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Statement

Setting (p. 397, Section 1). A $(p,t)$-arrangement is a set of $p$ points and $t$ lines in the euclidean or the real projective plane such that each of the $t$ lines contains exactly $3$ of the points; $t(p)$ is the largest $t$ for which a $(p,t)$-arrangement exists. The lines of an arrangement need not be all the lines through exactly three of the points, but those lines always form one, so $t(p)$ is the largest number of lines through exactly three points of a $p$-point set (an observation of this page).

Theorem 1 (p. 397). For every $p\ge3$,

$$t(p)\ge1+\Bigl\lfloor\frac{p(p-3)}{6}\Bigr\rfloor,$$

where the print writes $[x]$ for the integer part $\lfloor x\rfloor$.

The proof (pp. 397--401) is constructive: for each $p\ge3$ it exhibits $p$ points forming a $(p,t)$-arrangement with $t$ equal to the bound. The points lie on the real "odd circuit" of a non-singular cubic, and the one with parameter $0$ is a point at infinity of the cubic in the normal form used, so the arrangement is first obtained in the real projective plane; a projective map sending to infinity a line that misses the $p$ points carries it into the euclidean plane with the same collinear triples. Since a line meets a non-singular cubic in at most three points, no line contains four of the constructed points. Both remarks are observations of this page, not statements of the paper.

Read depth. Claims checked: the definitions, the statement and the counting step of the proof were read clause by clause on the page images of the print. The facts about cubics that the proof cites (the normal form and Abel's collinearity criterion) were not checked here. Nothing here is independently reviewed.

Proof pointer

Pp. 397--401. A non-singular real cubic is projectively equivalent to $y^2=4x^3-g_2x-g_3$, parametrized on its odd circuit by the Weierstrass function, $P(u)=(\wp(u),\wp'(u))$ for real $u$, with real period $2\omega$; three points $P(u),P(u'),P(u'')$ of the odd circuit are collinear exactly when $u+u'+u''\equiv0\pmod{2\omega}$ (the paper's (3), p. 400, cited to Abel through White, Hilton, Coolidge and Whittaker--Watson). The paper takes the $p$ points $P(2\omega k/p)$, $k=0,\ldots,p-1$, so the collinear triples correspond to the unordered triples of distinct residues mod $p$ with sum $0$. Counting the $p^2$ ordered solutions of $k+k'+k''\equiv0$, removing those with a repeated entry and correcting for the solutions of $3k\equiv0$ gives $1+\lfloor p(p-3)/6\rfloor$ triples (pp. 400--401). Figures 2 and 3 (pp. 401--402) draw the case $p=12$, the second after the projective change that puts the three collinear inflection points at infinity.

Remark (2) (p. 418) compares Sylvester's 1867--1868 constructions on cubics: on the paper's reading of Sylvester's choice of starting point, his arrangement is isomorphic to the one above when $3\nmid p$ and has one collinear triple fewer when $3\mid p$. Table I's footnote (p. 399) attributes the lower bounds for $\tilde t(p)$ to Theorem 1 and those for $t(p)$ to the observation preceding Theorem 9; read against the text, the two attributions appear interchanged (an observation of this page).

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 $t(n)=f_3(n)$, so the theorem gives $f_3(n)\ge1+\lfloor n(n-3)/6\rfloor$ for $n\ge3$. With the pair count $F_3(n)\le n(n-1)/6$ this gives both limits for $k=3$, as the problem's claim page for this paper records. The paper says nothing about $k\ge4$.
• Problem 101 and Problem 588: these ask about lines through four (respectively $k\ge4$) points when no line holds five (respectively $k+1$). The construction is the case $k=3$ of that setting, $n$ points with no four on a line and $n^2/6-O(n)$ lines through three of them, so for $k=3$ the count is not $o(n^2)$. The paper proves nothing about four-point lines.
• Problem 211: the paper does not discuss it. For $n\ge6$ the constructed set has at most $n-3$ points on a line and $\binom n2-2t=n(n+3)/6+O(1)$ lines through two or more of its points, where $t=1+\lfloor n(n-3)/6\rfloor$; with $k=n-3$ this is $(1/6+o(1))kn$, so the constant $1/6$ that the problem's page discusses could not be raised along this range. The deduction is the problem page's and this page's, not the paper's.