# Prove that there is a circle containing exactly $2018$ points

circlescombinatoricscontest-mathcurvespigeonhole-principle

Problem

Given a set $$\mathtt{E}$$ containing $$2017^{2019}$$ points on the plane. Prove that there is a circle containing exactly $$2018$$ points from the set $$\mathtt{E}$$ (these points are on the open disk).

[Maths Olympiad (Morocco $$2018$$)]

My Approach
It's enough to find a line dividing the plane into two parts, the first one contains $$2018$$ points of $$\mathtt{E}$$, and the other contains the rest of the points of $$\mathtt{E}$$. As we can take a circle tangent to this line and with a radius big enough. I wasn't able to prove this, but it seems true, and it seems that we can somehow use pigeonhole principle to prove it.

Your idea is good, and you just need to find the simplest rigorous way forward. One way is as follows: Take any ray $$L$$ with $$2018$$ points on its left. (I leave you to figure out how to rigorously prove the existence of such $$L$$.) Let $$P$$ be a point on $$L$$. Let $$C(k)$$ be the circle on the left of $$L$$ that is tangent to $$L$$ at $$P$$. Clearly $$C(k+1)$$ encircles at least as many points in $$E$$ as $$C(k)$$. But $$E$$ is finite, so $$C(k)$$ can only increase finitely many times as $$k$$ increases. Thus $$C(m)$$ for some $$m$$ encircles all the points in $$E$$.