HMMT February

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