In the sides of a 30-degree angle, we drew triangles as shown in the figure. The points $A_{i}$ are on one side, and the points $B_{i}$ are on the other side: the angles $O A_{i} B_{i}$ are 120 degrees, and the angles $O B_{i} A_{i+1}$ are right angles. Determine the ratio of the areas of the triangles $A_{1} B_{1} A_{2}$ and $A_{48} B_{48} A_{49}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5df2143229b99329f40cg-1.jpg?height=415&width=702&top_left_y=302&top_left_x=690)
Solution. In the triangle $O A_{i} B_{i}$, $\angle B_{i} O A_{i}=30^{\circ}, \angle O A_{i} B_{i}=120^{\circ}$, therefore $\angle O B_{i} A_{i}=180^{\circ}-\left(\angle B_{i} O A_{i}+\right.$ $\left.\angle O A_{i} B_{i}\right)=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e640800f789d55b90e3eg-1.jpg?height=501&width=857&top_left_y=270&top_left_x=623)

The angles $\angle O B_{i} A_{i+1}$ are right angles, so $\angle B_{i+1} B_{i} A_{i+1}=180^{\circ}-90^{\circ}=90^{\circ}$. Therefore, in the triangle $B_{i+1} B_{i} A_{i+1}$, there is a right angle and a 30-degree angle. Hence $\angle B_{i} A_{i+1} B_{i+1}=60^{\circ}$ and

$$
2 B_{i} A_{i+1}=A_{i+1} B_{i+1}
$$

Since $\angle O A_{i} B_{i}=120^{\circ}$, therefore $\angle O A_{i+1} B_{i}=\angle O A_{i+1} B_{i+1}-\angle B_{i} A_{i+1} B_{i+1}=120^{\circ}-60^{\circ}=60^{\circ}$, and $\angle B_{i} A_{i} A_{i+1}=$ $180^{\circ}-\angle O A_{i} B_{i}=180^{\circ}-120^{\circ}=60^{\circ}$, so in the triangle $B_{i} A_{i} A_{i+1}$, there are two 60-degree angles, meaning these triangles are equilateral. Therefore, $A_{i} A_{i+1}=B_{i} A_{i+1}$, and $A_{i+1} B_{i+1}=A_{i+1} A_{i+2}$, thus from (1) we get that

$$
\frac{A_{i+1} A_{i+2}}{A_{i} A_{i+1}}=2
$$

Therefore:

$$
\frac{A_{48} A_{49}}{A_{1} A_{2}}=\frac{A_{48} A_{49}}{A_{47} A_{48}} \cdot \frac{A_{47} A_{48}}{A_{46} A_{47}} \cdot \ldots \cdot \frac{A_{2} A_{3}}{A_{1} A_{2}}=2^{47}
$$

Since the triangles $A_{1} B_{1} A_{2}$ and $A_{48} B_{48} A_{49}$ are equilateral and similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Therefore, the desired ratio is:

$$
1:\left(2^{47}\right)^{2}=1: 2^{94} .
$$

Based on the work of Judit Nagy (Miskolc, Földes F. Gymnasium, 2nd year).