9. Given is a regular tetrahedron of volume 1 . We obtain a second regular tetrahedron by reflecting the given one through its center. What is the volume of their intersection?
Solution: $1 / 2$
Imagine placing the tetrahedron $A B C D$ flat on a table with vertex $A$ at the top. By vectors or otherwise, we see that the center is $3 / 4$ of the way from $A$ to the bottom face, so the reflection of this face lies in a horizontal plane halfway between $A$ and $B C D$. In particular, it cuts off the smaller tetrahedron obtained by scaling the original tetrahedron by a factor of $1 / 2$ about $A$. Similarly, the reflections of the other three faces cut off tetrahedra obtained by scaling $A B C D$ by $1 / 2$ about $B, C$, and $D$. On the other hand, the octahedral piece remaining remaining after we remove these four smaller tetrahedra is in the intersection of $A B C D$ with its reflection, since the reflection sends this piece to itself. So the answer we seek is just the volume of this piece, which is
$$
\begin{array}{l}
\text { (volume of } A B C D)-4 \cdot(\text { volume of } A B C D \text { scaled by a factor of } 1 / 2) \\
=1-4(1 / 2)^{3}=1 / 2
\end{array}
$$