Belarus2022

Answer: $\frac{m^2}{2}$. Denote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. It is easy to see that the perpendicularity of the lines $AC$ and $BD$ is equivalent to the equality $(a+c)(b+d)=-1$ and the condition about the difference of projections is equivalent to the equality $(a+c)-(b+d)=m$. The fact that the point $F$ belongs to the lines $AC$ and $BD$ means that $ac=bd=-\frac{1}{4}$. Let $a+c=p$ and $b+d=q$, then the numbers $a$ and $c$ are the roots of the quadratic equation $x^2-px-\frac{1}{4}=0$ and the numbers $b$ and $d$ are the roots of the quadratic equation $x^2-qx-\frac{1}{4}=0$. Hence $c-a=\sqrt{p^2+1}$ and $d-b=\sqrt{q^2+1}$. Since the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ are perpendicular, its area $S$ is equal to half the product of the lengths of the diagonals. So, $$ $$\begin{array}{rl}S& =\frac{1}{2}(c-a)(d-b)\sqrt{1+(c+a{)}^2}\sqrt{1+(d+b{)}^2}=\\ & =\frac{1}{2}(1+p^2)(1+q^2)=\frac{(q+q{p}^2)(p+p{q}^2)}{2pq}=\frac{-(p-q{)}^2}{-2}=\frac{m^2}{2}.\end{array}$$