The line with only two points on it

Take a handful of points on a page, not all sitting on one line, and draw every line that passes through at least two of them. Most of those lines will carry exactly two points; a few, if the points were placed carefully, will carry three or four. The question Sylvester asked in 1893 is whether the careful placing can be pushed all the way: can every line be made to carry at least three?

The answer is no, and the interesting part is not the answer but its proof, which is four sentences long, uses no coordinates, and was not found for forty years.

What is being counted #

Fix a finite set of points in the plane, not all on a single line. A connecting line is a line through at least two of them, taken together with every point of the set that happens to lie on it. That last clause is doing real work: on the grid above, the top row is one line carrying three points, not three separate lines each carrying two, and a count that made the second mistake would find no ordinary lines anywhere.

An ordinary line is a connecting line carrying exactly two. Sylvester’s question is whether a configuration can have none.

Three points give three ordinary lines and no choice about it. Four points in general position give six, again with no choice. The first configuration with any freedom in it is the near-pencil: put all but one of the points on a single line, and the last one off it.

The near-pencil is extremal in the opposite sense from the one Sylvester was asking about. What is wanted is a configuration with few ordinary lines, ideally none, and the near-pencil has almost nothing but.

The configuration that comes closest #

There is a seven-point arrangement that gets remarkably near. Take a triangle, add the midpoints of its three sides, and add the centroid. Every side of the triangle now carries three points — two corners and a midpoint — and every median carries three as well, since a median runs from a corner through the centroid to the opposite midpoint.

Seven points make twenty-one pairs. The six rich lines account for eighteen of them, three pairs each, so exactly three pairs are left over — and those three are the sides of the middle triangle formed by the midpoints. Three ordinary lines from seven points, which is fewer ordinary lines than points, and is the best known for that size.

It is very nearly a counterexample, and that is what makes Sylvester’s question feel open until it is settled. One more rich line and the count would be down to nought.

The proof, which is a minimisation #

Here is the whole argument, due to Leroy Kelly in 1948.

Among all pairs consisting of a point $P$ of the set and a connecting line $\ell$ not passing through $P$, choose one for which the distance from $P$ to $\ell$ is smallest. Such a pair exists, because there are finitely many points and finitely many lines and at least one pair to choose from — that last clause is where “not all in one line” is used, and it is used only there.

Claim: that line $\ell$ is ordinary.

Suppose it were not. Then $\ell$ carries at least three points of the set. Drop a perpendicular from $P$ to $\ell$, meeting it at a foot $F$. The three points on $\ell$ cannot be spread with at most one on each side of $F$, so two of them lie strictly on the same side — call the nearer one $B$ and the farther one $C$. Now look at the line through $P$ and $C$. The distance from $B$ to that line is strictly less than the distance from $P$ to $\ell$, which contradicts the choice of the minimum.

The last step is the one worth pausing on, because it is the only geometry in the argument. Triangle $PFC$ has a right angle at $F$, and $B$ sits between $F$ and $C$. The distance from $B$ to line $PC$ is a leg of a triangle similar to $PFC$ but scaled down by the ratio $BC/FC$, which is less than one because $B$ is strictly inside the segment. So the new distance is a genuine shrink, not a tie.

That is the entire proof. It has no coordinates, no equations, no case analysis beyond “two on one side”, and it produces an ordinary line rather than merely proving one exists.

Why the first proofs were harder #

Sylvester posed the problem in 1893 and it went unanswered. Gallai proved it in 1933 by projecting the configuration into a picture where one point is sent to infinity and arguing about the resulting parallel families — a real proof, but one that needs projective machinery to state.

Kelly’s argument arrived fifteen years later and is short enough to hold in the head. What it needed was a willingness to minimise something that is not obviously the right quantity. Nothing about the statement suggests measuring distances; the statement is about incidence, which is a notion that survives stretching the page. The proof leaves the category the problem is stated in — and that is exactly why it works, because a shortest distance is a thing a finite set of points genuinely has, and incidence alone offers nothing to be extreme about.

There is a moral in this for the rest of the site. Three colours force a triangle is settled by a pigeonhole count; six people at a party by the same; this one is settled by picking the smallest of finitely many real numbers, which is the pigeonhole principle’s cousin — the observation that a finite non-empty set of numbers has a least element, and that assuming otherwise gives something to contradict.

Where the proof fails, and it does fail #

The argument uses distance, which the real plane has and other planes do not. That is not a technicality: the theorem is false over the complex numbers.

The Hesse configuration is nine points and twelve lines in the complex projective plane, arranged so that every one of the twelve lines carries exactly three of the nine points, and every one of the thirty-six pairs of points lies on one of them. There are no ordinary lines at all. It is the configuration of the nine inflection points of a smooth cubic curve, and it cannot be drawn on a real page — the nine-point grid is the closest thing to it that fits in the real plane, and the grid has twelve ordinary lines rather than none.

Seven points and seven lines builds the Fano plane over the field of two elements and finds it entirely consistent — but that consistency is combinatorial. Sylvester’s theorem says that no such structure can be laid out with real coordinates and straight edges, so the drawing’s curved seventh line is not a failure of draughtsmanship. It is the theorem, seen from the other side.

So the theorem is best read as a statement about what the real numbers do that other fields do not. Incidence structures with no ordinary line are common; realisable ones do not exist.

How few is few #

Once no configuration can have zero ordinary lines, the question becomes how close it can get. The answer is known and took a long time.

Kelly and Moser proved in 1958 that $n$ points always determine at least $3n/7$ ordinary lines, and that the seven-point configuration above shows $3n/7$ is exactly right at $n = 7$. The true answer for large $n$ is $n/2$, conjectured by Dirac in 1951 and proved by Green and Tao only in 2013 — sixty years later, with the proof running to nearly two hundred pages and settling the question for all sufficiently large $n$.

Two configurations achieve $n/2$ for even $n$, both due to Böröczky, built from the vertices of a regular $m$-gon together with points at infinity in the directions of its sides. They are the reason the bound cannot be improved, and their existence is why the general proof is hard: the extremal examples are not close to the near-pencil, so no argument that merely rules out degenerate configurations can succeed.

The opposite question, which is much harder #

Sylvester’s problem asks for few ordinary lines. The orchard problem, which is older, asks for many rich ones: given $n$ trees, how many rows of exactly three can be planted?

The two questions are not opposites in any formal sense, and the second is far worse behaved. For nine points the answer is ten rows of three, and finding an arrangement takes some ingenuity; for twelve it is nineteen. The general answer was conjectured by Burr, Grünbaum and Sloane in 1974 to be $\lfloor n(n-3)/6 \rfloor + 1$ and proved for large $n$ by Green and Tao in the same paper that settled Dirac’s conjecture — the two questions turn out to need the same machinery, which is the sort of coincidence that says the machinery is about something more general than either.

What makes the orchard problem hard is that the extremal configurations are not built out of grids or polygons at all. They come from cubic curves: take an elliptic curve, and use the fact that three of its points are collinear exactly when they add to zero in the curve’s own group law. Choosing $n$ points forming a subgroup makes an enormous number of collinear triples appear at once, and no arrangement built from ordinary symmetry gets close.

That is a connection worth registering. A question about dots on a page is answered by the arithmetic of a curve, and the reason is that the curve supplies a way of manufacturing collinearity to order. Nothing in the statement of the orchard problem suggests looking for an algebraic structure whose addition law is collinearity, and that is exactly what the answer needs.

The dual statement, which is the same theorem #

Every statement about points and lines in the plane has a partner obtained by swapping the two words, and this one is worth writing out because it sounds like a different theorem.

Take finitely many lines in the plane, not all through a single point and not all parallel. Then some point lies on exactly two of them. In this form the result is about an arrangement of lines and the vertices it creates, and the vertices where exactly two lines cross are called simple. The dual of the near-pencil is a family of lines all through one point plus one more; the dual of the Kelly–Moser configuration is an arrangement of seven lines with only three simple vertices.

The two statements are the same statement, transported. The map that trades circles for lines is a different transport of the same kind, and the habit is worth naming: a theorem is often two theorems that a reader would not have thought to connect, and the connection is a dictionary rather than an argument.

Where it came from #

Sylvester published the question in the Educational Times in 1893 as a problem, not a conjecture, and no solution appeared. He seems not to have returned to it.

The problem was rediscovered by Erdős in 1943, who could not prove it and passed it on; Gallai supplied a proof within weeks, and Erdős published it with attribution. Kelly’s proof came in 1948 and was itself published by Coxeter in a paper making the point that the theorem is not projective — that any proof must use something the real numbers have and the projective plane over an arbitrary field does not.

That point is the durable one. A great deal of plane geometry survives being flattened into incidence structure, and this theorem is a marker for where that flattening stops.

What the pictures cannot show #

The figures here draw configurations at particular sizes, and the drawings are honest about those sizes and about nothing beyond them. Nine points, twenty lines, twelve ordinary — all computed. What no drawing can show is that no configuration escapes, and the enumeration above does not show it either: it covers the sets that fit on a four-by-four grid, which is finitely many out of infinitely many.

The Kelly–Moser configuration’s coordinates here are whole numbers because they were chosen to be. Nothing about the configuration requires that, and the collinearity of a median with the centroid is a fact about ratios rather than about integers; the integer coordinates are a convenience that lets the collinearity test be exact rather than a property of the object.

And the complex counterexample cannot be drawn at all. The Hesse configuration is nine points in a plane over the complex numbers, which is four real dimensions, and every picture of it that appears in print is a picture of something else — usually the nine inflection points of a cubic, three of which are real and six of which are not.

The ladder from here #

Below: seven points and seven lines, the incidence structure that has no ordinary line and no realisation, and more things than boxes, the finiteness argument this proof is a version of. Sideways: three colours force a triangle, another statement that something unavoidable happens in every configuration; the plane, divided by whoever is nearest, where a finite point set again forces a structure nobody put there; and sixteen trees on four points, a count over configurations rather than a property forced on all of them. Above: the Dirac–Motzkin bound and its proof, the dual theorem about arrangements of lines, and the classification of configurations that achieve the minimum.

What is worth carrying away #

The statement is about incidence and the proof is about distance, and there is no way to fix that. Every attempt to prove Sylvester’s theorem inside the language it is stated in fails, because the statement is false in that language — the Fano and Hesse configurations are there, consistent, waiting.

So the thing the proof really establishes is a boundary. It says that the real plane is not merely an incidence structure with coordinates attached; it carries an ordering and a metric, and those leak into facts that look purely combinatorial. A theorem that needs a metric to prove a combinatorial statement is saying the statement was never combinatorial.