2002 AIME II Problems/Problem 15

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

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

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

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

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

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 $12+221+49=\boxed{282}$ .

2002 AIME II (Problems • Answer Key • Resources)
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2002 AIME II Problems/Problem 15

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