Difference between revisions of "2005 AMC 12A Problems/Problem 22"
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<cmath>(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400</cmath> | <cmath>(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400</cmath> | ||
<cmath>\sqrt{l^2 + w^2 +h^2} = 20 = diameter</cmath> | <cmath>\sqrt{l^2 + w^2 +h^2} = 20 = diameter</cmath> | ||
| − | < | + | <cmath>r=\frac{20}{2} = 10</cmath> |
== See also == | == See also == | ||
Revision as of 19:05, 19 June 2017
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 
.
Surface area = ![\[2lw+2lh+2wl=384\]](http://latex.artofproblemsolving.com/5/f/9/5f9ca6c64954614c2d3c725223a35917292f446a.png)
Sum of all edges = ![\[4l+4w+4h=112 \Longrightarrow l + w + h = 28\]](http://latex.artofproblemsolving.com/5/3/c/53c45dbe2b23530ae4dc076796e311ba0f25cd26.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/6/6/0/6607e6c38ab202cd07d54df0d03e98af0083c131.png)
![\[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400\]](http://latex.artofproblemsolving.com/6/7/d/67dd3bb33bf380d8846978dd1087e25720f5eb12.png)
![\[\sqrt{l^2 + w^2 +h^2} = 20 = diameter\]](http://latex.artofproblemsolving.com/9/5/7/957c4c0ce0286ed1e8c64ac7f178cb9f833f4518.png)
![\[r=\frac{20}{2} = 10\]](http://latex.artofproblemsolving.com/b/0/2/b02e0fe4b5a02b2a38e0bd6c43b7a4de9b99acef.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. 
