2007 AMC 10A Problem 21

P21-25~3 minPrint

Solid Geometry Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

One length is passed along: the outer cube's edge equals the sphere's diameter, which equals the long diagonal through the inner cube, corner to opposite corner.

Solution

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

Why this works

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 $\frac{2}{\sqrt3}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 15: Curved Surfaces