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 $\boxed{169}$ .

2005 AIME II (Problems • Answer Key • Resources)
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2005 AIME II Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See also