# Difference between revisions of "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 1

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

## Solution 2 (Alcumus)

Let $r_1$ and $r_2$ be the radii of the circles. Then the centers of the circles are of the form $(kr_1,r_1)$ and $(kr_2,r_2)$ for the same constant $k,$ since the two centers are collinear with the origin. Since $(9,6)$ lies on both circles, $$(kr - 9)^2 + (r - 6)^2 = r^2,$$where $r$ represents either radius. Expanding, we get $$k^2 r^2 - (18k + 12) r + 117 = 0.$$We are told the product of the circles is 68, so by Vieta's formulas, $\frac{117}{k^2} = 68.$ Hence, $k^2 = \frac{117}{68},$ and $k = \sqrt{\frac{117}{68}}.$ $[asy] unitsize(0.25 cm); pair[] O; real[] r; pair P; r = 4.096; r = 16.6; O = (r/(2/3*sqrt(17/13)),r); O = (r/(2/3*sqrt(17/13)),r); P = reflect(O,O)*(9,6); draw(Circle(O,r)); //draw(Circle(O,r)); draw(arc(O,r,130,300)); draw((0,0)--(8,12*sqrt(221)/49*8)); draw((0,0)--(30,0)); draw((0,0)--O--(O.x,0)); draw(O--(O + reflect((0,0),(10,12*sqrt(221)/49*10))*(O))/2); label("y = mx", (8,12*sqrt(221)/49*8), N); dot("(9,6)", (9,6), NE); dot("(kr,r)", O, N); dot(P,red); [/asy]$

Since the circle is tangent to the line $y = mx,$ the distance from the center $(kr,r)$ to the line is $r.$ We can write $y = mx$ as $y - mx = 0,$ so from the distance formula, $$\frac{|r - krm|}{\sqrt{1 + m^2}} = r.$$Squaring both sides, we get $$\frac{(r - krm)^2}{1 + m^2} = r^2,$$so $(r - krm)^2 = r^2 (1 + m^2).$ Since $r \neq 0,$ we can divide both sides by 0, to get $$(1 - km)^2 = 1 + m^2.$$Then $1 - 2km + k^2 m^2 = 1 + m^2,$ so $m^2 (1 - k^2) + 2km = 0.$ Since $m \neq 0,$ $$m(1 - k^2) + 2k = 0.$$Hence, $$m = \frac{2k}{k^2 - 1} = \frac{2 \sqrt{\frac{117}{68}}}{\frac{117}{68} - 1} = \boxed{\frac{12 \sqrt{221}}{49}}.$$

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