Four cubes with edge lengths $2,3,5,10$ are stacked in increasing order from top to bottom, with their back-left vertical edges aligned. Point $X$ is the top-back-left corner of the smallest cube, and point $Y$ is the bottom-front-right corner of the largest cube. What is the length of the portion of $\overline{XY}$ inside the cube with edge length $5$?

1. The cube edges are $2,3,5,10$, so the stack is $20$ units high. Taking the common back-left vertical edge as the $z$-axis gives $X=(0,0,20)$ and $Y=(10,10,0)$.
2. At height $z$, the segment $\overline{XY}$ has coordinates $x=y=\frac{20-z}{2}$. The edge-$5$ cube occupies heights $10$ through $15$. There, $x=y$ decreases from $5$ to $\frac52$, so the segment stays inside this cube's footprint throughout the interval.
3. The segment changes vertically by $5$ and changes by $\frac52$ in each horizontal coordinate while inside the cube. Its length is therefore $\sqrt{5^2+\left(\frac52\right)^2+\left(\frac52\right)^2}=\frac{5\sqrt6}{2}$.