Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 12B Problems/Problem 14"

(→‎Problem 14)
Line 28: Line 28:
  
 
<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>
 +
 +
==Solution 2==
 +
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.
 +
~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}}

Revision as of 09:15, 11 October 2021

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:

\[144 =(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+xz)\]

\[144 = x^2+y^2+z^2 + 94\]

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

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

Solution 2

Let $a, b, c$ be the side lengths of the rectangular prism. So we have $2(ab + ac + bc) = 94$ and $a + b + c= 12$. Note that the interior diagonal is just $\sqrt{a^2 + b^2 + c^2}$. Since $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + ac + bc) = 144 - 94 = 50$. So the length of the interior diagonal is just $5\sqrt{2}$. Since we have 4 of the same interior diagonals that answer is $20\sqrt{2}$ which is D. ~coolmath_2018

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_14&oldid=163407"