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

Revision as of 15:37, 1 July 2022

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

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

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 $\boxed{43}$ .

Solution 4

Extend $\overline{AB}$ past $B$ to point $X$ so that $CPTX$ is cyclic. Then, by Power of a Point on $CPTX$ , $(CQ)(QB) = (PQ)(QX)$ . By Power of a Point on $CATB$ , $(CQ)(QT) = (AQ)(QB) = 42$ . Thus, $(PQ)(QX) = 42$ , so $BX = 8$ .

By the Inscribed Angle Theorem on $CPTX$ , $\angle SCT = \angle BXT$ . By the Inscribed Angle Theorem on $ASTC$ , $\angle SCT = \angle SAT$ , so $\angle BXT = \angle SAT$ . Since $ASTB$ is cyclic, $\angle AST = \angle TBX$ . Thus, $\triangle AST \sim \triangle XBT$ , so $AS/XB = ST/BT$ . Solving for $ST$ yields $ST = \frac{35}{8}$ , for a final answer of $35+8 = \boxed{043}$ .

~ Leo.Euler

Solution 5 (5 = 2 + 3)

By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find $AS\cdot BT+AB\cdot ST=AT\cdot BS.$ Projecting through $C$ we have $\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}.$ Therefore $AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies$ $\left(\frac {AQ \cdot PB}{PQ} - AB\right)\times ST = AS \cdot BT \implies$ $ST = \frac {AS \cdot BT \cdot PQ}{AQ \cdot PB – AB \cdot PQ}$ $ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed {43}.$

Shelomovskii, vvsss, www.deoma-cmd.ru

@@ Line 41: / Line 41: @@
 ==Solution 2 (Projective Geometry)==
+[[File:2016 AIME II 10c.png|400px|right]]
 Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}</math>

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