2005 AMC 12A Problem 22

Attempt Problem 22 of the 2005 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12A solutions, or check the answer key.

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22.

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?

88

1010

1212

1414

1616

Answer: B
Concepts:rectangular prismspherealgebraic manipulation
Difficulty rating: 1990
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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