Belarusian Mathematical Olympiad

Let $A_1(0;a_1)$, $B_1(b_1;0)$, $C_1(c_1;0)$, $D_1(0;d_1)$, $A(a;1/a)$, $B(b;1/b)$, $C(c;1/c)$, $D(d;1/d)$ be the marked points. Since any vertical and any horizontal line meets the hyperbola $y=1/x$ at most at one point we see that the numbers $a$, $b$, $c$, $d$ are pairwise distinct.



The equation of the line $AB$ has the form $y-\frac{1}{a}=k_1(x-a)$. Since point $B$ belongs to the line $AB$ its coordinates must satisfy the line equation, i.e. $\frac{1}{b}-\frac{1}{a}=k_1(b-a)$, so $k_1=-\frac{1}{ab}$. Similarly, for the line $CD$ we have $y-\frac{1}{c}=k_2(x-c)$, and so $$k_2=-\frac{1}{cd}.\text{ Since }AB\parallel CD\text{ we have }{k}_1=k_2,\text{ so}$$ $$ab=cd⟹\frac{a}{c}=\frac{b}{d}.(1)$$

The equation of the line $AC$ has the form $y-\frac{1}{a}=k(x-a)$. Setting $x=0$ in this equation, we find the ordinate of $A_1$: $a_1=\frac{1}{a}-ka$, and setting $y=0$ in the equation we find the abscissa of $C_1$: $c_1=a-\frac{1}{ka}$. Since point $C$ belongs to the line $AC$ its coordinates must satisfy the line equation, i.e. $\frac{1}{c}-\frac{1}{a}=k(c-a)$, so $k=-\frac{1}{ac}$. Therefore $a_1=\frac{1}{a}+\frac{1}{c}$ and $c_1=a+c$. Similarly, from the equation of the line $BD$ we find $d_1=\frac{1}{b}+\frac{1}{d}$ and $b_1=b+d$. Therefore, $$S(A_{1}O{C}_1)=\frac{1}{2}|a_1{c}_1|=\frac{1}{2}\cdot \frac{(a+c{)}^2}{|ac|}=\frac{1}{2}\cdot \frac{(a/c+1{)}^2}{|a/c|},$$ $$S(D_{1}O{B}_1)=\frac{1}{2}|b_1{d}_1|=\frac{1}{2}\cdot \frac{(b+d{)}^2}{|bd|}=\frac{1}{2}\cdot \frac{(b/d+1{)}^2}{|b/d|}.$$ Now from (1) it follows that $S(A_{1}O{C}_1)=S(D_{1}O{B}_1)$.