Example 1 Try to find the volume of the edge-tangent sphere (a sphere that is tangent to all edges) of a regular tetrahedron with edge length $a$.


Solution 1: As shown in Figure 1, let $O$ be the center of the regular tetrahedron $ABCD$. In the right triangle $\triangle OBF$, by the Pythagorean theorem, the circumradius of this regular tetrahedron $AO=BO=\frac{\sqrt{6}}{4} a$. Therefore, the radius of the edge-tangent sphere $r=OP=\frac{\sqrt{2}}{4} a$ (the inradius $OP=\frac{\sqrt{6}}{12} a$, and the center $O$ is the point dividing the altitude $AF$ into four equal parts). Thus, the volume of the edge-tangent sphere $V=\frac{4}{3} \pi r^{3}=\frac{\sqrt{2}}{24} \pi a^{3}$.

Solution 2: Construct a plane parallel to the opposite edge through each edge of the regular tetrahedron, forming a cube with edge length $\frac{\sqrt{2}}{2} a$. Then, the edge-tangent sphere of the regular tetrahedron is the inscribed sphere of the cube, hence $r=\frac{\sqrt{2}}{4} a$. The rest follows as in Solution 1.