Difference between revisions of "2005 AIME II Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | {{solution}} | + | Rewrite the given equations as |
| + | <math>(x+5)^2 + (y-12)^2 = 256</math> | ||
| + | <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 | ||
| + | |||
| + | <math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math> | ||
| + | <math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}.</math> | ||
| + | |||
| + | Solving for <math>r</math> in both equations and setting them equal yields | ||
| + | <math>20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}</math> | ||
| + | |||
| + | Squaring both sides, canceling common terms, and rearranging yields | ||
| + | |||
| + | <math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math> | ||
| + | |||
| + | Squaring again and canceling yields | ||
| + | |||
| + | <math>1 = \frac{x^2}{100} + {(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. | ||
| + | |||
| + | 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> | ||
| + | |||
| + | Expanding yields | ||
| + | <math>(3+4a^2)x^2 - 96ax + 276 = 0</math> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is 169. | ||
== See also == | == See also == | ||
Revision as of 02:50, 14 March 2007
Problem
Let 
and 
denote the circles 
and 
respectively. Let 
be the smallest possible value of 
for which the line 
contains the center of a circle that is externally tangent to and internally tangent to 
Given that 
where 
and 
are relatively prime integers, find 
Solution
Rewrite the given equations as
Let 
have center 
and radius 
. Now, if two circles with radii 
and 
are externally tangent, then the distance between their centers is 
, and if they are internally tangent, it is 
. So we have
Solving for in both equations and setting them equal yields
Squaring both sides, canceling common terms, and rearranging yields
Squaring again and canceling yields
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 
, we substitute for 
:
Expanding yields
We want the value of that makes the line tangent to the ellipse, which will mean that for that choice of there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is 0. So
Solving yields 
, so the answer is 169.
