Exercise 3. Let $x, y, z$ be non-zero real numbers such that $x+y+z=0$. Suppose that

$$
\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x}{z}+\frac{z}{y}+\frac{y}{x}+1
$$

Determine the value of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$.
Solution to Exercise 3 Let \( a \) be the common value of \( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \) and \( \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 \) so that

\[
\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = a
\]

and

\[
\frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 = a
\]

Let's add these two equations:

\[
2a = \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 = 1 + \frac{x+z}{y} + \frac{x+y}{z} + \frac{y+z}{x} = 1 - \frac{y}{y} - \frac{z}{z} - \frac{x}{x} = -2
\]

In particular, \( a = -1 \), so \( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \) is -1.

Graders' Comment: The exercise is very well solved. Some students propose different proofs from the solution, but often these require more calculations.