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 "2013 AIME I Problems/Problem 7"

(→‎Solution)
(→‎Solution)
Line 9: Line 9:
 
900 = <math>\frac{1}{2}</math>((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2)((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2 - 10)((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2 - <math>\sqrt{(x/2)^2 + 64}</math>)((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2 - <math>\sqrt{(x/2)^2 + 36}</math>).
 
900 = <math>\frac{1}{2}</math>((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2)((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2 - 10)((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2 - <math>\sqrt{(x/2)^2 + 64}</math>)((10 + <math>\sqrt{(x/2)^2 + 64}</math> + <math>\sqrt{(x/2)^2 + 36}</math>)/2 - <math>\sqrt{(x/2)^2 + 36}</math>).
  
−
Solving, we get <math>\boxed{041}</math>.
+
Solving, we get <math>\boxed{041}</math>.
 +
 
 +
P.S.
 +
No expressions cancel out. Good luck.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2013|n=I|num-b=6|num-a=8}}

Revision as of 17:08, 18 March 2013

Problem 7

A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$.

Solution

After using the pythagorean formula three times, we can quickly see that the sides of the triangle are 10, $\sqrt{(x/2)^2 + 64}$, and $\sqrt{(x/2)^2 + 36}$. Therefore, we can use Heron's formula to set up an equation for the area of the triangle.

The semi perimeter is (10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2

900 = $\frac{1}{2}$((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2 - 10)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2 - $\sqrt{(x/2)^2 + 64}$)((10 + $\sqrt{(x/2)^2 + 64}$ + $\sqrt{(x/2)^2 + 36}$)/2 - $\sqrt{(x/2)^2 + 36}$).

Solving, we get $\boxed{041}$.

P.S. No expressions cancel out. Good luck.

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_7&oldid=51856"