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

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Revision as of 10:00, 7 April 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

  • $\frac{\sqrt{a^2+b^2+c^2}}{2}=10=r$

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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