The Sylvester–Gallai theorem in geometry states that, given a finite number of points in the Euclidean plane, either

  1. All the points are collinear; or

  2. There is a line which contains exactly two of the points.

I think the theorem is also true for circle, plane version and curve version, can you give a proof of these versions?

Conjecture: (circle version):

Given a finite number of points in the Euclidean plane, either

  1. All the points are concyclic; or

  2. There is a circle which contains exactly three points.

Conjecture: (Plane version)

Given a finite number of points in the three-dimensional space, either

  1. All the points in a plane; or

  2. There is a plane which contains exactly three points.

Conjecture: (Curve version)

Let $n, N$ be the natural number with $(\frac{3n+n^2}{2} < N < \infty)$, give $N$ points in the Euclidean plane, either

  1. All the points lie on a curve of degree $n$; or

  2. There is a curve of degree $n$ which contains exactly $\frac{3n+n^2}{2}$ of the points.

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    $\begingroup$ Motzkin, The lines and planes connecting the points of a finite set, Trans Amer Math Soc 70 (1951) 451-464, MR0041447 (12, 849c), proved some generalizations of Sylvester-Gallai. I think both of your conjectures are there. Gupta, Non-concyclic sets of points, Proc Nat Inst Sci India 19 (1953) 315-316, MR0056941 (15, 149d) proved the circle conjecture, but the review by Blumenthal says it's already in the Motzkin paper. $\endgroup$ – Gerry Myerson Jun 8 '17 at 23:20
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    $\begingroup$ See also Wiseman and Wilson, A Sylvester theorem for conic sections, Discrete Comput Geom 3 (1988) no. 4 295-305, MR0947218 (89f:52024). $\endgroup$ – Gerry Myerson Jun 8 '17 at 23:24

The plane version is false as stated. Choose two skew lines $\ell_1$ and $\ell_2$ in $\mathbb{R}^3$ and place $\geq 3$ points on each of them. Any three of these points will either all lie on the same $\ell_i$ (and thus be collinear) or two of them will lie on the same $\ell_i$, so the plane they span contains that $\ell_i$ and thus contains three collinear points.

A true version is that there is a plane containing precisely $k$ of the points for which $k-1$ of them are collinear. This is due to Motzkin (Trans. AMS, 1951), with a higher dimensional generalization by Hansen (Math. Scand. 1965). I took the above counterexample from Hansen's paper.

I think the curve version should be false, but I haven't broken it yet. I have now discovered two papers: Wiseman and Wilson, (Disc. and Comp. Geom 1988) and Czapliński et. al., (Rendiconti del Seminario Matematico della Università di Padova 2016) which pose the curve version as an open question, so it probably isn't easy to resolve.

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    $\begingroup$ An earlier version of this answer claimed the opposite, and tried to reduce the problem to the SG-theorem in $\mathbb{R}^2$. The error was that multiple points of the original configuration could become the same point in $\mathbb{R}^2$. $\endgroup$ – David E Speyer Jun 9 '17 at 13:12

For the circles part: it's true.

Assume that each circle defined by our set of points $S$ contains $\geq 4$ points; in particular, there are $\geq 4$ points in total. Fix one of them, say $p$, and consider all the circles determined by our points and passing through $p$; each such circle passes through $\geq3$ points from $S-\{p\}$.

Now apply inversion with center $p$ - under this, the above observation shows that $S-\{p\}$ goes to a set of points with the property that any line through two of them contains a third. Applying the standard version of the theorem, we see that all the images of $S-\{p\}$ under the inversion are collinear - hence, $S-\{p\}$ all lie on a circle through $p$.

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  • $\begingroup$ For this argument, are you assuming that three collinear points define a "circle" (i.e., the line through the points)? $\endgroup$ – Timothy Chow Jun 8 '17 at 20:26
  • $\begingroup$ you kind of have to right? otherwise the original question is false too: e.g. if all the points are on a line $\endgroup$ – amakelov Jun 9 '17 at 7:34
  • $\begingroup$ A line is special case of a circle, a line is a circle with radius equal $\infty$ math.stackexchange.com/questions/82220/… $\endgroup$ – Cố Gắng Lên Jun 9 '17 at 11:37

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