, 2006

Solution:

The answer is $27$. Let $H$ be the orthogonal projection of $D$ onto $AB$. From the data of the problem it follows immediately that $AD = 3$ and $DH = \frac{3 \sqrt{3}}{2}$. Let $Q, R$ be the orthogonal projections of $P$ onto $CD$ and $AB$, respectively.

Since $P$ is the centroid of $ACD$, and since the diagonals of a parallelogram bisect each other, we have $PD = \frac{1}{3} BD$. From the similarity of triangles $BDH, BPR$ it follows that $PR = \frac{2}{3} DH = \sqrt{3}$. The area of $ABP$ is therefore $S(ABP) = \frac{1}{2} AB \cdot PR = 3 \sqrt{3}$.

The area of the quadrilateral $ACPD$ can be obtained as the difference between the area of $ACD$ and the area of $PCD$. We observe that $QR$ is the height of $ACD$ and $PQ$ the height of $PCD$.

Then $S(ACPD) = S(ACD) - S(PCD) = \frac{1}{2} CD \cdot QR - \frac{1}{2} CD \cdot PQ = \frac{1}{2} CD \cdot (QR - PQ) = \frac{1}{2} CD \cdot PR = 3 \sqrt{3}$. (We note, incidentally, that $ABP$ and $ACPD$ are equivalent; this remains true whatever the position of $P$.) The product of the two areas is therefore $(3 \sqrt{3})^2 = 27$.