Difference between revisions of "2005 AMC 12A Problems/Problem 22"
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Box P has dimensions <math>l</math>, <math>w</math>, and <math>h</math>. | Box P has dimensions <math>l</math>, <math>w</math>, and <math>h</math>. | ||
Its surface area is <cmath>2lw+2lh+2wl=384,</cmath> | Its surface area is <cmath>2lw+2lh+2wl=384,</cmath> | ||
| − | + | and the sum of all its edges is <cmath>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.</cmath> | |
The diameter of the sphere is the space diagonal of the prism, which is <cmath>\sqrt{l^2 + w^2 +h^2}.</cmath> | The diameter of the sphere is the space diagonal of the prism, which is <cmath>\sqrt{l^2 + w^2 +h^2}.</cmath> | ||
Notice that <cmath>(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,</cmath> so the diameter is | Notice that <cmath>(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,</cmath> so the diameter is | ||
| − | <cmath>\sqrt{l^2 + w^2 +h^2} = 20</cmath>. The radius is half of the diameter, so | + | <cmath>\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20</cmath>. The radius is half of the diameter, so |
<cmath>r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.</cmath> | <cmath>r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.</cmath> | ||
Revision as of 18:00, 30 October 2018
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 
.
Its surface area is ![\[2lw+2lh+2wl=384,\]](http://latex.artofproblemsolving.com/b/e/6/be6f4fdd31b4b9eb2042e2256288b5ca1e0d73c5.png)
and the sum of all its edges is ![\[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\]](http://latex.artofproblemsolving.com/8/6/3/86391c1d6be352d4b06253140f65e260ee99b750.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/3/1/3/31384be5c4dd31ee79fefadd8f754b0e543415a6.png)
Notice that ![\[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\]](http://latex.artofproblemsolving.com/2/0/e/20ea50479e3eb41006e5cba6901f12d434e6abda.png)
so the diameter is
![\[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20\]](http://latex.artofproblemsolving.com/2/8/4/284c40611d19d2a3218cf5a74a1592b80cc6fc97.png)
. The radius is half of the diameter, so
![\[r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.\]](http://latex.artofproblemsolving.com/8/a/0/8a0ac99e9fe30e6457565990dbbf925afcada5c3.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. 
