# Packing tetrahedra

Last spring I saw a great colloquium talk on packing regular tetrahedra in space by Jeffrey Lagarias. He pointed out that in some sense the problem goes back to Aristotle, who apparently claimed that they tile space. Since Aristotle was thought to be infallible, this was repeated throughout the ages until someone (maybe Minkowski?) noticed that they actually don’t.

John Conway and Sal Torquato considered various quantitative questions about packing, tiling, and covering, and in particular asked about the densest packing of tetrahedra in space. They optimized over a very special kind of periodic packing, and in the densest packing they found, the tetrahedra take up about 72% of space.

Compare this to the densest packing of spheres in space, which take up about 74%. If Conway and Torquato’s example was actually the densest packing of tetrahedra, it would be a counterexample to Ulam’s conjecture that the sphere is the worst case scenario for packing.

But a series of papers improving the bound followed, and as of early 2010 the record is held by Chen, Engel, and Glotzer with a packing fraction of 85.63%.

I want to advertise two attractive open problems related to this.

(1) Good upper bounds on tetrahedron packing.

At the time of the colloquium talk I saw several months ago, it seemed that despite a whole host of papers improving the lower bound on tetrahedron packing, there was no upper bound in the literature. Since then Gravel, Elser, and Kallus posted a paper on the arXiv which gives an upper bound. This is very cool, but the upper bound on density they give is something like $1- 2.6 \times 10^{-25}$, so there is still a lot of room for improvement.

(2) Packing tetrahedra in a sphere.

As far as I know, even the following problem is open. Let’s make our lives easier by discretizing the problem and we simply ask how many tetrahedra we can pack in a sphere. Okay, let’s make it even easier: the edge length of each of the tetrahedra is the same as the radius of the sphere. Even easier: every one of the tetrahedra has to have one corner at the center of the sphere. Now how many tetrahedra can you pack in the sphere?

It is fairly clear that you can get 20 tetrahedra in the sphere, since the edge length of the icosahedron is just slightly longer than the radius of its circumscribed sphere. By comparing the volume of the regular tetrahedron to the volume of the sphere, we get a trivial upper bound of 35 tetrahedra. But by comparing surface area instead, we get an upper bound of 22 tetrahedra.

There is apparently a folklore conjecture that 20 tetrahedra is the right answer, so proving this comes down to ruling out 21 or 22. To rule out 21 seems like a nonlinear optimization problem in some 63-dimensional space.

I’d guess that this is within the realm of computation if someone made some clever reductions. Oleg Musin settled the question of the kissing number in 4-dimensional space in 2003. To rule out kissing number of 25 is essentially optimizing some function over a 75-dimensional space. This sounds a little bit daunting, but it is apparently much easier than Thomas Hales’s proof of the Kepler conjecture. (For a nice survey of this work, see this article by Pfender and Ziegler.)

## 4 thoughts on “Packing tetrahedra”

1. Anonymous says:

if the edges of tetrahedra is 2 the area of the equilateral
triangles has area sqrt(3). the area of the circumsphere is 16\pi and you can fit 16\pi/sqrt(3) = 29.02. i am wondering
how are you are getting about 22?

• matthewkahle says:

It is true that the area of the equilateral triangle is sqrt(3), but the surface area is larger once it is projected out on to the sphere. The formula for solid angle for a regular tetrahedron is given here: http://en.wikipedia.org/wiki/Tetrahedron

2. Anonymous says:

thanks for blogging about geometry. i think i have it now. the dihedral angle of tetrahedra is $cos^{-1}(1/3)$ and the area projected by one face is $(3cos^{-1}(1/3)-\pi$ so $\frac{4\pi}{3\cos^{-1}(1/3) – pi)$ is about 22.795.

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