System of Equations: Distances from the Torricelli Point to the Vertices of a Triangle I am calculating the length between the Torricelli point and the triangle apex. With the law of cosines I get this system:
$$\left\{\begin{array}{cc}
x^2+y^2+xy&=a^2\,,\\x^2+z^2+xz&=b^2\,,\\z^2+y^2+zy&=c^2\,.\end{array}\right.$$
Please help me solve it.

In order for me to remember which variable corresponds to which geometric object. I will relabel them as follows:
Let $A,B,C$ be the 3 vertices and $a,b,c$ be the 3 sides of a triangle in usual notation. Let $u, v, w$ be the distance 
between the Torricelli point and $A,B,C$ respectively. 
Let $\Delta$ be the area of the triangle. The problem at hand becomes: 

Given $$\begin{align} \alpha &\stackrel{def}{=} a^2 = v^2 + vw + w^2\\
\beta  &\stackrel{def}{=} b^2 = w^2 + wu + u^2\\ \gamma
&\stackrel{def}{=} c^2 = u^2 + uv + v^2 \end{align}$$ 
  How to solve for $u,v,w$?

Let $\delta = uv + vw + wu$, it is not hard to verify
$$\begin{array}{rcrcrcrl} \alpha &+& \beta &+& \gamma &+& 3\delta &= 2(u+v+w)^2\\
 -\alpha &+& \beta &+& \gamma &+& \delta  &= 2u(u+v+w)\\
  \alpha &-& \beta &+& \gamma &+& \delta  &= 2v(u+v+w)\\
  \alpha &+& \beta &-& \gamma &+& \delta  &= 2w(u+v+w)
\end{array}$$
This implies
$$(u,v,w) = \left(\frac{P-2a^2}{Q}, \frac{P-2b^2}{Q}, \frac{P-2c^2}{Q}\right)$$
where $\;P = a^2 + b^2 + c^2 + \delta\;$ and $\;Q = \sqrt{2(a^2 + b^2 + c^2 + 3\delta)}\;$. 
Everything comes down to the computation of $\delta$.
Notice the area of triangle can be expressed in terms of $\delta$,
$$\Delta = \frac12(uv + vw + wu)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4}\delta$$ 
With help of Heron's formula, $\delta$ can be computed as follows:
$$\delta = \frac{4}{\sqrt{3}}\Delta =
\frac{1}{\sqrt{3}}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$
