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Difference between revisions of "2002 AIME II Problems/Problem 15"

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(Took me a couple years, but I solved it.)
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== Solution ==
 
== Solution ==
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{{solution}}
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Let the smaller angle between the <math>x</math>-axis and the line <math>y=mx</math> be <math>\theta</math>. Note that the centers of the two circles lie on the angle bisector of the angle between the <math>x</math>-axis and the line <math>y=mx</math>. Also note that if <math>(x,y)</math> is on said angle bisector, we have that <math>\frac{y}{x}=\tan{\frac{\theta}{2}}</math>.  Let <math>\tan{\frac{\theta}{2}}=m_1</math>, for convenience. Therefore if <math>(x,y)</math> is on the angle bisector, then <math>x=\frac{y}{m_1}</math>. Now let the centers of the two relevant circles be <math>(a/m_1 , a)</math> and <math>(b/m_1 , b)</math> for some positive reals <math>a</math> and <math>b</math>. These two circles are tangent to the <math>x</math>-axis, so the radii of the circles are <math>a</math> and <math>b</math> respectively. We know that the point <math>(9,6)</math> is a point on both circles, so we have that
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<cmath>(9-\frac{a}{m_1})^2+(6-a)^2=a^2</cmath>
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<cmath>(9-\frac{b}{m_1})^2+(6-b)^2=b^2</cmath>
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Expanding these and manipulating terms gives
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<cmath>\frac{1}{m_1}^2a^2-[(18/m_1)+12]a+117=0</cmath>
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<cmath>\frac{1}{m_1}^2b^2-[(18/m_1)+12]b+117=0</cmath>
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It follows that <math>a</math> and <math>b</math> are the roots of the quadratic
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<cmath>\frac{1}{m_1}^2x^2-[(18/m_1)+12]x+117=0</cmath>
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It follows from Vieta's Formulas that the product of the roots of this quadratic is <math>117m_1^2</math>, but we were also given that the product of the radii was 68. Therefore <math>68=117m_1^2</math>, or <math>m_1^2=\frac{68}{117}</math>. Note that the half-angle formula for tangents is
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<cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}</cmath>
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Therefore
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<cmath>\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}</cmath>
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Solving for <math>\cos{\theta}</math> gives that <math>\cos{\theta}=\frac{49}{185}</math>. It then follows that <math>\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}</math>.
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It then follows that <math>m=\tan{\theta}=\frac{12\sqrt{221}}{49}</math>. Therefore <math>a=12</math>, <math>b=221</math>, and <math>c=49</math>. The desired answer is then <math>12+221+49=\boxed{282}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=14|after=Last Question}}
 
{{AIME box|year=2002|n=II|num-b=14|after=Last Question}}

Revision as of 16:44, 27 November 2011

Problem

Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$, and the product of the radii is $68$. The x-axis and the line $y = mx$, where $m > 0$, are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$, where $a$, $b$, and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$.

Solution

Let the smaller angle between the $x$-axis and the line $y=mx$ be $\theta$. Note that the centers of the two circles lie on the angle bisector of the angle between the $x$-axis and the line $y=mx$. Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$. Let $\tan{\frac{\theta}{2}}=m_1$, for convenience. Therefore if $(x,y)$ is on the angle bisector, then $x=\frac{y}{m_1}$. Now let the centers of the two relevant circles be $(a/m_1 , a)$ and $(b/m_1 , b)$ for some positive reals $a$ and $b$. These two circles are tangent to the $x$-axis, so the radii of the circles are $a$ and $b$ respectively. We know that the point $(9,6)$ is a point on both circles, so we have that

\[(9-\frac{a}{m_1})^2+(6-a)^2=a^2\]

\[(9-\frac{b}{m_1})^2+(6-b)^2=b^2\]

Expanding these and manipulating terms gives

\[\frac{1}{m_1}^2a^2-[(18/m_1)+12]a+117=0\]

\[\frac{1}{m_1}^2b^2-[(18/m_1)+12]b+117=0\]

It follows that $a$ and $b$ are the roots of the quadratic

\[\frac{1}{m_1}^2x^2-[(18/m_1)+12]x+117=0\]

It follows from Vieta's Formulas that the product of the roots of this quadratic is $117m_1^2$, but we were also given that the product of the radii was 68. Therefore $68=117m_1^2$, or $m_1^2=\frac{68}{117}$. Note that the half-angle formula for tangents is

\[\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\]

Therefore

\[\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}\]

Solving for $\cos{\theta}$ gives that $\cos{\theta}=\frac{49}{185}$. It then follows that $\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}$.

It then follows that $m=\tan{\theta}=\frac{12\sqrt{221}}{49}$. Therefore $a=12$, $b=221$, and $c=49$. The desired answer is then $12+221+49=\boxed{282}$.

See also

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