Difference between revisions of "2000 AIME I Problems/Problem 4"

m (Solution)
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a_2 + a_3 + a_5 &= a_7\
 
a_2 + a_3 + a_5 &= a_7\
 
a_2 + a_7 &= a_8\
 
a_2 + a_7 &= a_8\
a_1 + a_4 + a_6 &= a_9\ \intertext{and}
+
a_1 + a_4 + a_6 &= a_9\
a_6 + a_9 &= a_7 + a_8\end{align*}</cmath>
+
a_6 + a_9 &= a_7 + a_8.\end{align*}</cmath>
  
 
With a bit of trial and error and some arithmetic, we can use these equations to find that <math>5a_1 = 2a_2</math>; we can guess that <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>.  These numbers are relatively prime, as desired.  (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.)  The perimeter is <math>2(61)+2(69)=\boxed{260}</math>.
 
With a bit of trial and error and some arithmetic, we can use these equations to find that <math>5a_1 = 2a_2</math>; we can guess that <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>.  These numbers are relatively prime, as desired.  (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.)  The perimeter is <math>2(61)+2(69)=\boxed{260}</math>.

Revision as of 19:31, 1 March 2015

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]

Solution

Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle.

The picture shows that \begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a_9 &= a_7 + a_8.\end{align*}

With a bit of trial and error and some arithmetic, we can use these equations to find that $5a_1 = 2a_2$; we can guess that $a_1 = 2$. Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) The perimeter is $2(61)+2(69)=\boxed{260}$.

See also

2000 AIME I (ProblemsAnswer KeyResources)
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

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