Difference between revisions of "2016 AIME II Problems/Problem 10"
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Edit: Note that the finish is much simpler. Once you get, <math>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}</math>, so <math>ST=\dfrac{AS*\sin(\beta)}{\sin(\alpha)}=7*(15/24)=35/8</math>. | Edit: Note that the finish is much simpler. Once you get, <math>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}</math>, so <math>ST=\dfrac{AS*\sin(\beta)}{\sin(\alpha)}=7*(15/24)=35/8</math>. | ||
− | ==Solution 2(Projective Geometry)== | + | ==Solution 2 (Projective Geometry)== |
Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}</math> | Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}</math> | ||
Revision as of 20:25, 5 March 2020
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, , so .
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 .
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.