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.
Best Answer
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.