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

Page

Toolbox

Search

2000 AIME I Problems/Problem 4

Problem

Solution

See also