For a regular triangle $A B C$, inscribe a circle and then inscribe a regular triangle $A^{\prime} B^{\prime} C^{\prime}$ of the same circumference. The intersection points of the corresponding sides form another regular triangle (see the 3rd problem of the first round of this year's Arany Dániel Junior Competition, problems 41-42 in our previous issue). The second triangle can be labeled in three ways, and thus we generally arrive at three regular triangles. How do the sides of these three triangles relate to each other?
The figure shows the regular triangle $ABC$, the regular triangle $A^{\prime} B^{\prime} C^{\prime}$ rotated by an angle $\alpha$, and the common circumscribed circle around them.

![](https://cdn.mathpix.com/cropped/2024_05_02_9ae8c5534dbcf37b0c4eg-1.jpg?height=989&width=1033&top_left_y=259&top_left_x=535)

Considering different pairs of sides, the resulting regular triangles are $A_{1} B_{1} C_{1}$, $A_{2} B_{2} C_{2}$, and $A_{3} B_{3} C_{3}$. As proven in the text, the center of the regular triangle formed by the intersection points of the sides is the same as the center $O$ of the original triangles. This applies to each of the resulting regular triangles.

According to one of the well-known forms of the Law of Sines, in any triangle, the ratio of a side to the sine of the opposite angle is equal to the diameter of the circumscribed circle. Since in each of the regular triangles every angle is $60^{\circ}$, the ratio of the sides of the three triangles to be examined will be equal to the ratio of the radii of the circumscribed circles of the triangles. Let's calculate the radii $r_{1}, r_{2}, r_{3}$ of the respective circumscribed circles.

If we drop a perpendicular from point $O$ to the segment $BC$, the length of the resulting segment $OT$ is half the radius $r$ of the circumscribed circle of the original triangle $ABC$, because $OT$ is one-third of the entire median. If we draw a perpendicular to the rotated segment $B^{\prime} C^{\prime}$, the angle between the resulting segment $OT^{\prime}$ and the previous segment $OT$ will be $\alpha$.

If we examine the corresponding vertices $A_{1}-A_{2}-A_{3}$, $B_{1}-B_{2}-B_{3}$, and $C_{1}-C_{2}-C_{3}$, they are located on the symmetry axes of the figure, and point $O$ is on each of these lines.

Due to symmetry, $\angle T O C_{2} = \angle C_{2} O T^{\prime} = \frac{\alpha}{2}$. The angle $\angle B_{1} O A_{1}$ is $120^{\circ}$, because in the circumscribed circle of $\Delta A_{1} B_{1} C_{1}$, the corresponding inscribed angle is $60^{\circ}$. Thus, by symmetry, $\angle B_{1} O C_{2} = \angle C_{2} O A_{1} = 60^{\circ}$.

From the right triangle $B_{1} O T$:

$$
r_{1} = \frac{\frac{r}{2}}{\cos \left(60^{\circ} - \frac{\alpha}{2}\right)}
$$

From the triangle $O T C_{2}$:

$$
r_{2} = \frac{\frac{r}{2}}{\cos \frac{\alpha}{2}}
$$

And similarly from the triangle $O T A_{3}$:

$$
r_{3} = \frac{\frac{r}{2}}{\cos \left(60^{\circ} + \frac{\alpha}{2}\right)}
$$

The ratio of the sides of the regular triangles is the same as the ratio of the radii of the circumscribed circles, which is a function of the rotation angle $\alpha$:

$$
\frac{1}{\cos \left(60^{\circ} - \frac{\alpha}{2}\right)} : \frac{1}{\cos \frac{\alpha}{2}} : \frac{1}{\cos \left(60^{\circ} + \frac{\alpha}{2}\right)}
$$