, 2023

Solution:
Let $O$ and $r$ be the center and the circumradius of $\triangle ABC$. Let $T$ be the midpoint of the chord in question.

Note that $AO = \frac{AB}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$. Additionally, we have that $AT$ is half the distance from $A$ to $BC$, i.e. $AT = \frac{\sqrt{3}}{2}$. This means that $TO = AO - AT = \frac{\sqrt{3}}{6}$.

By the Pythagorean Theorem, the length of the chord is equal to:
$$2 \sqrt{r^{2} - OT^{2}} = 2 \sqrt{\frac{4}{3} - \frac{1}{12}} = 2 \sqrt{\frac{5}{4}} = \sqrt{5}$$

Solution:
Let the chord be $XY$, and the midpoints of $AB$ and $AC$ be $M$ and $N$, respectively, so that the chord has points $X, M, N, Y$ in that order. Let $XM = NY = x$. Power of a point gives
$$1^{2} = x(x+1) \Longrightarrow x = \frac{-1 \pm \sqrt{5}}{2}$$
Taking the positive solution, we have $XY = 2x + 1 = \sqrt{5}$.