2007 AMC 10A Problem 21

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 $\boxed{\textbf{(C)}\ 8}$.

Nested inscribed solids pass a single length along the chain: cube edge $\to$ sphere diameter $\to$ inner cube's space diagonal. Since surface area only needs $s^2$, there is no reason to extract $s$ itself. Note the surface area ratio $\frac{8}{24} = \frac13$ is exactly $\left(\frac{1}{\sqrt3}\right)^2$, the square of the edge ratio.

• Using the face diagonal $s\sqrt2$ instead of the space diagonal $s\sqrt3$ for the inscribed cube, which gives $12$.
• Taking the sphere's radius ($1$) rather than its diameter as the inner cube's diagonal, which gives surface area $2$ and no matching choice.