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 "2000 AIME I Problems/Problem 4"

(→‎Solution)
(→‎Solution)
Line 13: Line 13:
  
 
The picture shows that:
 
The picture shows that:
−
<math></math>\begin{eqnarray*}
+
<cmath>\begin{eqnarray*}
 
a_1+a_2 &=& a_3\\
 
a_1+a_2 &=& a_3\\
 
a_1 + a_3 &=& a_4\\
 
a_1 + a_3 &=& a_4\\
Line 21: Line 21:
 
a_2 + a_7 &=& a_8\\
 
a_2 + a_7 &=& a_8\\
 
a_1 + a_4 + a_6 &=& a_9\\
 
a_1 + a_4 + a_6 &=& a_9\\
−
a_6 + a_9 &=& a_7 + a_8\\
+
a_6 + a_9 &=& a_7 + a_8\end{eqnarray*}</cmath>
  
 
With a bit of trial and error and some arithmetic, we can use the last equation to find that <math>5a_1 = 2a_2</math>; [[without loss of generality]], let <math>a_1 = 2</math>. Then solving gives <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, which gives us <math>l=61,w=69</math> (relatively prime), and the perimeter is <math>2(61)+2(69)=\boxed{260}</math>.
 
With a bit of trial and error and some arithmetic, we can use the last equation to find that <math>5a_1 = 2a_2</math>; [[without loss of generality]], let <math>a_1 = 2</math>. Then solving gives <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, which gives us <math>l=61,w=69</math> (relatively prime), and the perimeter is <math>2(61)+2(69)=\boxed{260}</math>.

Revision as of 23:23, 24 March 2011

Problem

The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.

[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]

Solution

Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle.

The picture shows that: \begin{eqnarray*} a_1+a_2 &=& a_3\\ a_1 + a_3 &=& a_4\\ a_3 + a_4 &=& a_5\\ a_4 + a_5 &=& a_6\\ a_2 + a_3 + a_5 &=& a_7\\ a_2 + a_7 &=& a_8\\ a_1 + a_4 + a_6 &=& a_9\\ a_6 + a_9 &=& a_7 + a_8\end{eqnarray*}

With a bit of trial and error and some arithmetic, we can use the last equation to find that $5a_1 = 2a_2$; without loss of generality, let $a_1 = 2$. Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$ (relatively prime), and the perimeter is $2(61)+2(69)=\boxed{260}$.

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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=2000_AIME_I_Problems/Problem_4&oldid=37761"