Belorusija 2012

Answer: $12$. We use the following obvious inequalities. If the quadrilateral $ABCD$ has the sides $AB=a$, $BC=b$, $CD=c$, $DA=d$, and the area $S$, then $$S\le \frac{ab+cd}{2}\text{ because}$$ $$\begin{gathered} S=S(ABC)+S(ACD)= \\ =\frac{1}{2}ab\sin \beta +\frac{1}{2}cd\sin \delta \le \frac{ab+cd}{2}. \end{gathered}$$ Similarly, $S\le \frac{ad+bc}{2}$. Summing these inequalities, we easily obtain $S\le \frac{(a+c)(b+d)}{4}$. Hence $$S\le \frac{6\cdot 8}{4}=12.$$ It is easy to see that equality occurs for a rectangle.