Difference between revisions of "2005 AIME II Problems/Problem 8"
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and the answer is <math>m+n+p=\boxed{405}</math>. | and the answer is <math>m+n+p=\boxed{405}</math>. | ||
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+ | ==Solution 2== | ||
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+ | Call our desired length <math>x</math>. Note for any <math>X</math> on <math>\overline{AB}</math> and <math>Y</math> on <math>\overline{O_1O_2}</math> such that <math>\overline{XY}\perp\overline{AB}</math> that the function <math>f</math> such that <math>f(\overline{O_1Y})=\overline{XY}</math> is linear. Since <math>(0,4)</math> and <math>(14,10)</math>, we can quickly interpolate that <math>f(10)=\overline{O_3T}=\frac{58}{7}</math>. Then, extend <math>\overline{O_3T}</math> until it reaches the circle on both sides; call them <math>P,Q</math>. By Power of a Point, | ||
+ | <math>\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}</math>. | ||
+ | Since <math>\overline{AT}=\overline{TB}=\frac{1}{2}x</math>, | ||
+ | <cmath>(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2</cmath> | ||
+ | <cmath>\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2</cmath> | ||
+ | After solving for <math>x</math>, we get <math>x=\frac{8\sqrt{390}}{7}</math>, so our answer is <math>8+390+7=\boxed{405}</math> | ||
== See also == | == See also == |
Revision as of 21:51, 14 August 2011
Contents
Problem
Circles and are externally tangent, and they are both internally tangent to circle The radii of and are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of is also a common external tangent of and Given that the length of the chord is where and are positive integers, and are relatively prime, and is not divisible by the square of any prime, find
Solution
Let be the centers and the radii of the circles . Let be the points of tangency from the common external tangent of , respectively, and let the extension of intersect the extension of at a point . Let the endpoints of the chord/tangent be , and the foot of the perpendicular from to be . From the similar right triangles ,
It follows that , and that . By the Pythagorean Theorem on , we find that
and the answer is .
Solution 2
Call our desired length . Note for any on and on such that that the function such that is linear. Since and , we can quickly interpolate that . Then, extend until it reaches the circle on both sides; call them . By Power of a Point, . Since , After solving for , we get , so our answer is
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |