AIME 2016 II · 第 10 题

Solution 1

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

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

$$\begin{aligned}35+13ST&=\frac{21\cdot 35}{8}\\13ST&=\frac{13\cdot 35}{8}\\ST&=\frac{35}{8}.\end{aligned}$$ 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)

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