<s>[INST] 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.
 [/INST] Here is some progress.  This might be useful if your real task is not about finding the distances from the Torricelli point to the vertices, but the sum of the distances. 
Let $ABC$ be a triangle whose internal angles are smaller than $\dfrac{2\pi}{3}$ (which means the Torricelli point is not one of its vertices, otherwise the problem is trivial).  Suppose that $a=BC$, $b=CA$, and $c=AB$.  The point $O$ is the Torricelli point of the triangle $ABC$.  Write $p=OA$, $q=OB$, and $r=OC$.  (Note that $x=r$, $y=q$, and $z=p$ in your notations.)  Then,
$$a^2=q^2+qr+r^2\,,$$
$$b^2=r^2+rp+p^2\,,$$
and
$$c^2=p^2+pq+q^2\,.$$

Draw an equilateral triangle $BCB'$ such that $B'$ and $A$ are on the opposite sides of $BC$.  Suppose that $O'$ is on the line segment $OB'$ such that $O'O=r$.   It follows that $O'B'=q$, making
$$p+q+r=AB'=\sqrt{AC^2+B'C^2-2\cdot AC\cdot B'C\cos(\angle ACB')}\,.$$
We have $AC=b$, $B'C=a$, and $\angle ACB'=\angle ACB+\dfrac{\pi}{3}$, whence
$$\cos(\angle ACB')=\cos\left(\angle ACB+\frac{\pi}{3}\right)=\frac{\cos(\angle ACB)-\sqrt{3}\,\sin(\angle ACB)}{2}\,.$$
Since
$$\cos(\angle ACB)=\frac{a^2+b^2-c^2}{2ab}\text{ and }\sin(\angle ACB)=\frac{2\,\Delta}{ab}\,,$$
where $\Delta$ is the area of the triangle $ABC$, we get
$$p+q+r=\sqrt{\frac{a^2+b^2+c^2}{2}+2\sqrt{3}\,\Delta}\,.\tag{*}$$
Because
$$a^2+b^2+c^2=2\,(p+q+r)^2-3\,(rp+pq+qr)\,,$$
we get
$$rp+pq+qr=\dfrac{4}{\sqrt{3}}\,\Delta\,.\tag{#}$$
(This equality can be found using an argument relating the areas of the triangles $BOC$, $COA$, $AOB$, and $ABC$.)
We also have
$$a^2(q-r)+b^2(r-p)+c^2(p-q)=(q^3-r^3)+(r^3-p^3)+(p^3-q^3)=0\,.$$
According to WolframAlpha, the expressions for $p$, $q$, and $r$ are terrible.  Here is an expression for $p$:
$$p = \small\sqrt{\frac{-a^6+2b^6+2c^6 + 4 a^4( b^2  +c^2) - 5 a^2( b^4+c^4) +b^2c^2(b^2+c^2)- 3 a^2 b^2 c^2-\sqrt{3\,\Xi_a}}{6\,(a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2)}}\,,$$
where
$$\begin{align}\Xi_a&:=\small-a^{12} + 4 a^10 (b^2+c^2) - 6 a^8( b^4+c^4) - 10 a^8 b^2 c^2 + 4 a^6 (b^6+c^6)\\&\small\ \ \ \ \ + 10 a^6 b^2 c^2(b^2+c^2) - a^4 (b^8+c^8) - 6 a^4 b^2 c^2(b^4+c^4) - 3 a^4 b^4 c^4 \\&\small\ \ \ \ \ \ \ \ \ \ + 2 a^2 b^2 c^2 (b^8+c^6)- b^4 c^4(b^4+c^4) + 2 b^6 c^6\,.\end{align}$$
For $q$ and $r$, the expressions are similar.  Apparently,
$$\Xi_a=16\Delta^2(a^2-b^2)^2(a^2-c^2)^2\,.$$
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