Let $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 = a$. Then: $$-3 = -1 -1 -1 = \frac{-x}{x}+ \frac{-y}{y} +\frac{-z}{z} = \frac{y+z}{x}+ \frac{x+z}{y} +\frac{y+x}{z} = (\frac{x}{y} + \frac{y}{z} + \frac{z}{x}) + (\frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1) -1 = 2a -1 \Longrightarrow a = -1$$

pggp wrote: Let $x$, $y$, $z$ be non-zero real numbers, such that $x + y + z = 0$ and the numbers $$ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \quad \text{and} \quad \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 $$are equal. Determine their common value. Can you share the other problems in the Final？ Thank you very much

gnoka wrote: pggp wrote: Let $x$, $y$, $z$ be non-zero real numbers, such that $x + y + z = 0$ and the numbers $$ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \quad \text{and} \quad \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 $$are equal. Determine their common value. Can you share the other problems in the Final？ Thank you very much https://artofproblemsolving.com/community/c1403865_2019_polish_junior_mo_finals