Problem 92 — Visual explainer (undergrad)
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Problem statement
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Let $f(n)$ be maximal such that there exists a set $A$ of $n$ points in $\mathbb{R}^2$ in which every $x\in A$ has at least $f(n)$ points in $A$ equidistant from $x$.

Is it true that $f(n)\leq n^{o(1)}$? Or even $f(n) < n^{O(1/\log\log n)}$?

Picture
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Plane picture:

y ^
  |    o      o
  |        o
  | o
  +-----------------> x

Question: what to look at
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- Restate the problem as: given the constraint, what asymptotic bound or
  structure must follow?
- Use the comments as benchmarks (best known bounds / constructions).

Context
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Area hint: incidence geometry / combinatorial geometry.

Notation (if it appears above)
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- `o(g(n))` means "grows strictly smaller than g(n)" (ratio → 0 as n→∞).
- `O(g(n))` means "at most a constant times g(n)" for large n.

Benchmarks / known results (from comments)
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- This is a stronger form of the unit distance conjecture (see [90]).
- The set of lattice points imply $f(n) > n^{c/\log\log n}$ for some constant
  $c>0$.
- Erd\H{o}s offered \$500 for a proof that $f(n) \leq n^{o(1)}$ but only \$100
  for a counterexample.
- This latter prize is downgraded to \$50 in.