# Difference between revisions of "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 $$\sqrt{l^2 + w^2 +h^2}$$ $$(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400$$ $$\sqrt{l^2 + w^2 +h^2} = 20 = diameter$$ $r=\frac{20}{2} = 10$