# 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 $h$. Its surface area is $$2lw+2lh+2wh=384,$$ *I fixed a typo here* and the sum of all its edges is $$l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.$$

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

## Solution 2

As in the previous solution, we have that $2lw+2lh+2wh=384$ and $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28$, and the diameter of the sphere is the space diagonal of the prism, $\sqrt{l^2 + w^2 + h^2}$.

Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that $h=0$. (This essentially means that we have an infinitesimally thin box.) We now have that $2lw = 384$ and $l + w = 28$, and we are solving for $\sqrt{l^2 + w^2}$. Because $$(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,$$ this means that $$l^2 + w^2 = 28^2 - 384 = 400,$$ so the space diagonal is $\sqrt{400} = 20$. Since the diameter of the sphere is $20$, the radius is $\boxed{\textbf{(B) } 10}$. ~emerald_block