On a 2-dimensional plane, 2019 points are given such that, for every 3 points picked, there are 2 points whose distance is smaller than 1.

Is there a circle with radius 1 that can contain at least 1010 of the 2019 points?

Yes, of course
No, there isn't

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Suppose the statement is false. Take any one point $P$ and construct a unit circle $C_P$ with center $P$ . By our assumption, $C_P$ contains no more than 1009 points, so there are at least 1010 points outside of $C_P$ .

Take any three points outside $C_P$ , call them $A, B$ and $C$ . Then $AP, BP$ and $CP$ are all greater than 1, so $AB, BC$ and $AC$ must all be less than 1.

If we "push" $A, B$ and $C$ as far apart as possible under this constraint, they will be at the vertices of an equilateral triangle of side length 1 (technically just under 1 but that should not affect the proof.)

Now consider any other point $X$ also outside $C_P$ . Since $XP$ is greater than 1, $XA, XB$ and $XC$ must all be less than 1, i.e $X$ must lie in the intersection of the unit circles centered at $A, B$ and $C$ (the shaded portion in the left image below). Since $X$ could be any point outside of $C_P$ , this means that

allpoints outside of $C_P$ must lie in that intersection. But clearly that intersection can itself be contained within a unit circle, as shown in the right image.So all points outside of $C_P$ must lie within that unit circle, and there are at least 1010 such points.