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 $$\begin{array}{rlr}AB=BC=CA& \iff (b-a{)}^2+(1/b-1/a{)}^2=& =(c-b{)}^2+(1/c-1/b{)}^2=(a-c{)}^2+(1/a-1/c{)}^2.\end{array}(1)$$

From (1) we obtain $$\begin{array}{rlr}{b}^2-2ab+a^2+1/b^2-2/(ab)+1/a^2& =c^2-2bc+b^2+1/c^2-2/(bc)+1/b^2& ⟹(a^2-c^2)-2b(a-c)+\frac{c^2-a^2}{a^2{c}^2}-\frac{2(c-a)}{abc}=0.\end{array}$$

Since $a\ne c$ we have $$a+c-2b-\frac{a+c}{a^2{c}^2}+\frac{2}{abc}=0.$$ In a similar way from (1) we obtain two more equalities $$b+a-2c-\frac{b+a}{b^2{a}^2}+\frac{2}{abc}=0\text{and}c+b-2a-\frac{c+b}{c^2{b}^2}+\frac{2}{abc}=0.$$ Summing all three equalities, we obtain $\frac{6}{abc}-\frac{a+c}{a^2{c}^2}-\frac{b+a}{b^2{a}^2}-\frac{c+b}{c^2{b}^2}=0$, so $\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}=6$. Thus $$T=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1=3+6=9.$$