Difference between revisions of "2014 AMC 12B Problems/Problem 14"
(Created page with "==Solution== Let the side lengths of the rectangular box be <math> x, y</math> and <math>z</math>. From the information we get <cmath> 4(x+y+z) = 48 \Rightarrow x+y+z = 12 <...") |
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Revision as of 17:53, 20 February 2014
Solution
Let the side lengths of the rectangular box be 
and 
.
From the information we get
![\[4(x+y+z) = 48 \Rightarrow x+y+z = 12\]](http://latex.artofproblemsolving.com/b/c/8/bc86401a844711e88def93692937422d75fb2fea.png)
![\[2(xy+yz+xz) = 94\]](http://latex.artofproblemsolving.com/9/e/7/9e781c0ada050703109cb29ab2e9696f544177b8.png)
The sum of all the lengths of the box's interior diagonals is
![\[4 \sqrt{x^2+y^2+z^2}\]](http://latex.artofproblemsolving.com/e/a/7/ea7e1a232129089dd72f040de7f6db3dad14b926.png)
Squaring the first expression, we get:
![\[144 =(x+y+z)^2 = x^2+y^2+x^2 + 2(xy+yz+xz)\]](http://latex.artofproblemsolving.com/8/f/d/8fdc1165a4a7da1576c47d879ed69a47cb8306c5.png)
![\[144 = x^2+y^2+x^2 + 94\]](http://latex.artofproblemsolving.com/5/2/d/52d64add03f1a2c4b8acad534a803526f558b07d.png)
Hence ![\[x^2+y^2+x^2 = 50\]](http://latex.artofproblemsolving.com/6/6/c/66c06f28ada26f9aefaaefe78c2a8b2f6e067b2b.png)
![\[4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2}\]](http://latex.artofproblemsolving.com/8/3/5/835d55911dfd8ed143f7905470abcd933e4bde66.png)
