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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 6 (p. 410, quoted). "A (10, 13)-arrangement is impossible."

Theorems 3 and 4 give $t(10)\le13$ (an observation of this page, evaluating their formulas at $p=10$), and an arrangement with more than $13$ lines would contain one with exactly $13$, by dropping lines; so the theorem gives $t(10)\le12$. With the lower bound $12$ of Theorem 1 this determines $t(10)=12$, 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. 410--412. Here $\Gamma(\mathcal A)$ has $10$ nodes, $6$ edges and odd valences, which forces a star on one point $A$ with three leaves plus three disjoint edges. Deleting $A$ and its three lines leaves a $(9,10)$-arrangement $\mathcal B$ whose lines, according to whether three named points are collinear, are one of two explicit lists $(A_1)$, $(A_2)$; a relabelling maps $(A_2)$ to $(A_1)$. With two points of $(A_1)$ sent to infinity, the case analysis over three orderings of the resulting pencils (Figures 12 and 13) rules out each; the third case (Figure 12(c)) is left to the reader.

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 $f_3(10)=12$. A single value of $n$; it does not affect the limits the problem asks for.