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 it's 12 edges is 112. What is $r$ ?

Solution

The box P has dimensions a, b, and c. Therefore,

$2ab+2ac+2bc=384$
$4a+4b+4c=112$
$a+b+c=28$

Now we make a formula for r. Since the diameter of the sphere is the space diagonal of the box,

$r=\frac{\sqrt{a^2+b^2+c^2}}{2}$

We square a+b+c:

$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784$

We get that

$\frac{\sqrt{a^2+b^2+c^2}}{2}=10=r$

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2005 AMC 12A Problems/Problem 22

Problem

Solution

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