Annual Harvard-MIT Mathematics Tournament

Solution:
Answer: $\frac{1}{4}$

Without loss of generality, assume that $x \geq y$ — half the volume of the solid is on this side of the plane $x = y$. For each value of $c$ from $0$ to $\frac{1}{2}$, the region of the intersection of this half of the solid with the plane $y = c$ is a trapezoid. The trapezoid has height $1 - 2c$ and average base $\frac{1}{2}$, so it has an area of $\frac{1}{2} - c$.

The total volume of this region is $\frac{1}{2}$ times the average area of the trapezoids, which is $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$. Double that to get the total volume, which is $\frac{1}{4}$.