2005 AMC 12A Problem 22

Below is the professionally curated solution for Problem 22 of the 2005 AMC 12A, 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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Concepts:rectangular prism sphere algebraic manipulation

Difficulty rating: 1990

22.

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 its $12$ edges is $112.$ What is $r?$

$8$

$10$

$12$

$14$

$16$

Solution:

Let the dimensions be $x, y, z.$ The $12$ edges give $4(x + y + z) = 112,$ so $x + y + z = 28,$ and the surface area gives $2xy + 2yz + 2xz = 384.$

The space diagonal is a diameter of the sphere, so $(2r)^2 = x^2 + y^2 + z^2 = (x + y + z)^2 - (2xy + 2yz + 2xz) = 28^2 - 384 = 400.$

Thus $2r = 20$ and $r = 10.$

Thus, the correct answer is B.

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