San José State University |
---|
applet-magic.com Thayer Watkins Silicon Valley & Tornado Alley USA |
---|
Regularities of the Platonic and Archimedean Polyhedra |
The Platonic polyhedra are those with a single type of polygonal face and with the same number of edges meeting at each vertex. This latter number is called the degree of a vertex. The Archimedean polyhedra have more than one type of polygonal faces but the vertices are all of a single degree. They are sometimes referred to as convex, by which is meant that they are the surface of convex polyhedral solids. The data for the Platonic and Archimedean polyhedra are given in the table below ordered first by the number of vertices and then by the degree of the vertices.
v | k | n | f_{n} | m | f_{m} | p | f_{p} | e | Name |
4 | 3 | 3 | 4 | 6 | tetrahedron | ||||
6 | 4 | 3 | 8 | 12 | octahedron | ||||
8 | 3 | 4 | 6 | 12 | cube | ||||
12 | 3 | 3 | 4 | 6 | 4 | 18 | truncated tetrahedron | ||
12 | 4 | 3 | 8 | 4 | 6 | 24 | cuboctahedron | ||
12 | 5 | 3 | 20 | 30 | icosahedron | ||||
20 | 3 | 5 | 12 | 30 | dodecahedron | ||||
24 | 3 | 3 | 8 | 8 | 6 | 36 | truncated cube | ||
24 | 3 | 4 | 6 | 6 | 8 | 36 | truncated octahedron | ||
24 | 4 | 3 | 8 | 4 | 18 | 48 | rhombicuboctahedron | ||
24 | 5 | 3 | 32 | 4 | 6 | 60 | snub cuboctahedron | ||
30 | 4 | 3 | 20 | 5 | 12 | 60 | icosidodecahedron | ||
48 | 3 | 4 | 12 | 6 | 8 | 8 | 6 | 72 | truncated cuboctahedron |
60 | 3 | 3 | 20 | 10 | 12 | 90 | truncated dodecahedron | ||
60 | 3 | 5 | 12 | 6 | 20 | 90 | truncated icosahedron | ||
60 | 4 | 3 | 20 | 4 | 30 | 5 | 12 | 120 | rhombicosidodecahedron |
60 | 5 | 3 | 80 | 5 | 12 | 150 | snub icosidodecahedron | ||
120 | 3 | 4 | 30 | 6 | 20 | 10 | 12 | 180 | truncated icosidodecahedron |
where:
There are alternative names for some of the polyhedra. The truncated octahedron is sometimes called the greater rhombicuboctahedron and the truncated icosidodecahedron is also called the greater rhombicosidodecahedron. The cube, to be consistent with the naming of the other polyhedra, could be called the hexahedron.
The most notable regularities in the above table are that the numbers of edges are multiples of six and the numbers of faces and vertices are all even. However there is the triangular prism which has two triangles and three squares which has nine edges, a number not divisible by six or two. The polygonal prisms are polyhedra that satisfy the conditions defining the Archimedean polyhedra but are generally not included simply because they are so simple and constitute an infinite set. The cube happens to be a member of that set. For more on such polyhedra see the Prosaic Polyhedra.
The Euler-Poincare characteristic Χ for a geometric structure is f-e+v, where f, e and v are the number of its faces, edges and vertices, respectively. The faces are two dimensional, the edges one dimensional and the vertices zero dimensional. Letting f_{q} denote the number of faces with q edges and v_{j} the number of vertices of degree j, then
The value of q must be greater than or equal to 3.
The regular convex polyhedra all have an Euler-Poincare characteristic of 2. Thus one condition satisfied by them is
A face-by-face count of the edges yields a value of Σ_{q}qf_{q}, but this is equal to 2e because each edge is counted exactly twice since it is the edge of two faces. Thus another condition satified by the polyhedra is
A vertex-by-vertex count of the edges yields a value of Σ_{j}jv_{j}, but this is equal to 2e because each edge is counted exactly twice since it impinges upon exactly two vertices. Thus another condition satified by the polyhedra is
The first condition involves the term Σ_{j}v_{j}−e. This term can be constructed from the third condition as
If the first equation is multiplied by two and the above expression substituted in, the result is
The second and third condition can be combined to yield the equation
Thus the quantity e has be eliminated from the equations. Subtracting the first of the above two equations from the second yields
But Σ_{j}v_{j} is just v so the above equation can be put into the form
This means that if the number of vertices is changed by an amount Δv then the changes in the number of faces Δf_{q} have to satisfy the condition
For two polyhedra with the same number of vertices but a different degree for the vertices, the changes in the number of faces have to satisfy the condition
For example, consider the change from an icosahedron which has 20 triangles and 0 hexagons to a cuboctahedron which has 8 triangles and 6 squares. The changes are then Δf_{3}=−12 and Δf_{4}=+6. Then
For polyhedra made up of n-gons and m-gons
Thus the tradeoff, so to speak, between n-gons and m-gons is (m-2)/(n-2). Thus for m=4 and n=3 that tradeoff is 2/1 and hence to get 6 more squares the number of triangles has to be reduced by 2 times 6 or 12.
For polyhedra having vertices of a single degree k it holds that
This means that for polyhedra whose vertices are of an odd degree (3 or 5) it must be that the number of vertices is an even number. It also happens that even for the Platonic and Archimedean polyhedra with vertices of an even degree the number of vertices are even, but it would be desirable to have a mathematical proof rather than the results of examining all of the cases.
As of yet there is no proof that the number of edges of the Platonic and Archimedean polyhedra must be divisible by six; they just are. As was noted previously the triangular prism illustrates that the divisibility of the number edges by six does not hold for all polyhedra, even those with just two types of faces and vertices all of a single degree.
(To be continued.)
If k edges meet at a vertex then k polygon faces meet at that same vertex. Let g_{q} be the number of q-gons impinging upon a vertex. Then
The interior angle θ_{q} of a q-gon, measured as fraction of a full circle, is (1/2−1/q). The constraint on the g_{q}'s is that the sum of the interior angles of the polygons meeting at a vertex must be less than a full circle; i.e.,
A vertex-by-vertex count of the q-gons will give a figure of g_{q}v_{k}. But each q-gon has q vertices so it is counted q times. Thus
Since g_{q} must be an integer these equations impose a constraint upon the values of the f_{q}'s.
The values of the g_{q} for the Platonic and Archimedean polyhedra are given in the table below. The values are determined from the above formula, but can be verified by a simple count in the image which can be accessed by clicking on the name of the polyhedron.
v | k | n | g_{n} | m | g_{m} | p | g_{p} | e | Name |
4 | 3 | 3 | 3 | 6 | tetrahedron | ||||
6 | 4 | 3 | 4 | 12 | octahedron | ||||
8 | 3 | 4 | 1 | 12 | cube | ||||
12 | 3 | 3 | 1 | 6 | 2 | 18 | truncated tetrahedron | ||
12 | 4 | 3 | 2 | 4 | 2 | 24 | cuboctahedron | ||
12 | 5 | 3 | 5 | 30 | icosahedron | ||||
20 | 3 | 5 | 3 | 30 | dodecahedron | ||||
24 | 3 | 3 | 1 | 8 | 2 | 36 | truncated cube | ||
24 | 3 | 4 | 1 | 6 | 2 | 36 | truncated octahedron | ||
24 | 4 | 3 | 1 | 4 | 3 | 48 | rhombicuboctahedron | ||
24 | 5 | 3 | 4 | 4 | 1 | 60 | snub cuboctahedron | ||
30 | 4 | 3 | 2 | 5 | 2 | 60 | icosidodecahedron | ||
48 | 3 | 4 | 1 | 6 | 1 | 8 | 1 | 72 | truncated cuboctahedron |
60 | 3 | 3 | 1 | 10 | 2 | 90 | truncated dodecahedron | ||
60 | 3 | 5 | 1 | 6 | 2 | 90 | truncated icosahedron | ||
60 | 4 | 3 | 1 | 4 | 2 | 5 | 1 | 120 | rhombicosidodecahedron |
60 | 5 | 3 | 4 | 5 | 1 | 150 | snub icosidodecahedron | ||
120 | 3 | 4 | 1 | 6 | 1 | 10 | 1 | 180 | truncated icosidodecahedron |
For polyhedra of no more than two types of faces, n-gons and m-gons, and vertices all of the same degree k the constraining equations can be reduced to a set of two equations in three unknowns, f_{n}, f_{m} and v.
These two equations can be solved for f_{n} and f_{m} as linear functions of v. These take the form of
where r, s and t are integers which are a function of m, n and k. Specifically s=2(m-n), r=mk-2(m+k) and t=−nk+2(n+k). Expressions of the form 2(m+k)−mk occur often enough to define a function φ(x,y)=2(x+y)−xy.
Thus the solutions take the form
It is presumed that m is the larger of m and n. Since the number of m-gons at a vertex, g_{m}=mf_{m}/v, must be greater than or equal to zero
It is notable that this lower limit for v depends only upon n and k and is independent of m.
Let M be the maximum number of m-gons that can meet at a vertex. Then
So, for any combination of m, n and k the allowable number of vertices is limited to a particular range from 4n/φ(n,k) to 4n[(M/m)−φ(n,k)].
Within that range the allowable values of v are those such that
This is equivalent to
This reduces to
It is also requiered that f_{n} and f_{m} be nonnegative.
Consider the case of n=3, m=5 and k=4. For j=0, 1, 2 the values of v are 6, 10 and 30. These correspond to the octahedron, the skewed pentagonal prism and the icosidodecahedron, respectively. For all higher values of j the computed values of v are negative. Thus the only allowable values of v are 6, 10 and 30.
For the case of n=3, m=5 and k=3 the only nonnegative values of v generated for integral values of j are v=4 and v=20. These are the case of the tetrahedron and the dodecahedron.
For n=3, m=6 and k=3 the allowable values of v are 4, 6 and 12. The value of v=6 corresponds to the cases of the tetrahedron and v=12 to the truncated tetrahedron, but the case of v=6 is a puzzle. It would have to have 9 edges, which is not an even number. It should have 4 triangles and one hexagon. The one hexagon would require all of the vertices and 6 of the edges. A figure with those specificatitions can be constructed but it is simply a flat hexagon with three pairs of vertices connected and not a polyhedron.
The table below gives the allowable number of vertices for the various possible combinations of n, m and k. The symbol {. . .} denotes a set and {_} denotes the empty set.
n | m | k | Allowable v |
3 | 4 | 3 | {4, 6, 8} |
3 | 4 | 4 | {6, 8, 12, 24} |
3 | 4 | 5 | {12, 24} |
3 | 5 | 3 | {4, 20} |
3 | 5 | 4 | {6, 10, 30} |
3 | 5 | 5 | {12, 60} |
3 | 6 | 3 | {4, 6, 12} |
3 | 6 | 4 | {6, 12} |
3 | 6 | 5 | {12} |
3 | 8 | 3 | {4, 24} |
3 | 10 | 3 | {4, 60} |
4 | 5 | 3 | {8, 10, 20} |
4 | 5 | 4 | {_} |
4 | 6 | 3 | {8, 12, 24} |
4 | 6 | 4 | {_} |
5 | 6 | 3 | {20, 30, 60} |
5 | 6 | 4 | {_} |
As illustration of the analysis behind the above table take the case of n=3, m=4 and k=4.
Integer j | Vertices v_{4} | f_{3} | f_{4} | e | polyhedron |
0 | 6 | 8 | 0 | 12 | octahedron |
1 | 8 | 8 | 2 | 16 | twisted cube |
2 | 12 | 8 | 6 | 24 | cuboctahedron |
3 | 24 | 8 | 18 | 48 | rhombicuboctahedron |
The twisted cube has a square at the bottom and a square at the top which is turned 45° with respect to the bottom square and with the top and bottom squares connected with triangles. The fact that there are exactly 8 triangles for each polyhedron illustrates an analog of Euler's Five Pentagon Theorem . In this case, any polyhedron whose faces are either triangles or squares and whose vertices are all of degree four must have exactly eight triangular faces.
Another interesting case of this sort is the polyhedra whose faces are either squares or hexagons and with degree 3 vertices.
Integer j | Vertices v_{3} | f_{4} | f_{6} | e | polyhedron |
0 | 8 | 6 | 0 | 12 | cube |
1 | 12 | 6 | 2 | 18 | hexagonal prism |
2 | 24 | 6 | 8 | 36 | truncated octahedron |
Such polyhedra must have exactly 6 square faces.
To show that the integers generating the number of vertices are not necessarily sequential consider the case of n=3, m=5 and k=3.
Integer j | Vertices v_{3} | f_{3} | f_{5} | e | polyhedron |
0 | 4 | 4 | 0 | 6 | tetrahedron |
3 | 20 | 0 | 12 | 30 | dodecahedron |
The other values of the integer j generate nonintegral values for all the variables.
Here is one last illustration, that of polyhedra having faces that are either triangles or pentagons and vertices all of degree 4.
Integer j | Vertices v_{4} | f_{3} | f_{5} | e | polyhedron |
0 | 6 | 8 | 0 | 12 | octahedron |
1 | 10 | 10 | 2 | 20 | twisted pentagonal prism |
2 | 30 | 20 | 12 | 60 | icosidodecahedron |
The numbers and types of faces of the Platonic and Archimedean polyhedra are the nonnegative integer solutions of a set of equations. Some solutions of this sort correspond to the polyhdera which are polygonal prisms. Such prismatic figures have the same regularity as the Archimedean polyhedra. There is only one case in which such a solution did not correspond to a polyhedron. Thus the geometric problem of the existence of particular arrangements of polygonal faces, edges and vertices can be reduced to a numerical one.
HOME PAGE OF Thayer Watkins |