Example 11 Four spheres, each with a radius of 1, are mutually tangent, three on the floor, and the fourth above them. A tetrahedron with side length $S$ has faces that are tangent to all these spheres. Find $S$.
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Figure $23-18$
Solution The centers of the four spheres are the vertices of a regular tetrahedron with side length 2. Let this tetrahedron be $A B C D$, and let the height from $A$ intersect the base $B C D$ at $G$, then $G$ is the centroid of $\triangle B C D$. Taking the midpoint of $B C$ as $E$, we extract the plane figure $\triangle A E G$, as shown in Figure $23-18$.

The distance from each face of the tetrahedron that is externally tangent to the spheres to the corresponding face of the tetrahedron $A B C D$ is 1, and it is still a regular tetrahedron. Therefore, we extend $\triangle A E G$ to the corresponding $\triangle A_{1} E_{1} G_{1}$ in this tetrahedron, which only requires drawing $A_{1} E_{1} \parallel A E, E_{1} G_{1} \parallel E G$, with the distance between the parallel lines being 1, thus obtaining $\triangle A_{1} E_{1} G_{1}$. By finding the sides of $\triangle A_{1} E_{1} G_{1}$ through $\triangle A E G$, we can then find $S$.

In fact, it is easy to see that $A E=\frac{\sqrt{3}}{2} A C=\sqrt{3}, E G=\frac{1}{3} A E=\frac{\sqrt{3}}{3}, A G=\sqrt{A E^{2}-E G^{2}}=\frac{2}{3} \sqrt{6}, \frac{A_{1} A}{A E}=\frac{1}{E G}, A_{1} A=\frac{A E}{E G}=3, A_{1} G_{1}=A_{1} A+A G+G G_{1}=4+\frac{2}{3} \sqrt{6}$.

Also, $\frac{A_{1} G_{1}}{A_{1} E_{1}}=\frac{A_{1} G_{1}}{\frac{\sqrt{3}}{2} S}=\frac{A G}{A E}$, we get $S=\frac{A_{1} G_{1} \cdot A E}{\frac{\sqrt{3}}{2} A G}=2+2 \sqrt{6}$.