Difference between revisions of "2016 AIME II Problems/Problem 10"
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==Solution== | ==Solution== | ||
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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)); | ||
− | + | </asy> | |
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
.