49th Mathematical Olympiad in Ukraine

From the first equality we can get that $a-b=\frac{1}{c}-\frac{1}{b}$ or that $(a-b)=\frac{b-c}{bc}$. Analogously $b-c=\frac{c-a}{ac}$ and $c-a=\frac{a-b}{ab}$. Thus we obtain: $a-b=\frac{a-b}{(abc{)}^2}$. Since $a$, $b$ are distinct, $(abc{)}^2=1$, which means that $abc=\pm 1$. Let us show that there exist numbers $a$, $b$, $c$ for which both of the values can be achieved.

Take $a=1$. Then we have two equations on $b$, $c$: $1+\frac{1}{b}=c+1$ and $1+\frac{1}{b}=b+\frac{1}{c}$. From the first one we get that $c=\frac{1}{b}\Rightarrow 2b=1+\frac{1}{b}$ or $2b^2-b-1=0$. This equation has two roots $-b_1=-\frac{1}{2}$ and $b_2=1$. The second root coincides with $a$, thus we take $b=-\frac{1}{2}$ and $c=-2$. For these $a$, $b$, $c$ we have $abc=1$ and it is easy to check that they satisfy the problem conditions.

Now take $a=-1\Rightarrow -1+\frac{1}{b}=c-1$ and $-1+\frac{1}{b}=b+\frac{1}{c}\Rightarrow c=\frac{1}{b}$ and $2b=\frac{1}{b}-1$. So $2b^2+b-1=0$. The roots of this equation are $b_1=\frac{1}{2}$ and $b_2=-1$. Again we choose only one value $b=\frac{1}{2}\Rightarrow c=2$ and $abc=-1$. A direct check shows that these $a$, $b$, $c$ fulfill the conditions.