Semiregular polyhedron

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A semiregular polyhedron is a geometric shape constructed from a finite number of regular polygon faces with every face edge shared by one other face, and with every vertex containing the same sequence of faces, and, moreover, for every two vertices there is an isometry mapping one into the other.

There are 3 classes of semiregular polyhedra:

• Archimedean solids - 13 polyhedra with more than one polygon face type.
• Prisms - infinite set: 2 N-gons and N-squares.
• Antiprisms - infinite set: 2 N-gons and 2N triangles.

The Platonic solids are a set of 5 regular polyhedra.

These 4 sets compose the convex polyhedra, along with a set of 53 nonconvex forms compose the larger set of uniform polyhedra.

There's one example of a polyhedron with every vertex containing the same sequence of faces, that fails to have, for every pair of vertices, an isometry mapping the first vertex into the second, is the elongated square gyrobicupola.

The regular and semiregular polyhedra are related to the regular and semiregular tilings of the plane. See List of uniform planar tilings.

[edit] Existence of the semiregular polyhedra

The existence of these polyehedra can be enumerated by looking at their vertex configuration and the angle defect: A set of regular faces must have internal angles less than 360 degees.

NOTE: The vertex figure can represent a regular or semiregular tiling on the plane if equal to 360. It can represent a tiling of the hyperbolic plane if greater than 360 degrees.

For uniform polyhedra, the angle defect can be used to compute the number of vertices. (The angle defect is defined as 360 degrees minus the sum of all the internal angles of the polygons that meet at the vertex.) Descartes' theorem states that the sum of all the angle defects in a topological sphere must add to 4*π radians or 720 degrees.

Since uniform polyhedra have all identical vertices, this relation allows us to compute the number of vertices: Vertices = 720/(angle-defect).

Example: A truncated cube 3.8.8 has an angle defect of 30 degrees. Therefore it has 720/30=24 vertices.

In particular it follows that {a,b} has 4/(2-b(1-2/a)) vertices.

Every enumerated vertex configuration potentially uniquely defines a semiregular polyhedron. However not all configurations are possible.

Topological requirements limit existence. Specifically p.q.r implies that a p-gon is surrounded by alternating a q-gons and r-gons, so either p is even or q=r. Similarly q is even or p=r. Therefore potentially possible triples are 3.3.3, 3.4.4, 3.6.6, 3.8.8, 3.10.10, 3.12.12, 4.4.n (for any n>2), 4.6.6, 4.6.8, 4.6.10, 4.6.12, 4.8.8, 5.5.5, 5.6.6, 6.6.6. In fact, all these configurations with three faces meeting at each vertex turn out to exist.

Similarly when four faces meet at each vertex, p.q.r.s, if one number is odd its neighbors must be equal.

[edit] A complete enumeration of convex uniform polyhedra

The number in parentheses is the number of vertices, determined by the angle defect.

Triples

• Platonic solids 3.3.3 (4), 4.4.4 (8), 5.5.5 (20)
• prisms 3.4.4 (6), 4.4.4 (8; also listed above), 4.4.n (2n)
• Archimedean solids 3.6.6 (12), 3.8.8 (24), 3.10.10 (60), 4.6.6 (24), 4.6.8 (48), 4.6.10 (120), 5.6.6 (60).
• regular tiling 6.6.6
• semiregular tilings 3.12.12, 4.6.12, 4.8.8

Quadruples

• Platonic solid 3.3.3.3 (6)
• antiprisms 3.3.3.3 (6; also listed above), 3.3.3.n (2n)
• Archimedean solids 3.4.3.4 (12), 3.5.3.5 (30), 3.4.4.4 (24), 3.4.5.4 (60)
• regular tiling 4.4.4.4
• semiregular tilings 3.6.3.6, 3.4.6.4

Quintuples Finally configurations with five and six faces meeting at each vertex:

• Platonic solid 3.3.3.3.3 (12)
• Archimedean solids 3.3.3.3.4 (24), 3.3.3.3.5 (60) (both chiral)
• semiregular tilings 3.3.3.3.6 (chiral), 3.3.3.4.4, 3.3.4.3.4 (note that the two different orders of the same numbers give two different patterns)

Sextuples

• regular tiling 3.3.3.3.3.3