2007 AMC 10A Problems/Problem 21

Problem

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

Solution 1

$[asy] import three; draw(((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0))^^((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1))^^((0,0,0)--(0,0,1))^^((0,1,0)--(0,1,1))^^((1,1,0)--(1,1,1))^^((1,0,0)--(1,0,1))); draw(shift((0.5,0.5,0.5))*scale3(1/sqrt(3))*shift((-0.5,-0.5,-0.5))*rotate(aTan(sqrt(2)),(0,0,0.5),(1,1,0.5))*(((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0))^^((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1))^^((0,0,0)--(0,0,1))^^((0,1,0)--(0,1,1))^^((1,1,0)--(1,1,1))^^((1,0,0)--(1,0,1)))); dot((0.5,0.5,1)^^(0.5,0.5,0)); [/asy]$

We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.

Let $S$ be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length $s\sqrt{3}$ where $s$ is a side of the cube, the ratio of a side of the inner square to that of the outer square (and the side of the outer square = the diagonal of the inner square), we have $\frac{S}{24} = \left(\frac{1}{\sqrt{3}}\right)^2$ . Thus $S = 8\Rightarrow \mathrm{(C)}$ .

Solution 2 (computation)

The area of each face of the outer cube is $\frac {24}{6} = 4$ , and the edge length of the outer cube is $2$ . This is also the diameter of the sphere, and thus the length of a long diagonal of the inner cube.

A long diagonal of a cube is the hypotenuse of a right triangle with a side of the cube and a face diagonal of the cube as legs. If a side of the cube is $x$ , we see that $2 = \sqrt {x^{2} + (\sqrt {2}x)^{2}}\Rightarrow x = \frac {2}{\sqrt {3}}$ .

Thus the surface area of the inner cube is $6x^{2} = 6\left(\frac {2}{\sqrt {3}}\right )^{2} = 8$ .

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2007 AMC 10A Problems/Problem 21

Problem

Contents

Solution

Solution 1

Solution 2 (computation)

See also