2005 AMC 12A Problems/Problem 22

Problem

A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$ ?

$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$

Solution

Box P has dimensions $l$ , $w$ , and $h$ . Surface area = $2lw+2lh+2wl=384$ Sum of all edges = $4l+4w+4h=112 \Longrightarrow l + w + h = 28$

The diameter of the sphere is the space diagonal of the prism, which is $\sqrt{l^2 + w^2 +h^2}$ $(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400$ $\sqrt{l^2 + w^2 +h^2} = 20 = diameter$ $r=\frac{20}{2} = 10$

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2005 AMC 12A Problems/Problem 22

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