Difference between revisions of "2000 AIME I Problems/Problem 4"
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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 =</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>. | 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 =</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>. | ||
− | With a '''lot''' of algebra, <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, and the perimeter is <math>2(61)+2(69)=\boxed{398}</math>. | + | With a '''lot''' of algebra, <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, <math>l=61,w=69</math>, and the perimeter is <math>2(61)+2(69)=\boxed{398}</math>. |
Revision as of 15:31, 24 November 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 length and the width . Call the squares' side lengths from smallest to largest .
The picture shows that $a_1_+a_2= a_3$ (Error compiling LaTeX. ), , , , , ,, and .
With a lot of algebra, , , , , 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 |