4. Let $A_{1}, A_{2}, A_{3}, \ldots, A_{11}, A_{12}$ be the consecutive vertices of a regular dodecagon with side $a=2 \text{~cm}$. Let the diagonals $A_{1} A_{10}, A_{2} A_{11}$, and $A_{3} A_{12}$ intersect at points $A, B$, and $C$. Find the area $P_{A B C}$.
Solution. Let $O$ be the center of the circumscribed circle around the regular dodecagon. Let $A_{1} A_{10}$ and $A_{2} A_{11}$ intersect at $A$, $A_{2} A_{11}$ and $A_{3} A_{12}$ intersect at $B$, and $A_{1} A_{10}$ and $A_{3} A_{12}$ intersect at $C$. Using the properties of equal peripheral angles over the same arc, the property of central and peripheral angles over the same arc, and the fact that $\measuredangle A_{1} O A_{2}=30^{\circ}$, it is proven that

![](https://cdn.mathpix.com/cropped/2024_06_05_777e4c6fa99d478c3e87g-4.jpg?height=551&width=570&top_left_y=1700&top_left_x=912)

$$
\measuredangle A_{1} A_{10} A_{2}=\measuredangle A_{12} A_{3} A_{10}=\measuredangle A_{3} A_{12} A_{1}=\measuredangle A_{10} A_{1} A_{12}=\measuredangle A_{12} A_{1} A_{2}=\measuredangle A_{1} A_{2} A_{11}=30^{\circ}
$$

Since $A_{1} A_{12}, A_{2} A_{11}$, and $A_{3} A_{10}$ are parallel, it follows that $\measuredangle C A B=\measuredangle C B A=30^{\circ}$, which means that $P_{A B C}=\overline{A C}^{2} \frac{\sqrt{3}}{4}$. Let $M$ and $N$ be points on $A_{1} A_{12}$ and $A_{2} A_{11}$, respectively, such that $M, N, C$ are collinear and $M N$ is perpendicular to $A_{1} A_{12}$, and thus to $A_{2} A_{11}$. Since $C N$ and $C M$ are heights of isosceles triangles, it follows that $M$ and $N$ are midpoints of $A_{1} A_{12}$ and $A B$, respectively. Then, in sequence, it is concluded that:

$1=\overline{A_{1} M}=\overline{A_{1} C} \frac{\sqrt{3}}{2}, \overline{M C}=\frac{\overline{A_{1} C}}{2}=\frac{1}{\sqrt{3}}, \overline{M N}=\frac{\overline{A_{1} A_{12}}}{2}=1, \overline{N C}=\frac{\sqrt{3}-1}{\sqrt{3}}$ and $\overline{A C}=2 \overline{N C}=2 \cdot \frac{\sqrt{3}-1}{\sqrt{3}}$ Therefore,

$$
P_{A B C}=\frac{4(\sqrt{3}-1)^{2} \sqrt{3}}{3 \cdot 4}=\frac{(\sqrt{3}-1)^{2} \sqrt{3}}{3} \mathrm{~cm}^{2}
$$

## III year