Difference between revisions of "2000 AIME I Problems/Problem 4"
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a_6 + a_9 &= a_7 + a_8.\end{align*}</cmath> | a_6 + a_9 &= a_7 + a_8.\end{align*}</cmath> | ||
− | + | Notice that <math>a_7 + a_9 = a_6 + a_8</math>. Expressing all terms 3 to 9 in terms of <math>a_1</math> and <math>a_2</math> and substituting their expanded forms into the previous equation will give the expression <math>5a_1 = 2a_2</math>. | |
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+ | We can guess that <math>a_1 = 2</math>. (If we started with <math>a_1</math> odd, the resulting sides would not be integers and we would need to scale up by a factor of <math>2</math> to make them integers; if we started with <math>a_1 > 2</math> even, the resulting dimensions would not be relatively prime and we would need to scale down.) 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>. These numbers are relatively prime, as desired. The perimeter is <math>2(61)+2(69)=\boxed{260}</math>. | ||
== See also == | == See also == |
Revision as of 00:16, 30 July 2018
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
Notice that . Expressing all terms 3 to 9 in terms of and and substituting their expanded forms into the previous equation will give the expression .
We can guess that . (If we started with odd, the resulting sides would not be integers and we would need to scale up by a factor of to make them integers; if we started with even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives , , , which gives us . These numbers are relatively prime, as desired. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.