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

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== Solution ==
 
== Solution ==
  
{{solution}}
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Rewrite the given equations as
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<math>(x+5)^2 + (y-12)^2 = 256</math>
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<math>(x-5)^2 + (y-12)^2 = 16</math>
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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
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<math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math>
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<math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}.</math>
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Solving for <math>r</math> in both equations and setting them equal yields
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<math>20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}</math>
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Squaring both sides, canceling common terms, and rearranging yields
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<math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math>
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Squaring again and canceling yields
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<math>1 = \frac{x^2}{100} + {(y-12)^2}{75}.</math>
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So the locus of points that can be the center of the circle with the desired properties is an ellipse.
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Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:
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<math>1 = \frac{x^2}{100} + {(ax-12)^2}{75}</math>
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Expanding yields
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<math>(3+4a^2)x^2 - 96ax + 276 = 0</math>
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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
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<math>(-96a)^2 - 4(3+4a^2)(276) = 0</math>
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Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is 169.
  
 
== See also ==
 
== See also ==

Revision as of 03:50, 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

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

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 $y$:

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

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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