Difference between revisions of "2000 AIME I Problems/Problem 4"
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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. | 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. | ||
− | The picture shows that <math>a_1+a_2= a_3</math> | + | The picture shows that: |
+ | <math></math>\begin{eqnarray*} | ||
+ | <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> | ||
+ | <math>a_6 + a_9 = a_7 + a_8</math> | ||
With a bit of trial and error and some arithmetic, we can use the last equation to find that <math>5a_1 = 2a_2</math>; [[without loss of generality]], let <math>a_1 = 2</math>. 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> (relatively prime), and the perimeter is <math>2(61)+2(69)=\boxed{260}</math>. | With a bit of trial and error and some arithmetic, we can use the last equation to find that <math>5a_1 = 2a_2</math>; [[without loss of generality]], let <math>a_1 = 2</math>. 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> (relatively prime), and the perimeter is <math>2(61)+2(69)=\boxed{260}</math>. |
Revision as of 00:19, 25 March 2011
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: $$ (Error compiling LaTeX. ! Missing $ inserted.)\begin{eqnarray*}
With a bit of trial and error and some arithmetic, we can use the last equation to find that ; without loss of generality, let . Then solving gives , , , which gives us (relatively prime), 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 |