Christmas Streak 76/88: Geometry Rush #6

Notice that $\rm P,~Q$ are centroids of $\rm \triangle BCD$ and $\rm \triangle ACD.$

From it we note $\rm\overline{OQ}:\overline{QD}=\overline{OP}:\overline{PC}=1:2.$

Then we infer that $\rm \overline{PQ}//\overline{CD}.$

Now look at $\triangle \rm OPQ.$

Since $\rm \triangle OPQ\sim\triangle OCD,$ with a ratio of $1:3,$ we figure out that $\triangle \rm OPQ=\dfrac{1}{9}\triangle OCD=\dfrac{1}{9}\times6\times4\times\dfrac{1}{2}=\dfrac{4}{3}.$

Extend $\overline{\rm DM}$ until it meets with $\overrightarrow{\rm AB}$ and call the intersection $\rm S.$

Since $\rm \triangle RAS \sim \triangle RND,$ drawing perpendicular line from $\rm R$ to $\rm \overline{AB}(\overline{CD}),$

we figure out that $\rm \triangle RAS=12\times8\times\dfrac{4}{5}\times\dfrac{1}{2}=\dfrac{192}{5}.$

Also note that $\rm\overline{AR}:\overline{RN}=4:1,~\overline{AQ}:\overline{QN}=2:1~\Rightarrow~\overline{AQ}:\overline{QR}=5:1.$

Then $\rm \triangle PRQ \sim \triangle SRA,$ with a ratio of $1:6.$

Therefore $\rm \triangle PRQ=\dfrac{1}{36}\triangle SRA=\dfrac{1}{36}\times\dfrac{192}{5}=\dfrac{16}{15},$

showing $\rm \square OPRQ=\triangle OPQ + \triangle PRQ = \dfrac{4}{3}+\dfrac{16}{15}=\dfrac{12}{5}$

$\therefore~p+q=\boxed{17}.$

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Other methods exist:

1. Use $\rm \triangle DOP = \dfrac{1}{6}\triangle BCD$ and find the ratio of $\rm \overline{AQ}:\overline{QR}:\overline{RN}=10:2:3$ to figure out $\rm \triangle DQR.$ ( $\rm \square OPRQ = \triangle DOP - \triangle DQR$ )

2. Find the respective areas of $\rm \triangle ARD,~\triangle AQB,~\triangle BOC,~\triangle CPD,$ and subtract them all from $\rm \square ABCD.$

etc.