jmc

Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$ Note that the slope of $\overline{AC}$ is $\tan 60^{\circ }=\sqrt{3}.$ Hence, the equation of the line containing $\overline{AC}$ is$$y=x\sqrt{3}+1.$$This will intersect the ellipse when\begin{eqnarray}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\ & = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x(13x+8\sqrt 3)=0\implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray}We ignore the $x=0$ solution because it is not in quadrant 3. Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\left(\frac{8\sqrt{3}}{13},y_0\right)$ and $\left(\frac{-8\sqrt{3}}{13},y_0\right),$ respectively, for some value of $y_0.$ It is clear that the value of $y_0$ is irrelevant to the length of $BC$. Our answer is$$BC=2\ast \frac{8\sqrt{3}}{13}=\sqrt{4{\left(\frac{8\sqrt{3}}{13}\right)}^2}=\sqrt{\frac{768}{169}}⟹m+n=\boxed{937}.$$