1. Let the regular hexagon be $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$. Points $P_{1}$ and $P_{2}$ are the midpoints of sides $\overline{A_{4} A_{5}}$ and $\overline{A_{3} A_{4}}$. What is the ratio of the areas of triangle $\Delta P_{1} A_{1} P_{2}$ and the regular hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$?
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_8091e4f2781742023a76g-17.jpg?height=817&width=722&top_left_y=734&top_left_x=290)

Let: $\left|P_{1} P_{2}\right|=m,|F E|=x,|E S|=y,\left|A_{5} A_{4}\right|=a$.

Then the area of the hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ is $P_{6}=6 \cdot \frac{a^{2} \sqrt{3}}{4}$, and the area of the triangle $\Delta A_{1} P_{2} P_{1}$ is $P_{3}=\frac{m \cdot(a+x+y)}{2}$.

$\left|P_{1} P_{2}\right|=m=\left|A_{5} E\right|=v$ and $|F E|=x=\frac{1}{4} a$, because $\overline{P_{1} P_{2}}$ is the midline of the triangle $\triangle A_{5} A_{3} A_{4}$.

$|E S|=y=\frac{1}{2} a$, because $y$ is the length of the leg opposite the angle of $30^{\circ}$ in the right triangle $\triangle A_{5} E S$ with the hypotenuse of length $a$. It follows that

$P_{3}=\frac{m \cdot(a+x+y)}{2}=\frac{v \cdot\left(a+\frac{1}{4} a+\frac{1}{2} a\right)}{2}=\frac{a \sqrt{3}}{2} \cdot \frac{\frac{7}{4} a}{2}=\frac{7 a^{2} \sqrt{3}}{16}$, so the desired ratio is:

$P_{3}: P_{6}=\left(\frac{7 a^{2} \sqrt{3}}{16}\right):\left(6 \cdot \frac{a^{2} \sqrt{3}}{4}\right)=\frac{7}{16}: \frac{6}{4}=\frac{7}{24}$.