A sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube? $\text{(A)}\ 3 \qquad$ $ \text{(B)}\ 6 \qquad$ $ \text{(C)}\ 8 \qquad$ $ \text{(D)}\ 9 \qquad$ $ \text{(E)}\ 12$
Solution
$6a^2=24\Rightarrow a=2$, so the inscribed sphere has diameter $2$.
This diameter is the inner cube's space diagonal: $b\sqrt3=2\Rightarrow b^2=\frac43$.
Inner surface area is $6b^2=8$. The answer is (\textbf{C}).
The vertices of a regular octahedron are the centers of the six faces of a cube with edge length $6$. A sphere is inscribed in the octahedron. What is the radius of the sphere?
$\text{(A)}\ 3 \qquad$ $ \text{(B)}\ 6 \qquad$ $ \text{(C)}\ 8 \qquad$ $ \text{(D)}\ 9 \qquad$ $ \text{(E)}\ 12$ 