# Difference between revisions of "2016 AIME II Problems/Problem 10"

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import cse5; | import cse5; | ||

pathpen = black; pointpen = black; | pathpen = black; pointpen = black; | ||

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MP("5",(B+T)/2,dir(140)); | MP("5",(B+T)/2,dir(140)); | ||

MP("7",(A+S)/2,dir(40)); | MP("7",(A+S)/2,dir(40)); | ||

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Let <math>\angle ACP=\alpha</math>, <math>\angle PCQ=\beta</math>, and <math>\angle QCB=\gamma</math>. Note that since <math>\triangle ACQ\sim\triangle TBQ</math> we have <math>\tfrac{AC}{CQ}=\tfrac56</math>, so by the Ratio Lemma <cmath>\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.</cmath>Similarly, we can deduce <math>\tfrac{PC}{CB}=\tfrac47</math> and hence <math>\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}</math>. | Let <math>\angle ACP=\alpha</math>, <math>\angle PCQ=\beta</math>, and <math>\angle QCB=\gamma</math>. Note that since <math>\triangle ACQ\sim\triangle TBQ</math> we have <math>\tfrac{AC}{CQ}=\tfrac56</math>, so by the Ratio Lemma <cmath>\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.</cmath>Similarly, we can deduce <math>\tfrac{PC}{CB}=\tfrac47</math> and hence <math>\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}</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>. | 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>. |

## Revision as of 21:08, 17 March 2016

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

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 .