AIME 2005 II · 第 15 题

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

$$\begin{aligned} r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{aligned}$$ Solving for $r$ in both equations and setting them equal, then simplifying, yields

$$\begin{aligned} 20 - \sqrt{(x+5)^2 + (y-12)^2} &= \sqrt{(x-5)^2 + (y-12)^2} \\ 20+x &= 2\sqrt{(x+5)^2 + (y-12)^2} \end{aligned}$$ 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.

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.

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 $\boxed{169.}$