Difference between revisions of "2005 AMC 12A Problems/Problem 22"
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We get that | We get that | ||
| − | <cmath>r=\frac{\sqrt{a^2+b^2+c^2}}{2}=B)10</cmath> | + | <cmath>r=\frac{\sqrt{a^2+b^2+c^2}}{2}=\boxed{\textbf{(B) }10}</cmath> |
== See also == | == See also == | ||
Revision as of 19:35, 7 November 2013
Problem
A rectangular box 
is inscribed in a sphere of radius 
. The surface area of is 384, and the sum of the lengths of it's 12 edges is 112. What is ?
Solution
The box P has dimensions 
, 
, and 
. Therefore,
![\[2ab+2ac+2bc=384\]](http://latex.artofproblemsolving.com/2/c/9/2c9e5b2fa39b10af037e22f59acdd80581ac9eb6.png)
![\[4a+4b+4c=112 \Longrightarrow a + b + c = 28\]](http://latex.artofproblemsolving.com/1/0/c/10ce527b25a26f4473ef7e090e27abfce7c47d1b.png)
Now we make a formula for . Since the diameter of the sphere is the space diagonal of the box,
![\[r=\frac{\sqrt{a^2+b^2+c^2}}{2}\]](http://latex.artofproblemsolving.com/c/6/1/c6196f7d49c0d0d35b9b565c5ce071f727b9ac6d.png)
We square 
:
![\[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784\]](http://latex.artofproblemsolving.com/c/6/7/c679ae0653aadd1d56e493a2b252d9277e338078.png)
We get that
![\[r=\frac{\sqrt{a^2+b^2+c^2}}{2}=\boxed{\textbf{(B) }10}\]](http://latex.artofproblemsolving.com/5/f/e/5fe5ee459ca4b6ed2b7cd468d101d8f2f7a69d90.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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
