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Statement

Setting (pp. 53--54). An angle of a planar configuration is the angle, at most $\pi$, formed at one of its points by two others. $\alpha(m)$ is the greatest number such that every configuration of $m$ points in the plane contains an angle $\beta\ge\alpha(m)$ (inequality (1)); the paper asks for its exact value and whether (1) can be replaced by the strict inequality (3), $\beta>\alpha(m)$. Szekeres (1941, the paper's reference [3]) proved that (i) every configuration of $N=2^n+1$ points has an angle greater than $(1-1/n+1/nN^2)\pi$, and (ii) for every $\varepsilon>0$ some configuration of $2^n$ points has every angle less than $(1-1/n)\pi+\varepsilon$; hence (2) for $2^n<m\le2^{n+1}$, $[1-1/n+1/n(2^n+1)^2]\pi\le\alpha(m)\le[1-1/(n+1)]\pi$.

Theorem 1 (p. 54, quoted). "Every plane configuration of $2^n$ points ($n\geqq3$) contains an angle greater than $(1-1/n)\pi$."

Section 4 restates it (p. 59) as Theorem 1*: a set of $2^n$ points in the plane is not $P_n$, where a set is $P_n$ when every angle formed by three of its points is at most $(1-1/n)\pi$; the proof takes $n>2$.

Consequence (p. 54). With (ii), Theorem 1 gives $\alpha(2^n)=(1-1/n)\pi$ for $n\ge3$, and the strict inequality (3) holds for $m=2^n$, $n\ge3$. The paper says the problem is thus completely settled for $m=2^n$, $n\ge2$; for $n=2$ this rests on the value $\alpha(4)=\pi/2$, attained by the square (see the small values).

Proof pointer

Sections 3 and 4, pp. 57--60. A partition of the complete graph $C^{(N)}$ into $n$ edge classes is even when no class contains an odd circuit. Lemma 2 (p. 57, from [3]) bounds $N\le2^n$ for an even partition into $n$ classes, and Lemma 3.1 (p. 58) adds that when $N=2^n$ every vertex meets every class. Splitting the plane, from a chosen direction, into $2n$ sectors of angle $\pi/n$ and putting the edge $p_\mu p_\nu$ into class $i$ when its direction lies in sector $T_i$ or $T_{n+i}$ gives a partition of the configuration's complete graph; Lemma 5 (p. 59, from [3]) says it is even when the set is $P_n$. For a $P_n$ set of $2^n$ points, a hull angle below $(1-1/n)\pi$ lets the sectors be aligned with a hull side so that a hull vertex misses a class, against Lemma 3.1. Otherwise the hull has $2n$ vertices $p_1q_1\cdots p_nq_n$, all its angles equal to $(1-1/n)\pi$. If the two $n$-gons through alternate hull vertices have all their angles equal to $(1-2/n)\pi$, then $p_1\cdots p_n$ is a regular $n$-gon; the set has at least two points inside the hull (as $2^n-2n\ge2$ for $n\ge3$), one of them not the centre, and it sees two hull vertices at an angle above $(1-1/n)\pi$, directly if it lies in a triangle $p_iq_ip_{i+1}$ and by Lemma 6 (p. 59) otherwise. If some angle of those $n$-gons is below $(1-2/n)\pi$, Lemma 3 (p. 57) with $i=2$ gives the contradiction.

Dependencies

Lemmas 2 and 5 are proved in G. Szekeres, On an extremum problem in the plane, Amer. J. Math. 63 (1941), 208--210 (the paper's reference [3]); Lemma 2 is reproved on p. 57. Lemma 1 is the bipartiteness of graphs without odd circuits (König). Lemma 6 is credited to Problem 4086 of the Amer. Math. Monthly 54 (1947), p. 117. The configurations (ii) behind the upper bound $\alpha(2^n)\le(1-1/n)\pi$ are from [3] and are not reconstructed in this paper.

Read depth. Claims checked: the setting, Theorem 1, Theorem 1* and the consequence were read clause by clause on the page images of the print, and the proof (pp. 57--60) was followed. Szekeres's (i) and (ii) were not read in their source. 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: the problem's $\alpha_n$ is the paper's $\alpha(m)$; Theorem 1 with Szekeres's configurations determines it at $m=2^n$, $n\ge3$, as $(1-1/n)\pi$, and shows every such configuration has an angle strictly above it. It determines no other value.