## Task A-1.2.

Outside the regular polygon $A_{1} A_{2} \ldots A_{n}$, there is a point $B$ such that the triangle $A_{1} A_{2} B$ is equilateral. Determine all $n$ for which the points $B, A_{2}$, and $A_{3}$ are consecutive vertices of some regular polygon.
## Solution.

Let the new regular polygon have $m$ vertices. There are two cases:

## 1. Case

The new polygon lies outside the given polygon. In other words, the $m$-gon and the $n$-gon are on opposite sides of the line $A_{2} A_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_7166a15529367baa5955g-02.jpg?height=554&width=619&top_left_y=891&top_left_x=707)

Then we have $\varangle B A_{2} A_{3}+\varangle A_{1} A_{2} A_{3}+60^{\circ}=360^{\circ}$.

The internal angle of a regular $k$-gon is $\frac{k-2}{k} \cdot 180^{\circ}$, so we have:

$$
\begin{gathered}
\frac{n-2}{n} \cdot 180^{\circ}+\frac{m-2}{m} \cdot 180^{\circ}+60^{\circ}=360^{\circ} \\
3 m(n-2)+3 n(m-2)+m n=6 m n \\
m n-6 m=6 n \\
m=\frac{6 n}{n-6}=6+\frac{36}{n-6} .
\end{gathered}
$$

Clearly, $n-6 \in \mathbb{N}$ and $n-6$ must divide 36.

By checking all possibilities

| $n-6$ | 1 | 2 | 3 | 4 | 6 | 9 | 12 | 18 | 36 |
| :---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| $n$ | 7 | 8 | 9 | 10 | 12 | 15 | 18 | 24 | 42 |
| $m$ | 42 | 24 | 18 | 15 | 12 | 10 | 9 | 8 | 7 |

we find nine solutions.

## 2. Case

The observed $m$-gon and $n$-gon are on the same side of the line $A_{2} A_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_7166a15529367baa5955g-03.jpg?height=542&width=665&top_left_y=366&top_left_x=684)

In this case, $\varangle B A_{2} A_{3}=\varangle A_{1} A_{2} A_{3}+60^{\circ}$

$$
\begin{gathered}
\frac{m-2}{m} \cdot 180^{\circ}=60^{\circ}+\frac{n-2}{n} \cdot 180^{\circ} \\
3 n(m-2)=m n+3 m(n-2) \\
6 n+n m=6 m \\
n=\frac{6 m}{m+6}=6-\frac{36}{m+6}
\end{gathered}
$$

Since $n \geqslant 3$, $\frac{36}{m+6}$ must be in the set $\{1,2,3\}$.

Then we have

$$
\begin{array}{c|rrr}
m+6 & 36 & 18 & 12 \\
m & 30 & 12 & 6 \\
n & 5 & 4 & 3
\end{array}
$$

so we have found three more solutions.

Thus, the values of $n$ that satisfy the conditions of the problem are $n \in\{3,4,5,7,8,9,10,12,15,18,24,42\}$.