## Virtual insane inverse hole

Okay, so if I’m not doing J91/Z4 in Antiprism anytime soon, how come I’ve gone ahead and done Z4?

off_align -F J2,5,0,5 J63 > piece1.off
off_align -F piece1.off,5,6,9 J91 | off_util -M b -M a -x V | \
off_align -F piece1.off,8,6,6 | off_util -M b -M a -x V | antiview

I checked out Adventures Among the Toroids again. Not sure when the library stopped stamping due dates in books but the last and (I think) only stamped date after the book’s acquisition in 1982 was June 2, 2008, which presumably was me. It’s mine again for the next year if it’s not recalled.

## Virtual regular hole

Way way back, before there was a Mathematrec, on my other blog I posted about a book titled Adventures Among the Toroids: a study of Quasi-Convex, Aplanar, Tunneled Orientable Polyhedra of Positive Genus having Regular Faces with Disjoint Interiors, being an elaborate Description and Instructions for the Construction of an enormous number of new and fascinating Mathematical Models of interest to Students of Euclidean Geometry and Topology, both Secondary and Collegiate, to Designers, Engineers and Architects, to the Scientific Audience concerned with Molecular and other Structural Problems, and to Mathematicians, both professional and dilettante with hundreds of Exercises and Search Projects many completely outlined for Self-Instruction (Revised Second Edition), by B. M. Stewart, and I showed a picture of a paper model I’d made of one of Stewart’s toroidal polyhedra, designated Q32/S3S3. Today after much trial and even more error, I figured out how to draw that same polyhedron using the open source Antiprism software. The command is

off_align -F oct,0,0,0 oct | off_align -F J27,2,0,3 | \
off_util -D f19,8 | antiview


(which stacks and merges two octahedra, stacks them inside a triangular orthobicupola and subtracts them from it, removes the two additional coincident faces which off_align doesn’t, and then displays the result). If you don’t have Antiprism but you have something that’ll display OFF files, here it is: https://drive.google.com/open?id=1M_a474kL95XMdTl4qHHfH6AfCrm2Tnfw. And if you have neither, here’s a static image of it:And no, I am not planning on doing the same for J91/Z4 anytime soon. Feel free to take that on yourself.

## How’d you get there, Archimedes

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.

For instance, start with a regular tiling of the plane — one with congruent convex regular polygons and congruent vertices, such as:

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.

What happens if you truncate or rectify 884? Under rectification the octagons become smaller, re-oriented octagons; the squares become smaller, re-oriented squares, and the vertices become triangles:

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:

Of course there is also 333333:

but that’s dual to 666 and it turns out we can’t get anything from 333333 we can’t get from 666.

Truncating 666 turns the hexagons into dodecagons, with the vertices replaced by triangles:

Two dodecagons and a triangle at each vertex, so it’s 12-12-3.

If we rectify 666, we still get triangles at the vertices but the hexagons become smaller, re-oriented hexagons:

Which is 6363.

Let’s start with that, now, and try truncating it. (The resulting polygons get pretty small on the same scale, so I’ll enlarge it.)

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.

Likewise if you rectify 6363 the vertices become rectangles, but you can move stuff around to make them into squares, like this:

6434.

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.

You can do a similar thing with 12-12-3. Alternate the dodecagons to make them hexagons, leaving the triangles alone; divide the remaining space into triangles; adjust to make the polygons regular:

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.

There’s one more uniform tiling and you could think of it as a sort of half snub: Take 4444 and pull it apart in the vertical direction only, filling in the horizontal gaps with triangles:

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.

## Getting rid of triangles, and other ways to Euler

That proof that any convex polyhedron made of pentagons and hexagons must contain at least 12 pentagons rests on Euler’s polyhedron formula: If a convex polyhedron has $V$ vertices, $E$ edges, and $F$ faces, then

$V-E+F=2$

So how is that proved? Lots of ways, some of which rely on the Jordan curve theorem, so then you need a proof of that. An outline of one that doesn’t use the Jordan curve theorem, Triangle Removal, is as follows.

First, transform the polyhedron into a planar graph. You can think of this as removing one face of the polyhedron, and then stretching the boundaries of that face and deforming the other edges until all the edges lie in a plane. Or you can think of it as putting a light source near one face and a plane on the opposite side: the shadow cast onto the plane is the planar graph. This graph has the same number of vertices and edges as the polyhedron, and apparently one fewer face — until you realize the entire plane exterior to the graph can be thought of as a face too.

Okay, we have our graph. Now do the following: Find any interior face that has more than three edges, and draw a diagonal through it from one vertex to another. Doing that increases the number of edges by 1, and divides one face into two, but leaves the vertices unchanged. That means $V-E+F$ is unchanged. Repeat that until all the interior faces are triangles.

Now do this:

1. Identify a triangle that has two edges on the border of the graph. If you can’t find one, go to step 2. If you can find one, remove those two edges. This eliminates two edges, one face, and one vertex, so again $V-E+F$ is unchanged. Now repeat step 1.
2. Identify a triangle that has one edge on the border of the graph. Remove that edge. This eliminates one edge, one face, and no vertices, so again $V-E+F$ is unchanged. Now repeat step 1.

Keep repeating as long as you can. Here I get hand wavy, because I’m less than clear on the argument, but: You don’t want to do step 2 if you can do step 1, because that leads to dead ends — specifically, a graph broken up into two disconnected graphs. Always do step 1 if you can, step 2 only if you can’t do step 1, and the graph will stay connected.

For instance, starting from a (kinda irregular) square pyramid. Here it’s turned into a planar graph with one quadrilateral and three triangular faces (the fourth triangle mapping to the exterior of the graph). The quadrilateral is divided into two triangles, and then triangles are removed one at a time.

Since the graph stays connected, the process continues until there’s only one triangle left. At this point we stop and take a tally: There are 3 vertices, 3 edges, and 2 faces (the interior of the triangle, and the exterior). So $V-E+F = 2$, and since every step we took preserved the value of $V-E+F,$ this proves $V-E+F=2$ for the original polyhedron.

Something about that proof appeals to me, even though there’s that one part I’m hazy on.

The Sum of Angles proof I find clearer, but less cute. Here again you turn the polyhedron into a planar graph with straight edges. Now add up the interior angles of all the faces (including the exterior face). For face $i$, which has $k_i$ edges, the sum of the interior angles is $(k_i-2)\pi$. Summing over all the faces,

$\Sigma_{i=1}^F(k_i-2)\pi = \left(\Sigma_{i=1}^Fk_i-\Sigma_{i=1}^F2\right) = (2E-2F)\pi$

But we get the same result if we sum the angles meeting at each vertex. For each interior vertex, the sum is $2\pi$, adding up to $2\pi(V-k)$ where $k$ is the number of vertices on the border. For the vertices on the border, it’s $2(\pi-\theta_i)$ where $\theta_i$ is the exterior angle at the $i$th vertex of the exterior polygon, and the 2 is because we include the (equal) contributions of the internal and external polygons. But the sum of the exterior angles of a polygon is $2\pi$, so the exterior vertices contribute a total of $2\pi k-4\pi$. Altogether, then, the sum is $2\pi V-4\pi$.

So $(2E-2F)\pi = 2\pi V-4\pi$, or $E-F = V-2$,

Either way, no Jordan curve theorem required! Yay.

## Hexagonal filth

I was amused by this report of a petition to change the representation of soccer balls on UK signs. At present the image is of a ball divided into brown and white hexagons. At least on one side.

The problem is, if it’s hexagons all the way around, it’s a geometric impossibility. You can’t have a convex polyhedron consisting solely of hexagons, regular or otherwise. In fact: If a convex polyhedron has only hexagons and/or pentagons as faces, then there must be at least 12 pentagons.

The proof is based on Euler’s polyhedron formula: If a convex polyhedron has $V$ vertices, $E$ edges, and $F$ faces, then

$V-E+F=2$.

For our hex-and-pent-agon polyhedron, consisting of $h$ hexagons and $p$ pentagons, $F = p+h$, of course. What’s $E$? Well, each hexagon has 6 edges and each pentagon has 5, but that counts each edge twice since each edge belongs to two faces. So $E = (5p+6h)/2$.

Now, at each vertex, at least 3 edges meet. If you add up the number of edges at each vertex over all vertices you get at least $3V$; but since each edge contributes 2 to that total, $3V \le 2E$.

But from Euler’s formula,

$3V = 6+3E-3F \le 2E$

or

$6-3F \le -E$

and substituting our formulas for $F$ and $E$,

$6-3p-3h \le -(5p+6h)/2$

or

$12-6p-6h \le -5p-6h$

or

$12-p \le 0$

or

$p \ge 12$.

QED.

(The equality holds if every vertex has 3 edges — or equivalently, 3 faces. In particular, if you’re using only regular hexagons and pentagons, then there isn’t room at a vertex for more than 3 faces to meet there. So for a polyhedron made of regular hexagons and pentagons, or any polyhedron with exactly three hexagons and/or pentagons meeting at each vertex, $p = 12$.)

Hence the government must be petitioned. As The Aperiodical says, “Ban this hexagonal filth!”