All three vertices of an equilateral triangle are on the parabola $y=2x^2$, and one of its sides has slope $\frac12$. The $x$-coordinates of the three vertices have a sum of $m/n$, where $m$ and $n$ are relatively prime positive integers. What is the value of $m+n$?

1. For two points $(u,2u^2)$ and $(v,2v^2)$ of the parabola, the chord has slope $\frac{2u^2-2v^2}{u-v}=2(u+v)$. So if the vertices have $x$-coordinates $x_1,x_2,x_3$, the sides have slopes $2(x_1+x_2),\ 2(x_2+x_3),\ 2(x_3+x_1)$, and these add up to $4(x_1+x_2+x_3)$.
2. The sides of an equilateral triangle make $60^\circ$ angles. If one side has slope $\frac12=\tan\theta$, the other two have slopes $\tan(\theta\pm60^\circ)=\dfrac{\frac12\pm\sqrt3}{1\mp\frac{\sqrt3}2}=\dfrac{1\pm2\sqrt3}{2\mp\sqrt3}$.
3. Their sum is $\dfrac{(1+2\sqrt3)(2+\sqrt3)+(1-2\sqrt3)(2-\sqrt3)}{(2-\sqrt3)(2+\sqrt3)}=\dfrac{16}1=16$.
4. The three slopes add up to $\frac12+16=\frac{33}2$, so $x_1+x_2+x_3=\frac{33}{2}\cdot\frac14=\frac{33}8$ and $m+n=33+8=\boxed{41}$.