Belarus2022

Denote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. The perpendicularity of the lines $AB$ and $CD$ is equivalent to $acbd=-1/4$ and the fact that the point $F$ belongs to the lines $AC$ and $BD$ is equivalent to the equalities $a+c=2ac+1$ and $b+d=2bd+1$. Denote $ac=p$ and $bd=q$, then the numbers $a$ and $c$ are the roots of the quadratic equation $x^2-(2p+1)x+p=0$ and the numbers $b$ and $d$ are the roots of the quadratic equation $x^2-(2q+1)x+q=0$. Hence $c-a=\sqrt{4p^2+1}$ and $d-b=\sqrt{4q^2+1}$. Note that $$\frac{1}{2a}-\frac{1}{2c}=\frac{c-a}{2ac}=\frac{\sqrt{4p^2+1}}{2p}=-2q\cdot \sqrt{\frac{1}{4q^2}+1}=\sqrt{4q^2+1}=d-b.$$ Similarly, $\frac{1}{2d}-\frac{1}{2b}=c-a$. Thus the projections of the segment $AC$ on the abscissa and ordinate axes are equal, respectively, to the projections of the segment $BD$ on the ordinate and abscissa axes, hence $AC=BD$. Since the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ are perpendicular, its area $S$ is equal to half the product of the diagonals. Therefore $$S=\frac{1}{2}\cdot AC\cdot BD=\frac{1}{2}A{C}^2=\frac{1}{2}((c-a{)}^2+(b-d{)}^2)=\frac{1}{2}(4p^2+4q^2+2).$$