Difference between revisions of "2016 AIME II Problems/Problem 10"
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By Ptolemy's Theorem applied to quadrilateral <math>ASTB</math>, we find | By Ptolemy's Theorem applied to quadrilateral <math>ASTB</math>, we find | ||
<cmath>5\cdot 7+13\cdot ST=AT\cdot BS.</cmath> | <cmath>5\cdot 7+13\cdot ST=AT\cdot BS.</cmath> | ||
− | Therefore, in order to find <math>ST</math>, it suffices to find <math>AT\cdot BS</math>. We do this using similar triangles. | + | Therefore, in order to find <math>ST</math>, it suffices to find <math>AT\cdot BS</math>. We do this using similar triangles, which can be found by using Power of a Point theorem. |
As <math>\triangle APS\sim \triangle CPB</math>, we find | As <math>\triangle APS\sim \triangle CPB</math>, we find |
Revision as of 18:28, 13 March 2017
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 .
Contents
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, , so .
Solution 2
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 .
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.