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
Notice that AS is parallel to BC, and BT is parallel to AC. So, AT = BC = 63/4, and BS = AC = 35/6. Then, by Ptolemy's, we have 63/4 * 35/6 = 735/8 = 13ST + 35 => ST = 35/8 => 043. - asdf334 (please texify)
Solution 2
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 3 (Projective Geometry)
Projecting through we have which easily gives
Solution 4
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.