A complete post is here: blog.geomblog.org/2026/05/the-... The unit distance problem: Let P be a set of n points in the plane (amen). What is the maximum number of unit distances that can be achieved? (the amen is a running joke in my old field of computational geometry) 2/n

The unit distances problem

Note that this is really about duplicate distances (because if you get a bunch of pairs of points at distance d, you can scale the point set so that d = 1). It's also trivial to see that the maximum is at least n-1 (points evenly spaced on the line), and that the number is at most n-choose-2. 3/n

So what's the answer? Erdos showed using a fairly complicated construction that you can get a set of points that has $n^{1 + 1/\log\log n}$ unit distances. On the other hand, a famous result from 1983 by Spencer, Szemederi and Trotter showed that you can't get more than $O(n^{4/3})$ distances. 4/n

Erdos himself used a really elegant but weaker argument to show that you can't get more than $O(n^{3/2})$ distances. And the argument was cool and deceptive (in retrospect). To get a distance pair of 1, a circle of size 1 around a point must touch another point (and vice versa). 5/n

Draw an edge between those two points when this happens. So here's a fun fact: you can't create a situation where two points are connected by edges to each of three other points. Formally you can't create a situation where there's a $K_{2,3}$ bipartite graph hidden inside the set of edges. 6/n

By a well known result in graph theory this means this graph can't have too many edges in it (because if it did you'd eventually find one of these special graphs): specifically no more than $O(n^{3/2})$ edges. 7/n

So there's a limit. But what's the real limit? A long standing conjecture was that you could NOT get anything nontrivially more than n pairs. Specifically that you couldn't get $\Omega(n^{1 + \epsilon})$ pairs for any $\epsilon$. This is sad because the gap between this, and $n^{4/3}$ is huge. 8/n