HMMT February

Observe that $$\begin{aligned} & \triangle A C X \sim \triangle Y C B \Longrightarrow \frac{A C}{A X}=\frac{C Y}{B Y} \\ & \triangle A C Y \sim \triangle X C B \Longrightarrow \frac{A C}{A Y}=\frac{C X}{B X} \end{aligned}$$ Mulitplying these two equations together, we get that $$A C^{2}=\frac{(C X \cdot C Y)(A X \cdot A Y)}{B X \cdot B Y}=\frac{24}{5}$$ Analogously, we obtain that $$B C^{2}=\frac{(C X \cdot C Y)(B X \cdot B Y)}{A X \cdot A Y}=\frac{10}{3}$$ Hence, we have $$A B=A C+B C=\sqrt{\frac{24}{5}}+\sqrt{\frac{10}{3}}=\frac{11 \sqrt{30}}{15}$$ implying the answer.