2005 AMC 12A Problems/Problem 22

Revision as of 21:08, 30 January 2015 by Swe1 (talk | contribs) (Grammar)

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

The box P has dimensions $a$, $b$, and $c$. Therefore, \[2ab+2ac+2bc=384\] \[4a+4b+4c=112 \Longrightarrow a + b + c = 28\]

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 \[r=\frac{\sqrt{a^2+b^2+c^2}}{2}=\boxed{\textbf{(B) }10}\]

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS