49th Mathematical Olympiad in Ukraine

Consider the first equations of the system. It is easy to see that $xy\ne 0$, and we $$\text{get }\frac{9}{2(x+y)}=\frac{x+y}{xy}\iff 9xy=2x^2+4xy+2y^2\iff 2y^2-5xy+2x^2=0$$

$y=\frac{5x\pm \sqrt{25x^2-16x^2}}{4}=\frac{5x\pm 3x}{4}$. It follows that all couples $(x,2x)$ and $(x,\frac{1}{2}x)$ without $(0,0)$ are solutions of the first equation. We are putting these solutions in the second equation. From this equation we have $x^2+y^2=5$, $|x|\ge \sqrt{2}$ and $y\le \sqrt{3}$. Further, putting $(x,2x)$ the first condition becomes $5x^2=5\iff x=\pm 1$. And we have solutions $(1,2)$ and $(-1,-2)$ but these couples don't satisfy the condition $|x|\ge \sqrt{2}$. Analogously putting $(x,\frac{1}{2}x)$ the first condition becomes $x^2+\frac{1}{4}{x}^2=5\iff x^2=4$. Here we have solutions $(2,1)$ and $(-2,-1)$ which satisfy our conditions.