Consider any $3$ points.We must have at least two points which are at distance less then $sqrt(3)$.Now,if we consider all triplets,we will have at least $111*110*109/6/109=2035>1998$,cause every pair of vertices is in $109$ triplets.

The $1998$ segments formed by the $111$ points must be of lengths strictly less than $\sqrt{3}$; this is how it was given in the exam.

Here is a solution for the actual problem (it is still too trivial to be on a TST, in my opinion). First, we can pick an arbitrary large number of triplets on the circle, which divide the circle into $3$ equal arcs and are pairwise disjoint, so we can pick one that has no points on the circle from these points. Now, divide the circle with these $3$ points; if two points are in the same area, their distance is less than $\sqrt{3}$, so let $a,b,c$ be the numbers of points in this areas. We have at least $a(a-1)/2+b(b-1)/2+c(c-1)/2$, and this is minimum when $a=b=c$, so we have at least $3\cdot 37\cdot 36/2=1998$ segments.

Prove the stronger version: Prove that we can replace $1998$ by $3025.$

This is only true if we go back from $< \sqrt{3}$ to $\leq \sqrt{3}$, otherwise three bunches of $37$ coinciding points, situated at the vertices of an equilateral triangle, realize just (precisely) $1998$ such segments. Consider an edge between two vertices (points given) if their mutual distance is strictly larger than $\sqrt{3}$. Then there is obviously no $K_3$ (induced triangle), so by Turán's theorem there are at most $\lfloor 111/2 \rfloor \cdot \lceil 111/2 \rceil = 55\cdot 56 = 3080$ such edges. It means there are at least $\binom {111}{2} - 3080 = 3025$ edges missing, corresponding to segments of length at most $\sqrt{3}$. A simple model realizing this minimum is given by the same Turán's theorem; two bunches of $55$, respectively $56$ points, round antipodal points on the circle.

mavropnevma wrote: t $\lfloor 111/2 \rfloor \cdot \lceil 111/2 \rceil = 55\cdot 56 = 3080$ such edges. It means there are at least $\binom {111}{2} - 3080 = 3025$ edges missing, corresponding to segments of length at most $\sqrt{3}$. A simple model realizing this minimum is given by the same Turán's theorem; two bunches of $55$, respectively $56$ points, round antipodal points on the circle. First question: Why is there no $ K_3 $ Second question : Why are there at most $\lfloor 111/2 \rfloor \cdot \lceil 111/2 \rceil = 55\cdot 56 = 3080$? Thank you in advance