Archimedean Solids

Archimedes appreciated the beauty and perfection of the platonic solids, but he grew tired of seeing the same five shapes. He wanted a bit more variety. So he developed the Archimedean solids, also called the semiregular 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 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 around. This is called a gonal prism. Each vertex joins an n-gon and two squares. When n = 4 you have the cube.

Next, take the aforementioned 2 n-gons, floor and ceiling, and twist the ceiling just a bit, so that the points of the ceiling float above the edges of the floor. Upward pointing equilateral triangles connect the edges on the floor with the points above. Similarly, downward pointing equilateral triangles connect the edges above with the points below. This is the anti-gonal prism. Every corner joins an n-gon with three triangles, hence the shape is semiregular. An example is shown below for n = 9. The top and bottom rings don't quite line up with the wall, because of the gaps between triangles. Thus the rings are a bit inside, but that's ok.

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. There are 12 pentagons and 20 hexagons. Since there are no triangles, and too many faces to extend (within the limits of 800 buckyballs), I cannot build this shape.

Some semiregular solids come from regular solids under various transformations. Consider the process of truncation. Given a regular solid, take a file and file down the corners until all edges have the same, shorter length. 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 [8,8,3].

Apply this process to the 8 points of an octahedron, giving the truncated octahedron, with formula [6,6,4], wherein each vertex conects two hexagons and a square. An example is shown below. This uses extended hexagons for stability, and is thus a rather large shape. In fact it barely holds its own weight. I had to build it in the air, since a ¾ constructed shape would collapse in on itself. Once complete, I very gently set it down on one of its square faces, and there it is.

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 both the regular solid and its dual. 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 cuboctahedron again.

Perform this operation on the dodecahedron, or the icosahedron, and get the icosadodecahedron. A picture is shown below, using indented cups as pentagons.

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 original 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 you can perform the same operation to the octahedron and get the same shape, hence it is the rhombic cuboctahedron.

Finally we have the snub operation. Start with the cuboctahedron, which consists of squares and triangles. Pull these faces apart, leaving gaps between. 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 more 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. Here is the snub cuboctahedron with indented squares for stability.

There is also a snub icosadodecahedron with 2×30+32 = 92 faces. Here is ¾ of the shape at T9 (looking a bit like a bowl), then the complete shape at T6, probably the roundest thing you can make out of buckyballs.

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. 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? Put various n-gons together at a corner and find out. Start with a lemma.

Suppose each vertex joins a pentagon, a triangle, and two squares, in that order: [5,3,4,4]. Walk around the pentagon and the edges border a triangle, a square, a triangle, a square, a triangle, and then we're in trouble, because the sixth edge is the first edge, and it must border a triangle and a square simultaneously. This is the walk around test, and it forces the two numbers on either side of an odd number to be the same. Thus [5,3,4,4] is invalid, from the perspective of the pentagon or the triangle, but [5,4,3,4] is fine.

Specifically, [x,a…b] is invalid if x is odd, and a ≠ b, and x is bracketed by a and b wherever it appears, as in [x,a,x,b], or [x,x,b], or [x,x,b,b]. [7,3,3,3] is fine, since 3 is bracketed by 7 and 3, or by 3 and 3, and in fact it makes a perfectly good solid, an antigonal prism. Another walkaround failure is [x,x,a…b], where x a and b are distinct, as gons must alternate x, a or b, x, a or b, etc.

To get things started, assume each vertex meets something higher than an octagon. The high shape x consumes 140° or more. Note that x cannot join four other shapes at a vertex, (140 + 4×60 = 380), 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 by the walkaround test. This has positive curvature for x < 12. Yet if x is odd, and greater than 3, x 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 [10,4,6] is valid. 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. (No room for a triangle and two squares.) 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 second shape is a square then so is the third. 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], the cuboctahedron. One square joins 4 triangles to make a snub solid.

Bring in the platonics, and that's all the semiregular solids. They are presented in the table below. You can compute the number of vertices from the pattern, using the Angular Defect Theorem. Thus you can tell in advance whether you have enough buckyballs to build the shape.

I haven't addressed the quasiregular solids, because I don't have enough buckyballs to make the interesting ones, such as the triangle spoked icosadodecahedron.