Belarusian Mathematical Olympiad

Answer: $\frac{16}{3}$. Let $A(a;1/a)$, $B(b;1/b)$, $C(c;1/c)$, $D(d;1/d)$ be the marked points (see Fig. 1). 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. Since the opposite sides of the parallelogram are equal, we have $$\begin{array}{rlr}AB=CD& ⟺(b-a{)}^2+(1/b-1/a{)}^2=(d-c{)}^2+(1/d-1/c{)}^2& ⟺(b-a{)}^2\left(1+\frac{1}{(ab{)}^2}\right)=(d-c{)}^2\left(1+\frac{1}{(cd{)}^2}\right).(1)\end{array}$$ Moreover, since opposite sides of the parallelogram are equal and parallel we see that their projections on any line are equal, in particular, their projections on $Ox$-axis are equal, i.e. $|b-a|=|d-c|$. Then from (1) it follows that $$1+\frac{1}{a^2{b}^2}=1+\frac{1}{c^2{d}^2}⟹a^2{b}^2=c^2{d}^2.$$ Similarly, from $BC=AD$ we obtain $b^2{c}^2=a^2{d}^2$. Therefore, $a^2=c^2$ and $b^2=d^2$, and so, taking into consideration $a\ne c$ and $b\ne d$, we obtain $a=-c$, $d=-b$. Without loss of generality we suppose that $a<0<b$, and then $b<c$. Since $AB=2BC$ we have $$(b+c{)}^2\left(1+\frac{1}{(bc{)}^2}\right)=(b-a{)}^2\left(1+\frac{1}{(ab{)}^2}\right)=A{B}^2=(2BC{)}^2=$$ $$=4(c-b{)}^2\left(1+\frac{1}{(bc{)}^2}\right),$$ so $(b+c{)}^2=4(b-c{)}^2$. Then $b+c=2(c-b)$, so we have $c=3b$. Fig. 1 Fig. 2 Consider the pentagon $B_{1}BC{C}_{2}O$ (see Fig. 2). Since $B_1(0,1/b)$, $B_2(b,0)$, $C_1(0,1/c)$, $C_2(c,0)$ we have $$\begin{array}{rl}S(B_{1}BC{C}_{2}O)& =S(B_{1}B{B}_{2}O)+S(BC{C}_2{B}_2)=b\cdot \frac{1}{b}+\frac{1/b+1/c}{2}\cdot (c-b)=\\ & =[c=3b]=1+\frac{4}{3b}\cdot \frac{1}{2}\cdot 2b=\frac{7}{3}.\end{array}$$ On the other hand, $$\begin{array}{rl}S(B_{1}BC{C}_{2}O)& =S(O{B}_{1}B)+S(OBC)+S(OC{C}_2)=\\ & =\frac{1}{2}\cdot \frac{1}{b}\cdot b+S(OBC)+\frac{1}{2}\cdot \frac{1}{c}\cdot c=1+S(OBC),\end{array}$$ so $S(OBC)=\frac{4}{3}$. Since $S(ABCD)=4S(OBC)$ the required area of the parallelogram $ABCD$ is equal $\frac{16}{3}$.