San José State University
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The Algebraic Determination
of the Structures of Polyhedra

This material deals with the problem of determining which combinations of faces, edges and vertices correspond to actual polyhedra. It does so by considering three types of constraints on these combinations. One type of constraint is equations that the numbers of the components must satisfy. Another is inequalities that they have to satisfy. The equations and inequalities can be satisfied by non-integer values so the requirement that the numbers of the components must be postive integers is another type of constraint. The analysis here will be limited to polyhedra having faces which consist of no more than two types of polygons and vertices of a single degree. The degree of a vertex is the number of edges terminating at that vertex. This set of polyhedra includes the Platonic polyhedra and most of the Archimedean polyhedra.

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-gonal and m-gonal faces, respectively, of the polyhedron. The number of degree k vertices is denoted as vk. The number of edges is denoted as e.

The Equational Constraints

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 in four unknowns to be satisfied; 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 might be 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. These are the specifications of prism-like polyhedron with pentagons at the top and bottom which are connected by triangular faces. (It is not a prism because one pentagonal face is rotated with respect to the other.)

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 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 come 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,

qfq = vkgq
for q = {n, m}.

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 q-gon is counted exactly q times. Thus multiplying the equation Σqgq = k by vk gives Σqqfq. This equation is the same as the count of edges face-by-face count of the edges. Therefore the equation Σqgq = 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 in the constraint on the number of polygons which can meet at a vertex introduces two new unknowns, the gn's, 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 qfq=vkgq it follows that

must be nonnegative integers

Now reconsider the case of n=3, m=5 and k=4 with v4=8. The solution was f3=9 and f1=1. The quantities corresponding to the number of polygons at each vertex for this case would be 3*9/8 and 5*1/8, which are not integers. Thus the integer constraint eliminates this solution.

For this case with v4=10 the solution is f3=10 and f5=2. The numbers of the polygons coming together at a vertex would be 3*10/10=3 and 5*2/10=1. These are integers and a prism-like pentagonal polyhedron does exist.

The general solution for the case n=3, m=5 and k=4 was found previously to be

f3 = v4/2 + 5
f5 = v4/2 - 3

These equations indicate that v4 must be an even number. Furthermore

g3 = 3f3/v4 = 3/2 + 15/v4
g5 = 5f5/v4 = 5/2 − 15/v4

must be nonnegative integers. For the second equation this means

5/2 − 15/v4 ≥ 0
and hence
5/2 ≥ 15/v4
and thus
v4 ≥ 6

But there can be no more than two pentagons comming together at a vertex so g5 ≤ 2 and thus

5/2 − 15/v4 ≤ 2
and hence
1/2 ≤ 15/v4
and therefore
v4 ≤ 30

The integer constraint is sastified v4=6 and v4=30. It is also satisfied for v4=10, but for no other value of v4.

The Integer Constraints Imposed
Upon the Number of Vertices

The general solutions for the numbers of faces were found to be

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

These mean that

gn = nfn/vk = [n(m(k-2)-2k) + 4nm/vk]/(2(m-n))
gm = mfm/vk = [m(n(k-2)-2k) + 4nm/vk]/(2(n-m))

which reduce to

gn = n(m(k-2)-2k)/(2(m-n)) + (2nm/(m-n))/vk
gm = −m(n(k-2)-2k)/(2(m-n)) − (2nm/(m-n))/vk
or, equivalently
gm = m(2k-n(k-2))/(2(m-n)) − (2nm/(m-n))/vk

The Lower Limit to the Number of Polygonal Faces
with the Largest Number of Sides

The requirement that gm≥0 reduces to

vk ≥ 4n/[2k−n(k-2)]

For example, if k=4 and n=3 then v4 ≥ 4*3/(2*4-3*2)=6, as was found previously when m=5. But the lower limit of 6 is independent of m. Thus any polyhedron with triangular faces and all vertices of degree 4 must have 6 or more triangular faces.

on vk
for fm=0

The Upper Limit to the Number of Polygonal Faces
with the Largest Number of Sides

The interior angle of a polygon with m sides when measured as a fraction of a full circle is θm=(1/2−1/m). The maximum number of these that could come together at a vertex is floor(1/θm) where floor(z) denotes the largest integer less than or equal to z. Let this value be denoted as M.

The requirement that gm≤M reduces to

m(2k-n(k-2))/(2(m-n)) − M ≤ (2nm/(m-n))/vk
or, equivalently
vk ≤ [(2nm/(m-n))]/[m(2k-n(k-2))/(2(m-n))−M]
which can be reduced to
vk ≤ 4nm/[m(2k-n(k-2))−2M(m-n)]

For m=5, M=2. Thus for m=5, n=3 and k=4

vk ≤ 4*3*5/[5*(2*4-3*2)-2*2*2]=60/(5*2-8)=60/2=30

As was found previously v4≤30.

Now consider the case of k=5 and n=3, m=5 and M=2 as before. The limits for v5 are:

v5 ≥ 4*3/[2*5-3*2]=12/4=3
v5 ≤ 4*3*5/[5*(2*5-3*3)-2*2*2]=60/1=60

The analysis does not indicate that there are polyhedra corresponding to the derived limits. In this there is no such polyhedron corresponding to the lower limit but there is one corresponding to the upper limit. It is called the snub dodecahedron and has 12 pentagonal faces, 80 triangular faces and 150 edges for its 60 vertices of degree 5.

snub dodecahedron

on vk
563260truncated icosahedron


There are constraints imposed on the number of faces, edges and vertices of polyhedra having two types of faces and vertices all of the same degree by:

These constraints appear to be sufficient to limit the solutions to those values that correspond to the specification of actual polyhedra.

(To be continued.)

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