2018 AIME II Problems/Problem 14
Contents
Problem
The incircle of triangle is tangent to at . Let be the other intersection of with . Points and lie on and , respectively, so that is tangent to at . Assume that , , , and , where and are relatively prime positive integers. Find .
Solution 1
Let the sides and be tangent to at and , respectively. Let and . Because and are both tangent to and and subtend the same arc of , it follows that . By equal tangents, . Applying the Law of Sines to yields Similarly, applying the Law of Sines to gives It follows that implying . Applying the same argument to yields from which . The requested sum is .
Solution 2 (Projective)
Let the incircle of be tangent to and at and . By Brianchon's theorem on tangential hexagons and , we know that and are concurrent at a point . Let . Then by La Hire's lies on the polar of so lies on the polar of . Therefore, also passes through . Then projecting through , we have Therefore, . Since we know that and . Therefore, and . Since , we also have . Solving for , we obtain .
-Vfire
Solution 3 (Combination of Law of Sine and Law of Cosine)
Let the center of the incircle of be . Link and . Then we have
Let the incircle of be tangent to and at and , let and .
Use Law of Sine in and , we have
therefore we have
Solve this equation, we have
As a result, , , , ,
So,
Use Law of Cosine in and , we have
And we have
So
Solve this equation, we have
As a result,
So, the final answer of this question is
~Solution by (Frank FYC)
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.