Difference between revisions of "2014 AMC 12B Problems/Problem 14"
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<cmath>4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2} </cmath> | <cmath>4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2} </cmath> | ||
| + | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2014|ab=B|num-b=13|num-a=15}} | ||
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Revision as of 12:24, 22 February 2014
Problem 14
A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?
$\textbf{(A)}\ 8\sqrt{3}\qquad\textbf{(B)}\ 10\sqrt{2}\qquad\textbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2}$ (Error compiling LaTeX. Unknown error_msg)
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)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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. 
