# Tetrahedron packing in Cube

I'm thinking about following solid geometry problem.

Q: Suppose you have a box of "cube" shape with edge length 1. Then, How many regular tetrahedrons(with edge length 1) can be in the box?

So, this is kind of packing problem inside cube. I guess the answer is 3. but I don't know how to prove 3 is the maximum number. Is there any rigorous way to show this?

Thanks for any help in advance.

• Here's a Wiki article that may be of interest: en.wikipedia.org/wiki/Tetrahedron_packing The results, which are based off experiment and simulation, hint that a 0.7 to 0.85 packing fraction might be reasonable. Maybe higher. Given that your tetrahedra have a $V\approx0.12$, then the answer could be up to $7$! Edit: I say up to seven, since the article does not make it clear when it's considering finite volumes and even then with what sorts of boundaries. Feb 24, 2016 at 2:22
• @SunTaek Yes, I was completely looking at the wrong item in the table -- oops! Feb 24, 2016 at 2:25
• @zahbaz Thanks for interesting article. hmm...but I cannot construct mental picture of 5 or 6 regular tetrahedron in the unit cube... Feb 24, 2016 at 2:27
• @SunTaek. Agreed. I'll stay tuned to this post. This is an interesting problem. Feb 24, 2016 at 2:28
• I can see some ways to pack two tetrahedra in a cube, same edge length. i would be interested in a diagram or calculations that prove that three is possible. Feb 24, 2016 at 18:32

1. "3" is possible, as shown in following diagram.
The vertices of the red tetrahedron are $$\left(\frac12,-\frac12,\frac12\right),\;\; \left(\frac12,-\frac12,-\frac12\right),\;\; \left(\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}},0\right)\;\;\text{ and }\;\; \left(-\frac{1}{2\sqrt{2}},-\frac{1}{2\sqrt{2}},0\right)$$ The green and blue tetrahedrons can be obtained from the red one by rotating it along the $(1,1,1)$ diagonal for $120^\circ$ and $240^\circ$ respectively.

2. I believe "3" is the maximum number. Following is a heuristic argument:

Let's say we have $n$ tetrahedrons inside a cube. There are $\frac{n(n-1)}{2}$ ways of picking a pair of tetrahedron $A, B$ among them.

Pick a point $a$ from $A$, a point $b$ from $B$ such that the distance $|a-b|$ is maximized. No matter how I place $A$ and $B$, I always get $|a - b| \ge 2\sqrt{\frac23} \approx 1.633$.

Since this value is very close to $\sqrt{3} \approx 1.732$, the diameter of the cube, the points $a, b$ will be very close to the two end points of a diagonal. This means each pair of tetrahedron will occupy at least one diagonals of the cube.

Since a cube has $4$ diagonals and it seems impossible for different pairs of tetrahedron to share a diagonal, we find: $$\frac{n(n-1)}{2} \le 4 \implies n \le 3$$

• amazing! rough, but convincing idea. Thank you So much! By the way, may I ask you which program you used to draw that picture? Feb 25, 2016 at 17:48
• @SunTaek I have some script to generate the figure in X3D format, I then use the javascript library x3dom to display and convert it to a picture inside a web browser (firefox). Feb 25, 2016 at 18:06