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Posers and Puzzles

Posers and Puzzles

  1. 03 Jun '08 23:19
    Prove that if X is a finite set of points on the plane (not all in one line) then there is a line passing through exactly 2 points of X.
  2. 04 Jun '08 00:01
    Extremal Principle+PHP+Contradiction
    Let me work out the details...
  3. 04 Jun '08 00:11
    Suppose for a contradiction that any line passing through 2 points of X also passes through a least 3, and that not all points of X are on the same line.

    Take all pairs of {lines passing through 2 points of X} and {a point not on that line}. There is at least one such pair becuase not all points of X are on a line. Call the line L and the point y. Choose the pair that minimizes the distance from L to y.

    Drop the perpendicular from y to the L. By the PHP at least 2 points on L lie on the same side of the perpendicular. Let these two points be a and b, with a closer to the perpendicular than b. Then the distance from a to line yb is less than the distance from L to y, contradicting its minimality. QED
  4. Standard member TheMaster37
    Kupikupopo!
    04 Jun '08 13:44 / 1 edit
    Suppose for a contradiction that any line passing through 2 points of X also passes through a least 3, and that not all points of X are on the same line.

    Take all pairs of {lines passing through 2 points of X} and {a point not on that line}. There is at least one such pair becuase not all points of X are on a line. Call the line L and the point y. Choose the pair that minimizes the distance from L to y.


    I suppose you want {lines passing through 3 points of X}?

    Drop the perpendicular from y to the L. By the PHP at least 2 points on L lie on the same side of the perpendicular.

    I am not familiar with PHP but if a line contains exactly three points and the perpendicular goes through one of those three, this statement is false...in the convention I am used to (a point on a line is on neither side of the line).

    Let these two points be a and b, with a closer to the perpendicular than b. Then the distance from a to line yb is less than the distance from L to y, contradicting its minimality. QED
  5. 04 Jun '08 21:45
    Since every line passing through 2 points also passes through 3 (as i supposed), it doesn't matter whether i say 2 or 3.

    Oops about the second one, i should have mentioned it. Fortunatly, the proof doesn't break down because of that, if we take a to be the base of the perpendicular a to by is still less that y to L.
  6. Standard member TheMaster37
    Kupikupopo!
    05 Jun '08 13:20
    Since every line passing through 2 points also passes through 3 (as i supposed), it doesn't matter whether i say 2 or 3.

    Right you are, my bad!

    Oops about the second one, i should have mentioned it. Fortunatly, the proof doesn't break down because of that, if we take a to be the base of the perpendicular a to by is still less that y to L.

    Very true, now we have a very elegant proof

    I don't supose there is an intuistionistic proof...