Difference between revisions of "2005 AIME II Problems/Problem 15"
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== Problem == | == Problem == | ||
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Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math> | Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math> | ||
== Solution == | == Solution == | ||
− | + | Rewrite the given equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>. | |
− | 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 | 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> | <math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math> | ||
− | <math>16 - r = \sqrt{(x+5)^2 + (y-12)^2} | + | <math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}</math> |
Solving for <math>r</math> in both equations and setting them equal yields | Solving for <math>r</math> in both equations and setting them equal yields | ||
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<math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math> | <math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math> | ||
− | Squaring again and canceling yields | + | Squaring again and canceling yields <math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math> |
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− | <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. | ||
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<math>(-96a)^2 - 4(3+4a^2)(276) = 0</math> | <math>(-96a)^2 - 4(3+4a^2)(276) = 0</math> | ||
− | Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is 169. | + | Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>169</math>. |
== See also == | == See also == | ||
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{{AIME box|year=2005|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2005|n=II|num-b=14|after=Last Question}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 13:30, 19 April 2008
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 and .
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 .
See also
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 |