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Statement

Here $t(p)$ is the largest number of lines through exactly three points of a $p$-point set, as defined on the Theorem 1 page, and $\Gamma(\mathcal A)$ is the graph of the pairs of points on no line of the arrangement, defined on the Theorem 3 page.

Theorem 7 (p. 412, quoted). "A (12, 20)-arrangement is impossible."

Theorems 3 and 4 give $t(12)\le20$ (an observation of this page, evaluating their formulas at $p=12$), and an arrangement with more than $20$ lines would contain one with exactly $20$, by dropping lines; so the theorem gives $t(12)\le19$. With the lower bound $19$ of Theorem 1 this determines $t(12)=19$, as Table I (p. 399) records.

A note added in proof (p. 422) states that the theorem had already been established by J. Novák (1970).

Read depth. Claims checked: the statement was read on the page images of the print, and the case analysis was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pp. 412--414. Here $\Gamma(\mathcal A)$ is a perfect matching on $12$ nodes. Sending the ends $A$, $B$ of one edge to infinity gives two pencils of five parallel lines, and the other ten points form a set $\Pi$ of their $25$ crossings with two on each line. The ten remaining ("skew") lines must join each point of $\Pi$ to six others, and no point may be unjoined to two others; the paper turns these into conditions $(C_1)$, $(C_2)$ and rules out every placement by a case analysis on the boundary of the grid (Figures 14--19).

Source. S. A. Burr, B. Grünbaum and N. J. A. Sloane, The orchard problem, Geometriae Dedicata 2 (1974), 397--424, DOI 10.1007/BF00147569 (source card).

Bears on

• Problem 669: in the problem's notation the theorem, with Theorems 1, 3 and 4, gives $f_3(12)=19$. A single value of $n$; it does not affect the limits the problem asks for.