Problem 1353

In the following, we set $n=2019$ and, for any vertex $P_{i}$, we denote $a_{i}$ as the angle $\widehat{P}_{i} / n$. Let $P_{k}$, $P_{k+1}$, $P_{k+2}$, $P_{k+3}$, and $P_{k+4}$ be four consecutive vertices, with the indices of the vertices considered modulo $n+2$.
The sum of the angles in a triangle is $180^{\circ}$. By considering the triangles $P_{i} P_{i+1} P_{i+2}$, $P_{i} P_{i+1} P_{i+3}$, and $P_{i} P_{i+1} P_{i+4}$, we observe that

$$\begin{aligned} a_{i} + n a_{i+1} + a_{i+2} & = 180^{\circ} \\ 2 a_{i} + (n-1) a_{i+1} + a_{i+3} & = 180^{\circ} \\ 3 a_{i} + (n-2) a_{i+1} + a_{i+4} & = 180^{\circ} \end{aligned}$$

We will then subtract these equations two by two:
$\triangleright (1)-(2)$ indicates that $a_{i+1} + a_{i+2} = a_{i} + a_{i+3}$, which means $a_{i+1} - a_{i} = a_{i+3} - a_{i+2}$;
$\triangleright (2)-(3)$ indicates that $a_{i+1} + a_{i+3} = a_{i} + a_{i+4}$, which means $a_{i+1} - a_{i} = a_{i+4} - a_{i+3}$.
We observe that $a_{i+4} - a_{i+3} = a_{i+3} - a_{i+2}$. The difference between two consecutive angles $a_{j}$ and $a_{j+1}$ is therefore equal to a constant $\mathbf{c}$.
This means that $a_{i+k} = a_{i} + k \mathbf{c}$ for all integers $i$ and $k$. In particular, $a_{1} = a_{n+3} = a_{1} + (n+2) \mathbf{c}$, so $\mathbf{c} = 0^{\circ}$. Thus, our polygon has all its vertices equal.
The triangle $P_{i} P_{i+1} P_{i+2}$ is therefore isosceles at $P_{i+1}$, so that $P_{i} P_{i+1} = P_{i+1} P_{i+2}$. Since two consecutive sides are of the same length, our polygon has all its sides of the same length, and it is indeed a regular polygon.
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