
3. A dissection of a convex polygon into finitely many triangles by segments is called a trilateration if no three vertices of the created triangles lie on a single line (vertices of some triangles might lie inside the polygon). We say that a trilateration is good if its segments can be replaced with one-way arrows in such a way that the arrows along every triangle of the trilateration form a cycle and the arrows along the whole convex polygon also form a cycle. Find all $n \geq 3$ such that the regular $n$-gon has a good trilateration.

(Josef Greilhuber, Austria)


Solution. We show that the regular $n$-gon has a good trilateration if and only if $3 \mid n$.

Given a regular $n$-gon and its good trilateration, color the triangles whose arrows go clockwise in black and the other ones in white. In this way, any two triangles sharing an edge have received different colors and all the triangles sharing an edge with the perimeter of the whole $n$-gon have received the same color (wlog black). We say that a segment in the trilateration is interior if it is not one of the sides of the $n$-gon. Let $x$ be the number of interior segments. Since each interior segment is a side of precisely one white triangle and the sides of white triangle are all different interior segments, we have $3 \mid x$. Arguing likewise for the black triangles, we obtain $3 \mid x+n$. Hence $3 \mid n$.

It remains to show that when $3 \mid n$ then the regular $n$-gon has a good trilateration. This is straightforward by mathematical induction.
![](https://cdn.mathpix.com/cropped/2024_04_17_07439ea9f51f17c31e5fg-2.jpg?height=236&width=954&top_left_y=2150&top_left_x=550)

## Czech-Polish-Slovak Match

## IST Austria, 23-26 June 2019

(Second day - 25 June 2019)
