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 possible 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

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

$r + 4 = \sqrt{(x-5)^2 + (y-12)^2}$

$16 - r = \sqrt{(x+5)^2 + (y-12)^2}$

Solving for $r$ in both equations and setting them equal yields $20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}$

Squaring both sides, canceling common terms, and rearranging yields

$20+x = 2\sqrt{(x+5)^2 + (y-12)^2}$

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 $y$ :

$1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75}$

Expanding yields

$(3+4a^2)x^2 - 96ax + 276 = 0$

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 $169$ .

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

Page

Toolbox

Search

2005 AIME II Problems/Problem 15

Problem

Solution

See also