Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.

If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.

I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.

To my surprise, this problem can be solved using geometry.
Identify the Euclidean plane with complex plane $\mathbb{C}$.
and let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity.
Consider triangle $\triangle ABC$ with vertices at $A = x$, $B = y\omega$ and $C = z\omega^2$.
The sides of the triangle equal to
$$\begin{align}
a^2 = BC^2 &= | y\omega - z\omega^2|^2 = y^2 + yz+ z^2 = 1\\
b^2 = AC^2 &= | x - z\omega^2|^2 = x^2 + xz + z^2 = 4\\
c^2 = AB^2 &= | x - y\omega|^2 = x^2+xy+y^2 = 3
\end{align}
$$
Since $a^2 + c^2 = b^2$, this is a right angled triangle with area $\mathcal{A} = \frac12 ac 
=\frac{\sqrt{3}}{2}$.
An alternate way to compute the area is cut the triangle into 3 pieces along the line $OA$, $OB$ and $OC$. This gives us
$$\mathcal{A}= \frac12 (xy + yz + xz)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4} (xy+yz+xz)$$
Combine these two results, we find:
$$xy+yz+xz = 2$$
Update
For a pure algebraic answer, one can substitute above expression of $a^2,b^2,c^2$ into Heron's formula for area of triangle, $$\mathcal{A} = \frac14\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$
simplify and obtain following algebraic identity
$$\begin{array}{rl}
3(xy+yz+zx)^2 =& \phantom{+0}((x^2+xz+y^2) + (y^2+yz+z^2) + (z^2+xz+x^2))^2\\
&-2((x^2+xz+y^2)^2 + (y^2+yz+z^2)^2 + (z^2+xz+x^2)^2)
\end{array}
$$
Using given values of $a, b, c$, we find
$$3(xy+yz+xz)^2 = (3+4+1)^2 - 2(3^2+4^2+1^2) = 64 -2(26) = 12 = 3(2)^2$$
This leads to the same conclusion $xy+yz+xz = 2$ as before.
