Erdős Problem #89 $500 Prize

The Distinct Distances Problem

Given n points in the plane, what is the minimum number of distinct distances between pairs of points? A deceptively simple question with deep connections to algebraic geometry.

The Problem

Let g(n) be the minimum number of distinct distances determined by n points in the plane. Erdős conjectured:

g(n) = Ω(n / √log n)

The best known lower bound is g(n) = Ω(n / log n), proved using algebraic methods.

Explore the problem
Part 1

What Are Distinct Distances?

Place n points anywhere in the plane. Now measure the distance between every pair of points. The distinct distances problem asks: what's the minimum number of different distances you can achieve?

Distinct Distances
Given n points P₁, P₂, ..., Pₙ in ℝ², count how many different values appear among all pairwise distances |PᵢPⱼ| for i ≠ j.

There are n(n-1)/2 pairs, but many distances might be equal.

Simple Example: 3 Points

With 3 points, you have 3 pairs. An equilateral triangle gives only 1 distinct distance (all sides equal). But can you always achieve so few? For larger n, minimizing distinct distances becomes much harder.

Part 2

Try It Yourself

Point Placer

Click to place points. Watch how the number of distinct distances changes!

Click to add points
Points (n)
0
Point Pairs
0
Distinct Distances
0
Ratio (Distinct/n)
-
Distances will appear here...
Part 3

Famous Configurations

Some point arrangements are especially good at minimizing distinct distances. The √n × √n grid is the classic example that achieves near-optimal bounds.

√n × √n Grid
For n = 9: only 5 distinct distances
Regular n-gon
For n = 6: only ⌊n/2⌋ distances
Triangular Lattice
Also achieves good bounds

The Grid is Special

A √n × √n integer lattice achieves about cn / √log n distinct distances — exactly what Erdős conjectured as the minimum! This suggests the conjecture might be tight.

Part 4

Mathematical Progress

Progress on the Conjecture 70% Resolved
Guth-Katz Theorem (2015)
Using algebraic geometry and the polynomial method, Guth and Katz proved:

g(n) = Ω(n / log n)

This is only a log factor away from Erdős's conjecture of n / √log n.

Why Is the Last Factor So Hard?

The gap between n / log n and n / √log n might seem small, but closing it has resisted all attempts. The algebraic methods that achieved the current bound seem to have reached their limits.

New ideas — perhaps from combinatorics, number theory, or algebraic geometry — will be needed to fully resolve Erdős's conjecture.

Part 5

Why Does It Matter?

The distinct distances problem connects to:

The Unit Distance Companion Problem
Erdős also asked about the maximum number of pairs at distance exactly 1. This is Problem #90 — the "unit distance problem" — another $500 prize!