6. (2000 National High School Mathematics League Question) A sphere is tangent to all six edges of a regular tetrahedron. If the edge length of the regular tetrahedron is $a$, then the volume of this sphere is $\qquad$ .
6. The center of a regular tetrahedron is the center of the sphere, and the radius can be calculated, thus allowing the volume of the sphere to be determined.

As shown in the figure, let the center of the sphere be $O$, the radius be $r$, the volume be $V$, the center of face $BCD$ be $O_{1}$, and the midpoint of edge $BC$ be $E$. Then,
$A O_{1}=\sqrt{a^{2}-O_{1} B^{2}}=\sqrt{a^{2}-\frac{1}{3} a^{2}}=\frac{\sqrt{6}}{3} a$,
From $O B^{2}=O_{1} O^{2}+O_{1} B^{2}=\left(O_{1} A-O B_{0}^{2}+O_{1} B^{2}\right)$
we get $\frac{2}{3} a^{2}-\frac{2 \sqrt{6}}{3} a \cdot O B+\frac{1}{3} a^{2}=0$,
Thus, $O B^{2}=\frac{3}{2 \sqrt{6}} a=\frac{\sqrt{6}}{4} a$,
Therefore, $r=O E=\sqrt{O B^{2}-B E^{2}}=\sqrt{\frac{3}{8} a^{2}-\frac{1}{4} a^{2}}=\frac{1}{2 \sqrt{2}} a$.
$$
V=\frac{4}{3} \pi r^{2}=\frac{4}{3} \pi \frac{1}{16 \sqrt{2}} a^{3}=\frac{\sqrt{2}}{24} \pi a^{3} .
$$

Note: This problem can be adapted as (8th IMO problem): Prove that the sum of the distances from the center of the circumscribed sphere of a regular tetrahedron to its four vertices is less than the sum of the distances from any other point in space to the four vertices.