calculating the Fermat point of a triangle

If the angle at A is 120 degrees or more, $a^2 \geq b^2 + bc + c^2$ and the Fermat point is at A. Check this for B and C, as well.

When all angles are less than 120 degrees, the Fermat point is number 13 in the list of triangle centers here:

http://faculty.evansville.edu/ck6/encyclopedia/ETC.html

where you can find barycentric coordinates of that point as a function of the sides of the triangle. Given barycentric coordinates for any point its Cartesian coordinates can be calculated from the Cartesian coordinates of the vertices of the triangle.

Here after 9 years. I had the same question but never found an answer on the web so I decided to compute the Fermat-Torricelli point barycentric coordinates myself.

Let $A,B,C$ be the angles of the triangle and $\alpha=A+\frac{\pi}{3},\beta=B+\frac{\pi}{3},\gamma=C+\frac{\pi}{3}$. The formula is:

$\overrightarrow{PT}=(\sin\alpha \sin\beta \sin C+\sin\alpha \sin\gamma \sin B+\sin\beta \sin\gamma \sin A)^{-1}(\sin\beta \sin\gamma \sin A\cdot\overrightarrow{PA}+\sin\alpha \sin\gamma \sin B \cdot \overrightarrow{PB} +\sin\alpha \sin\beta \sin C \cdot \overrightarrow{PC})$

So the ratio of the barycentric coordinates is : $(\sin\beta \sin\gamma \sin A:\sin\alpha \sin\gamma \sin B:\sin\alpha \sin\beta \sin C)$.

I hope that you see the symmetry.

I made a GeoGebra program that computes the point using the formula and, to check its validity, I've added the lines that intersect to the point. The notations are a bit different than here.

I hope that for those who are searching it now, this will be useful.