Square $ABCD$ has side length $45$. Point $P$ lies inside the square so that $AP=36$ and $BP=27$. The centroids of $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, $\triangle DAP$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?

1. Let $E,F,G,H$ be the midpoints of the sides. The centroid of each triangle lies on the median from $P$, two thirds of the way to the midpoint, so the centroid quadrilateral is the image of the square $EFGH$ under the dilation about $P$ with ratio $\frac23$. Since $[EFGH]=\frac12[ABCD]$, its area is $\frac49\cdot\frac12[ABCD]=\dfrac29[ABCD]$, independent of the position of $P$.
2. $\dfrac29(45)^2=\dfrac29(2025)=\boxed{450}$.