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Statement

Notation as on the Theorem 1 page.

Theorem 2 (p. 60, quoted). "In a plane configuration of $N=2^n-k$ points ($0<k<2^{n-1}$) there is an angle $\ge(1-1/n-k/2N)\pi$."

The introduction (p. 54) states it as $\alpha(2^n-k)\ge(1-1/n)\pi-k\pi/2(2^n-k)$ for $0<k<2^{n-1}$. No range for $n$ is printed; the range of $k$ is empty unless $n\ge2$.

The suggestion beside it (p. 54). The authors write that it is not impossible that $\alpha(m)=(1-1/n)\pi$ for $2^{n-1}<m<2^n$, $n\ge4$, and that (3) holds for every $m>6$, but that they can prove only Theorem 2.

Proof pointer

Pp. 60--61. Lemma 4 (p. 58) refines Lemma 3.1: for $N=2^n-k$, $0\le k<2^n$, and an even partition of $C^{(N)}$ into $n$ classes, writing $\nu(p)$ for the number of classes with no edge at $p$, $\sum_p(2^{\nu(p)}-1)\le k$. Assuming every angle is at most $(1-1/n)\pi-\tfrac12\delta-\delta'$ with $\delta=k\pi/N$ and $\delta'>0$, each point $p$ has a direction $\alpha(p)$ such that no segment from $p$ lies in the two opposite open sectors of that angle bounded by $\pm\alpha(p)$. Some open arc of length $\delta+\tfrac13\delta'$ contains $k+1$ of the $2N$ directions $\pm\alpha(p)$. A sector partition aligned with that arc, with the edges in two thin strips moved to other classes, stays even, and the $k+1$ points owning those directions have no edge in the first class, so the sum in Lemma 4 exceeds $k$.

Dependencies

Theorem 1's apparatus: Lemmas 1, 2 and 5 and the sector partitions of Section 4, with Lemma 4 of this paper.

Read depth. Claims checked: Theorem 2, its announcement and the suggestion on p. 54 were read clause by clause on the page images of the print; the proof (pp. 60--61) was followed for structure. 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: a lower bound for $\alpha_m$ when $2^{n-1}<m<2^n$; it determines no value. The suggestion that $\alpha(m)=(1-1/n)\pi$ throughout that range is a guess the paper does not prove; the problem's claim pages record what later became of it.