Given $n$ (pairwise) nonparallel lines in $\mathbb{R}^2 $. $\lbrace L_1,\ldots,L_n\rbrace $. The intersection of any two lines belongs to a third line in our set of lines. I would like to show that $\cap L_i\not = \emptyset.$
My best ideas suppose there is more than one point of intersection and to define a point of intersection of minimal distance (motivated by Kelly's proof of the Sylvester-Gallai Theorem http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem#Kelly.27s_proof ) from our original point of intersection then use the slopes of the lines to show that there is no such closest minimal intersection point by creating new points of intersection. However, there seems to be way too many cases.
Are there any nicer approaches to this problem? I was told this is an application of Sylvester-Gallai Theorem, but I don't see the connection. Any insight would be humbly appreciated.
I think you're on the right track. A possible proof is as follows:
By assumption every point of intersection must have at least three lines going through it. Consider all pairs of points of intersection and lines, and choose the pair such that the perpendicular distance from the point to the line is minimum. Let the point be
Aand the line closest to it belclose, and three lines throughAbel1,l2,l3clockwise as shown in the diagram attached.Now there are two possible cases for the intersection point
Coflclosewithl2we consider.Cis closer tol1thanAis tolclose, which is a contradiction. This is the case shown in the diagram.Cis closer tol3thanAis tolclose, which is again a contradiction.Edit: Note that this proof relies on the fact that the lines are pairwise nonparallel, for otherwise if
lcloseandl1were parallel the reasoning in the first case would not hold. As for what people mean by the question being an application of the Sylvester-Gallai Theorem, I think that the question is simply the projective dual of the Sylvester-Gallai Theorem.