Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Source. Conjecture (1) of problem 36, p. 101, of P. Erdős, Research problems, Period. Math. Hungar. 15 (1984), no. 1, 101--103, doi:10.1007/BF02109375. The edition read is named on the source card.
Statement
Setting (p. 101). is a set of points in the plane. has property when no line contains more than of its points; so says that the points are not all on one line.
Definition (p. 101). For with property , , is the maximal number of lines containing points of . Under such a line contains exactly of the points.
Conjecture (1) (p. 101). For fixed , as ,
Reported progress (p. 101). Erdős states that Kárteszi proved the first half, showing that is possible, and that Grünbaum improved this to display (2), ; the constant is printed with subscript . He suggests that (2) is perhaps best possible, writes that the second half of (1) is still open, and offers a prize for a proof or disproof of it.
The case (p. 101). In contrast, display (3) reports Sylvester's two-sided estimate , with the second constant printed as , citing Burr, Grünbaum and Sloane.
Proof pointer
None in the paper; the second half of (1) is posed as open. The note gives no argument for Kárteszi's or Grünbaum's bounds and cites Grünbaum's 1976 paper for (2).
Read depth
Claims checked: the definitions, (1), (2) and (3) were read clause by clause on the page image of p. 101. Nothing here is independently reviewed.
Dependencies
- Display (3) is cited to Burr, Grünbaum and Sloane, The orchard problem (card burr_1974_orchard_problem).
- Display (2) is cited to B. Grünbaum, New views on some old questions of combinatorial geometry, Colloquio Internazionale sulle Teorie Combinatorie (Rome, 1973), I, Accad. Naz. Lincei, 1976, 451--468, which the library does not hold.
Bears on
- Problem 588: the problem asks whether for , counting lines with at least points among points with no on a line; that is the second half of conjecture (1) for every . The note poses it and proves nothing towards it.
- Problem 101: the problem asks whether points with no five on a line determine lines with four points, which is the case of the second half of conjecture (1). The note proves nothing towards it.