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
$$k_2 = -\frac{1}{cd}. \text{ Since } AB \parallel CD \text{ we have } k_1 = k_2, \text{ so}$$
$$ab = cd \implies \frac{a}{c} = \frac{b}{d}. \qquad (1)$$

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,
$$S(A_1 OC_1) = \frac{1}{2} |a_1 c_1| = \frac{1}{2} \cdot \frac{(a+c)^2}{|ac|} = \frac{1}{2} \cdot \frac{(a/c+1)^2}{|a/c|},$$
$$S(D_1 OB_1) = \frac{1}{2} |b_1 d_1| = \frac{1}{2} \cdot \frac{(b+d)^2}{|bd|} = \frac{1}{2} \cdot \frac{(b/d+1)^2}{|b/d|}.$$
Now from (1) it follows that $S(A_1 OC_1) = S(D_1 OB_1)$.