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

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(Solution 6)
 
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== Problem ==
 
== Problem ==
Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math>
+
Let <math> w_1 </math> and <math> w_2 </math> denote the [[circle]]s <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positive value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math>
  
== Solution ==
+
== Solution 1 ==
 
Rewrite the given equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>.
 
Rewrite the given equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>.
  
 
Let <math>w_3</math> have center <math>(x,y)</math> and radius <math>r</math>.  Now, if two circles with radii <math>r_1</math> and <math>r_2</math> are externally tangent, then the distance between their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>.  So we have
 
Let <math>w_3</math> have center <math>(x,y)</math> and radius <math>r</math>.  Now, if two circles with radii <math>r_1</math> and <math>r_2</math> are externally tangent, then the distance between their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>.  So we have
  
<math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math>
+
<cmath>\begin{align*}
 +
r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\
 +
16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*} </cmath>
  
<math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}</math>
+
Solving for <math>r</math> in both equations and setting them equal, then simplifying, yields
  
Solving for <math>r</math> in both equations and setting them equal yields
+
<cmath>\begin{align*}
<math>20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}</math>
+
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*} </cmath>
  
Squaring both sides, canceling common terms, and rearranging yields
+
Squaring again and canceling yields <math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math>
 +
 
 +
So the locus of points that can be the center of the circle with the desired properties is an [[ellipse]].
 +
 
 +
<center><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));
  
<math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math>
+
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"));
 +
}
  
Squaring again and canceling yields <math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math>
+
bluecirc(-9.2); bluecirc(-4); bluecirc(3);
 +
</asy></center>
 +
 
 +
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math> and expand:
 +
<cmath>1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.</cmath>
 +
 
 +
We want the value of <math>a</math> that makes the line <math>y=ax</math> tangent to the ellipse, which will mean that for that choice of <math>a</math> there is only one solution to the most recent equation. But a quadratic has one solution [[iff]] its discriminant is <math>0</math>, so <math>(-96a)^2 - 4(3+4a^2)(276) = 0</math>.
 +
 
 +
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>\boxed{169}</math>.
 +
 
 +
== Solution 2 ==
 +
As above, we rewrite the equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>. Let <math>F_1=(-5,12)</math> and <math>F_2=(5,12)</math>. If a circle with center <math>C=(a,b)</math> and radius <math>r</math> is externally tangent to <math>w_2</math> and internally tangent to <math>w_1</math>, then <math>CF_1=16-r</math> and <math>CF_2=4+r</math>. Therefore, <math>CF_1+CF_2=20</math>. In particular, the locus of points <math>C</math> that can be centers of circles must be an ellipse with foci <math>F_1</math> and <math>F_2</math> and major axis <math>20</math>.
 +
 
 +
Clearly, the minimum value of the slope <math>a</math> will occur when the line <math>y=ax</math> is tangent to this ellipse. Suppose that this point of tangency is denoted by <math>T</math>, and the line <math>y=ax</math> is denoted by <math>\ell</math>. Then we reflect the ellipse over <math>\ell</math> to a new ellipse with foci <math>F_1'</math> and <math>F_2'</math> as shown below.
 +
<center><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></center>
 +
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 <math>F_1</math>, <math>T</math>, and <math>F_2'</math> are collinear, and similarly, <math>F_2</math>, <math>T</math> and <math>F_1'</math> are collinear. Therefore, <math>OF_1F_2F_2'F_1'</math> is a pentagon with <math>OF_1=OF_2=OF_1'=OF_2'=13</math>, <math>F_1F_2=F_1'F_2'=10</math>, and <math>F_1F_2'=F_1'F_2=20</math>. Note that <math>\ell</math> bisects <math>\angle F_1'OF_1</math>. We can bisect this angle by bisecting <math>\angle F_1'OF_2</math> and <math>F_2OF_1</math> separately.
 +
 
 +
We proceed using complex numbers. Triangle <math>F_2OF_1'</math> is isosceles with side lengths <math>13,13,20</math>. The height of this from the base of <math>20</math> is <math>\sqrt{69}</math>. Therefore, the complex number <math>\sqrt{69}+10i</math> represents the bisection of <math>\angle F_1'OF_2</math>.
 +
 
 +
Similarly, using the 5-12-13 triangles, we easily see that <math>12+5i</math> represents the bisection of the angle <math>F_2OF_1</math>. Therefore, we can add these two angles together by multiplying the complex numbers, finding
 +
<cmath>\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.</cmath>
 +
Now the point <math>F_1</math> is given by the complex number <math>-5+12i</math>. Therefore, to find a point on line <math>\ell</math>, we simply subtract <math>\frac{1}{2}\angle F_1'OF_1</math>, which is the same as multiplying <math>-5+12i</math> by the conjugate of <math>(\sqrt{69}+10i)(12+5i)</math>. We find
 +
<cmath>(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).</cmath>
 +
In particular, note that the tangent of the argument of this complex number is <math>\sqrt{69}/10</math>, which must be the slope of the tangent line. Hence <math>a^2=69/100</math>, and the answer is <math>\boxed{169}</math>.
 +
 
 +
== Solution 3 ==
 +
We use the same reflection as in Solution 2. As <math>OF_1'=OF_2=13</math>, we know that <math>\triangle OF_1'F_2</math> is isosceles. Hence <math>\angle F_2F_1'O=\angle F_1'F_2O</math>. But by symmetry, we also know that <math>\angle OF_1T=\angle F_2F_1'O</math>. Hence <math>\angle OF_1T=\angle F_1'F_2O</math>. In particular, as <math>\angle OF_1T=\angle OF_2T</math>, this implies that <math>O, F_1, F_2</math>, and <math>T</math> are concyclic.
  
So the locus of points that can be the center of the circle with the desired properties is an ellipse.
+
Let <math>X</math> be the intersection of <math>F_2F_1'</math> with the <math>x</math>-axis. As <math>F_1F_2</math> is parallel to the <math>x</math>-axis, we know that <cmath>\angle TXO=180-\angle F_1F_2T.\tag{1}</cmath> But <cmath>180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}</cmath> By the fact that <math>OF_1F_2T</math> is cyclic, <cmath>\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}</cmath> Therefore, combining (1), (2), and (3), we find that
 +
<cmath>\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}</cmath>
  
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:
+
By symmetry, we also know that
 +
<cmath>\angle F_1TO=\angle OTF_1'.\tag{5}</cmath>
 +
Therefore, (4) and (5) show by AA similarity that <math>\triangle F_1OT\sim \triangle OXT</math>. Therefore, <math>\angle XOT=\angle OF_1T</math>.
  
<math>1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75}</math>
+
Now as <math>OF_1=OF_2'=13</math>, we know that <math>\triangle OF_1F_2'</math> is isosceles, and as <math>F_1F_2'=20</math>, we can drop an altitude to <math>F_1F_2'</math> to easily find that <math>\tan \angle OF_1T=\sqrt{69}/10</math>. Therefore, <math>\tan\angle XOT</math>, which is the desired slope, must also be <math>\sqrt{69}/10</math>. As before, we conclude that the answer is <math>\boxed{169}</math>.
  
Expanding yields
+
==Solution 4==
 +
[[Image:2005_AIME_II_-15.png||center|800px]]
 +
First, rewrite the equations for the circles as <math>(x+5)^2+(y-12)^2=16^2</math> and <math>(x-5)^2+(y-12)^2=4^2</math>. 
 +
Then, choose a point <math>(a,b)</math> that is a distance of <math>x</math> from both circles.  Use the distance formula between <math>(a,b)</math> and each of <math>A</math> and <math>C</math> (in the diagram above).  The distances, as can be seen in the diagram above are <math>16-x</math> and <math>4+x</math>, respectively.
 +
<cmath>(a-5)^2+(b-12)^2=(4+x)^2</cmath>
 +
<cmath>(a+5)^2+(b-12)^2=(16-x)^2</cmath>
 +
Subtracting the first equation from the second gives <cmath>20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2</cmath>
 +
Substituting this into the first equation gives
 +
<cmath>a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4</cmath>
 +
<cmath>b^2-24b+69+\frac{3a^2}4=0</cmath>
 +
Now, instead of converting this to the equation of an eclipse, solve for <math>b</math> and then divide by <math>a</math>.
 +
<cmath>b=\frac{24\pm\sqrt{300-3a^2}}{2}</cmath>
 +
We take the smaller root to minimize <math>\frac b a</math>.
 +
<cmath>\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}</cmath>
 +
Now, let <math>10\cos\theta=a</math>.  This way, <math>\sqrt{100-a^2}=10\sin\theta</math>.
 +
Substitute this in.  <math>\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta</math>
 +
Then, take the derivative of this and set it to 0 to find the minimum value.
 +
<math>\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}</math>
 +
Then, use this value of <math>\sin\theta</math> to find the minimum of <math>\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta</math> to get <math>\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}</math>
  
<math>(3+4a^2)x^2 - 96ax + 276 = 0</math>
+
==Solution 5 (probably fastest)==
 +
Like before, notice that the distances from the centers of the given circles to the desired center are <math>4+r</math> and <math>16-r</math>, which add up to <math>20</math>. This means that the possible centers of the third circle lie on an ellipse with foci <math>(-5, 12)</math> and <math>(5, 12)</math>. Using the fact that the sum of the distances from the foci is <math>20</math>, we find that the semi-major axis has length <math>10</math> and the semi-minor axis has length <math>5\sqrt{3}</math>. Therefore, the equation of the ellipse is <cmath>\dfrac{x^2}{100}+\dfrac{(y-12)^2}{75} = 1,</cmath> where the numbers <math>100</math> and <math>75</math> come from <math>10^2</math> and <math>(5\sqrt{3})^2</math> respectively.
  
We want the value of <math>a</math> that makes the line <math>y=ax</math> tangent to the ellipse, which will mean that for that choice of <math>a</math> there is only one solution to the most recent equation.  But a quadratic has one solution iff its discriminant is 0.  So
 
  
<math>(-96a)^2 - 4(3+4a^2)(276) = 0</math>
+
We proceed to find <math>m</math> using the same method as Solution 1.
  
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>169</math>.
+
==Solution 6==
 +
First, obtain the equation of the ellipse as laid out in previous solutions. We now scale the coordinate plane in the <math>x</math> direction by a factor of <math>\frac{\sqrt{3}}{2}</math> centered at <math>x=0.</math> This takes the ellipse to a circle centered at <math>(0,12)</math> with radius <math>5\sqrt{3}</math> and takes the line <math>y=ax</math> to <math>y=\left( \frac{\sqrt{3}}{2} \right)^{-1} ax.</math> 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 <math>\sqrt{69}.</math> By similar triangles, the slope of this line is then <math>\frac{\sqrt{69}}{5\sqrt{3}}.</math> We multiply this by <math>\frac{\sqrt{3}}{2}</math> to get <math>a=\frac{\sqrt{69}}{10},</math> so our final answer is <math>\boxed{169.}</math>
  
 
== See also ==
 
== See also ==
Line 40: Line 132:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:17, 28 December 2020

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

2005 AIME II -15.png

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 $\boxed{169.}$

See also

2005 AIME II (ProblemsAnswer KeyResources)
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