2000 AIME I Problems/Problem 4

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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.

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

2000 AIME I (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2000 AIME I Problems/Problem 4

From AoPSWiki

Problem

Solution

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