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

Revision as of 07:55, 26 January 2012

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 $\begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ \intertext{and} a_6 + a_9 &= a_7 + a_8\end{align*}$

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 $2(61)+2(69)=\boxed{260}$ .

@@ Line 12: / Line 12: @@
 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:
+The picture shows that
-<cmath>\begin{eqnarray*}
+<cmath>\begin{align*}
-a_1+a_2 &=& a_3\\
+a_1+a_2 &= a_3\\
-a_1 + a_3 &=& a_4\\
+a_1 + a_3 &= a_4\\
-a_3 + a_4 &=& a_5\\
+a_3 + a_4 &= a_5\\
-a_4 + a_5 &=& a_6\\
+a_4 + a_5 &= a_6\\
-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\\
+a_1 + a_4 + a_6 &= a_9\\ \intertext{and}
-a_6 + a_9 &=& a_7 + a_8\end{eqnarray*}</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 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 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>.
 == 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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Difference between revisions of "2000 AIME I Problems/Problem 4"

Revision as of 07:55, 26 January 2012

Problem

Solution

See also