We draw the diagonals of the convex quadrilateral $A B C D$, and then the centroids of the 4 triangles thus formed. What fraction of the area of quadrilateral $A B C D$ is the area of the quadrilateral determined by the 4 centroids?
Given the convexity of quadrilateral $ABCD = N$, the diagonals $AC$ and $BD$ intersect at point $M$ inside $N$. The four triangles in question are $MAB$, $MBC$, $MCD$, and $MDA$. The centroids $S_A$, $S_B$, $S_C$, and $S_D$ of these triangles can be found by taking the midpoints $A_0$, $B_0$, $C_0$, and $D_0$ of the sides $AB$, $BC$, $CD$, and $DA$, respectively, and then marking the point that is two-thirds of the way along the segments $MA_0$, $MB_0$, $MC_0$, and $MD_0$ from $M$.

![](https://cdn.mathpix.com/cropped/2024_05_02_33803649ade60bfa54c7g-1.jpg?height=517&width=594&top_left_y=346&top_left_x=749)

According to this, the quadrilateral $N_S$ defined by the four centroids is a $2:3$ scaled-down version of the quadrilateral $A_0B_0C_0D_0 = N_0$ with $M$ as the center of scaling, so the ratio of their areas is $4:9$.

On the other hand, the ratio of the areas of $N_0$ and $N$ is $1:2$, because the four triangular parts of $N$ outside $N_0$ are equal to the corresponding (i.e., adjacent) parts of $N_0$ divided into four triangles by the medians $MA_0$, $MB_0$, $MC_0$, and $MD_0$. For example, the heights of the triangles $BA_0B_0$ and $MA_0B_0$ are equal, since $A_0B_0$ is the midline of triangle $ABC$, and $M$ lies on segment $AC$.

Therefore, the area of $N_S$ is

$$
\frac{1}{2} \cdot \frac{4}{9} = \frac{2}{9}
$$

of the area of $N$.