Belarusian Mathematical Olympiad

Answer: $a = 1$.

Let $A = x - \frac{y}{z} - \frac{z}{y}$, $B = y - \frac{z}{x} - \frac{x}{z}$, and $C = z - \frac{x}{y} - \frac{y}{x}$. By condition, it follows that
$$0 = A - B = (x - y) + \frac{x - y}{z} - \frac{z(x - y)}{xy} = (x - y) \left( 1 + \frac{1}{z} - \frac{z}{xy} \right) = \frac{(x - y)(xyz + xy - z^2)}{xyz}.$$
Since $x \neq y$, we obtain $z^2 - xy - xyz = 0$. Similarly, considering the difference $B - C$, we obtain $x^2 - yz - xyz = 0$. Therefore,
$$0 = (z^2 - xy - xyz) - (x^2 - yz - xyz) = (z - x)(x + y + z).$$
Since $x \neq z$, we have $x + y + z = 0$. Dividing this equality by $x$, $y$, and $z$ consequently, we have
$$1 + \frac{y}{x} + \frac{z}{x} = \frac{x}{y} + 1 + \frac{z}{y} = \frac{x}{z} + \frac{y}{z} + 1 = 0.$$
Therefore, $\frac{y}{z} + \frac{z}{y} + \frac{x}{z} + \frac{x}{y} + \frac{y}{x} = -3$. Since $3a = A + B + C$, we obtain
$$a = \frac{1}{3} \cdot \left( x + y + z - \frac{y}{z} - \frac{z}{y} - \frac{x}{z} - \frac{x}{y} - \frac{y}{z} \right) = \frac{1}{3} \cdot (0 + 3) = 1.$$
It is easy to verify that $x = 4$, $y = -2 - \frac{2}{\sqrt{5}}$, and $z = -2 + \frac{2}{\sqrt{5}}$ satisfy the problem condition for $a = 1$.