$2$ out of $4$ points, each of distance at most $1$ apart, are at most $1/\sqrt2$ apart Given $4$ points in the plane such that any pair of them are a distance of at most $1$ apart, show that some pair of them must be of distance at most $1/\sqrt{2}$  apart.
I figured the solution might involve a pigeonhole argument, but as my geometry is rusty I'm having trouble turning the "distance of at most $1$" condition into a condition that bounds all $4$ points in a region that can then be subdivided into $3$ regions in which every point must be within $1/\sqrt{2}$. Or that might be entirely on the wrong track.
 A: The given answers give excellent visual representations of what's going on, but I think I stumbled upon the kind of argument I was looking for:
Claim: Given $4$ points in the plane, $3$ of them must either be collinear or form either a right or an obtuse triangle. 
Proof: Consider the convex hull of the $4$ points.
If it is a segment, then there are $3$ collinear points.
If it is a triangle, say $ABC$, then the point in the interior of this triangle, say $D$, forms three different triangles with the other three points whose angles at $D$ add up to $360^\circ$, meaning that at least one of these triangles is right or obtuse.
Finally, if the convex hull is a quadrilateral, its angles also add up to $360^\circ$, so at least one of its angles must be right or obtuse, giving a right or obtuse triangle.

But if three of the points form a right or obtuse triangle $XYZ$ with obtuse angle $Y$, then $XY^2 + YZ^2 \leq XZ^2$. If $XY$ and $YZ$ are both greater than $1/\sqrt{2}$, this would make $XZ > 1$, a contradiction. So there is always a pair of points at most $1/\sqrt{2}$ apart.
A: By eyeballfrog's comment, if there's an arrangement of four points such that all pairwise distances lie in the interval $(1/\sqrt{2}, 1]$, then by scaling, we can guarantee that two are a distance $1$ apart. Let these points be $A$ and $B$. The diagram shows circles of radius $1/\sqrt{2}$ and $1$ with centers $A$ and $B$. The remaining two points must be in the curvilinear regions $CDEF$ and $C'D'E'F'$, excluding the inner arcs of radius $1/\sqrt{2}$.
They can't be in different regions, as $ACBC'$ is a square with side-length $1/\sqrt{2}$, so $CC' = 1$, and between any two points in different regions there is a vertical distance of at least $CC'$.
But they can't be in the same region either. The triangle $ABE$ is equilateral, so $EC = \sqrt{3}/{2} - 1/2$. Applying the law of cosines to the triangle $ABD$ gives $$AD^2 = AB^2 + BD^2 - 2 AB \cdot BD \cos \angle ABD.$$ As $AB = BD = 1$ and $AD = 1/\sqrt{2}$, this gives $\cos \angle ABD = BG = 3/4$ and thus $DF = 1/2$ (because $DF \parallel AB$). Therefore, $CDEF$ can be contained in a rectangle with sides through $C, D, E, F$ and parallel to $CE$ and $DF$. This rectangle has diagonal length $\sqrt{CE^2 + DF^2} < 1/\sqrt{2}$, so all points in $CDEF$ (and, by symmetry, $C'D'E'F'$) are closer than $1/\sqrt{2}$.

A: Bear with me.
Okay...  So two points $A$ and $D$ must be the furthest (or tied for furthest) apart. Say they are $d$ apart and $d \le 1$.  We might as well assume they are exactly $1$ apart; everything we calculate will be in terms of that distance.
I drew the following image Based on $A$ and $D$:

The Black circles are centered at $A$ and $D$ and have a radius of $1$ (or $d$).  And the red circles have a radius of $\frac 1{\sqrt 2}$.  (As $\frac 12 < \frac 1{\sqrt 2} < 1$ we know the red circles intersect.)
The points $e$ and $f$ are where the circles of radius $\frac 1{\sqrt 2}$ intersect.  And $g$ is are the lines $ef$ and $AD$ intersect.
The geometry you need to know is that $\triangle Aeg$ and $\triangle Afg$ are congruent right triangles and by the Pythagorean theorem that $eg^2 + Ag^2 = Ae^2$.  But $Ae=\frac 1{\sqrt 2}$ and $Ag = \frac 12$ so .... $eg = \frac 12$.
That means the distance from $e$ to $f$ is $1$.
Well... so?
Well where can we put $B$ and $C$?
Well we can't put them outside the black circles.  That'd mean they are more than $1$ away from both $A$ and $D$.
We can't put them in the blue areas because that would mean they are more than one away from either $A$ or from $D$.
We can't put one of them in a light green area the other in the other light green area because that would mean they are more than $1$ apart from each other.
If we put either $B$ or $C$ in a lavender or in the yellow area that would mean that point is within $\frac 1{\sqrt 2}$ of either $A$ or $D$ and our result would be fullfilled.
The only unconsidered option is that both $B$ and $C$ are place in the same light green area.  Then they'd both be more than $\frac 1{\sqrt 2}$ for $A$ and $B$ but they'd be pretty close to each other.
How close? 
