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

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== Solution ==
 
== Solution ==
{{solution}}
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Call the length <math>l</math> and the width <math>w</math>. Call the squares' side lengths from smallest to largest <math>a_1,\ldots,a_9</math>.
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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>.
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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>.
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== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2000|n=I|num-b=3|num-a=5}}

Revision as of 15:30, 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.

[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 length $l$ and the width $w$. Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$.

The picture shows that $a_1_+a_2= a_3$ (Error compiling LaTeX. ! Double subscript.), $a_1 + a_3 = a_4$, $a_3 + a_4 = a_5$, $a_4 + a_5 =$, $a_2 + a_3 + a_5 = a_7$, $a_2 + a_7 = a_8$,$a_1 + a_4 + a_6 = a_9$, and $a_6 + a_9 = a_7 + a_8$.

With a lot of algebra, $a_9 = 36$, $a_6=25$, $a_8 = 33$, and the perimeter is $2(61)+2(69)=\boxed{398}$.


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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