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Geometry Difficulty 7.5 National olympiad, round 2 Prove it Hong Kong

$ABCD$ is a parallelogram with $\angle B$ acute. A circle is tangent to $BC$ , $CD$ and $DA$ . The circle intersects $AC$ at $M$ and $N$ , where $M$ is closer to $A$ than $N$ . If $AM = 9$ , $MN = 16$ and $NC = 2$ , find the area of $ABCD$ .

Solution

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

$27^2 - 21^2 = CS^2 = (x+6)^2 - (x-6)^2,$

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

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

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.