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

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== Solution==
 
== Solution==
 
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>.  
Surface area = <cmath>2lw+2lh+2wl=384</cmath>
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Its surface area is <cmath>2lw+2lh+2wl=384,</cmath>
Sum of all edges = <cmath>4l+4w+4h=112 \Longrightarrow l + w + h = 28</cmath>
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while the sum of all 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>
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The diameter of the sphere is the space diagonal of the prism, which is <cmath>\sqrt{l^2 + w^2 +h^2}.</cmath>
<cmath>(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400</cmath>
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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 = diameter</cmath>
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<cmath>\sqrt{l^2 + w^2 +h^2} = 20</cmath>. The radius is half of the diameter, so
<cmath>r=\frac{20}{2} = \boxed{\textbf{(B)} 10}</cmath>
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<cmath>r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 18:58, 30 October 2018

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 $h$. Its surface area is \[2lw+2lh+2wl=384,\] while the sum of all edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28\]

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

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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