Belorusija 2012

Answer: $S(ABCD)=L^2/8$. Let the area of $ABCD$ be a maximum for some $AB=x$, $BD=y$, $CD=z$, $x+y+z=L$. Since $S(ABCD)=S(ABD)+S(DBC)=\frac{1}{2}AB\cdot BD\sin \angle ABD+\frac{1}{2}BD\cdot DC\sin \angle BDC$, 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, $$\begin{array}{rl}S(ABCD)& =\frac{1}{2}xy+\frac{1}{2}yz=\frac{1}{2}y(x+z)=\\ & =[x+y+z=L\Rightarrow x+z=L-y]=\frac{1}{2}y(L-y).\end{array}$$



It is easy to see that for $0<y<L$ the value of the product $y(L-y)$ is a maximum if $y=L/2$ and it is equal to $L^2/4$. Therefore, the maximal value of $ABCD$ with given sum of its sides $AB$, $CD$ and diagonal $BD$, $AB+BD+DC=L$, is equal to $L^2/8$.