imc

A regular unit tetrahedron can be split into eight tetrahedra that have lengths of $\frac{1}{2}$. The volume of a regular tetrahedron can be found using the formula for the area of a pyramid: $\frac{1}{3}\cdot B\cdot H$ where $B$ is the area of the base and $H$ is the height. For a tetrahedron of side length 1, its base area is $\frac{\sqrt{3}}{4}$, and its height can be found using the Pythagorean Theorem. Its height is $\sqrt{1^2-{\left(\frac{\sqrt{3}}{3}\right)}^2}=\frac{\sqrt{2}}{\sqrt{3}}$. Its volume is $\frac{1}{3}\times \frac{\sqrt{3}}{4}\times \frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}}{12}$. The tetrahedron actually has side length $\sqrt{2}$, so the actual volume is $\frac{\sqrt{2}}{12}\times {\sqrt{2}}^3=\frac{1}{3}$. On the eight small tetrahedra, the four tetrahedra on the corners of the large tetrahedra are not inside the other large tetrahedra. Thus, $\frac{4}{8}=\frac{1}{2}$ of the large tetrahedra will not be inside the other large tetrahedra. The intersection of the two tetrahedra is thus $\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}=\boxed{\frac{1}{6}}$.