Difference between revisions of "2016 AIME I Problems/Problem 6"
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+ | ==Problem== | ||
In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | ||
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+ | ==Solution== | ||
+ | It is well known that <math>DA = DI = DB</math> and so we have <math>DA = DB = 5</math>. Then <math>\triangle DLB \sim \triangle ALC</math> and so <math>\frac{AL}{AC} = \frac{3}{5}</math> and from the angle bisector theorem <math>\frac{CI}{IL} = \frac{5}{3}</math> so <math>CI = \frac{10}{3}</math> and our answer is <math>\boxed{013}</math> |
Revision as of 16:07, 4 March 2016
Problem
In let be the center of the inscribed circle, and let the bisector of intersect at . The line through and intersects the circumscribed circle of at the two points and . If and , then , where and are relatively prime positive integers. Find .
Solution
It is well known that and so we have . Then and so and from the angle bisector theorem so and our answer is