HMMT-Feb 2013 guts, Problem 27

Let $W$ be the hypercube $\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid 0 \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 1\right\}$. The intersection of $W$ and a hyperplane parallel to $x_{1}+x_{2}+x_{3}+x_{4}=0$ is a non-degenerate 3 -dimensional polyhedron. What is the maximum number of faces of this polyhedron?

Solution: The number of faces in the polyhedron is equal to the number of distinct cells (3dimensional faces) of the hypercube whose interior the hyperplane intersects. However, it is possible to arrange the hyperplane such that it intersects all 8 cells. Namely, $x_{1}+x_{2}+x_{3}+x_{4}=\frac{3}{2}$ intersects all 8 cells because it passes through $\left(0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ (which is on the cell $\left.x_{1}=0\right),\left(1, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}\right)$ (which is on the cell $x_{1}=1$ ), and the points of intersection with the other 6 cells can be found by permuting these quadruples. $\fbox{8}$.

Answer: 8