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Difference between revisions of "2014 AMC 12B Problems/Problem 14"
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Squaring the first expression, we get:  
 
Squaring the first expression, we get:  
  
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<cmath>144 =(x+y+z)^2 =  x^2+y^2+z^2 + 2(xy+yz+xz)</cmath>
+
\begin{align*}
−
 
+
144 =(x+y+z)^2 &=  x^2+y^2+z^2 + 2(xy+yz+xz) \\
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<cmath>144 =  x^2+y^2+z^2 + 94</cmath>
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&=  x^2+y^2+z^2 + 94.
 +
\end{align*}
  
 
Hence <cmath>x^2+y^2+z^2 = 50</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>
 
<cmath>4 \sqrt{x^2+y^2+z^2} =  \boxed{\textbf{(D)}\ 20\sqrt 2} </cmath>
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==Solution 2==
 
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Let <math>a, b, c</math> be the side lengths of the rectangular prism. So we have <math>2(ab + ac + bc) = 94</math> and <math>a + b + c= 12</math>. Note that the interior diagonal is just <math>\sqrt{a^2 + b^2 + c^2}</math>. Since <math>a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + ac + bc) = 144 - 94 = 50</math>. So the length of the interior diagonal is just <math>5\sqrt{2}</math>. Since we have 4 of the same interior diagonals that answer is <math>20\sqrt{2}</math> which is D.
 
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~coolmath_2018
 
  
 
== See also ==
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:27, 18 June 2024

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?

$\textbf{(A)}\ 8\sqrt{3}\qquad\textbf{(B)}\ 10\sqrt{2}\qquad\textbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2}$

Solution

Let the side lengths of the rectangular box be $x, y$ and $z$. From the information we get

\[4(x+y+z) = 48 \Rightarrow x+y+z = 12\]

\[2(xy+yz+xz) = 94\]

The sum of all the lengths of the box's interior diagonals is

\[4 \sqrt{x^2+y^2+z^2}\]

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 \[x^2+y^2+z^2 = 50\]

\[4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2}\]

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

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