Mongolian Mathematical Olympiad

Let $F_n$ denote the set of configurations of $n$ points joined by line segments in such a way that no two of the line segments intersect inside the circle. If $f\in F_n$ then the given $n$ points are said to be vertices of subconfiguration $f$. First we shall prove that it is possible to colour vertices of $f$ by 3 colours in such a way that the endpoints of any line segment joining vertices are of different colour.

Let us proceed by induction. It is trivial for $n\le 3$. Now suppose that it is possible to colour vertices of any subconfiguration of $f\in F_{n-1}$ by 3 colours in such a way that endpoints of any line segment joining vertices have different colour. If $f\in F_n$ then it is easy to show that there exists a vertex of $f$ in which no more than 2 line segments have endpoints. Then let $f^{\prime }$ denote a configuration formed after deleting this vertex and line segments with end in the vertex. Note that $f^{\prime }\in F_{n-1}$. By the induction hypothesis, it is possible to colour vertices of $f^{\prime }$ by 3 colours and considering the fact that no more than 2 line segments have endpoints in the deleted vertex, we conclude that it is possible to colour the deleted vertex differently from another end of line segments.

Now we consider $f\in F_{148}$. By the above proved statement, it is possible to colour vertices of $f$ by 3 colours and there exist 50 vertices with the same colour. Since these 50 vertices are not connected, we have the desired result.