HongKong 2022-23 IMO Selection Tests

Let $P$, $Q$, $R$ be the points where the circle touches $BC$, $CD$ and $DA$ respectively. Using power, we have $AR=\sqrt{AM\cdot AN}=15$ and $CP=CQ=\sqrt{CN\cdot CM}=6$. Let $DR=DQ=x$ and $S$ be the foot of the perpendicular from $C$ to $AD$. Then we have $DS=x-6$, $AS=21$ and $CD=x+6$. Using



$$27^2-21^2=C{S}^2=(x+6{)}^2-(x-6{)}^2,$$

we get $x=12$ and $CS=12\sqrt{2}$. The area of $ABCD$ is thus equal to

$$AD\cdot CS=(21+6)\cdot 12\sqrt{2}=324\sqrt{2}.$$