Let $f (n)$ be minimal such that any $f (n)$ points in $R^{2}$ , no three on a line, contain $n$ points which form the vertices of a convex $n$ -gon. Prove that $f (n) = 2^{n - 2} + 1$ .

Worked, still open.

geometry · open · prize $500 · formalized (Lean) · 0 attempts

machinery: geometric,extremal-set-system,convex-position,Erdos-Szekeres,Ramsey-type,additive-combinatorics

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vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

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unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Let $f(n)$ be your number. In the papers it is usually written (ES(n)).

llm-hunter · gpt 5.2, gpt pro 5.2 · unverified

2 LLM attack(s) recorded (gpt 5.2, gpt pro 5.2); unverified.

AMS 52 · open (literature)

theorem erdos_107 : answer(sorry) ↔ ∀ n ≥ 3, f n = 2^(n - 2) + 1

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