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

(Solution)
 
(Solution)
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==Solution==
 
==Solution==
[asy]
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<asy>
 
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]
+
</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 $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$, then $ST=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm);  pair A = origin, B = (13,0), P = (4,0), Q = (7,0), 	T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), 	S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10));  Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1)); DrawPathArray(A--S^^B--T,rgb(0,0.4,0)); D(S--T,rgb(1,0.2,0.4));  D("A",A,dir(215)); D("B",B,dir(330)); D("P",P,dir(240)); D("Q",Q,dir(240)); D("T",T,dir(290)); D("C",C,dir(120)); D("S",S,dir(250));  MP("4",(A+P)/2,dir(90)); MP("3",(P+Q)/2,dir(90)); MP("6",(Q+B)/2,dir(90)); MP("5",(B+T)/2,dir(140)); MP("7",(A+S)/2,dir(40)); [/asy] Let $\angle ACP=\alpha$, $\angle PCQ=\beta$, and $\angle QCB=\gamma$. Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$, so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\]Similarly, we can deduce $\tfrac{PC}{CB}=\tfrac47$ and hence $\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}$.

Now Law of Sines on $\triangle ACS$, $\triangle SCT$, and $\triangle TCB$ yields \[\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.\]Hence \[\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},\]so \[TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.\]Hence $ST=\tfrac{35}8$ and the requested answer is $35+8=\boxed{43}$.

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