HMMT February, 2023

Solution:
Observe that
$$\begin{aligned} & \triangle A C X \sim \triangle Y C B \Longrightarrow \frac{A C}{A X} = \frac{C Y}{B Y} \\ & \triangle A C Y \sim \triangle X C B \Longrightarrow \frac{A C}{A Y} = \frac{C X}{B X} \end{aligned}$$
Multiplying 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.