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

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==Solution 2 (Projective Geometry)==
 
==Solution 2 (Projective Geometry)==
 
[[File:2016 AIME II 10c.png|400px|right]]
 
[[File:2016 AIME II 10c.png|400px|right]]
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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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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{043}</math>.
  
 
==Solution 3==
 
==Solution 3==
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==Solution 4 ==
 
==Solution 4 ==
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Extend <math>\overline{AB}</math> past <math>B</math> to point <math>X</math> so that <math>CPTX</math> is cyclic. Then, by Power of a Point on <math>CPTX</math>, <math>(CQ)(QB) =  (PQ)(QX)</math>. By Power of a Point on <math>CATB</math>, <math>(CQ)(QT) = (AQ)(QB) = 42</math>. Thus, <math>(PQ)(QX) = 42</math>, so <math>BX = 8</math>.  
+
Extend <math>\overline{AB}</math> past <math>B</math> to point <math>X</math> so that <math>CPTX</math> is cyclic. Then, by Power of a Point on <math>CPTX</math>, <math>(CQ)(QT) =  (PQ)(QX)</math>. By Power of a Point on <math>CATB</math>, <math>(CQ)(QT) = (AQ)(QB) = 42</math>. Thus, <math>(PQ)(QX) = 42</math>, so <math>BX = 8</math>.  
  
 
By the Inscribed Angle Theorem on <math>CPTX</math>, <math>\angle SCT = \angle BXT</math>. By the Inscribed Angle Theorem on <math>ASTC</math>, <math>\angle SCT = \angle SAT</math>, so <math>\angle BXT = \angle SAT</math>. Since <math>ASTB</math> is cyclic, <math>\angle AST = \angle TBX</math>. Thus, <math>\triangle AST \sim \triangle XBT</math>, so <math>AS/XB = ST/BT</math>. Solving for <math>ST</math> yields <math>ST = \frac{35}{8}</math>, for a final answer of <math>35+8 = \boxed{043}</math>.
 
By the Inscribed Angle Theorem on <math>CPTX</math>, <math>\angle SCT = \angle BXT</math>. By the Inscribed Angle Theorem on <math>ASTC</math>, <math>\angle SCT = \angle SAT</math>, so <math>\angle BXT = \angle SAT</math>. Since <math>ASTB</math> is cyclic, <math>\angle AST = \angle TBX</math>. Thus, <math>\triangle AST \sim \triangle XBT</math>, so <math>AS/XB = ST/BT</math>. Solving for <math>ST</math> yields <math>ST = \frac{35}{8}</math>, for a final answer of <math>35+8 = \boxed{043}</math>.
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~blusoul
 
~blusoul
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==Solution 7 (no trig or projections)==
 +
 +
Note that since <math>\triangle SAP~\triangle BCP</math>, <math>\frac{9}{SP}=\frac{BC}{7}=\frac{PC}{4}</math>. Furthermore, since <math>\triangle ACQ~\triangle TBQ</math>, we have <math>\frac{7}{TQ}=\frac{AC}{5}=\frac{QC}{6}</math>. From Stewart's on triangle <math>BCP</math>, we have <math>25CQ+BC^2\cdot TQ=TQ\cdot CQ\cdot TC+36TC</math>, and since <math>TQ\cdot CQ=6\cdot7=42</math> by power of a point, this simplifies to <math>25CQ+BC^2\cdot TQ=78TC</math>. Similarly, <math>49CP+AC^2\cdot SP=52SC</math>. Finally, using Ptolemy's on quadrilateral <math>ACBS</math> yields <math>13SC=7BC+SB\cdot AC</math>, and using Ptolemy's on quadrilateral <math>ACBT</math> yields <math>13TC=5AC+TA\cdot BC</math>. From Ptolemy's on <math>ABTS</math>, we find <math>SB\cdot TA=13ST+35</math>, which is nice because it contains <math>ST</math>.
 +
We return to our first Stewart's equation: <math>25CQ+BC^2\cdot TQ=78TC</math>, and we notice that <math>CQ</math> and <math>TQ</math> can be related to <math>AC</math> using our similar triangle conditions. Substituting gives us <math>30AC+\frac{35BC^2}{AC}=78TC</math>, which by four times our first Ptolemy's equation also equals <math>30AC+6TA\cdot BC</math>. Thus, <math>\frac{35BC^2}{AC}=6TA\cdot BC</math> and <math>TA=\frac{35}{6}\cdot\frac{BC}{AC}</math>. Similarly, from our other Stewart's equation, we find <math>28BC+\frac{63AC^2}{BC}=52SC=28BC+4SB\cdot AC</math>, or <math>SB=\frac{63}{4}\cdot\frac{AC}{BC}</math>. Plugging this into our final Ptolemy's equation, we find <cmath>SB\cdot TA=13ST+35\Longrightarrow\frac{35\cdot63}{6\cdot4}=13ST+35\Longrightarrow ST=\frac{\frac{35\cdot21}{8}-35}{13}=\frac{35\cdot\frac{13}{8}}{13}=\frac{35}{8},</cmath>giving us our final answer of <math>\boxed{043}</math>.
 +
 +
~wuwang2002
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2016|n=II|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:49, 22 November 2024

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)

2016 AIME II 10c.png

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 $ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{043}$.

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 $CPTX$, $(CQ)(QT) = (PQ)(QX)$. 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)

2016 AIME II 10.png

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 \[AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies\] \[\left(\frac {AQ \cdot PB}{PQ} - AB\right)\times ST = AS \cdot BT \implies\] \[ST = \frac {AS \cdot BT \cdot PQ}{AQ \cdot PB – AB \cdot PQ}\] \[ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed {43}.\]

vladimir.shelomovskii@gmail.com, vvsss

Solution 6

Connect $AT$ and $\angle{SCT}=\angle{SAT}, \angle{ACS}=\angle{ATS}, \frac{ST}{\sin \angle{SAT}}=\frac{AS}{\sin \angle{ATS}}$

So we need to get the ratio of $\frac{\sin \angle{ACS}}{\sin \angle{SCT}}$

By clear observation $\triangle{CAQ}\sim \triangle{BTQ}$, we have $\frac{CQ}{AC}=\frac{6}{5}$, LOS tells $\frac{AC}{\sin \angle{CPA}}=\frac{4}{\sin \angle{ACS}}; \frac{CQ}{\sin \angle{CPQ}}=\frac{3}{\sin \angle{PCQ}}$ so we get $\frac{\sin \angle{PCQ}}{\sin \angle{ACS}}=\frac{5}{8}$, the desired answer is $7\cdot \frac{\sin \angle{SAT}}{\sin \angle{ATS}}=\frac{35}{8}$ leads to $\boxed{043}$

~blusoul

Solution 7 (no trig or projections)

Note that since $\triangle SAP~\triangle BCP$, $\frac{9}{SP}=\frac{BC}{7}=\frac{PC}{4}$. Furthermore, since $\triangle ACQ~\triangle TBQ$, we have $\frac{7}{TQ}=\frac{AC}{5}=\frac{QC}{6}$. From Stewart's on triangle $BCP$, we have $25CQ+BC^2\cdot TQ=TQ\cdot CQ\cdot TC+36TC$, and since $TQ\cdot CQ=6\cdot7=42$ by power of a point, this simplifies to $25CQ+BC^2\cdot TQ=78TC$. Similarly, $49CP+AC^2\cdot SP=52SC$. Finally, using Ptolemy's on quadrilateral $ACBS$ yields $13SC=7BC+SB\cdot AC$, and using Ptolemy's on quadrilateral $ACBT$ yields $13TC=5AC+TA\cdot BC$. From Ptolemy's on $ABTS$, we find $SB\cdot TA=13ST+35$, which is nice because it contains $ST$. We return to our first Stewart's equation: $25CQ+BC^2\cdot TQ=78TC$, and we notice that $CQ$ and $TQ$ can be related to $AC$ using our similar triangle conditions. Substituting gives us $30AC+\frac{35BC^2}{AC}=78TC$, which by four times our first Ptolemy's equation also equals $30AC+6TA\cdot BC$. Thus, $\frac{35BC^2}{AC}=6TA\cdot BC$ and $TA=\frac{35}{6}\cdot\frac{BC}{AC}$. Similarly, from our other Stewart's equation, we find $28BC+\frac{63AC^2}{BC}=52SC=28BC+4SB\cdot AC$, or $SB=\frac{63}{4}\cdot\frac{AC}{BC}$. Plugging this into our final Ptolemy's equation, we find \[SB\cdot TA=13ST+35\Longrightarrow\frac{35\cdot63}{6\cdot4}=13ST+35\Longrightarrow ST=\frac{\frac{35\cdot21}{8}-35}{13}=\frac{35\cdot\frac{13}{8}}{13}=\frac{35}{8},\]giving us our final answer of $\boxed{043}$.

~wuwang2002

See also

2016 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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