2005 AMC 12A Problems/Problem 22
Problem
A rectangular box 
is inscribed in a sphere of radius 
. The surface area of is 384, and the sum of the lengths of its 12 edges is 112. What is ?
Solution
Box P has dimensions 
, 
, and 
.
Its surface area is ![\[2lw+2lh+2wl=384,\]](http://latex.artofproblemsolving.com/b/e/6/be6f4fdd31b4b9eb2042e2256288b5ca1e0d73c5.png)
while the sum of all its edges is ![\[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28\]](http://latex.artofproblemsolving.com/5/6/5/565c9380cb323ca34042ce94b4604c175bbfd1c9.png)
The diameter of the sphere is the space diagonal of the prism, which is ![\[\sqrt{l^2 + w^2 +h^2}.\]](http://latex.artofproblemsolving.com/3/1/3/31384be5c4dd31ee79fefadd8f754b0e543415a6.png)
Notice that ![\[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\]](http://latex.artofproblemsolving.com/2/0/e/20ea50479e3eb41006e5cba6901f12d434e6abda.png)
so the diameter is
![\[\sqrt{l^2 + w^2 +h^2} = 20\]](http://latex.artofproblemsolving.com/1/f/e/1fe9c17cffdbdb97c24b028e0c17da1c7dafeb3e.png)
. The radius is half of the diameter, so
![\[r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.\]](http://latex.artofproblemsolving.com/8/a/0/8a0ac99e9fe30e6457565990dbbf925afcada5c3.png)
See also
| 2005 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
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