Harvard-MIT November Tournament

Solution:

The answer is $\frac{1}{2}$.

Label the vertices of the cube $A, B, C, D, E, F, G, H$, such that $ABCD$ is the top face of the cube, $E$ is directly below $A$, $F$ is directly below $B$, $G$ is directly below $C$, and $H$ is directly below $D$.

We can obtain a volume of $\frac{1}{2}$ by taking the vertices $A, B, C, F$, and $H$.

To compute the volume of $ABCFH$, we will instead compute the volume of the parts of the cube that are not part of $ABCFH$. This is just the three tetrahedrons $CFGH$, $AEFH$, and $ACDH$, which each have volume $\frac{1}{6}$ (by using the $\frac{1}{3}bh$ formula for the volume of a pyramid).

Therefore, the volume not contained in $ABCFH$ is $3 \cdot \frac{1}{6} = \frac{1}{2}$, so the volume contained in $ABCFH$ is $1 - \frac{1}{2} = \frac{1}{2}$.