HMMT February, 2023

Solution:

Let three points be $A$, $B$, and $C$ with $x$-coordinates $a$, $b$, and $c$, respectively. Let the circumcircle of $\triangle ABC$ meet the line $y = b$ at point $P$. Then, we have $\angle BPC = 60^{\circ} \Longrightarrow PC = \frac{2}{\sqrt{3}}(c-b)$. Similarly, $AP = \frac{2}{\sqrt{3}}(b-a)$. Thus, by the Law of Cosines,
$$\begin{aligned} AC^{2} & = AP^{2} + PC^{2} - 2 \cdot AP \cdot PC \cos 120^{\circ} \\ & = \frac{4}{3}\left((c-b)^{2} + (b-a)^{2} + (c-b)(b-a)\right) \\ & = \frac{4}{3}\left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right) \\ & = \frac{4}{3}\left((a+b+c)^{2} - 3(ab + bc + ca)\right) \end{aligned}$$
By Vieta's we have $a + b + c = 9$ and $ab + bc + ca = 10$, so we have $AC^{2} = \frac{4}{3}(81 - 30) = 68$, implying that the answer is $\sqrt{68} = 2\sqrt{17}$.