HMMT-Nov 2009 guts, Problem 24

Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figure whose vertices are the 5 chosen points?

Solution: Label the vertices of the cube $A, B, C, D, E, F, G, H$, such that $A B C D$ 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 $A C B F H$, we will instead compute the volume of the parts of the cube that are not part of $A B C F H$. This is just the three tetrahedrons $C F G H, A E F H$, and $A C D H$, which each have volume $\frac{1}{6}$ (by using the $\frac{1}{3} b h$ formula for the area of a pyramid). Therefore, the volume not contained in $A C B F H$ is $3 \cdot \frac{1}{6}=\frac{1}{2}$, so the volume contained in $A C B F H$ is $1-\frac{1}{2}=\frac{1}{2}$. $\fbox{\frac{1}{2}}$.

Answer: \frac{1}{2}