Hong Kong Preliminary Selection Contest

We use $[X_1{X}_2\cdots X_k]$ to denote the area of the $k$-sided polygon $X_1{X}_2\cdots X_k$. Let $K$ and $L$ be the midpoints of $AD$ and $BC$ respectively. Then $△DKN\sim △DAC$ with side-length ratio $1:2$ and so $[DKN]=\frac{1}{4}[DAC]$. Similarly, we have $[AMK]=\frac{1}{4}[ABD]$, $[BLM]=\frac{1}{4}[BCA]$ and $[CNL]=\frac{1}{4}[CDB]$.

Summing up these equations, we get $$[DKN]+[AMK]+[BLM]+[CNL]=\frac{1}{2}[ABCD].$$ This shows $[ABCD]=2[KMLN]=4[NML]$. By the midpoint theorem, we have $NL=\frac{1}{2}BD=4=NM$ and $ML=\frac{1}{2}AC=3$. Hence the height of $△NML$ from $N$ has length $\sqrt{4^2-{\left(\frac{3}{2}\right)}^2}=\frac{\sqrt{55}}{2}$.



Thus $[NML]=\frac{3\sqrt{55}}{4}$ and so $[ABCD]=3\sqrt{55}$.