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),
Lets assume we have such a set of points obeying the conditions, and we call each set of 3 points lying on a straight line a 3set. Lets count the number of 3sets P in two different ways:
According to the problem, each choice of 2 points lie in a 3set. And also, each 3set contains 3 distinct pairs. Thus
P=C(n,2)/3=n(n-1)/6
Now consider any given point Q and all 3sets that Q lies in. Since all points lies in any 3set containing Q, and there are 2 points different from Q in each 3 set, we get that that the number of 3sets containing Q is (n-1)/2. Next consider another point R and we count all 3sets R lies in. But we have already counted one of the sets in the Qpart so the number for R is (n-3)/2. Continuing in this way shows that P=Sum(n+1-2j)/2=(n+1)(n-1)/8, j goes from 1 to (n-1)/2 (after we reach (n-1)/2 the terms become non positive, so thus we must have counted all 3sets by then).
so: (n+1)(n-1)/8=n(n-1)/6 ->3(n+1)=4n->n=3
thus it is only possible for 3 points, so the answer to the question is no.
Edited on November 7, 2008, 8:27 pm
Proof
