Question

7. Suppose there are 10 lines in the plane and no two of them are parallel. Suppose that at every point in the plane there are at most 3 lines passing through that point. Show that if we fix one of the 10 lines Lo, then there are at least 5 points on Lo where other lines intersect it.

Answer 1

Suppose there are 10 lines in the plane and no two of them are parallel. Also suppose that at every point in the plane there are at most 3 lines passing through that point. So let one of the lines is fixed and denoted by Lo.Now Lo is not parallel to the rest 9 lines, and thus when they intersect Lo there can be maximum 9 points on Lo possible where other lines intersect Lo. So if the 9 other lines intersect at different points then clearly there will be 9 points on Lo If the lines who cut the Lo at a single point they can be put as a pair, for example L1, L2 both intersect Lo at a single point, then we make the pair (L, L2). Now maximum such pairs possible is 4 as there are 9 line not counting Lo in this list. Thus 8 lines were considered in pairing they will intersect Lo at 4 points. Finally, there remain a single line and this also will cut Lo at a point different from the previous four points, as otherwise the number of lines paasing through a given point will become more than 3, which is not allowed. Therefore, this point becomes the 5th point. Thus there are at least 5 points on Lo where the remaining 9 lines intersect.