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

(Solution)
(Solution)
Line 14: Line 14:
 
The picture shows that:
 
The picture shows that:
 
<math></math>\begin{eqnarray*}
 
<math></math>\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\\
a_3 + a_4 = a_5
+
a_3 + a_4 &=& a_5\\
a_4 + a_5 = a_6
+
a_4 + a_5 &=& a_6\\
a_2 + a_3 + a_5 = a_7
+
a_2 + a_3 + a_5 &=& a_7\\
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\\
  
 
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 00:22, 25 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: $$ (Error compiling LaTeX. ! Missing $ inserted.)\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\\

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 (ProblemsAnswer KeyResources)
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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