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2005 AMC 12A Problem 22

Problem 22 of 25HarderAlgebraGeometry

A rectangular box PP is inscribed in a sphere of radius r.r. The surface area of PP is 384,384, and the sum of the lengths of its 1212 edges is 112.112. What is r?r?

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Solution

Let the dimensions be x,y,z.x, y, z. The 1212 edges give 4(x+y+z)=112,4(x + y + z) = 112, so x+y+z=28,x + y + z = 28, and the surface area gives 2xy+2yz+2xz=384.2xy + 2yz + 2xz = 384. The space diagonal is a diameter of the sphere, so (2r)2=x2+y2+z2=(x+y+z)2(2xy+2yz+2xz)=282384=400. \begin{aligned} &(2r)^2 = x^2 + y^2 + z^2 \\ &= (x + y + z)^2 \\ &\quad {}- (2xy + 2yz + 2xz) \\ &= 28^2 - 384 = 400. \end{aligned} Thus 2r=202r = 20 and r=10.r = 10. Thus, the correct answer is B.

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Concepts: rectangular prism · sphere · algebraic manipulation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.