Difference between revisions of "2005 AMC 12A Problems/Problem 22"
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== Problem == | == Problem == | ||
A rectangular box <math> P </math> is [[inscribe]]d in a [[sphere]] of [[radius]] <math>r</math>. The [[surface area]] of <math>P</math> is 384, and the sum of the lengths of it's 12 edges is 112. What is <math>r</math>? | A rectangular box <math> P </math> is [[inscribe]]d in a [[sphere]] of [[radius]] <math>r</math>. The [[surface area]] of <math>P</math> is 384, and the sum of the lengths of it's 12 edges is 112. What is <math>r</math>? | ||
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| + | <math>\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16</math> | ||
== Solution == | == Solution == | ||
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* <math>4a+4b+4c=112</math> | * <math>4a+4b+4c=112</math> | ||
* <math>a+b+c=28</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, | + | Now we make a formula for <math>r</math>. 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> | * <math>r=\frac{\sqrt{a^2+b^2+c^2}}{2}</math> | ||
| − | We square a+b+c: | + | We square <math>a+b+c</math>: |
* <math>(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784</math> | * <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 | We get that | ||
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== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}} | |
| − | + | ||
| − | + | [[Category:Intermediate Geometry Problems]] | |
Revision as of 16:00, 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 . Since the diameter of the sphere is the space diagonal of the box,
We square 
:
We get that
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 | |
