Difference between revisions of "2005 AMC 12A Problems/Problem 22"

Revision as of 20:20, 3 July 2013

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 it's 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

The box P has dimensions $a$ , $b$ , and $c$ . Therefore,

$2ab+2ac+2bc=384$
$4a+4b+4c=112 \Longrightarrow a + b + c = 28$

Now we make a formula for $r$ . Since the diameter of the sphere is the space diagonal of the box,

$r=\frac{\sqrt{a^2+b^2+c^2}}{2}$

We square $a+b+c$ :

$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+384=784$

We get that

$\frac{\sqrt{a^2+b^2+c^2}}{2}=10=r$

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 12 Problems and Solutions

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

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 [[Category:Intermediate Geometry Problems]]
 [[Category:3D Geometry Problems]]
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Difference between revisions of "2005 AMC 12A Problems/Problem 22"

Revision as of 20:20, 3 July 2013

Problem

Solution

See also