62nd Belarusian Mathematical Olympiad

Answer: $S(ABCD)=\frac{L^2}{8}$. Let the area of $ABCD$ be a maximum for some $AB=x$, $BD=y$, $CD=z$, $x+y+z=L$. Since $S(ABCD)=S(ABD)+S(DBC)=\frac{1}{2}AB\cdot BD\sin \angle ABD+\frac{1}{2}BD\cdot BC\sin \angle BDC$, we see that the area of the quadrilateral with fixed values of $x$, $y$, $z$ is maximum if $\angle ABD=\angle CDB=90^{\circ }$. Therefore, $$S(ABCD)=\frac{1}{2}xy+\frac{1}{2}yz=\frac{1}{2}y(x+z)=\frac{1}{2}y(L-y).$$ It is easy to see that for $0<y<L$ the value of the product $y(L-y)$ is a maximum if $y=\frac{L}{2}$ and it is equal to $\frac{L^2}{4}$. Therefore, the maximal value of $ABCD$ with given sum of its sides $AB$, $CD$ and diagonal $BD$, $AB+BD+DC=L$, is equal to $\frac{L^2}{8}$.