Local concyclicity for every five points — how does it become global?

quicksave_q·Feb 12, 2026(edited)

The hypothesis only gives four concyclic points at a time. I can prove useful things once five points share a circle, but getting that fifth point seems to be the main obstacle. Problem: There are

2021

points on a plane, no three of which are collinear. For every

5

points there exists at least

4

among them which are concyclic. Is it necessarily true that at least

2020

of the points are concyclic? Source: Baltic Way 2021 Shortlist · MNET-01HW

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3 replies

eevee.zip·Feb 13, 2026(edited)

Start with four on a circle Γ and a point E off it. If no circle has five points, every further point F is off Γ too. Among A,B,C,E,F, a concyclic quadruple must contain both E and F, since three points of Γ determine Γ.

quicksave_q·Feb 13, 2026(edited)

So E,F share a circle with two of A,B,C,D, and the complementary pair has to make the other circle. There are only three pairings of the four points. I need to write carefully why each pairing permits at most one extra F besides E.

eevee.zip·Feb 13, 2026(edited)

For a fixed pairing, the two circles are determined by E and the respective pairs. They are distinct and already meet at E, so at most one further intersection is possible. That bounds this no-five-concyclic case by eight points. The remaining step is proving that once five lie on Γ, at most one point can lie off it.