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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 8 (p. 414, quoted). "A (14, 28)-arrangement is impossible."

Theorems 3 and 4 give $t(14)\le28$ (an observation of this page, evaluating their formulas at $p=14$), and an arrangement with more than $28$ lines would contain one with exactly $28$, by dropping lines; so the theorem gives $t(14)\le27$. With the lower bound $26$ of Theorem 1, Table I (p. 399) records $26\le t(14)\le27$. For pseudolines, Theorem 10 gives $\tilde t(14)\ge27$, and Theorem 9 carries this theorem's bound to pseudolines, so $\tilde t(14)=27$, as Table I marks.

Read depth. Claims checked: the statement was read on the page images of the print. The proof is not in the paper, so the theorem rests on the authors' word. Nothing here is independently reviewed.

Proof pointer

P. 415. The paper prints no proof. It states that the proof is analogous to that of Theorem 7, with $\Gamma(\mathcal A)$ now seven disjoint edges and many subcases, and offers to send the complete proof to interested readers. The corpus holds no copy of that proof.

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 Theorem 1, gives $26\le f_3(14)\le27$, the upper bound resting on an unprinted proof. A single value of $n$; it does not affect the limits the problem asks for.