2016 AIME II Problems/Problem 10
Contents
Problem
Triangle is inscribed in circle . Points and are on side with . Rays and meet again at and (other than ), respectively. If and , then , where and are relatively prime positive integers. Find .
Solution 1
Let , , and . Note that since we have , so by the Ratio Lemma Similarly, we can deduce and hence .
Now Law of Sines on , , and yields Hence so Hence and the requested answer is .
Edit: Note that the finish is much simpler. Once you get , you can solve quickly from there getting .
Solution 2 (Projective Geometry)
Projecting through we have which easily gives .
Solution 3
By Ptolemy's Theorem applied to quadrilateral , we find Therefore, in order to find , it suffices to find . We do this using similar triangles, which can be found by using Power of a Point theorem.
As , we find Therefore, .
As , we find Therefore, .
As , we find Therefore, .
As , we find Therefore, . Thus we find But now we can substitute in our previously found values for and , finding Substituting this into our original expression from Ptolemy's Theorem, we find Thus the answer is .
Solution 4
Extend past to point so that is cyclic. Then, by Power of a Point on , . By Power of a Point on , . Thus, , so .
By the Inscribed Angle Theorem on , . By the Inscribed Angle Theorem on , , so . Since is cyclic, . Thus, , so . Solving for yields , for a final answer of .
~ Leo.Euler
Solution 5 (5 = 2 + 3)
By Ptolemy's Theorem applied to quadrilateral , we find Projecting through we have Therefore
vladimir.shelomovskii@gmail.com, vvsss
Solution 6
Connect and
So we need to get the ratio of
By clear observation , we have , LOS tells so we get , the desired answer is leads to
~blusoul
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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