Please reference this problem.

In the above referenced problem a debate was raised as to the validity of given solution.  This debate caused me to ponder the following problem.

If you place 3 points in a line then obviously every line through any 2 points will intersect exactly 3 points.  Now the real question is, is it possible to place more than 3 points in a plane such that all lines between any 2 points intersect exactly 3 points, no more no less.

If the answer to the above question is in the negative, prove it. Otherwise, derive the appropriate example(s),

The is a more beautiful proof to a wider problem (at least in my opinion). It was an arab mathematician who thought of it first (in recorded history), but I cannot remember his name :(

Problem:

Given N points in plane, there is at least a line connecting 2 of the points, that does not pass through a third.

Proof:

Construct all lines passing through 2 points, and all non-zero distances from a point to a line. Because N is finite, there must be a minimum such distance. let A be the point and BC be the 2 points that determine the line. thus we have:

   A

/_|__\

B H    C

H is an artificial point, where the perpendicular from A intersects BC. It may not be part of the N points.

Now, if line BC does not contain another point D, problem is already proven.

If D is to the right of C, then the distance from C to the line AD is shorter than the minimum distance [which we established to be the distance from A to BC]

If D is between H anc C, then the distance from D to the line AC is shorter than the minimum distance

If D is between H anc B, then the distance from D to the line AB is shorter than the minimum distance

If D is to the left of B, then the distance from B to the line AD is shorter than the minimum distance