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

(Solution)
Line 23: Line 23:
 
Squaring again and canceling yields
 
Squaring again and canceling yields
  
<math>1 = \frac{x^2}{100} + {(y-12)^2}{75}.</math>
+
<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.
 
So the locus of points that can be the center of the circle with the desired properties is an ellipse.
Line 29: Line 29:
 
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:
 
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:
  
<math>1 = \frac{x^2}{100} + {(ax-12)^2}{75}</math>
+
<math>1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75}</math>
  
 
Expanding yields
 
Expanding yields

Revision as of 03:51, 14 March 2007

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 $(x+5)^2 + (y-12)^2 = 256$ $(x-5)^2 + (y-12)^2 = 16$

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.

See also

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