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 $x\ne y$, 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 $x\ne z$, 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 $a=1$.