<s>[INST] System and triangle If $a,b$ and $c$ are the lengths of sides of the triangle, find $x,y$ and $z$ such that
$$\begin{array}{l}
x^2y^2+x^2z^2=axyz\\
y^2x^2+y^2z^2=bxyz\\
z^2x^2+z^2y^2=cxyz
\end{array}$$
I see that $x=y=z=0$ is a solution. Moreover, $x=y=0$ and  $z \in \mathbb{R}$ is solution, or $x=z=0$ and  $y \in \mathbb{R}$ is a solution, $y=z=0$ and  $x \in \mathbb{R}$ too is a solution and if $a=b=c$ $x=y=z=a/2$ is solution. Are There other solutions?  This problem was proposed in the RPM- Revista do professor de Matemática in Brazil.
 [/INST] If $a, b, c$ are the side length of a triangle then
$$  
  u = \frac 12(b+c-a) \, , \quad v = \frac 12(c+a-b) \, ,\quad w = \frac 12(a+b-c)
$$
are positive real numbers such that
$$
  a = v + w \, , \quad b = w + u \, ,\quad c = u + v \, .
$$
(This is the so-called "Ravi substitution", see for example 
Ravi Substitution explained.)
Then your system of equations is equivalent to
$$
 y^2 z^2 = u x y z \\
 z^2 x^2 = v x y z \\
 x^2 y^2 = w x y z 
$$
Now multiply the first two equations. This gives
$x^2 y^2 z^4 = uv (xyz)^2$ and therefore $z^2 = uv$ or $xyz = 0$.
It follows that positive solutions $x, y, z$ must satisfy
$$
  x = \sqrt{vw}  \, , \quad  y = \sqrt{wu}  \, , \quad  z = \sqrt{uv}
$$
and conversely one can verify that this is indeed a solution.
</s>