Question:
Suppose 2016 points of the circumference of a circle are coloured red and the remaining points are coloured blue. Given any natural number $n \geq 3$, prove that there is a regular n-sided polygon all of whose vertices are blue. (Source: INMO 2016)
Solution:
Suppose 2016 points of the circumference of a circle are coloured red and the remaining points are coloured blue. Given any natural number $n \geq 3$, prove that there is a regular n-sided polygon all of whose vertices are blue. (Source: INMO 2016)
Solution:
Consider a regular $(2017 \times n)$-gon on the circle, say with vertices $A_1, A_2, A_3,\dots, A_{2017 \times n}$. For each $j$, such that $1 \leq j \leq 2017$, consider the set of points $\{A_k : k \equiv j \mod 2017\}$. Note that these are the vertices of a regular $n$-gon, say $S_j$. Thus, we get 2017 regular $n$-gons: $S_1, S_2,\dots, S_{2017}$. Since there are only 2016 red points on the circumference, by pigeonhole principle there must be at least one $n$-gon among these 2017 which does not contain any red vertex. But then, all the vertices of this regular $n$-gon must be blue. Hence proved.