Difference between revisions of "2000 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
− | The diagram shows a rectangle that has been dissected into nine non-overlapping | + | The diagram shows a [[rectangle]] that has been dissected into nine non-overlapping [[square]]s. Given that the width and the height of the rectangle are relatively prime positive integers, find the [[perimeter]] of the rectangle. |
<center><asy>draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); | <center><asy>draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); | ||
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== Solution == | == Solution == | ||
− | + | Call the squares' side lengths from smallest to largest <math>a_1,\ldots,a_9</math>, and let <math>l,w</math> represent the dimensions of the rectangle. | |
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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>. | ||
+ | [[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>. | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=3|num-a=5}} | {{AIME box|year=2000|n=I|num-b=3|num-a=5}} | ||
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+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 17:11, 31 December 2007
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 , and let represent the dimensions of the rectangle.
The picture shows that , , , , , , , and .
Without loss of generality, let . With a bit of trial and error and some arithmetic, , , , which gives us , and the perimeter is .
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 |