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Difference between revisions of "2002 AIME I Problems/Problem 2"
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== Problem ==
 
== Problem ==
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The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as <math>\dfrac{1}{2}(\sqrt{p}-q)</math> where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.
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The diagram shows twenty congruent [[circle]]s arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The [[ratio]] of the longer dimension of the rectangle to the shorter dimension can be written as <math>\dfrac{1}{2}(\sqrt{p}-q)</math> where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.
  
 
<center>[[Image:AIME_2002I_Problem_02.png]]</center>
 
<center>[[Image:AIME_2002I_Problem_02.png]]</center>
  
 
== Solution ==
 
== Solution ==
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Let the radius of the circles is <math>r</math>. The longer dimension can be written as <math>14r</math>, and by Pythagorean Theorem, we can find that the shorter dimension: <math>2r\sqrt{3}+2r</math>.
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Let the [[radius]] of the circles be <math>r</math>. The longer dimension of the rectangle can be written as <math>14r</math>, and by the [[Pythagorean Theorem]], we find that the shorter dimension is <math>2r\left(\sqrt{3}+1\right)</math>.
  
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Therefore, <math>\frac{14r}{2r\sqrt{3}+2r}=\frac{1}{2}(\sqrt{p}-q)</math>
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Therefore, <math>\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)</math>. Thus we have <math>p=147</math> and <math>q=7</math>, so <math>p+q=\boxed{154}</math>.
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After simplifying, we get <math>7\sqrt{3}-7=\sqrt{p}-q</math>, which gives <math>p=147</math> and <math>q=7</math>. So, <math>p+q=154</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2002|n=I|num-b=1|num-a=3}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 13:50, 10 June 2008

Problem

The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$.

AIME 2002I Problem 02.png

Solution

Let the radius of the circles be $r$. The longer dimension of the rectangle can be written as $14r$, and by the Pythagorean Theorem, we find that the shorter dimension is $2r\left(\sqrt{3}+1\right)$.

Therefore, $\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)$. Thus we have $p=147$ and $q=7$, so $p+q=\boxed{154}$.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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