2000 AIME I Problems/Problem 4
Contents
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 1
Call the squares' side lengths from smallest to largest , and let represent the dimensions of the rectangle.
The picture shows 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 .
Solution 1.2 (more detail)
We can just list the equations: We can then write each in terms of and as follows Since Since the side lengths of the rectangle are relatively prime, we can see that and Therefore, ~peelybonehead
Solution 2 Length-chasing (Angle-chasing but for side lengths)
We set the side length of the smallest square to 1, and set the side length of square in the previous question to a. We do some "side length chasing" and get . Solving, we get and the side lengths are and . Thus, the perimeter of the rectangle 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 | ||
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