Difference between revisions of "2016 AIME II Problems/Problem 10"
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Now Law of Sines on <math>\triangle ACS</math>, <math>\triangle SCT</math>, and <math>\triangle TCB</math> yields <cmath>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.</cmath>Hence <cmath>\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},</cmath>so <cmath>TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.</cmath>Hence <math>ST=\tfrac{35}8</math> and the requested answer is <math>35+8=\boxed{43}</math>. | Now Law of Sines on <math>\triangle ACS</math>, <math>\triangle SCT</math>, and <math>\triangle TCB</math> yields <cmath>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.</cmath>Hence <cmath>\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},</cmath>so <cmath>TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.</cmath>Hence <math>ST=\tfrac{35}8</math> and the requested answer is <math>35+8=\boxed{43}</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{ | + | 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== | ==Solution 2== |
Revision as of 19:14, 5 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.
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.