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

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== Solution==
 
== Solution==
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KEEP IN MIND, THIS IS THE SAME SOLUTION AS THE OTHER ONE ON THIS PAGE HOWEVER THERE WAS A MATHEMATICAL TYPO IN ONE EQUATION. I do not take credit for the rest of the work.
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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>.  
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Its surface area is <cmath>2lw+2lh+2wh=384,</cmath>
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Its surface area is <cmath>2lw+2lh+2wh=384,</cmath> **here is where I fixed the typo**
 
and the sum of all its edges is <cmath>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.</cmath>
 
and the sum of all its edges is <cmath>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.</cmath>
  

Revision as of 08:28, 26 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

KEEP IN MIND, THIS IS THE SAME SOLUTION AS THE OTHER ONE ON THIS PAGE HOWEVER THERE WAS A MATHEMATICAL TYPO IN ONE EQUATION. I do not take credit for the rest of the work.

Box P has dimensions $l$, $w$, and $h$. Its surface area is \[2lw+2lh+2wh=384,\] **here is where I fixed the typo** 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 (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

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