HMMT February, 2020

Solution:
Let $O$ be the circumcenter of $\triangle CDE$. Observe that $\triangle ABC \sim \triangle XYC$. Moreover, $\triangle ABC$ is a right triangle because $1^{2} + (4\sqrt{3})^{2} = 7^{2}$, so the length $XY$ is just equal to $2r$, where $r$ is the radius of the circumcircle of $\triangle CDE$. Since $D$ and $E$ are on the angle trisectors of angle $C$, we see that $\triangle ODE$, $\triangle XDO$, and $\triangle YEO$ are equilateral. The length of the altitude from $C$ to $AB$ is $\frac{4\sqrt{3}}{7}$. The distance from $C$ to $XY$ is $\frac{XY}{AB} \cdot \frac{4\sqrt{3}}{7} = \frac{2r}{7} \cdot \frac{4\sqrt{3}}{7}$, while the distance between lines $XY$ and $AB$ is $\frac{r\sqrt{3}}{2}$. Hence we have
$$\frac{4\sqrt{3}}{7} = \frac{2r}{7} \cdot \frac{4\sqrt{3}}{7} + \frac{r\sqrt{3}}{2}$$
Solving for $r$ gives that $r = \frac{56}{65}$, so $XY = \frac{112}{65}$.