Disjoint sets of vertices of a polygon This is the question:
Suppose that the vertices of a regular polygon of 20 sides are coloured with three colours – red, blue and green – such that there are exactly three red vertices. Prove that there are three vertices A,B,C of the polygon having the same colour such that triangle ABC is isosceles.
And this is the official solution to this question:
Since there are exactly three vertices, among the remaining 17 vertices there are nine of them of the same colour, say blue. We can divide the vertices of the regular 20-gon into four disjoint sets such that each set consists of vertices that form a regular pentagon. Since there are nine blue points, at least one of these sets will have three blue points. Since any three points on a pentagon form an isosceles triangle, the statement follows.
What do we mean by the "four disjoint sets"? What does it signify geometrically?
Also is the statement "among the remaining 17 vertices there are nine of them of the same colour" a consequence of the Pigeonhole principle?

For your second question; answer is yes. Pigeonhole principle repeatedly used to reach the answer.
By "four disjoint sets", answerer means that "four sets such that no two of them contains same element". Call the vertices $1$, $2$, $3$, $4$, etc. such that adjacent vertices differ by one, except vertices $20$ and $1$. Partition set of all vertices to four set:
$$
A=\{1,5,9,13,17\}\\
B=\{2,6,10,14,18\}\\
C=\{3,7,11,15,19\}\\
D=\{4,8,12,16,20\}\\
$$
It is easily seen that all vertices belong to same set creates pentagon. Apply pigeonhole principle again to reach the conclusion that one of them at least three blue points.
