In a planar 6-point set of diameter at most 2, how many distances can be greater than $\sqrt{2}$? I found a tough combinatorial geometry problem. Any discussion or advice is helpful.
6 points are on the plane such that any 2 points are at most distance 2 apart. What is the most number of pairs of points which are strictly greater than √2 apart?
 A: There is a way to get $12$ pairs, take an equilateral triangle of side length $1.5$, for each vertex add an extra point right next to it.(The picture shows the arrangement)

We now prove there can't be more than $12$ pairs, if there where more than $12$ pairs we would have two points $a,b$ such that the distance between each of these points and another is in the interval $(\sqrt{2},2]$.Why? For each point let $d(p)$ be the number of points that are at distance greater than $\sqrt2$. then the number of pairs is half of the sum of all of these numbers. So this sum must be at least $26$. If there aren't two points which satisfy all points are at distance greater than $\sqrt2$ this sum is at most $5+4+4+4+4+4=25$
We now prove the distance between any other two points $c,d$ is $\sqrt{2}$ or less. Suppose not, then consider the convex cuadrilateral with vertices $a,b,c,d$. One of its angles must be $90^\circ$ or more. Suppose without loss of generality it is $\angle ABC$. Then by law of cosines $AC^2=AB^2+BC^2-2\cos(\angle ABC)(AB\cdot BC)>AB^2+BC^2>2+2>4$.
So $AC>2$. Which is a contradiction, since the points where all at distance at most $2$.
A: We are dealing with a circle her such that maximum distance between any 2 points is 2, ie a circle of diameter 2
Mathematically, x^2 + y^2 <=1.
circle
Any 2 point in this circle has atmost 2units between them.
For strictly greater than √2, we take the most extreme condition, that is all points form the corners of a hexagon along the circumference. 
Hexagon
sides of hexagon are all 1 units. and diagonals are strictly greater than √(2). There are 9 diagonals hence 9 is the answer
