How to solve such a system of quadratic equations: $x^2+y^2-xy=a^2, x^2+z^2-xz=b^2, y^2+z^2-yz=c^2$ I don't know how to solve this this system:
$$x^2+y^2-xy=a^2\\x^2+z^2-xz=b^2\\y^2+z^2-yz=c^2$$ The system of quadratic equation in symmetry form has many geometric meaning,
this seems to be a triangular pyramid, three adjacent angles with a common point are $\frac{\pi}{3}$, and opposite to this point is a triangle with $a,b,c$.

But this geometric meaning makes no sense to me, it can't simplify the problem. I don't have other idea about it.


Solve the first and second equations as quadratic in $y$ and $z$ to get four pairs of solutions
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x-\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x+\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x+\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x-\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x+\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x+\sqrt{4 b^2-3 x^2}\right)$$
Suppose that the last is the "good" one, replace in the third equation which is
$$(a^2+b^2-c^2)-\frac{5 x^2}{4}+\frac x 4\left(\sqrt{4 a^2-3 x^2}+\sqrt{4 b^2-3 x^2} \right)-\frac{1}{4} \sqrt{4 a^2-3 x^2} \sqrt{4 b^2-3 x^2}=0$$
I suppose that multiple squaring steps would lead to an octic equation in $x$ which, I hope, could be factorized.
However, for given values of $(a,b,c)$, I would prefer to graph the function to locate more or less the solution and after to use Newton method for the solution in $x$ and then $y$ and $z$.
