9. As shown in Figure 7, the areas of equilateral triangles $\triangle ABC$ and $\triangle PQR$ are both equal to 1, and the center $M$ of $\triangle PQR$ lies on the boundary of $\triangle ABC$. If the area of the overlapping part of these two triangles is $S$, find the minimum value of $S$.
9. At the three vertices of the equilateral $\triangle PQR$, three congruent equilateral triangles are cut off, resulting in a regular hexagon with an area of $\frac{2}{3}$. $M$ is the center of this regular hexagon.

If point $M$ coincides with one of the vertices of $\triangle ABC$ (as shown in Figure 13(a)), it is easy to see that the overlapping area between the regular hexagon and $\triangle ABC$ is $\frac{1}{9}$. In Figure 13(b), when $\triangle ABC$ is rotated clockwise around point $M$, the triangle swept by the initial side and the triangle swept by the final side are congruent. Therefore, the overlapping area between the regular hexagon and $\triangle ABC$ remains constant.

If point $M$ is on the side of $\triangle ABC$, let's assume it is on $BC$ and closer to point $C$ (as shown in Figure 13(c)), draw a line $MN$ parallel to $AC$ intersecting side $AB$ at point $N$, then $\triangle BMN$ is an equilateral triangle. Since $MN = BM > CM$, both $BM$ and $MN$ intersect with the regular hexagon, so the overlapping area between $\triangle BMN$ and the regular hexagon is $\frac{1}{9}$.

When the regular hexagon is restored to the original equilateral triangle, the overlapping area does not decrease, so the minimum overlapping area between the two triangles is $\frac{1}{9}$ (as shown in Figure 13(a)).