I was thinking about the Archimedean solids, and pondering how the heck did Archimedes come up with, say, the snub cube or the truncated icosidodecahedron?(Images from Wikimedia Commons, created using Robert Webb’s Stella software. <– Mandatory fine print) Or more generally, how did he come up with all thirteen of the Archimedean solids?
And what are the Archimedean solids? They’re polyhedra, as you can presumably figure out from the pictures; like the Platonic solids (or regular polyhedra), they are convex polyhedra, their faces are regular polygons, and all the vertices are congruent. Unlike the Platonic solids, their faces are not congruent. The prisms and antiprisms (take two congruent polygons with unit length sides, place them in parallel planes a unit apart, and connect each vertex of one to the corresponding vertex of the other, making squares all around; that’s a prism. Do the same but rotate one of the polygons and put it a little closer, so you can connect each vertex of one polygon to two vertices of the other with unit length lines, making equilateral triangles all around; that’s an antiprism. There’s an infinite number of each, corresponding to all possible polygons — though a square prism is just a cube, and a triangular antiprism is an octahedron.) satisfy these requirements but are regarded separately. And the (deep breath) elongated square gyrobicupola or pseudorhombicuboctahedron satisfies them too, but isn’t usually counted: Archimedes didn’t (he may not have known about it), and its vertices are congruent but not transitive — which means it lacks the kind of rotational symmetry the other Archimedean solids (and the prisms and antiprisms) have. There turn out to be thirteen Archimedean solids, and Archimedes enumerated them all.
How? Well, he was a friggin genius, wasn’t he?
But also, you don’t have to pull them out of thin air. You can construct them all using the regular polyhedra as a starting point. I’m not saying it’s easy, but you can.
What’s easier, and gives you the general idea of how it would work for polyhedra, is to do the analogous thing with tilings of the plane — which, in a way, are like degenerate polyhedra.
This tiling has 4 squares around each vertex, so we’ll call it 4444. Now we truncate the vertices. What that means is to remove each vertex and replace it with a polygon whose vertices lie on the edges that met at that vertex, like so:
In this case since there are 4 edges at each vertex, the polygon is a square. If you make the new squares just the right size, then what’s left of the original squares is regular octagons. The result is:
A uniform, but not regular, tiling using octagons and squares — all convex regular polygons, and all the vertices are congruent, but the polygons aren’t. There are 2 octagons and a square at each vertex, so we call this 884.
If you make the new squares larger and larger, then the diagonal edges of the octagons get longer and longer and the vertical and horizontal edges get shorter and shorter, and eventually (when the new squares’ vertices are the midpoints of the edges of the original squares), the vertical and horizontal edges of the octagons vanish and the octagons become squares… which are congruent to the new squares, so you end up with:
Which is just the 4444 tiling we started out with, rescaled and reoriented. This operation is known as rectification.
This is a tiling, but not a uniform one. For one thing the triangles are not regular — they aren’t equilateral. For another thing the vertices aren’t all congruent. Truncation similarly doesn’t work out.
So let’s move on. Another regular tiling is the beastly 666:
but that’s dual to 666 and it turns out we can’t get anything from 333333 we can’t get from 666.
Two dodecagons and a triangle at each vertex, so it’s 12-12-3.
Which is 6363.
Well, that’s not good. The hexagons become regular dodecagons, but then the triangles become irregular hexagons, and the vertices become rectangles, not squares. But! Unlike the rectified 884, this one can be salvaged: push things around a bit and the rectangles can be made square, the hexagons regular:
And that’s uniform, 12-4-6.
Now for something completely different. Look at 884. What happens if you remove every other vertex of each octagon, thereby turning the octagons into squares? Leave the original squares alone. Like so:
You create lozenge-shaped gaps between the squares. They could be divided on a diagonal to make two very non equilateral triangles. That would be a tiling but not a uniform one.
But again, you can adjust the positions and sizes and angles of the squares to make the triangles come out equilateral. The result is this:
This is 43433.
This is 63333. Alone of the uniform tilings, it is not mirror symmetric. That is, there’s another form of it which can’t be obtained from the first by translation or rotation:43433 and 63333 are called snub tilings. Another way to think of them is you start with the regular tilings 4444 or 666 and pull them apart in all directions, creating gaps between the squares or hexagons, which you then fill with triangles, twisting the squares or hexagons as needed to make them fit.
It’s 44333. You can also think of this as 333333 pulled apart in the vertical direction with the gaps filled in with squares. It’s a planar analog of antiprisms and prisms, kind of.
That’s the lot. There are other transformations you can apply to create a new tiling from an existing one, and we haven’t tried applying all the transformations discussed here to all the tilings, but if you do you get either non uniform tilings or uniform tilings we’ve already found.
Left to the reader as an exercise: Do the same thing starting with regular polyhedra. Construct all the Archimedean solids. Extra credit: Uniform honeycombs of 3-space and uniform 4-polytopes.