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, | + | The box P has dimensions <math>a</math>, <math>b</math>, and <math>c</math>. Therefore, |
| − | * <math> 2ab+2ac+2bc=384</math> | + | * <math>2ab+2ac+2bc=384</math> |
| − | * <math>4a+4b+4c=112 | + | * <math>4a+4b+4c=112 \Longrightarrow a + b + c = 28</math> |
| − | + | ||
Now we make a formula for <math>r</math>. 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 <math>a+b+c</math>: | 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 | ||
* <math>\frac{\sqrt{a^2+b^2+c^2}}{2}=10=r</math> | * <math>\frac{\sqrt{a^2+b^2+c^2}}{2}=10=r</math> | ||
Revision as of 20:56, 21 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 
, 
, and 
. 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 | |
