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Is there a geometric object with finite numbers of points and lines (or curves) and the following 3 properties:

  1. Each line contains exactly 4 points.
  2. For every three points, there is a line containing these three points.
  3. There are 4 points that do not lie on any line.

Note that neither Euclidean spaces nor projective geometries over a finite field satisfy these conditions.

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  • $\begingroup$ If there are four points that don't lie on any line – if there is even one point that doesn't lie on any line – then how can it be that for every three points, there is a line containing those three points? Do you mean that there's a set of four noncollinear points? $\endgroup$ Oct 27, 2017 at 21:04
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    $\begingroup$ Yes, this is surely what they mean, and it seems to me natural enough phrasing for it. $\endgroup$ Oct 27, 2017 at 21:06
  • $\begingroup$ Do you presume that there is no more than one line through given two points? $\endgroup$ Oct 28, 2017 at 18:00
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    $\begingroup$ If you ask that any three points are contained in a unique block, this is precisely the definition of a Steiner Quadruple System. They are a well understood class of designs: it is known for example, that there exists an SQS on $v$ points whenever $v \equiv 2, 4 \mod 6$ and for no other values of $v$. $\endgroup$ Oct 28, 2017 at 18:27
  • $\begingroup$ @Sridhar, considering the discussion engendered by the answer you posted, it is not clear that you are justified in being so confident as to what exactly OP means. $\endgroup$ Oct 29, 2017 at 10:51

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Sure, it's easy. For example, let there be 5 points overall, and let each 4-element subset of those 5 points constitute a line… except for one particular 4-element subset of those 5 points which does not constitute a line. This satisfies all the required properties.

If you want UNIQUENESS of the line in property 2, then this becomes a trickier problem. But even then, it can be done: take the n-dimensional vector space over the field with two elements. Take points to correspond to vectors in this space, and take lines to correspond to sets of four points which sum to 0. Note there will be a unique line containing any three given points. What's more, as long as n is at least 3, there will be some four points which are not contained in any line.

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  • $\begingroup$ @sridhr Ramesh: This is not right.If i do what you say then the "one particular 4-element subset of those 5 points" will contradict the property two. Is this clear or do you need an example? $\endgroup$
    – ruhi
    Oct 27, 2017 at 22:03
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    $\begingroup$ For any three points you pick, there will be some line containing them. If the five elements overall are A,B,C,D,E, and the one four-tuple which doesn't make a line is {A,B,C,D}, and you pick the three elements, say, A,B,C, well they are still contained in the line {A, B, C, E}. And B,C,D are contained in the line {B,C,D,E}, and A, C, D are contained in the line {A, C, D, E} and so on. Every three elements is contained in some line, but A, B, C, D are not collinear. $\endgroup$ Oct 27, 2017 at 23:11
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    $\begingroup$ I'm sorry, I can't follow what you just said. Are you saying BCE is not on a line? BCE is on the line {B, C, D, E}, as well as also on the line {A, B, C, E}. $\endgroup$ Oct 28, 2017 at 0:42
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    $\begingroup$ Can you clarify to me how my first example does not satisfy the required properties? Either by 1. Pointing out a line that does not contain precisely four points, or by 2. Pointing out three points that do not lie on a line, or by 3. Somehow demonstrating that, contra my assertion, there IS a line which contains A, B, C, and D. Or, alternatively, by clarifying to me how I have mis interpreted what you intended with your descriptions of the required properties. $\endgroup$ Oct 28, 2017 at 0:45
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    $\begingroup$ I'll also note, I gave not just one example but two. The second one in some ways even nicer, because of its uniqueness element. Do you have a problem with the second example I gave? $\endgroup$ Oct 28, 2017 at 0:46
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Here is another presentation. I have not checked if it duplicates an example already given.

We use the six vertices of the regular octahedron, and we color the faces red and blue so that each edge of the octahedron lies next to a red face and to a blue face. Note that any two vertices either share an edge or are antipodal.

Now take three vertices of the octahedron. If they are not on a face, there is a plane passing through them that intersects a fourth vertex: call this collection a line. Otherwise take the color of the face of the vertices and call that a point and thus we have an eight point set and have 11 candidates for lines, eight of them coming from the faces. Finally, any antipodal pair matched with a color point gets the other color placed on that line, for a total of fourteen four point lines on eight points. This system satisfies the three stated conditions, as no line consists of (three vertices on) a face and a noncolor point.

Gerhard "Better Living Through Colored Polytopes" Paseman, 2017.10.28.

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  • $\begingroup$ Note that this also satisfies the stronger condition that every three points lie on a unique line. Indeed, this is the same as my second example, specialized to the case n = 3. To see this, label the three points of some face 001, 010, and 011, label that face's color 000, and label the three antipodal points 110, 101, and 100, respectively, with this antipodal face's color labelled 111. Now four points comprise a line if and only if their bitwise XOR comes out to 000. $\endgroup$ Oct 28, 2017 at 18:06

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