2005 AMC 12A Problem 22

Question

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

Let the dimensions be

x, y, z.

The edge sum gives

x + y + z

and the surface area gives

2xy + 2yz + 2xz

The sphere’s diameter is the space diagonal:

(2r)^2

= x^2 + y^2 + z^2

= (x+y+z)^2

- (2xy + 2yz + 2xz)

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

\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 = 20

and

r = 10.

Thus, the correct answer is B.

Problem 22 in other years