9. Take a unit sphere $S$, i.e., a sphere with radius 1. Circumscribe a cube $C$ about $S$, and inscribe a cube $D$ in $S$, so that every edge of cube $C$ is parallel to some edge of cube $D$. What is the shortest possible distance from a point on a face of $C$ to a point on a face of $D$ ?
Answer: $\frac{3-\sqrt{3}}{3}=1-\frac{\sqrt{3}}{3}$, or equivalent
Solution: Using the Pythagorean theorem, we know that the length of a diagonal of a cube of edge length $s$ is $s \sqrt{3}$. Since $D$ is inscribed in a sphere that has diameter 2 , this means that its side length is $2 / \sqrt{3}$.
The distance from a face of $D$ to a face of $C$ will be the distance between them along any line perpendicular to both of them; take such a line passing through the center of $S$. The distance from the center to any face of $D$ along this line will be half the side length of $D$, or $1 / \sqrt{3}$. The distance from the center to the edge of $C$ is the radius of $S$, which is 1 . Therefore the desired distance is $1-1 / \sqrt{3}=1-\frac{\sqrt{3}}{3}$.