13. Let $A B C D$ be a parallelogram. It is known that side $A B$ measures 6, angle $B \widehat{A} D$ measures $60^{\circ}$, and angle $A \widehat{D} B$ is a right angle. Let $P$ be the centroid of triangle $A C D$. Calculate the value of the product of the areas of triangle $A B P$ and quadrilateral $A C P D$.
13. The answer is 27. Let $H$ be the orthogonal projection of $D$ onto $AB$. From the data in 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.

![](https://cdn.mathpix.com/cropped/2024_04_17_5df6f2fb8919533f8a6eg-10.jpg?height=237&width=422&top_left_y=1715&top_left_x=1411)

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 $BDH$ and $BPR$, 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 by the difference between the area of $ACD$ and the area of $PCD$. Note that $QR$ is the height of $ACD$ and $PQ$ is the height of $PCD$.

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