A triangle has vertices $(0,0)$, $(1,1)$, and $(10m,0)$, and the line $y=mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$?

1. The cevian from $(0,0)$ must hit the midpoint of the opposite side, $\left(\frac{1+10m}2,\frac12\right)$, which must lie on $y=mx$: $\frac12=m\cdot\frac{1+10m}2$, i.e. $10m^2+m-1=0$.
2. The discriminant $1+40>0$, so there are two real nonzero roots (one positive, one negative), each giving a genuine triangle. By Vieta's, the sum of roots is $-\dfrac1{10}$.
3. The answer is $\boxed{-\dfrac1{10}}$.