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.
![[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]](http://latex.artofproblemsolving.com/d/8/1/d8140d628cc0905f333aa8deaface1b89ee6c983.png)
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. Unknown error_msg)\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 |