Let $A B C$ be an equilateral triangle with side length 2 that is inscribed in a circle $\omega$. A chord of $\omega$ passes through the midpoints of sides $A B$ and $A C$. Compute the length of this chord.
Solution 1: Let $O$ and $r$ be the center and the circumradius of $\triangle A B C$. Let $T$ be the midpoint of the chord in question. Note that $A O=\frac{A B}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$. Additionally, we have that $A T$ is half the distance from $A$ to $B C$, i.e. $A T=\frac{\sqrt{3}}{2}$. This means that $T O=A O-A T=\frac{\sqrt{3}}{6}$. By the Pythagorean Theorem, the length of the chord is equal to: \[ 2 \sqrt{r^{2}-O T^{2}}=2 \sqrt{\frac{4}{3}-\frac{1}{12}}=2 \sqrt{\frac{5}{4}}=\sqrt{5} \] Solution 2: Let the chord be $X Y$, and the midpoints of $A B$ and $A C$ be $M$ and $N$, respectively, so that the chord has points $X, M, N, Y$ in that order. Let $X M=N Y=x$. Power of a point gives \[ 1^{2}=x(x+1) \Longrightarrow x=\frac{-1 \pm \sqrt{5}}{2} \] Taking the positive solution, we have $X Y=2 x+1=\sqrt{5}$. $\fbox{\sqrt{5}}$.