Regular Patterns, Archimedean Solids

Semiregular (Archimedean) Solids

A semiregular solid consists of n-gons pasted together so that vertices are indistinguishable. This means each corner consists of the same n-gons, in the same clockwise order. The regular solids are semiregular, but there are many more.

Take two copies of an n-gon and make one the floor and one the ceiling. Connect them via n square walls. This is called a gonal prism. Each vertex joins an n-gon and two squares. When n = 4 you have the cube.

Notice that the dual of a semiregular solid need not be semiregular. The dual of a gonal prism produces n triangles on top and n triangles below. The top and bottom vertices connect n triangles; the rest connect four triangles.

Now take the aforementioned n-gons, floor and ceiling, and twist the ceiling just a bit, so that the points of the ceiling are above the edges of the floor. Let equilateral triangles connect the edges on the floor with the points above. Similarly, let equilateral triangles conect the edges above with the points below. This is the anti-gonal prism. Pull the floor and ceiling apart and it looks like the mouth of a wild animal. The triangle teeth point up and down, and mesh perfectly. Every corner joins an n-gon with three triangles, hence the shape is semiregular.

The soccer ball is a semiregular solid with formula 5,6,6. If you have one in your house, go get it and look at it. Every vertex joins a pentagon and two hexagons. Let's compute the number of faces using the earlier procedure. There are v vertices, 3v/2 edges, and v/5+2v/6 faces. Apply Euler's formula and get 60 vertices, 90 edges, and 32 faces - thus 12 pentagons and 20 hexagons.

Just as semiregular tilings often come from regular tilings, so semiregular solids often come from regular solids. Consider the process of truncation. Given a regular solid, take a file and file down the corners until the edges of the new faces are the same length as the edges of the original faces. Let's illustrate with the cube. A cube has 8 corners; file them down to make 8 triangles. The six faces of the cube, which use to be square, have become octagons. This is called a truncated cube. Each vertex connects two octagons and one triangle.

File a regular solid further, until the original edges disappear. The original faces remain, preserving the structure of the regular solid, and the new faces, corresponding to the vertices, reflect the structure of the dual. This is a hybrid shape, named for the regular solid and its dual.

To illustrate, file the corners of the cube down until the original edges shrink to zero. Each corner has become a triangle and each face has become a square standing on its point. This is the cuboctahedron. Start with the octahedron and apply the same process, and get the same shape, a cuboctahedron.

File the corners and edges down simultaneously to get a rounder, or rhombic form of the original shape. File the corners of a cube down to triangles, and the edges of the cube to squares, whence the faces become smaller squares. This shape is 3,4,4,4, and is called the rhombic cuboctahedron. It isn't called the rhombic cube, because we can perform the same operation to the octahedron and get the same shape, hence it is the rhombic cuboctahedron. We don't have to worry about the rhombictetrahedron; that is the same as the cuboctahedron.

Finally we have the snub operation. Start with the cuboctahedron, which consists of squares and triangles. Pull these faces apart, leaving gaps between. Now turn all the faces clockwise just a bit. In other words, each face turns clockwise as you look down on it from outside the shape. If two squares and two triangles use to meet at a vertex, they turn in synchrony to create a diamond gap. Fill this gap with two triangles. The cuboctahedron had 12 vertices, and each creates a diamond gap, thus there are 24 new triangle faces. The original shape had 14 faces, hence the snub cuboctahedron has 24+14 = 38 faces. There is also a snub icosidodecahedron with 2×30+32 = 92 faces. These are presented in the table below.

The snub shapes are the only semiregular solids that exhibit chirality. In other words, they come in left and right handed versions. Turn the faces clockwise, as described above, and find the right handed version (an arbitrary convention). Turn the faces counterclockwise and find the left handed version. Reflect one version through a plane, or through the origin, to find the other.

Have we described all the semiregular solids? Let's put various n-gons together at a corner and see what we can find. But first, a lemma.

The walkaround principle still holds. Thus 7,5,3 is impossible, because an odd shape, meeting two others, forces the others to be equal. Here is a generalization of that rule.

Let each corner have the pattern x,y,z,y where all three variables are different and y is odd. Walk around y back to start and arrive at a contradiction. The same is true of x,y,z,z and x,y,y,x and x,y,y,z. Thus 6,3,3,4 is impossible, even though the angles sum to less than 360°.

When building semiregular solids, we don't have to worry about anything beyond the octagon. Here's why. Let x be a 9-gon or higher, hence it consumes 140° or more. Note that x cannot join four other shapes at a vertex, and if it joins three other shapes, two of them must be triangles. If the third is a triangle we have the anti-gonal prism. If the third is a square, x,3,4,3 and x,4,3,3 both fail the walkaround test. If the third is a pentagon the angles exceed 360°. Thus x meets two other shapes. If one is a triangle the other must be x, which is valid for x < 12. Yet if x is odd, and greater than 3, it fails the walk around test. Therefore the semiregular solid x,3,x is valid for x = 3, 4, 6, 8, 10. That takes care of x meeting a triangle.

Without a triangle, x meets squares or higher. If x meets two squares we have the gonal prism. The shape x,4,5 fails the walkaround test, and x,4,6 fails the walk around test when x = 9 or 11. The pattern x,4,6 is valid when x = 10, another use for the decagon. These shapes consume too much angle when x is 12 or higher. Finally, x,5,5 fails the walk around test, and x,5,6 is too big. Hereinafter, we can restrict attention to octagons and below.

If the octagon meets three other shapes, two of them are triangles. Let the third be a square, to avoid the anti-gonal prism. Yet 8,3,4,3 and 8,3,3,4 fail the walkaround test, so the octagon meets two other shapes. If one of them is odd then the other is an octagon, and that only leaves room for a triangle. This is 8,8,3, the truncated cube. If the octagon meets two even shapes, one of them is a square. To avoid the gonal prism, the other is a hexagon. This is another valid shape. That takes care of the octagon.

If a heptagon, 7 sides, meets three shapes, two of them are triangles. The third creates the anti-gonal prism, or fails the walkaround test, hence the heptagon meets two other shapes. Since 7 is odd, these shapes are equal. They must be even, else we fail the walkaround test. Skip the squares (gonal prism) and move to hexagons, but that introduces too much angle.

If a pentagon meets two other shapes they too are pentagons, or they are copies of an even shape, a square (gonal prism) or a hexagon (soccer ball). If the pentagon meets four other shapes they are triangles, and that is valid. If the pentagon meets three other shapes, at least one is a triangle. If the second is a triangle, the third should not be a triangle, else we have an anti-gonal prism. Note that 5,3,3,4 5,3,4,3 5,3,3,5 5,3,3,6 5,3,6,3 all fail the walkaround test. That leaves 5,3,5,3, which is valid. If the third shape is a square then so is the fourth. Since 5,3,4,4 fails, we are left with 5,4,3,4, which is valid. That's it for the pentagon.

Join two hexagons with a triangle or a square; both are valid. Hereinafter there is at most one hexagon; everything else being squares or triangles.

If a hexagon meets two shapes, one cannot be a triangle, so we're talking about two squares, the gonal prism. If the hexagon meets 3 other shapes, the first two are triangles. Make the third a square, avoiding the anti-gonal prism. Yet 6,3,3,4 and 6,3,4,3 both fail the walkaround test. That's it for the hexagon.

Three squares and a triangle - that works. Two squares and two triangles makes the shape 4,3,4,3. One square joins 4 triangles to make a snub solid.

That's all the semiregular solids. They are presented in the table below.