Difference between revisions of "2005 AMC 12A Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | {{ | + | The box P has dimensions a, b, and c. Therefore, |
| + | * <math> 2ab+2ac+2bc=384</math> | ||
| + | * <math>4a+4b+4c=112</math> | ||
| + | * <math>a+b+c=28</math> | ||
| + | Now we make a formula for r. Since the diameter of the sphere is the space diagonal of the box, | ||
| + | * <math>r=\frac{\sqrt{a^2+b^2+c^2}}{2}</math> | ||
| + | We square a+b+c: | ||
| + | * <math>(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784</math> | ||
| + | We get that | ||
| + | * <math>\frac{\sqrt{a^2+b^2+c^2}}{2}=10=r</math> | ||
| + | |||
== See also == | == See also == | ||
* [[2005 AMC 12A Problems/Problem 21 | Previous problem]] | * [[2005 AMC 12A Problems/Problem 21 | Previous problem]] | ||
* [[2005 AMC 12A Problems/Problem 23 | Next problem]] | * [[2005 AMC 12A Problems/Problem 23 | Next problem]] | ||
* [[2005 AMC 12A Problems]] | * [[2005 AMC 12A Problems]] | ||
Revision as of 08:37, 9 September 2007
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 a, b, and c. Therefore,
Now we make a formula for r. Since the diameter of the sphere is the space diagonal of the box,
We square a+b+c:
We get that
