Belarusian Mathematical Olympiad

Answer 9.
Let $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$ be the marked points. Then the required product is equal to
$$T = (a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right).$$

Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a, b, c$ are pairwise distinct. By condition, the triangle $ABC$ is regular so
$$\begin{aligned} AB = BC = CA &\iff (b-a)^2 + (1/b - 1/a)^2 = \\ &= (c-b)^2 + (1/c - 1/b)^2 = (a-c)^2 + (1/a - 1/c)^2. \end{aligned} \quad (1)$$
From (1) we obtain
$$\begin{aligned} b^2 - 2ab + a^2 + 1/b^2 - 2/(ab) + 1/a^2 &= c^2 - 2bc + b^2 + 1/c^2 - 2/(bc) + 1/b^2 \\ &\implies (a^2 - c^2) - 2b(a-c) + \frac{c^2 - a^2}{a^2c^2} - \frac{2(c-a)}{abc} = 0. \end{aligned}$$
Since $a \neq c$ we have
$$a+c-2b-\frac{a+c}{a^2c^2}+\frac{2}{abc}=0.$$
In a similar way from (1) we obtain two more equalities
$$b+a-2c-\frac{b+a}{b^2a^2}+\frac{2}{abc}=0 \quad \text{and} \quad c+b-2a-\frac{c+b}{c^2b^2}+\frac{2}{abc}=0.$$
Summing all three equalities, we obtain $\frac{6}{abc} - \frac{a+c}{a^2c^2} - \frac{b+a}{b^2a^2} - \frac{c+b}{c^2b^2} = 0$, so $\frac{b}{c} + \frac{b}{a} + \frac{c}{a} + \frac{c}{b} + \frac{a}{b} + \frac{a}{c} = 6$. Thus
$$T = (a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 = 3 + 6 = 9.$$