This question is Q.13 of International Mathematical Olympiad Preliminary Selection Contest - Hong Kong 2019.
$A$, $B$, $C$ are three points on a circle while $P$ and $Q$ are two points on $AB$. The extensions of $CP$ and $CQ$ meet the circle at $S$ and $T$ respectively. If $AP=2$, $AQ=7$, $AB=11$, $AS=5$ and $BT=2$, find the length of $ST$.
My approach:
Let $BC=y$ and $AC=z$. It can be found that $CP=\dfrac{2y}5$, $PS=\dfrac{45}y$, $CQ=2z$ and $QT=\dfrac{14}z$. By applying cosine formula on $\triangle BCQ$, $\triangle BCT$, $\triangle ACP$ and $\triangle ACS$, I know that $y^2=\dfrac{1620}{11}$ and $z^2=\dfrac{994}{55}$.
Then I can find $\cos\angle PCQ$ and hence deduce that $ST=\dfrac{25}4$.
The calculation is tedious and I am quite sure that I have missed something. Does anybody have a better method?
Edit:
Michael Rozenberg gave a quick solution using cross-ratio.
From $\triangle BPS \sim \triangle CPA$ one obtains $$BS = \frac{AC\cdot BP}{CP} = \frac{45z}{2y}$$ and from $\triangle AQT \sim \triangle CQB$ one gets $$AT = \frac{BC\cdot AQ}{CQ} = \frac{7y}{2z}.$$ Hence $$BS \cdot AT = \frac{315}{4}.$$ We then get by Ptolemy theorem $$BS \cdot AT = AS \cdot BT + AB \cdot ST,$$ or equivalently $$\frac{315}{4}=5\cdot 2+11\cdot ST.$$ It follows that $$ST = \frac{\frac{315}{4}-10}{11} = \frac{25}{4}.$$