Difference between revisions of "2005 AMC 12A Problems/Problem 22"

(Solution)
(Solution)
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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>
<math>r=\frac{20}{2} = 10</math>
+
<cmath>r=\frac{20}{2} = 10</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 20:05, 19 June 2017

Problem

A rectangular box $P$ is inscribed in a sphere of radius $r$. The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$?

$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$

Solution

Box P has dimensions $l$, $w$, and $h$. Surface area = \[2lw+2lh+2wl=384\] Sum of all edges = \[4l+4w+4h=112 \Longrightarrow l + w + h = 28\]

The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}\] \[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400\] \[\sqrt{l^2 + w^2 +h^2} = 20 = diameter\] \[r=\frac{20}{2} = 10\]

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
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

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