Difference between revisions of "2014 AMC 12B Problems/Problem 14"
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| − | ==Solution== | + | ==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? | ||
| + | |||
| + | <math> \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} </math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
| + | |||
| + | ==Solution 1== | ||
Let the side lengths of the rectangular box be <math> x, y</math> and <math>z</math>. | Let the side lengths of the rectangular box be <math> x, y</math> and <math>z</math>. | ||
| Line 5: | Line 12: | ||
<cmath> 4(x+y+z) = 48 \Rightarrow x+y+z = 12 </cmath> | <cmath> 4(x+y+z) = 48 \Rightarrow x+y+z = 12 </cmath> | ||
| − | |||
<cmath>2(xy+yz+xz) = 94 </cmath> | <cmath>2(xy+yz+xz) = 94 </cmath> | ||
| Line 15: | Line 21: | ||
Squaring the first expression, we get: | Squaring the first expression, we get: | ||
| − | + | \begin{align*} | |
| + | 144 =(x+y+z)^2 &= x^2+y^2+z^2 + 2(xy+yz+xz) \\ | ||
| + | &= x^2+y^2+z^2 + 94. | ||
| + | \end{align*} | ||
| − | <cmath> | + | Hence <cmath>x^2+y^2+z^2 = 50</cmath> |
| − | + | <cmath>4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2} </cmath> | |
| + | ==Solution 2 (Observation)== | ||
| − | <cmath>4 \sqrt{ | + | Observe that a rectangular prism with side lengths of <math>3,4,5</math> fits all the necessary conditions. Thus, the answer is <cmath>4\sqrt{3^2+4^2+5^2} = \boxed{\textbf{(D)}\ 20\sqrt 2}</cmath> |
| + | ~LuisFonseca123 | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2014|ab=B|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:57, 1 September 2025
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?
Solution 1
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:
\begin{align*} 144 =(x+y+z)^2 &= x^2+y^2+z^2 + 2(xy+yz+xz) \\ &= x^2+y^2+z^2 + 94. \end{align*}
Hence ![\[x^2+y^2+z^2 = 50\]](http://latex.artofproblemsolving.com/8/4/3/84312496748f2bd8594ce025b657bf34646b97f2.png)
![\[4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2}\]](http://latex.artofproblemsolving.com/8/3/5/835d55911dfd8ed143f7905470abcd933e4bde66.png)
Solution 2 (Observation)
Observe that a rectangular prism with side lengths of 
fits all the necessary conditions. Thus, the answer is ![\[4\sqrt{3^2+4^2+5^2} = \boxed{\textbf{(D)}\ 20\sqrt 2}\]](http://latex.artofproblemsolving.com/2/2/f/22f98a1663c626f582ca6c86305eadf3e1b9297a.png)
~LuisFonseca123
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
