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

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<cmath>\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 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>
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== Solution 2 ==
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As in the previous solution, we have that <math>2lw+2lh+2wl=384</math> and <math>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28</math>, and the diameter of the sphere is the space diagonal of the prism, <math>\sqrt{l^2 + w^2 + h^2}</math>.
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Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that <math>h=0</math>. (This essentially means that we have an infinitesimally thin box.) We now have that <math>2lw = 384</math> and <math>l + w = 28</math>, and we are solving for <math>\sqrt{l^2 + w^2}</math>. Because <cmath>(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,</cmath> this means that <cmath>l^2 + w^2 = 28^2 - 384 = 400,</cmath> so the space diagonal is <math>\sqrt{400} = 20</math>. Since the diameter of the sphere is <math>20</math>, the radius is <math>\boxed{\textbf{(B)} 10}</math>. ~[[User:emerald_block|emerald_block]]
  
 
== See also ==
 
== See also ==

Revision as of 12:44, 25 March 2020

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 \[2lw+2lh+2wl=384,\] and the sum of all its 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 \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so \[r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.\]

Solution 2

As in the previous solution, we have that $2lw+2lh+2wl=384$ and $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28$, and the diameter of the sphere is the space diagonal of the prism, $\sqrt{l^2 + w^2 + h^2}$.


Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that $h=0$. (This essentially means that we have an infinitesimally thin box.) We now have that $2lw = 384$ and $l + w = 28$, and we are solving for $\sqrt{l^2 + w^2}$. Because \[(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,\] this means that \[l^2 + w^2 = 28^2 - 384 = 400,\] so the space diagonal is $\sqrt{400} = 20$. Since the diameter of the sphere is $20$, the radius is $\boxed{\textbf{(B)} 10}$. ~emerald_block

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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