2005 AMC 12A Problems/Problem 22

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 its 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

Box P has dimensions $l$ , $w$ , and $h$ . Its surface area is $2lw+2lh+2wl=384,$ and the sum of all its edges is $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.$

The diameter of the sphere is the space diagonal of the prism, which is $\sqrt{l^2 + w^2 +h^2}.$ Notice that $(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,$ so the diameter is $\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.$ The radius is half of the diameter, so $r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.$

2005 AMC 12A (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2005 AMC 12A Problems/Problem 22

Problem

Solution

See also