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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Notation as on the Theorem 1 page.

Theorem 3 (p. 61, in the note added in proof, quoted). "Every plane configuration of 2n−12^n-1 points (n≥2n\ge2) contains an angle not less than (1−1/n)π(1-1/n)\pi."

The paper adds (p. 61) that the theorem shows in particular α(2n−1)=(1−1/n)π\alpha(2^n-1)=(1-1/n)\pi, and that it cannot decide whether the strict inequality (3) holds for m=2n−1m=2^n-1.

The printed range is wrong at n=2n=2. There the theorem says that every three points of the plane form an angle of at least π/2\pi/2, which the equilateral triangle refutes, and the paper itself gives α(3)=π/3\alpha(3)=\pi/3 (p. 54). The proof needs a point of the configuration inside its convex hull: when every angle is below (1−1/n)π(1-1/n)\pi the hull has at most 2n−12n-1 vertices, which leaves an interior point exactly when 2n−1>2n−12^n-1>2n-1, that is, when n≥3n\ge3. The statement and the value α(2n−1)=(1−1/n)π\alpha(2^n-1)=(1-1/n)\pi hold for n≥3n\ge3.

Proof pointer

Pp. 61--62. Suppose every angle is below (1−1/n)π(1-1/n)\pi and take qq inside the hull, with largest angle (1−1/n)π−δ(1-1/n)\pi-\delta at qq. A sector partition aligned with that angle gives qq no edge in the first class, so Lemma 4 with k=1k=1 and Lemma 5 give every other point an edge in every class. As each hull angle is below (1−1/n)π(1-1/n)\pi, the hull has at most 2n−12n-1 vertices, fewer than the 2n2n sectors, so some hull vertex pip_i has its incoming side in a sector Tk−1T_{k-1} and its outgoing side outside TkT_k; such a vertex has no edge in class kk, a contradiction.

Dependencies

Lemmas 4 and 5 and the sector partitions, as on the Theorem 1 and Theorem 2 pages; the upper bound α(2n−1)≤α(2n)≤(1−1/n)π\alpha(2^n-1)\le\alpha(2^n)\le(1-1/n)\pi comes from Szekeres's configurations (ii).

Read depth. Claims checked: Theorem 3 and the sentence after it were read clause by clause on the page images of the print, and the proof (pp. 61--62) was followed; the failure at n=2n=2 is checked here against the paper's own value α(3)=π/3\alpha(3)=\pi/3. Nothing here is independently reviewed.

Source. P. Erdős and G. Szekeres, On some extremum problems in elementary geometry, Ann. Univ. Sci. Budapest. Eötvös Sect. Math. 3--4 (1960/1961), 53--62; the edition read is named on the source card.

Bears on

  • Problem 504: for n≥3n\ge3 it determines α2n−1=(1−1/n)π\alpha_{2^n-1}=(1-1/n)\pi; whether every configuration of 2n−12^n-1 points has an angle strictly above that value is left open in the paper.