1. Joined
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    03 Jun '08 12:46
    I don't have a solution...

    Is there a set of points S in the plane, such that every straight line in the plane contains exactly 2 points of S?
  2. Joined
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    03 Jun '08 15:30
    Originally posted by David113
    I don't have a solution...

    Is there a set of points S in the plane, such that every straight line in the plane contains exactly 2 points of S?
    My first instinct is to say no. I think I can prove that no continuous curve works, and clearly no bounded set works. But there could be some bizarre collection of discrete points that does the job - I can't see how to disprove that quite yet.
  3. Joined
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    03 Jun '08 15:411 edit
    What about the points that define a circle with a radius of infinity?
    Then every line should intersect this circle in exactly two points, at the polar coordinates r1=r2=inf, and phi1=f and phi2=f+pi.
    Is this the answer?
  4. Standard memberforkedknight
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    03 Jun '08 15:41
    Originally posted by David113
    I don't have a solution...

    Is there a set of points S in the plane, such that every straight line in the plane contains exactly 2 points of S?
    The closest thing that exists would probably be the Euclidean axes, which would work for all lines except y=0 and x=0
  5. Joined
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    03 Jun '08 16:15
    Originally posted by forkedknight
    The closest thing that exists would probably be the Euclidean axes, which would work for all lines except y=0 and x=0
    And all lines through the origin if that does not count as 2. Me thinks.
  6. In Christ
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    03 Jun '08 16:39
    What about the points that define a circle with a radius of infinity?

    That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
  7. Standard memberPalynka
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    03 Jun '08 16:52
    Originally posted by Jirakon
    What about the points that define a circle with a radius of infinity?

    That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
    A circumference is the immediate intuitive answer, but I fail to see what sense it makes to consider these intersections at infinity.

    Give me a functional form taking the limit of r to infinity and I'll give you an epsilon that proves that there is at least a line that doesn't intersect on two points.
  8. Joined
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    03 Jun '08 17:19
    Originally posted by Jirakon
    What about the points that define a circle with a radius of infinity?

    That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
    Care to mention one of the points in that set then?
  9. Joined
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    03 Jun '08 17:20
    Originally posted by Tera
    And all lines through the origin if that does not count as 2. Me thinks.
    And all lines parallel to either the x- or y- axis.
  10. Joined
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    03 Jun '08 17:54
    Originally posted by FabianFnas
    What about the points that define a circle with a radius of infinity?
    Then every line should intersect this circle in exactly two points, at the polar coordinates r1=r2=inf, and phi1=f and phi2=f+pi.
    Is this the answer?
    A point is an ordered pair of two REAL numbers. No infinity.
  11. Standard memberPalynka
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    03 Jun '08 18:013 edits
    Originally posted by mtthw
    My first instinct is to say no. I think I can prove that no continuous curve works, and clearly no bounded set works. But there could be some bizarre collection of discrete points that does the job - I can't see how to disprove that quite yet.
    I don't think it can be discrete points. Imagine if you have a set of discrete points and pick a line that intersects only two points. If you pick one of those points as center and keep rotating that line infinitesimally, then I don't see how it's possible to cover all possibilities with any mesh of discrete points.

    Edit - So I'd say no, there is no such set S.

    Edit 2 - Which probably means I'm wrong and missing an ingenious answer.
  12. Standard memberwolfgang59
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    03 Jun '08 19:20
    I agree there is no such set.
    ,,
    and I can kinda explain

    but a rigorous proof is alluding me.🙄
  13. Backside of desert
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    03 Jun '08 21:26
    Originally posted by Palynka
    A circumference is the immediate intuitive answer, but I fail to see what sense it makes to consider these intersections at infinity.
    .


    if you do include any trans finite numbers in the plane; then ithink that you need to include all trans finite numvbers.

    another quwstion is: can you define a plane, and lines on that plane, so that set s exists.
  14. In Christ
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    03 Jun '08 21:33
    A point is an ordered pair of two REAL numbers. No infinity.

    If that's the case, then there is no solution:

    Suppose such a set exists in which all points have finite coordinates. Take the point(s) with the highest y-value (y_max). Now imagine the line y = y_max + 1. This line does not intersect any of the points. Therefore no such set exists.
  15. Joined
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    03 Jun '08 21:47
    Originally posted by Jirakon
    What about the points that define a circle with a radius of infinity?

    That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
    Well, it was a wild guess, the closest I can get to an answer.

    Is it possible to construct a plane where inf really is included? As you can with R, i.e. R* ?
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