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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>, <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>, and <math>a_6 + a_9 = a_7 + a_8</math>.
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The picture shows that
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<cmath>\begin{align*}
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a_1+a_2 &= a_3\\
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a_1 + a_3 &= a_4\\
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a_3 + a_4 &= a_5\\
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a_4 + a_5 &= a_6\\
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a_2 + a_3 + a_5 &= a_7\\
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a_2 + a_7 &= a_8\\
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a_1 + a_4 + a_6 &= a_9\\
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a_6 + a_9 &= a_7 + a_8.\end{align*}</cmath>
  
[[Without loss of generality]], let <math>a_1 = 1</math>. With a bit of trial and error and some arithmetic, <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, which gives us <math>l=61,w=69</math>, and the perimeter is <math>2(61)+2(69)=\boxed{398}</math>.
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Expressing all terms 3 to 9 in terms of <math>a_1</math> and <math>a_2</math> and substituting their expanded forms into the previous equation will give the expression <math>5a_1 = 2a_2</math>.
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We can guess that <math>a_1 = 2</math>. (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.) 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. The perimeter is <math>2(61)+2(69)=\boxed{260}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:37, 1 August 2018

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*}

Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$.

We can guess that $a_1 = 2$. (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.) 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. 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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