Let $Q$ be a unit cube. We say a tetrahedron is "good" if all its edges are equal and all its vertices lie on the boundary of $Q$. Find all possible volumes of "good" tetrahedra.
Clearly, the volume of a regular tetrahedron contained in a sphere reaches its maximum value if and only if all four vertices of the tetrahedron lie on the surface of the sphere. Therefore, a "good" tetrahedron with maximum volume must have its vertices at the vertices of the cube (for a proof, inscribe the cube in a sphere). There are exactly two such tetrahedra, their volume being equal to $1-4 \cdot \frac{1}{6}=\frac{1}{3}$. On the other hand, one can find arbitrarily small "good" tetrahedra by applying homothety to the maximal tetrahedron, with the centre of the homothety in one of its vertices.