TITLE: Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$
QUESTION [44 upvotes]: 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.

REPLY [18 votes]: I would like to mention an other method, more algorithmic but also much more general to this problem. 
First compute a Gröbner basis (say, using the lexicographic order) of the three polynomials $$x^2 + xy + y^2 - 3 \\ y^2 + yz + z^2 - 1 \\ x^2 + xz + z^2 - 4$$ One obtains $\left\{7 z^3-4 z,2 y z-z^2,2
   y^2+3 z^2-2,x+2 y-3 z\right\}$. 
We want to evaluate $xy+yz+zx$, which is presumably a constant, under the given conditions. This means that $xy+yz+zx-a$ belongs to the ideal generated by the three above polynomials, for some constant $a$ to be determined. Let's evaluate the remainder of the Euclidean division of $xy+yz+zx-a$ by the Gröbner basis given above: one finds $$-2-a+7z^2 \, . $$ This should be zero, if our relation is to be a consequence of the original relations. So we deduce $a=7z^2-2$. 
What is the value of $z^2$? Well, the first term of the Gröbner basis tells us that $z(7z^2-4)=0$, so $z^2 = 0 , 4/7$, and therefore $a= \pm 2$. 
To fix the sign, we need the last piece of information, namely that $x,y,z \geq 0$. Clearly, if $z=0$ then the last term of the basis tells us that $x+2y=0$, which is forbidden (since $x=y=z=0$ is not a solution). Therefore, $z^2 = 4/7$ and $a=2$.