## Task B-4.5.

An equilateral triangle of area $P$ is rotated in its plane by $30^{\circ}$ about its centroid. What is the area of the intersection of the original triangle and the triangle obtained by the rotation?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8b541dbeede02a7e62f7g-19.jpg?height=702&width=697&top_left_y=454&top_left_x=682)

Let the vertices of the original triangle be $A, B, C$, the vertices of the rotated triangle be $A_{1}, B_{1}, C_{1}$, the centroid be $O$, and the intersection points of the sides of the two triangles be $D, E, F, G, H, I$. The figure obtained by the intersection of the two triangles is the hexagon DEFGHI.

Due to the rotation, $\varangle A O A_{1}=30^{\circ}$, and $\varangle O A A_{1}=\varangle O A_{1} A=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$. $OA$ is the angle bisector of $\varangle C A B$, and $O A_{1}$ is the angle bisector of $\varangle C_{1} A_{1} B_{1}$. It follows that $\varangle D A A_{1}=\varangle D A_{1} A=45^{\circ}$.

Thus, $|A D|=\left|A_{1} D\right|$, and $\varangle A D I=\varangle A D A_{1}=90^{\circ}$. Point $D$ lies on the perpendicular bisector of side $\overline{A A_{1}}$, or the angle bisector of $\varangle A O A_{1}$, since triangle $A O A_{1}$ is isosceles. Similarly, it can be shown that points $F$ and $H$ lie on the angle bisectors of $\varangle B O B_{1}$ and $\varangle C O C_{1}$, respectively, and that the angles at $F$ and $H$ are right angles.

By the SAS (Side-Angle-Side) congruence theorem, $\triangle O A D \cong \triangle O A_{1} D \cong \triangle O B F \cong \triangle O B_{1} F \cong \triangle O C H \cong \triangle O C_{1} H$.

(The angles at vertex $O$ are $15^{\circ}$, the angles at vertices $A, B, C, A_{1}, B_{1}, C_{1}$ are $30^{\circ}$, and the sides $|O A|,\left|O A_{1}\right|$, $|O B|,\left|O B_{1}\right|,|O C|$, and $\left|O C_{1}\right|$ are all equal to the radius of the circumscribed circle.)

We conclude that $|A D|=\left|A_{1} D\right|=|B F|=\left|B_{1} F\right|=|C H|=\left|C_{1} H\right|$.

From this, by the SAS congruence theorem, it follows that the triangles $\triangle A D I, \triangle A_{1} D E, \triangle B F E, \triangle B_{1} F G, \triangle C H G$, and $\triangle C_{1} H I$ are congruent (the angles are $30^{\circ}, 60^{\circ}, 90^{\circ}$).

Thus, $|D E|=|E F|=|F G|=|G H|=|H I|=|I D|$.

Therefore, the hexagon $D E F G H I$ is equilateral and consists of 6 triangles with the same height $r$ (the radius of the inscribed circle of the triangle) and the same base (a side of the hexagon). Consider one such triangle, $\triangle D E O$. The area of the hexagon is

$$
P_{6}=6 \cdot P_{\triangle}(D E O)=6 \cdot \frac{|D E| r}{2}=3|D E| r
$$

Let the side length of the original triangle be $a$. Then,

$$
\begin{gathered}
|A D|+|D E|+|E B|=|A B|=a \\
|D E|=|D I|=|A D| \sqrt{3},|E B|=2|B F|=2|A D| \\
|A D|+|A D| \sqrt{3}+2|A D|=a \Rightarrow|A D|=\frac{a}{3+\sqrt{3}} \\
|D E|=\frac{a}{3+3 \sqrt{3}} \sqrt{3}=\frac{a(\sqrt{3}-1)}{2}, r=\frac{a \sqrt{3}}{6} \\
P_{6}=3 \cdot \frac{a(\sqrt{3}-1)}{2} \cdot \frac{a \sqrt{3}}{6}=\frac{a^{2} \sqrt{3}}{4}(\sqrt{3}-1)=P(\sqrt{3}-1)
\end{gathered}
$$