Proportion of regular tetrahedron occupied by mutually tangent balls centred at its vertices

Question

The centres of four balls of radius $1$ are the vertices of a regular tetrahedron of side length $2$. What is the proportion of the tetrahedron occupied by the balls?

At first I thought it should just be the maximum packing density of spheres, $\frac{\pi}{3\sqrt2}\approx 0.74048$, but this is not true, because regular tetrahedrons do not tile space.

I considered the fact that the sphere kissing number is $12$, but it seems this fact would only be useful if regular tetrahedrons tiled space.

Answer 1

The solid angle of a regular tetrahedron is arccos(2327), so the volume inside the tetrahedron occupied by one ball is $\frac{1}{4\pi}\arccos\left(\frac{23}{27}\right)\times\frac{4\pi}{3}$. The volume of the tetrehedron is $\frac{\sqrt2}{12}\times 2^3$. So the answer to the question is $\sqrt2 \arccos(\frac{23}{27})\approx 0.7796$.