smc

Let the three points be at $A=(x_1,x_1^2)$, $B=(x_2,x_2^2)$, and $C=(x_3,x_3^2)$, such that the slope between the first two is $2$, and $A$ is the point with the least $y$-coordinate. Then, we have $\text{Slope of }AC=\frac{x_1^2-x_3^2}{x_1-x_3}=x_1+x_3$. Similarly, the slope of $BC$ is $x_2+x_3$, and the slope of $AB$ is $x_1+x_2=2$. The desired sum is $x_1+x_2+x_3$, which is equal to the sum of the slopes divided by $2$. To find the slope of $AC$, we note that it comes at a $60^{\circ }$ angle with $AB$. Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1+2i$ and $1+\sqrt{3}i$. What this does is generate the complex number that is at a $60^{\circ }$ angle with the complex number $1+2i$. Then, we can find the slope of the line between this new complex number and the origin: $$(1+2i)(1+\sqrt{3}i)$$ $$=1-2\sqrt{3}+2i+\sqrt{3}i$$ $$\text{Slope }=\frac{2+\sqrt{3}}{1-2\sqrt{3}}$$ $$=\frac{2+\sqrt{3}}{1-2\sqrt{3}}\cdot \frac{1+2\sqrt{3}}{1+2\sqrt{3}}$$ $$=\frac{8+5\sqrt{3}}{-11}$$ $$=\frac{-8-5\sqrt{3}}{11}.$$ The slope $BC$ can also be solved similarly, noting that it makes a $120^{\circ }$ angle with $AB$: $$(1+2i)(-1+\sqrt{3}i)$$ $$=-1-2\sqrt{3}-2i+\sqrt{3}i$$ $$\text{Slope }=\frac{\sqrt{3}-2}{-2\sqrt{3}-1}$$ $$=\frac{2-\sqrt{3}}{1+2\sqrt{3}}\cdot \frac{1-2\sqrt{3}}{1-2\sqrt{3}}$$ At this point, we start to notice a pattern: This expression is equal to $\frac{2+\sqrt{3}}{1-2\sqrt{3}}\cdot \frac{1+2\sqrt{3}}{1+2\sqrt{3}}$, except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out. Our sum is simply $2-2\cdot \frac{8}{11}=\frac{6}{11}$, and thus we can divide by $2$ to obtain $\frac{3}{11}$, which gives the answer $\boxed{14}$.