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Difference between revisions of "2000 AIME I Problems/Problem 4"

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The picture shows that <math>a_1+a_2= a_3</math>, <math>a_1 + a_3 = a_4</math>, <math>a_3 + a_4 = a_5</math>, <math>a_4 + a_5 = a_6</math>, <math>a_2 + a_3 + a_5 = a_7</math>, <math>a_2 + a_7 = a_8 </math>, <math> a_1 + a_4 + a_6 = a_9 </math>, and <math>a_6 + a_9 = a_7 + a_8</math>.
 
The picture shows that <math>a_1+a_2= a_3</math>, <math>a_1 + a_3 = a_4</math>, <math>a_3 + a_4 = a_5</math>, <math>a_4 + a_5 = a_6</math>, <math>a_2 + a_3 + a_5 = a_7</math>, <math>a_2 + a_7 = a_8 </math>, <math> a_1 + a_4 + a_6 = a_9 </math>, and <math>a_6 + a_9 = a_7 + a_8</math>.
  
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[[Without loss of generality]], let <math>a_1 = 1</math>. With a bit of trial and error and some arithmetic, <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, which gives us <math>l=61,w=69</math>, and the perimeter is <math>2(61)+2(69)=\boxed{398}</math>.
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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{398}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 10:40, 1 January 2008

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 $a_1+a_2= a_3$, $a_1 + a_3 = a_4$, $a_3 + a_4 = a_5$, $a_4 + a_5 = a_6$, $a_2 + a_3 + a_5 = a_7$, $a_2 + a_7 = a_8$, $a_1 + a_4 + a_6 = a_9$, and $a_6 + a_9 = a_7 + a_8$.

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 $2(61)+2(69)=\boxed{398}$.

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
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