Five lines can intersect in at most $(5^2-5)/2 =10$ ways.
So far I have found 1, 5, 6, 8, and 10 intersections.
Is there a reason why they only intersect in this many ways and why?
My approach is through combinations, $5C1$ for number of ways five lines can intersect at one point.
$(5C4*5C1)/5=5$ for number of ways 4 lines intersect at one point and the last line intersect the four.
and lastly $5C2=10$ for 10 intersections.
However I'm confused when I need to prove the 6 and 8 intersections.
[Math] Ways 5 lines can intersect
combinationscombinatoricsgeometrylinear algebra
Best Answer
One way to think about this is that $n$ lines usually have $\binom n2$ intersections. However, if a group of $k$ lines, $k \ge 3$ are concurrent (meet at a single point), their $\binom k2$ intersection points merge, reducing the total number of intersections by $\binom k2 - 1$. Let's call such a group a "bunch" just to have a name for it.
Further restricting the options, two different bunches can share at most one line. (If they shared two lines, then they'd have to be part of one bunch.) This limits the number of different bunches we can have.
In the case of $n=5$ lines total, our options are:
Parallel lines can also be handled with this method. In a "parallel bunch" of $k \ge 2$ lines, all $\binom k2$ intersection points are lost; a parallel bunch can share a line with a concurrent bunch, but no lines at all with another parallel bunch.
All the possible cases would get kind of annoying to list out, but the ones that get new counts of intersection points are:
We could mix and match parallel bunches and concurrent bunches, as well as use multiple smaller parallel bunches (e.g. a parallel bunch of $3$ and another parallel bunch of $2$, which is another way to get $6$ intersections) but in this case, it still doesn't help us get $3$, $2$, or $1$ intersections.
Another way to think about parallel lines is that they're equivalent to lines that intersect at a point at infinity. We can declare some intersection points to be points at infinity, and don't have to count them; however, at most one intersection point per line can be a point at infinity, otherwise the line is forced to be the line at infinity, which doesn't fit into our affine geometry framework.
In the case of $5$ lines, we can declare up to $2$ intersection points to be points at infinity, since each such point "uses up" two lines. This lets us reduce the number of intersection points by $1$ or by $2$. The exceptions are the final two configurations:
I think this approach is the more systematic one, but it requires thinking about points at infinity, which is an aspect of projective geometry that might make some people uncomfortable.