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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 5 (p. 409, quoted). "An (8, 8)-arrangement is impossible."

Theorems 3 and 4 give $t(8)\le8$ (an observation of this page, evaluating their formulas at $p=8$), and an arrangement with more than $8$ lines would contain one with exactly $8$, by dropping lines; so the theorem gives $t(8)\le7$. With the lower bound $7$ of Theorem 1 this determines $t(8)=7$, as Table I (p. 399) records.

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. 409--410. Here $\Gamma(\mathcal A)$ has $8$ nodes, $\binom82-24=4$ edges and odd valences, so it is a perfect matching. Sending the two ends $A$, $B$ of one edge to infinity, each lies on three lines of $\mathcal A$, two parallel pencils, and the other six points must sit among the nine crossings of those pencils, two on each line. Labelling the crossings as a $3\times3$ grid, the only candidates for the two remaining lines are the diagonals $159$ and $357$, which use only five points, and no sixth point completes an $(8,8)$-arrangement.

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 $f_3(8)=7$. A single value of $n$; it does not affect the limits the problem asks for.