The 2 pi/3 is fairly obvious (physics/small perturbations). Does that make it much easier to actually find the point? Probably. It suggests that doing things with equilateral triangles (pi/3) might be helpful.
Assuming there is a computable solution (does it exist and is it unique?) for the equilateral triangle T then transform the original triangle to T, compute the point and perform the inverse transform (like finding intersection of two ellipses by transforming them to circles and doing the radical axis trick).
There's several hidden assumption here, which is fine. They should be flagged and they might even break down.
Reducing the scope (initially) does no harm.
for an equilateral triangle with vertices at (0,0), (1,0) and (1/2,1/2*sqrt(3)) then (X,Y)=(1/2,1/(2*sqrt(3))) is the point that minimizes the sum of the distances
if you look at (d/dx) and (d/dy) at that point, and the second derivatives, it's a minimum
For the general case, kata's point about 2Pi/3 sounds right, so you can use geometry
If the vertices are at (x1,y1), (x2,y2) and (x3,y3) and the point is (X,Y) then the slope of the line from vertex 1 to the point is [$]\tan\theta_{1}=\frac{Y-y_{1}}{X-x_{1}}[$] and so on and then enforcing the angles between them leads to some messy equations for X and Y involving square roots.
Again, I'm not seeing nice neat expressions, just something that looks like it has to be solved numerically.
And indeed, I don't recall seeing an actual solution written down on that wiki page, just a description in words of what it was.
That page I linked contains several geometric methods. Maybe look them over and find the one you like best.