2007 AMC 10A Problem 21

Follow a single length from the outside in.

1. Outer cube. $6$ faces with total area $24$, so each face has area $4$ and each edge is $2$.
2. Sphere. It fits snugly inside, touching the middle of every face. Its diameter goes straight from one face to the opposite face, so it equals the edge: $2$.
3. Inner cube. Its $8$ corners all touch the sphere. Its long diagonal, from one corner through the center to the opposite corner, is a diameter of the sphere, so it is also $2$.
4. Long diagonal of a cube with edge $s$. A face diagonal is $\sqrt{s^2 + s^2} = s\sqrt2$. The long diagonal is the long side of a right triangle with legs $s$ (an edge going up) and $s\sqrt2$ (a face diagonal on the bottom):
$$\sqrt{s^2 + 2s^2} = \sqrt{3s^2} = s\sqrt3.$$
5. Find $s^2$. $s\sqrt3 = 2$, so $3s^2 = 4$ and $s^2 = \frac43$. (We only need $s^2$, since area uses $s^2$.)
6. Surface area. $6 s^2 = 6 \times \frac43 = 8$.

The answer is $\boxed{\textbf{(C)}\ 8}$.

When shapes fit inside each other, one length passes from each shape to the next: here cube edge, then sphere diameter, then the inner cube's long diagonal. Knowing that a cube with edge $s$ has long diagonal $s\sqrt3$ is the key fact. Area only needs $s^2$, so you never have to deal with $\frac{2}{\sqrt3}$.

The outer cube has six faces of area $\frac{24}{6} = 4$, so its edge is $2$.

A sphere inscribed in that cube touches all six faces, so its diameter equals the edge: $2$.

A cube inscribed in the sphere has all eight vertices on the sphere, so its space diagonal is a diameter: also $2$. For a cube of edge $s$, the space diagonal is $s\sqrt3$ (apply the Pythagorean theorem to an edge and a face diagonal: $s^2 + (s\sqrt2)^2 = 3s^2$). Hence
$$s\sqrt3 = 2 \quad\Longrightarrow\quad s^2 = \frac43,$$
and the inner cube's surface area is $6s^2 = 6 \cdot \frac43 = 8$.

The answer is $\boxed{\textbf{(C)}\ 8}$.