Difference between revisions of "2007 AMC 10A Problems/Problem 21"

Revision as of 15:13, 23 September 2019

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 $S = 8\Rightarrow \mathrm{\boxed{(C)}}$ .

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 $6x^{2} = 6\left(\frac {2}{\sqrt {3}}\right )^{2} = 8$ .

Solution 3, from AoPS

Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives $l^2 + l^2 + l^2 = 2^2 = 4.$ Hence each face has surface area $l^2 = \frac{4}{3} \ \text{square meters}.$ So the surface area of the inscribed cube is $6\cdot (4/3) = \boxed{8}$ square meters.

@@ Line 22: / Line 22: @@
 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 <math>x</math>, we see that <math>2 = \sqrt {x^{2} + (\sqrt {2}x)^{2}}\Rightarrow x = \frac {2}{\sqrt {3}}</math>.
 Thus the surface area of the inner cube is <math>6x^{2} = 6\left(\frac {2}{\sqrt {3}}\right )^{2} = 8</math>.
+===Solution 3, from AoPS===
+Since the [[surface area]] of the original [[cube]] is 24 square meters, each face of the cube has a surface area of <math>24/6 = 4</math> square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let <math>l</math> represent the side length of the inscribed cube. Applying the [[Pythagorean Theorem]] twice gives<cmath>
+l^2 + l^2 + l^2 = 2^2 = 4.
+</cmath>Hence each face has surface area<cmath>
+l^2 = \frac{4}{3} \ \text{square meters}.
+</cmath>So the surface area of the inscribed cube is <math>6\cdot (4/3) = \boxed{8}</math> square meters.
 == See also ==

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

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