Answer: $8+4 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a32a2f7b12fe1827bf2dg-3.jpg?height=637&width=1349&top_left_y=1037&top_left_x=369)
3. In a convex 12-sided polygon, all angles are equal. It is known that the lengths of some ten of its sides are 1, and the length of one more is 2. What can the area of this 12-sided polygon be? (M. A. Evdokimov) Solution. Consider the 12-sided polygon $A_{1} A_{2} \ldots A_{12}$, satisfying the condition of the problem. It has ten sides of length 1 and one side of length 2. Let $x$ be the length of the remaining side. Consider the vectors $\overline{A_{1} A_{2}}, \overline{A_{2} A_{3}}, \ldots, \overline{A_{12} A_{1}}$, as well as the unit vectors $\vec{e}_{1}, \vec{e}_{2}, \ldots, \vec{e}_{12}$ collinear with them. Then for some $i$ and $j$ the equality $\vec{e}_{1}+\ldots+2 \vec{e}_{i}+\ldots+x \vec{e}_{j}+\ldots+\vec{e}_{12}=\overrightarrow{0}$ holds. In addition, $\vec{e}_{1}+\vec{e}_{7}=\vec{e}_{2}+\vec{e}_{8}=\ldots=\vec{e}_{6}+\vec{e}_{12}=\overrightarrow{0}$, so $\vec{e}_{1}+\vec{e}_{2}+\ldots+\vec{e}_{12}=\overrightarrow{0}$. Subtracting the second obtained equality from the first, we get $\vec{e}_{i}+(x-1) \vec{e}_{j}=\overrightarrow{0}$. This is possible only if $\vec{e}_{i}=-\vec{e}_{j}$ and $x=2$. Therefore, in the original 12-sided polygon, there is a pair of parallel sides of length 2.

Due to the equality of all angles and corresponding sides, this 12-sided polygon has an axis of symmetry (see figure). To find the area, we will divide it into 4 trapezoids and a rectangle. We find $A_{3} A_{12}=A_{6} A_{9}=1+\sqrt{3}, A_{4} A_{11}=A_{5} A_{10}=2+\sqrt{3}$, so the desired area is

$$
S=2 \cdot(2+\sqrt{3})+\frac{\sqrt{3} \cdot(2+\sqrt{3}+1+\sqrt{3})}{2}+\frac{1+\sqrt{3}+1}{2}=8+4 \sqrt{3}
$$