Points in the plane - Pigeonhole Principle - Sharpen the result Thirteen points are given in the plane so that among any three of them there is a pair whose distance is less than $1$.  Prove that it is possible to select seven of the points so that they are all interior to a circle of radius $1$.
This problem can be solved by using the Pigeonhole Principle. Let $A$ be any given point, and consider the circle $C_A$ of radius $1$ around $A.$ If there are at least seven points in this circle, we are done. If not, let $B$ be a point outside $C_A$, and consider $C_B$, a circle of radius $1$ around $B.$
By the condition in the problem, all points are in $C_A$ or $C_B$, so by the Pigeonhole Principle at least seven of the points lie in the same circle.
Now the question is whether the result could be sharpened to assert that it is possible to select seven of the points so that they are all interior to a circle of radius $\frac{1}{\sqrt{3}}$. If so, how can this be shown?
 A: The picture shows an incorrect distribution of seven points in the circle, but helps us find proof of the statement:

Under given condition, it is possible to select seven of the points so that they are all inside a circle of radius $\frac{1}{\sqrt3}.$

Assume that seven of the points lie in a circle $C$ of radius $1,$ and the remaining six lie outside a circle $C'$ concentric to $C,$ radius of $C'$ being at least $3.$
Ignore for a while the six points that lie outside.
The picture shows the most stretched relative position of the seven points: one is a center of the circle, four points $1,4,5,3$ lie near the vertices of a regular hexagon, two other points are close to the arcs $(1,4)$ and $(3,5),$ respectively.
HOWEVER, the distribution of the thirteen points as decribed does not fulfill the condition. The problem only occurs if one of the three points lies outside $C'$ and the two inside $C,$ but they are too far apart. Therefore, the distance between any two points inside $C$ must be less than $1.$ Consequently, the seven points lie inside a curvilinear triangle (such as in the picture, with vertices $1,2,4$). As the circumradius of this triangle is $\frac{1}{\sqrt3},$ we are done.

