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I was trying to solve a problem and I got curious about this other one because it might give me some intuition:

Are there finitely many points in $\mathbb R^2$ such that they do not all lie on a straight line, and such that any straight line passing through two of them also passes through a third?

I tried to construct such points by hand and I couldn't do it. Anybody has an idea?

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  • $\begingroup$ My intuition (and just starting from a line and building out) is saying probably not in Euclidean Geometry. If we start with two points we need to add a third to that line, as well as another non-colinear point. That point needs another added point for each of the original points, and it keeps building out. There might be some interesting triangle fractals from just what I've been playing with $\endgroup$ – wjmccann Jun 12 at 15:43
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There aren't any such points. It is a statement of Sylvester–Gallai theorem which admits a very elegant and elementary proof (Kelly's proof) which is also available at Wikipedia.

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  • $\begingroup$ Also available at...? $\endgroup$ – user76284 Jun 12 at 19:48
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For a rough sketch of the proof of the theorem, we let the (finite) set of all lines formed by joining pairs of points be S and we consider pairs of (point, line), where the point lies in the set and the line lies in S. Oh, and the point doesn’t lie on the line. Obviously there are a finite number of such pairs.

Now if we assume that there are at least three points on each line, we consider the pair where the point line distance is minimal, and through a tiny bit of geometry (you should try it!) we see that there must be a smaller point line pair, which is a contradiction.

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