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2005 AMC 12A Problems/Problem 22

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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 • 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
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Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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