Prove that within a set of points distant less than 1 from each other, there cannot be 4 points all distant more than $1/\sqrt{2}$ Given a set of points in which the maximum distance between 2 points is no more than $1$, prove that there cannot be 4 points which all have distance $> 1/\sqrt{2}$ from each other.
What I have done so far:


*

*tried to draw 2 points, 4 circles originating from these 2 points (each of radius $1$ and $1/\sqrt{2}$, but failed to place the 3rd point somewhere from I could make clear argument abouth the 4th

*found this: Prove if there are 4 points in a unit circle then at least two are at distance less than or equal to $\sqrt2$, which is pretty much what I want (just different radius), but I can't prove that I can find a circle of diameter equal to $1$, covering my set of points; and I don't even know if that is true.

 A: Here is an approach.  It is not rigorously done, but I find it convincing.  
You can think of the contrapositive:  given four points in the plane that are at least $\frac 1{\sqrt 2}$ from each other, at least one pair has distance at least $1$.  We can enforce the pairwise distance by making each point the center of a disk of radius $\frac 1{2\sqrt 2}$ and demanding that the disks be disjoint.  They are not even allowed to be tangent to each other.  
Start by placing three of the disks at corners of an equilateral triangle.  There is clearly nowhere to place the fourth disk to have its center within $1$ of the other three.  The best we can do is make another equilateral triangle.  The two farthest centers are then $\sqrt 2$ apart, but the two on the shared side are only $\frac 1{\sqrt 2}$ apart.  Let the long axis of the figure be vertical.  We can spread the horizontal pair, sliding around the top disk to let the bottom one rise and approach the top one.  When the two side disks are $1$ apart they make (just under) a $\frac \pi 2$ angle at the top disk keeping the bottom disk (just over) $1$ from the top.  If we let the disks be tangent we get a square with side $\frac 1{\sqrt 2}$ and diagonal $1$.  We need to move them apart by some amount, which makes at least one diagonal increase above $1$.
