2016 AIME II Problems/Problem 10

Problem

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 1

Notice that AS is parallel to BC, and BT is parallel to AC. So, AT = BC = 63/4, and BS = AC = 35/6. Then, by Ptolemy's, we have 63/4 * 35/6 = 735/8 = 13ST + 35 => ST = 35/8 => 043. - asdf334 (please texify)

Solution 2

$[asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm); pair A = origin, B = (13,0), P = (4,0), Q = (7,0), T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10)); Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1)); DrawPathArray(A--S^^B--T,rgb(0,0.4,0)); D(S--T,rgb(1,0.2,0.4)); D("A",A,dir(215)); D("B",B,dir(330)); D("P",P,dir(240)); D("Q",Q,dir(240)); D("T",T,dir(290)); D("C",C,dir(120)); D("S",S,dir(250)); MP("4",(A+P)/2,dir(90)); MP("3",(P+Q)/2,dir(90)); MP("6",(Q+B)/2,dir(90)); MP("5",(B+T)/2,dir(140)); MP("7",(A+S)/2,dir(40)); [/asy]$ 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 $35+8=\boxed{43}$ .

Edit: Note that the finish is much simpler. Once you get $\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}$ , you can solve quickly from there getting $ST=\dfrac{AS \sin(\beta)}{\sin(\alpha)}=7\cdot \dfrac{15}{24}=\dfrac{35}{8}$ .

Solution 3 (Projective Geometry)

Projecting through $C$ we have $\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}$ which easily gives $ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}$

Solution 4

By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find $5\cdot 7+13\cdot ST=AT\cdot BS.$ Therefore, in order to find $ST$ , it suffices to find $AT\cdot BS$ . We do this using similar triangles, which can be found by using Power of a Point theorem.

As $\triangle APS\sim \triangle CPB$ , we find $\frac{4}{PC}=\frac{7}{BC}.$ Therefore, $\frac{BC}{PC}=\frac{7}{4}$ .

As $\triangle BQT\sim\triangle CQA$ , we find $\frac{6}{CQ}=\frac{5}{AC}.$ Therefore, $\frac{AC}{CQ}=\frac{5}{6}$ .

As $\triangle ATQ\sim\triangle CBQ$ , we find $\frac{AT}{BC}=\frac{7}{CQ}.$ Therefore, $AT=\frac{7\cdot BC}{CQ}$ .

As $\triangle BPS\sim \triangle CPA$ , we find $\frac{9}{PC}=\frac{BS}{AC}.$ Therefore, $BS=\frac{9\cdot AC}{PC}$ . Thus we find $AT\cdot BS=\left(\frac{7\cdot BC}{CQ}\right)\left(\frac{9\cdot AC}{PC}\right).$ But now we can substitute in our previously found values for $\frac{BC}{PC}$ and $\frac{AC}{CQ}$ , finding $AT\cdot BS=63\cdot \frac{7}{4}\cdot \frac{5}{6}=\frac{21\cdot 35}{8}.$ Substituting this into our original expression from Ptolemy's Theorem, we find $\begin{align*}35+13ST&=\frac{21\cdot 35}{8}\\13ST&=\frac{13\cdot 35}{8}\\ST&=\frac{35}{8}.\end{align*}$ Thus the answer is $\boxed{43}$ .

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

Page

Toolbox

Search