. This would mean f5=2 and f3=10. The number of edges would be 20. These conditions are fulfilled by prismatic polyhedron having pentagons at the top and bottom and triangles connecting them.

Before leaving this case consider a couple more values of v4. 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
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))
fm = [(n(k-2)-2k)vk + 4n]/(2(n-m))

These can be expressed as

fn = p*vk/q + 2m/(m-n)
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.


The Restriction on the Numbers of
Faces Imposed by the Constraint on
the Numbers of Polygons at a Vertex

Let gn be the number of n-gons at a vertex. The general constraint on the gn's is that

Σnθngn < 1

where θn is the interior angle of an n-gon, which in terms of a fraction of a full circle, is (½−1/n).

It is also required that

Σngn = k

where k is the degree of the vertex.

Consider the example of the octahedron. There are 8 triangles and 4 of these coming together at each of the 6 vertices. Each of the triangles at a vertex belongs to three vertices. So 4 times 6 is the number of triangles counted, but this is the same as 8 times 3 for the number of times triangles get counted. Thus when all of the vertices are of a single degree k,

nfn = vkgn

The quantity vkk, which is a vertex-by-vertex count of the edges, is equal to 2e because each edge is counted twice. This quantity is also a vertex-by-vertex count of the polygons but each n-gon is counted exactly n times. Thus multiplying the equation Σngn = k by vk gives Σnnfn. This equation is the same as the count of edges face-by-face count of the edges. Therefore the equation Σngn = k does not add to the number of independent equations.

The general problem in the previous analysis of the Archimedean polyhedra with two types of faces and vertices of only one type is that there ultimately were two equations in three unknowns. Bringing the constraint on the number of polygons which can meet at a vertex introduces two new unknowns but two new equations and one inequality. The new unknowns, the gn's, can easily be eliminated, but the inequality is retained so the system reduces to two equations and one inequality in three unknowns.

Now let us reconsider a solution to the equations for n=3, m=5 and k=4 that does not seem to correspond to any geometric entity. That solution is v4=8, f5=1, and f3=9. From the condition nfn=vkgn it follows that

gn = nfn/vk

The condition concerning the polyhedral angle at a vertex is

(1/6)g3 + (3/10)g5 < 1

If gn is replaced by nfn/vk the LHS of the above becomes

(1/6)(3*9/8) + (3/10)(5*1/8)

This evaluates to 0.75 so the constraint is not violated.

The problem with this solution is that with only one pentagon and each vertex of degree 4 there has to be three triangles at each of the five vertices of the pentagon. That requires 15 triangles but there is only 9. This is an impossible configuration.

The essential constraint is that the values of gn must be integers. For the solution of v4=8, f5=1 and f3=9 this is not the case; i.e., 3f3/v4=3*9/8 and 5f5/v4=5/8.

However for v4=10, f5=2 and f3=10. This would mean g3=3*10/10=3 and g5=5*2/10=1. As noted previously these conditions are fulfilled by prismatic polyhedron having pentagons at the top and bottom and triangles connecting them. Thus a polyhedron with these specifications does exist.


The equations which constrain the number of faces, edges and vertices of different types have 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. But if the constraint on the angle imposed at each vertex by the numbers of the different polygons then the solutions which cannot be achieved geometrically are eliminated.

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. It is the integral value requirements that impose the most severe restrictions on the solutions to the equations.

(To be continued.)

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