Find volume of the circumscribing tetrahedron Take four spheres each with a radius of 1.  Place them so they are mutually tangent.  Circumscribe a regular tetrahedron around the spheres.  Compute the volume of the tetrahedron.
 A: The tetrahedron $T_c$ formed by the centers of the four spheres has sidelength $2$, height $h=2\sqrt{2\over3}$, and volume $V_c={2\sqrt{2}\over3}$. The faces of $T_c$ have  distance
$$d={h\over4}={1\over\sqrt{6}}$$
from the center $C$ of $T_c$. The faces of the circumscribing tetrahedron $T$ have the distance $d+1$ from $C$. It follows that the volume $V$ of $T$ is given by
$$V=\left({d+1\over d}\right)^3 V_c={38\over3}\sqrt{2}+12\sqrt{3}\ .$$
A: Your spheres have a distance of $2$ from one another, so they form an inner tetrahedron of edge length $2$. The faces of the outer tetrahedron are a distance of $1$ away from those of the inner tetrahedron, since the outher face is tangent to the three spheres whose centers form the corners of the inner tetrahedron. So what's the dge length of that tetrahedron?
To answer that question, let's first consider a simpler situation. Start with a regular tetrahedron of edge length $a$. According to Wikipedia, it has volume $\frac1{12}\sqrt2a^3$. Now shift one of its planes by an offset of $d$ outwards, extending all faces accordingly. The result is a tetrahedron of edge length $b$ with a volume of $\frac1{12}\sqrt2b^2$. But you also know that the volume of a tetrahedron is $\frac13A_0h$ where $A_0$ is the area of its base and $h$ is the height orthogonal to that base. The original tetrahedron had a base area of $\frac14\sqrt3a^2$ so its height must have been $h=\frac13\sqrt6a$. Likewise for the new tetrahedron, so you have $d=\frac13\sqrt6b-\frac13\sqrt6a$ or $\frac12\sqrt6d=b-a$ or $b=a+\frac12\sqrt6d$. So at this point, you know that if you offset one face by $d$, the edge lengths increase by $\frac12\sqrt6d$.
Now back to the situation with your spheres. If you offset all four planes one after the other by a distance of $d=1$, then the edge length will increase by $\frac12\sqrt6$ at each step, so the final edge length is $2+2\sqrt6$. The volume for a polyhedron of that edge length is
$$V=\frac1{12}\sqrt2\bigl(2+2\sqrt6\bigr)^3=\frac{38}{3}\sqrt2+12\sqrt3$$
