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.
 A: *

*"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.

*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:
$$\frac{n(n-1)}{2} \le 4 \implies n \le 3$$
