# Difference between revisions of "2005 AIME II Problems/Problem 15"

## Problem

Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$

## Solution 1

Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$.

Let $w_3$ have center $(x,y)$ and radius $r$. Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$, and if they are internally tangent, it is $|r_1 - r_2|$. So we have

\begin{align*} r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*}

Solving for $r$ in both equations and setting them equal, then simplifying, yields

\begin{align*} 20 - \sqrt{(x+5)^2 + (y-12)^2} &= \sqrt{(x-5)^2 + (y-12)^2} \\ 20+x &= 2\sqrt{(x+5)^2 + (y-12)^2} \end{align*}

Squaring again and canceling yields $1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.$

So the locus of points that can be the center of the circle with the desired properties is an ellipse.

$[asy] size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d; pair A = (-5, 12), B = (5, 12), C = (0, 0); D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red); D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow(4)); void bluecirc (real x) { pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue); D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype("4 4")); } bluecirc(-9.2); bluecirc(-4); bluecirc(3); [/asy]$

Since the center lies on the line $y = ax$, we substitute for $y$ and expand: $$1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.$$

We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$, so $(-96a)^2 - 4(3+4a^2)(276) = 0$.

Solving yields $a^2 = \frac{69}{100}$, so the answer is $\boxed{169}$.

## Solution 2

As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$. Let $F_1=(-5,12)$ and $F_2=(5,12)$. If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$, then $CF_1=16-r$ and $CF_2=4+r$. Therefore, $CF_1+CF_2=20$. In particular, the locus of points $C$ that can be centers of circles must be an ellipse with foci $F_1$ and $F_2$ and major axis $20$.

Clearly, the minimum value of the slope $a$ will occur when the line $y=ax$ is tangent to this ellipse. Suppose that this point of tangency is denoted by $T$, and the line $y=ax$ is denoted by $\ell$. Then we reflect the ellipse over $\ell$ to a new ellipse with foci $F_1'$ and $F_2'$ as shown below.

$[asy] size(220); pair F1 = (-5, 12), F2 = (5, 12),C=(0,12); draw(circle(F1,16)); draw(circle(F2,4)); draw(ellipse(C,10,5*sqrt(3))); xaxis("x",Arrows); yaxis("y",Arrows); dot(F1^^F2^^C); real l(real x) {return sqrt(69)*x/10;} path g=graph(l,-7,14); draw(g); draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3))); pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10))); dot(T); pair F1P=reflect((0,0),(10,l(10)))*F1; pair F2P=reflect((0,0),(10,l(10)))*F2; dot(F1P^^F2P); dot((0,0)); label("F_1",F1,N,fontsize(9)); label("F_2",F2,N,fontsize(9)); label("F_1'",F1P,SE,fontsize(9)); label("F_2'",F2P,SE,fontsize(9)); label("O",(0,0),NW,fontsize(9)); label("\ell",(13,l(13)),SE,fontsize(9)); label("T",T,NW,fontsize(9)); draw((0,0)--F1--F2--F2P--F1P--cycle); draw(F1--F2P^^F2--F1P); [/asy]$

By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that $F_1$, $T$, and $F_2'$ are collinear, and similarly, $F_2$, $T$ and $F_1'$ are collinear. Therefore, $OF_1F_2F_2'F_1'$ is a pentagon with $OF_1=OF_2=OF_1'=OF_2'=13$, $F_1F_2=F_1'F_2'=10$, and $F_1F_2'=F_1'F_2=20$. Note that $\ell$ bisects $\angle F_1'OF_1$. We can bisect this angle by bisecting $\angle F_1'OF_2$ and $F_2OF_1$ separately.

We proceed using complex numbers. Triangle $F_2OF_1'$ is isosceles with side lengths $13,13,20$. The height of this from the base of $20$ is $\sqrt{69}$. Therefore, the complex number $\sqrt{69}+10i$ represents the bisection of $\angle F_1'OF_2$.

Similarly, using the 5-12-13 triangles, we easily see that $12+5i$ represents the bisection of the angle $F_2OF_1$. Therefore, we can add these two angles together by multiplying the complex numbers, finding $$\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.$$ Now the point $F_1$ is given by the complex number $-5+12i$. Therefore, to find a point on line $\ell$, we simply subtract $\frac{1}{2}\angle F_1'OF_1$, which is the same as multiplying $-5+12i$ by the conjugate of $(\sqrt{69}+10i)(12+5i)$. We find $$(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).$$ In particular, note that the tangent of the argument of this complex number is $\sqrt{69}/10$, which must be the slope of the tangent line. Hence $a^2=69/100$, and the answer is $\boxed{169}$.

## Solution 3

We use the same reflection as in Solution 2. As $OF_1'=OF_2=13$, we know that $\triangle OF_1'F_2$ is isosceles. Hence $\angle F_2F_1'O=\angle F_1'F_2O$. But by symmetry, we also know that $\angle OF_1T=\angle F_2F_1'O$. Hence $\angle OF_1T=\angle F_1'F_2O$. In particular, as $\angle OF_1T=\angle OF_2T$, this implies that $O, F_1, F_2$, and $T$ are concyclic.

Let $X$ be the intersection of $F_2F_1'$ with the $x$-axis. As $F_1F_2$ is parallel to the $x$-axis, we know that $$\angle TXO=180-\angle F_1F_2T.\tag{1}$$ But $$180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}$$ By the fact that $OF_1F_2T$ is cyclic, $$\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}$$ Therefore, combining (1), (2), and (3), we find that $$\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}$$

By symmetry, we also know that $$\angle F_1TO=\angle OTF_1'.\tag{5}$$ Therefore, (4) and (5) show by AA similarity that $\triangle F_1OT\sim \triangle OXT$. Therefore, $\angle XOT=\angle OF_1T$.

Now as $OF_1=OF_2'=13$, we know that $\triangle OF_1F_2'$ is isosceles, and as $F_1F_2'=20$, we can drop an altitude to $F_1F_2'$ to easily find that $\tan \angle OF_1T=\sqrt{69}/10$. Therefore, $\tan\angle XOT$, which is the desired slope, must also be $\sqrt{69}/10$. As before, we conclude that the answer is $\boxed{169}$.

## Solution 4

First, rewrite the equations for the circles as $(x+5)^2+(y-12)^2=16^2$ and $(x-5)^2+(y-12)^2=4^2$. Then, choose a point $(a,b)$ that is a distance of $x$ from both circles. Use the distance formula between $(a,b)$ and each of $A$ and $C$ (in the diagram above). The distances, as can be seen in the diagram above are $16-x$ and $4+x$, respectively. $$(a-5)^2+(b-12)^2=(4+x)^2$$ $$(a+5)^2+(b-12)^2=(16-x)^2$$ Subtracting the first equation from the second gives $$20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2$$ Substituting this into the first equation gives $$a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4$$ $$b^2-24b+69+\frac{3a^2}4=0$$ Now, instead of converting this to the equation of an eclipse, solve for $b$ and then divide by $a$. $$b=\frac{24\pm\sqrt{300-3a^2}}{2}$$ We take the smaller root to minimize $\frac b a$. $$\frac b a=\frac{24-\sqrt{300-3a^2}}{2a}=\frac{24-\sqrt3\cdot\sqrt{100-a^2}}{2a}=\frac{12}a-\frac{\sqrt3}{2a}\sqrt{100-a^2}$$ Now, let $10\cos\theta=a$. This way, $\sqrt{100-a^2}=10\sin\theta$. Substitute this in. $\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ Then, take the derivative of this and set it to 0 to find the minimum value. $\frac{6}{5}\sec\theta\tan\theta-\frac{\sqrt3}{2}\sec^2\theta=0\rightarrow\frac{6}{5}\sin\theta-frac{\sqrt3}{2}=0\rightarrow\sin\theta=\frac{5\sqrt3}{12}$ Then, use this value of $\sin\theta$ to find the minimum of $\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ to get $\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}$

## Solution 5 (probably fastest)

Like before, notice that the distances from the centers of the given circles to the desired center are $4+r$ and $16-r$, which add up to $20$. This means that the possible centers of the third circle lie on an ellipse with foci $(-5, 12)$ and $(5, 12)$. Using the fact that the sum of the distances from the foci is $20$, we find that the semi-major axis has length $10$ and the semi-minor axis has length $5\sqrt{3}$. Therefore, the equation of the ellipse is $$\dfrac{x^2}{100}+\dfrac{(y-12)^2}{75} = 1,$$ where the numbers $100$ and $75$ come from $10^2$ and $(5\sqrt{3})^2$ respectively.

We proceed to find $m$ using the same method as Solution 1.

## Solution 6

First, obtain the equation of the ellipse as laid out in previous solutions. We now scale the coordinate plane in the $x$ direction by a factor of $\frac{\sqrt{3}}{2}$ centered at $x=0.$ This takes the ellipse to a circle centered at $(0,12)$ with radius $5\sqrt{3}$ and takes the line $y=ax$ to $y=\left( \frac{\sqrt{3}}{2} \right)^{-1} ax.$ The tangent point of our line to the circle with positive slope forms a right triangle with the origin and the center of the circle. Thus, the distance from this tangent point to the origin is $\sqrt{69}.$ By similar triangles, the slope of this line is then $\frac{\sqrt{69}}{5\sqrt{3}}.$ We multiply this by $\frac{\sqrt{3}}{2}$ to get $a=\frac{\sqrt{69}}{10},$ so our final answer is $\boxed{169.}$

 2005 AIME II (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byLast Question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions