2000 AIME I Problems/Problem 4

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 $2(61)+2(69)=\boxed{260}$ .

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2000 AIME I Problems/Problem 4

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