We are given a regular tetrahedron $ABCD$ ($ABC$ is its` base and $D$ is its vertex) and we reflect it through the middle of its height (point reflexion) - and thus we obtain a congruent regular tetrahedron $A'B'C'D'$.
$D'$ lies in the center of $ABC$, and $D$ in the center of $A'B'C'$.
Planes $\pi (ABC) \ || \ \pi (A'B'C'), \ \ \ \pi (ABD) \ || \ \pi (A'B'D'), \ \ \ \pi (B'C'D') \ || \ \pi (BCD)$, $ \ \ \ \pi (A'C'D') \ || \ \pi (ACD)$.
I drew a picture and I think that the intersection of the two tetrahedrons is a parallelepiped, but I don't know how to prove it more formally (I mean, I know that the respective sides of the tetrahedrons are parallel, because we reflect $ABCD$ in a point, but I am not sure if that's enough).
Secondly, how can we calculate the volume of the intersection?
Could you help me with that?