The Pyraminx is shaped like a regular tetrahedron: it has four faces. The Rubik’s Cube, being a cube ‘n’ all, has six faces. Four faces… six faces… what about five faces? What would a five-faced puzzle look like?

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### A polyhedron with five faces?

We know that there are only five convex regular polyhedra — the Platonic solids. None of them have five faces. Nor do any of the thirteen Archimedean solids.

Does a polyhedron with five faces even exist? Yes of course. Starting with any polyhedron, you can always increase the number of faces by one, simply by snipping off one vertex. Snip a vertex off a tetrahedron, and what you’re left with is a polyhedron with five faces.

But it isn’t a very regular polygon. You want a degree of symmetry and regularity in your puzzles. This raises some interesting questions:

- What kinds of five-faced polyhedra are there?
- Which of them is the most regular?

In order to answer these questions, we need some understanding of topology.

### Topology

Start with a cube. Squash it or stretch it along one axis, and you end up with a square prism. Squash it or stretch it along two axes and you get a rectangular prism. Squash or stretch it off-axis in various ways, and you can end up with all sorts of rhombohedrons, parallopipeds and general cuboids.

You can make all sorts of shapes by distorting a cube, but they’ll all have six faces, twelve edges, and eight vertices. Each face will always be a quadrilateral, and there will always be three edges meeting at each vertex.

Like the cube, the pentagonal pyramid also has six faces, but no matter how you deform a cube, you can never turn it into a pentagonal pyramid, because a pentagonal pyramid has a completely different *topology*. Like the cube, it has six faces, but it has only ten edges and just six vertices. Its faces are triangles and pentagons, and five edges meet at its apex vertex.

You can’t change the number of edges or vertices by deforming a cube, so you can’t deform a cube into a pentagonal pyramid.

So when we ask how many five-sided polyhedra there are, there are two possible answers: we could count every possible deformation, in which case there would be an infinite number, amongst which there would be a great many notable variants.

Or we can do the sensible thing and count polyhedra with distinct topology.

### Remember duality

Before we start counting polyhedra, let’s remember the implications of polyhedron duality: every polygon with five faces is dual to a polygon with five vertices, and vice versa. The problem of finding a polygon with five faces is therefore equivalent to the problem of finding a polygon with five vertices. The latter, however, is in some ways easier to conceptualise. So for now, instead of trying to count polyhedra with five faces, we’ll count their duals: polyhedra with five vertices.

### Polyhedral graphs

The first step to counting polyhedron topologies is to find a way to abstract out the topological information we want — to capture the topological information without the geometrical details. The standard way to do this is to make a graph with a node for each vertex, and an edge between nodes wherever an edge runs between the corresponding vertices. Such a graph is known as a *1-skeleton*, but in this context we can just call it a *skeleton*. Here are the skeletons of a tetrahedron, an octahedron and a cube:

The cool thing about these skeletons is that the cube, the square prism, the rectangular prism, the parallelopiped, the rhombohedron, the cuboid, and every other deformation of the cube you can imagine, all have the same skeleton, but the skeleton of the pentagonal pyramid is different. To generalise: topologically equivalent polyhedra share the same skeleton, but topologically distinct polyhedra have distinct skeletons.

So we can count topologically distinct polyhedra simply by counting skeletons. Is it really that simple? The difficulty is, not every imaginable graph represents a polyhedron. We can’t just start drawing random graphs and counting them. We need to know which graphs to count. We need to know which graphs are the skeletons of polyhedra. This problem was solved in 1922 by a mathematician named Ernst Steinitz. Steinitz proved that the 1-skeletons of the polyhedra are precisely the 3-vertex-connected planar graphs.

Okay, that sounded like gobbledegook, but stick with me for a moment because this is pretty cool.

Let’s start with the “planar graph” bit. A planar graph is a graph that can be drawn on a piece of paper without any edges crossing. Did you get that? No matter how complicated your three-dimensional polyhedron, there is always some way to draw its 1-skeleton as a two-dimensional graph without any crossing edges. Is math awesome or what! Here’s an icosahedron and its skeleton:

A 3-vertex-connected graph is a graph that can lose up to two of its vertices (and all the edges connected to them) and it will still be in one piece. It might be possible to break it in two by deleting three or more of its vertices, but two is not enough.

Now put them together. What Steinitz proved was that every 3-vertex-connected planar graph is the 1-skeleton of a convex polyhedron, and *vice versa*. So if we want to count polyhedra with five vertices, we simply count the 3-vertex-connected planar graphs with five vertices (take my word for it, these are much easier to count).

I shall delay no more: there are two 3-vertex-connected planar graphs with five vertices. Here they are:

Therefore there are two topologically distinct convex polyhedra with five vertices. They are the square pyramid and the triangular bipyramid:

And finally, by duality, there are two topologically distinct convex polyhedra with five faces. These are the duals of the polyhedra with five vertices. The square pyramid is its own dual. The dual of the triangular bipyramid is the triangular prism:

So are there five-sized twisty puzzles? I’ve never seen a square pyramid puzzle but there is indeed a triangular prism puzzle going around: the Pentahedron.

### Counting polyhedra

So there are two topologically distinct polyhedra with five faces. Two is a pretty manageable number. It leaves us thinking maybe there aren’t all that many polyhedra. We might even start fantasizing about a twisty puzzle of every shape.

Sorry, but the number of polyhedra increases super-exponentially with the number of vertices or edges. There is only one polyhedron with four vertices and four edges: the tetrahedron. There are only two polyhedra with five vertices (and therefore also only two with five faces). But there are seven polyhedra with six vertices, and 34 with seven vertices. Thereafter the sequence goes 257, 2606, 32300, 440564, 6384639, 9626938… you get the point.

Thus we abandon the vain hope of a world with twisty puzzles of every shape.

### The “best” five-sided polyhedron

If I can’t have a puzzle of every shape, can I at least have a puzzle that is the “best” shape for each number of vertices or faces? If I can only have one five-sided twisty puzzle, what should it be? This is a new problem, and one that is not easy to frame.

One might be tempted to define the “best” polyhedron as the one that distributes the vertices as far apart as possible. Imagine the vertices are positive charges that repel each other but are constrained to remain on a sphere. Where would they settle? This framing of the problem is known as Thomson’s Problem, and it is extremely difficult to solve. It’s also not a very useful framing of the problem. Though it seems like an eminently sensible definition, it fails the smell test by declaring the square antiprism a superior puzzle shape to the cube. This is clearly not the case, and an attempt to understand why brings us once again back to isometries: the square antiprism has only eight isometries, whereas the cube has 24. This — the number of isometries — must be our true measure of the “best” *n*-sided polyhedron.