Building a perfect D20

2003-05-27 21:47

I have a favorite toy. Well, I have tons of them, but my recent love is GeoMags.

Too lazy to follow a link? Imagine two objects. One is a magnet, just about exactly an inch long. The other is a metal ball about half an inch in diameter. The magnet is in a plastic shell that makes the ball snap sharply into place on either end. Now, get a whole lot of these. They stick together. They make pretty shapes.

Among the shapes you can make with them are regular polyhedra – that’s polyhedrons (multi-sided objects) whose sides and faces are all identical. There’s five of them, in total. Tetrahedron, cube, octahedron, dodecahedron, icosahedron. As the names suggest, these are 4, 6, 8, 12, and 20 sided. And they make great dice… So, an icosahedron is sometimes called a d20, mostly by gamers.

You can, in fact, make any of these objects out of GeoMags. For a d20, you need twelve balls and thirty rods. (Freudians, stop here and get the snickering over with, then continue.) No problem; you build this out of GeoMags, and it sticks together, and it’s solid.

But wait! What if it’s too small? Well, you can always use two rods for each side. (Okay, I was wrong; now it’s time to stop snickering. Really.) So, you build an icosahedron using 60 connecting rods, but still 12 balls. Only, this one’s weak; it has a lot of junctions that aren’t very sturdy. So, we add another 60 rods and 30 balls, and we “fill in” the sides; each side has an inverted triangle in it, with balls at the junctions. It looks sorta like this:

 \ / \ /
   \ /

Now, there’s something interesting here. If you actually build one of these, you’ll find that it tends to wobble and fold until you’re almost done – and then it becomes pretty solid.

However, you will most likely discover that it’s still a little weak in spots. To understand why, we must remember that these are magnets. Ever play with magnets much? North ends attach to south ends; north ends repel north ends.

Now, let’s say we’ve got a couple of magnets attached to a ball:


This will actually work! Although the north ends are repelling each other, they’re actually still both able to glue on to the neutral metal ball… However, they won’t attach as strongly as they would if you kept the polarity straight:


Thus, the puzzle: Taking our icosahedron two units on a side, how do we maximize its strength? Casual observation tells us the following:

  1. The connections are strongest when they are perfectly balanced – the same number of north and south poles touching each sphere.
  2. The vertices of the icosahedron, being connected to five magnets, cannot be perfectly balanced.

How close can we come? Casual introspection suggests that the vertices should be able to be within one charge of balanced, and the others should be able to be perfectly balanced. Well, at least, that’s the best we could possibly hope for. Is it really possible?

The naive solution to this involves a 2^120 problem; just count charges on an icosahedron with magnets in a given orientation, then try flipping them at random.

You can do better.

Imagine, if you will, a hypothetical single-unit icosahedron in which each of the 12 vertices is nearly-balanced – it’s either got one excess north, or one excess south. If we extend it to two-unit sides, nothing has changed. If we put joints in the middle of each side, the whole thing becomes foldy and unstable, but the magnetism hasn’t changed, and in fact, each of the middle spheres is balanced… That gets us faces that look like this:

 \     /
  *   *
   \ /

Obviously, since they’re in the middle of pairs of magnets facing the same way, the center joints are balanced. Now, if we add three bars to the middle of this, and ensure that they are all facing the same way, we get back to our “correctly” faced side, and each point remains balanced.

So, if we can build a single-unit icosahedron which is maximally balanced, we can build a two-unit icosahedron which is maximally balanced.

On the single-unit icosahedron, we need each vertex to have two north and two south poles touching it, and then half of them need one extra north, and half of them need one extra south.

One way to pursue this is by construction. Make a pair of triangles in which all the magnets are facing the “same” way (clockwise or counterclockwise). Now, attach two magnets to each vertex of these triangles, one north pole, one south pole. (The orientation of these matters in a way I am too intellectually lazy to figure out.) Attach balls to the six magnets attached to one of the triangles, and join them with six magnets all facing the “same way” (counterclockwise or clockwise)… and attach the magnets from the bottom face to them. Every vertex now has four magnets attached to it, two north, two south. If it doesn’t, then take pairs of magnets attached to one of the two triangles you started with, and swap them. You should eventually succeed. :)

Now… You have an icosahedron missing exactly six magnets, each of which will connect two of the twelve vertices together. Since all the vertices are balanced before you do that, afterwards, each vertex will have precisely one extra north or south magnet attached.

That’s it! The rest is simple construction, and the result is a maximally-strongly-connected icosahedron, with two-unit sides.

If someone sends me a 184-piece white GeoMag set, I’ll write about how to do this with three-unit sides, which I believe to be possible.

Peter Seebach



  1. Remind me around Thanksgiving!
    --linsee (seebs' mom)

    — linsee · 2003-05-28 10:13 · #

  2. I got a cheap geomag imitation for my daughter yesterday.
    I Couldn't wait for her to go to school so I could do all the platonic solids.
    I quickly found out that simple cubes are impossible.
    The rest are not too hard and I am as proud as punch of my 1st icosahedron (d20).

    I used an internal support but was able to remove it leaving a stable object.

    But I can't repeat the trick with the dodecahedron
    . I know it can be done with support from 5 rods + 1 ball per face - I've seen pictures.
    The magnetic forces have to be balanced to stop the structure tearing itself apart. I've not managed to complete one yet.
    Is there another trick or just down to the balancing act or is it because my cheapo copy isn't up to the job.

    BTW a great place to study polyhedra is

    Thanks for an entertaining post.

    — Andrew Wilkins · 2004-09-23 06:56 · #