There are examples of sets of $n$ points with $\sim n^2/6$ many collinear triples and no four points on a line. Such constructions are given by Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84].

Grünbaum [Gr76] constructed an example with $\gg n^{3/2}$ such lines. Erdős speculated this may be the correct order of magnitude. This is false: Solymosi and Stojaković [SoSt13] have constructed a set with no five on a line and at least \[n^{2-O(1/\sqrt{\log n})}\] many lines containing exactly four points.

See also [102]. A generalisation of this problem is asked in [588].

Equivalently, one can ask the dual problem: given $n$ points in $\mathbb{R}^2$ such that there are no lines containing at least four points then there are three points such that the lines determined by them are ordinary ones (i.e. contain exactly two points each).

The Sylvester-Gallai theorem implies that there must exist a point where only two lines from $A$ meet. This problem asks whether there must exist three such points which form a triangle (with sides induced by lines from $A$). Füredi and Palásti [FuPa84] showed this is false when $d\geq 4$ is not divisible by $9$. Escudero [Es16] showed this is false for all $d\geq 4$.

Conjectured by Erdős and de Bruijn. The Sylvester-Gallai theorem states that $f(n)\geq 1$. The fact that $f(n)\geq 1$ was conjectured by Sylvester in 1893. Erdős rediscovered this conjecture in 1933 and told it to Gallai who proved it.

That $f(n)\to \infty$ was proved by Motzkin [Mo51]. Kelly and Moser [KeMo58] proved that $f(n)\geq\tfrac{3}{7}n$ for all $n$. This is best possible for $n=7$. Motzkin conjectured that for $n\geq 13$ there are at least $n/2$ such lines. Csima and Sawyer [CsSa93] proved a lower bound of $f(n)\geq \tfrac{6}{13}n$ when $n\geq 8$. Green and Tao [GrTa13] proved that $f(n)\geq n/2$ for sufficiently large $n$. (A proof that $f(n)\geq n/2$ for large $n$ was earlier claimed by Hansen but this proof was flawed.)

The bound of $n/2$ is best possible for even $n$, since one could take $n/2$ points on a circle and $n/2$ points at infinity. Surprisingly, Green and Tao [GrTa13] show that if $n$ is odd then $f(n)\geq 3\lfloor n/4\rfloor$.

In particular, given any $2n$ points with at most $n$ on a line there are $\gg n^2$ many lines formed by the points. Solved by Beck [Be83] and Szemerédi and Trotter [SzTr83].

In [Er84] Erdős speculates that perhaps there are $\geq (1+o(1))kn/6$ many such lines, but says 'perhaps [this] is too optimistic and one should first look for a counterexample'. The constant $1/6$ would be best possible here, since there are arrangements of $n$ points with no four points on a line and $\sim n^2/6$ many lines containing three points (see Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84]).

A generalisation of [101] (which asks about $k=4$).

The restriction to $k\geq 4$ is necessary since Sylvester has shown that $f_3(n)= n^2/6+O(n)$. (See also Burr, Grünbaum, and Sloane [BGS74] and Füredi and Palásti [FuPa84] for constructions which show that $f_3(n)\geq(1/6+o(1))n^2$.)

For $k\geq 4$, Kárteszi [Ka] proved \[f_k(n)\gg_k n\log n.\] Grünbaum [Gr76] proved \[f_k(n) \gg_k n^{1+\frac{1}{k-2}}.\] Erdős speculated this may be the correct order of magnitude, but Solymosi and Stojaković [SoSt13] give a construction which shows \[f_k(n)\gg_k n^{2-O_k(1/\sqrt{\log n})}\]

The trivial greedy algorithm gives $g(n)\gg n^{1/2}$. A similar question can be asked for a set with no $k$ points on a line, searching for a subset with no $l$ points on a line, for any $3\leq l<k$.

Erdős thought that $g(n) \gg n$, but in fact $g(n)=o(n)$, which follows from the density Hales-Jewett theorem proved by Furstenberg and Katznelson [FuKa91] (see [185]).