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

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== Solution ==
 
== Solution ==
 
The box P has dimensions <math>a</math>, <math>b</math>, and <math>c</math>. Therefore,
 
The box P has dimensions <math>a</math>, <math>b</math>, and <math>c</math>. Therefore,
−
* <math>2ab+2ac+2bc=384</math>
+
<cmath>2ab+2ac+2bc=384</cmath>
−
* <math>4a+4b+4c=112 \Longrightarrow a + b + c = 28</math>
+
<cmath>4a+4b+4c=112 \Longrightarrow a + b + c = 28</cmath>
  
 
Now we make a formula for <math>r</math>. Since the [[diameter]] of the sphere is the space diagonal of the box,
 
Now we make a formula for <math>r</math>. Since the [[diameter]] of the sphere is the space diagonal of the box,
−
* <math>r=\frac{\sqrt{a^2+b^2+c^2}}{2}</math>
+
<cmath>r=\frac{\sqrt{a^2+b^2+c^2}}{2}</cmath>
  
 
We square <math>a+b+c</math>:
 
We square <math>a+b+c</math>:
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* <math>(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784</math>
+
<cmath>(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784</cmath>
  
 
We get that
 
We get that
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* <math>\frac{\sqrt{a^2+b^2+c^2}}{2}=10=r</math>
+
<cmath>r=\frac{\sqrt{a^2+b^2+c^2}}{2}=B)10</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 19:34, 7 November 2013

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 it's 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

The box P has dimensions $a$, $b$, and $c$. Therefore, \[2ab+2ac+2bc=384\] \[4a+4b+4c=112 \Longrightarrow a + b + c = 28\]

Now we make a formula for $r$. Since the diameter of the sphere is the space diagonal of the box, \[r=\frac{\sqrt{a^2+b^2+c^2}}{2}\]

We square $a+b+c$: \[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784\]

We get that \[r=\frac{\sqrt{a^2+b^2+c^2}}{2}=B)10\]

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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