## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-249.jpg?height=338&width=396&top_left_y=261&top_left_x=176)

Let the regular triangles $\triangle ABC$ and $\triangle A_1B_1C_1$ have a common center $O$, and $\triangle A_1B_1C_1$ is rotated around $O$ relative to $\triangle ABC$ by $30^\circ$. Denote $M$ and $K$ as the points of intersection of $A_1B_1$ with $AB$ and $BC$, respectively. Let

$$
\begin{aligned}
& a = AB. \text{ Then } \angle MBO = \angle BOB_1 = 30^\circ \text{ (by the problem statement)} \\
& \Rightarrow AB \parallel OB_1; \angle MA_1O = \angle AOA_1 = 30^\circ \Rightarrow A_1B_1 \parallel AO.
\end{aligned}
$$

Thus, $AMOB_1$ is a parallelogram, and $A_1B_1 \perp BC$, since $AO \perp BC \Rightarrow$ $\triangle BKM$ is a right triangle, $BM = AB - AM = AB - OB_1 = a - \frac{a}{\sqrt{3}}, \Rightarrow$

$$
\begin{aligned}
& \Rightarrow S_{BKM} = \frac{BM \cdot BK \cdot \sin 30^\circ}{2} = \frac{BM^2 \sin 30^\circ \cdot \cos 30^\circ}{2} = \frac{a^2(\sqrt{3}-1)^2 \sqrt{3}}{24} = \\
& = \frac{S(\sqrt{3}-1)^2}{6}, \text{ where } S = \frac{a^2 \sqrt{3}}{4} - \text{ the area of } \triangle ABC
\end{aligned}
$$

Since the area of the figure $\Phi$, which is the intersection of $\triangle ABC$ and $\triangle A_1B_1C_1$, is equal to $S_{ABC}$ minus the area of three triangles equal to $\triangle BKM$, then $S_{\Phi} = S - 3S_{BKM} = S(\sqrt{3}-1)$.

Answer: $S(\sqrt{3}-1)$.
10.411 Two regular triangles with area $S$ are given, where the second triangle is obtained by rotating the first triangle around its center by an angle of $30^{\circ}$. Calculate the area of the intersection of these triangles.