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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.

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    $\begingroup$ Four parallel and one not will give $4$ intersections. $\endgroup$ – David Feb 14 '18 at 5:32
  • $\begingroup$ Three parallel and two not will give 7. Two parallel and three not will give 9 (all assuming no three intersect in the same point). $\endgroup$ – Arthur Feb 14 '18 at 6:06
  • $\begingroup$ I wonder whether OP is ruling out parallel lines. If you allow them, you can also have 0 intersections. $\endgroup$ – Gerry Myerson Feb 14 '18 at 6:09
  • $\begingroup$ Clearly, the OP does rule out 0 intersections, but you can have parallel lines (just not all of them) $\endgroup$ – Max Punck-Institut Feb 14 '18 at 6:56
  • $\begingroup$ Correct me if I am wrong but I think there are more ways. I find at least 7 ways for four lines to intersect<: $\endgroup$ – Max Punck-Institut Feb 14 '18 at 7:06
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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:

  • No bunches, for $\binom 52 = 10$ intersection points.
  • One bunch of $3$ lines, for $\binom 52 - \left(\binom32 - 1\right) = 10 - 2 = 8$ intersection points.
  • Two bunches of $3$ lines, for $\binom 52 - \left(\binom32 - 1\right)- \left(\binom32 - 1\right) = 10 - 2 - 2 = 6$ intersection points.
  • One bunch of $4$ lines, for $\binom 52 - \left(\binom42 - 1\right) = 10 - 5 = 5$ intersection points.
  • One bunch of $5$ lines, for $\binom 52 - \left(\binom52 - 1\right) = 10 - 9 = 1$ intersection point.

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:

  • One parallel bunch of $2$ lines, for $\binom 52 - \binom 22 = 9$ intersection points.
  • One parallel bunch of $3$ lines, for $\binom 52 - \binom 32 = 7$ intersection points.
  • One parallel bunch of $4$ lines, for $\binom 52 - \binom 42 = 4$ intersection points.
  • One parallel bunch of $5$ lines, for $0$ intersection points.

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:

  • When four lines meet at a point, we can put that point at infinity (and then no other intersection point can be at infinity, because it would lie on one of these four lines too) or else put at most one intersection point with the fifth line at infinity. So we can reduce the number of intersection points by at most $1$.
  • When all five lines meet at a point, we can either put that point at infinity or not. Once again, we can reduce the number of intersection points by at most $1$.

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.

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  • $\begingroup$ So, you are forbidding parallel lines, right? $\endgroup$ – Gerry Myerson Feb 14 '18 at 12:02
  • $\begingroup$ @GerryMyerson I guess I was, but there's really no reason to; this approach generalizes. $\endgroup$ – Misha Lavrov Feb 14 '18 at 15:03
  • $\begingroup$ This is perfect, thank you! $\endgroup$ – mroch241 Feb 15 '18 at 21:27
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Concerning the problem "In how many ways can $5$ lines intersect in the plane?" there are $18$ different configurations.

We can partition $5$ lines into subsets of parallel lines in $7$ ways: $$(5),\quad(4,1),\quad(3,2),\quad (3,1,1),\quad (2,2,1),\quad (2,1,1,1),\quad(1,1,1,1,1)\ .$$

For $(5)$, $(4,1)$, and $(3,2)$ we obtain $1$ configuration each.

For $(3,1,1)$ the two single lines can intersect on one of the parallels or in a point not on these; makes $2$ configurations.

For $(2,2,1)$ the single line can pass through $0$, $1$, or $2$ of the four intersection points of the other lines; making $3$ configurations.

For $(2,1,1,1)$ the three single lines can produce a triangle or intersect in one point. In the first case the two parallels can pass through $0$, $1$, or $2$ vertices of the triangle; in the second case zero or one of the parallels can pass through the point of intersection of the three single lines. Makes $5$ configurations.

For $(1,1,1,1,1)$ we can have (i) one point incident with five lines, (ii) one point incident with four lines, (iii) two points incident with three lines, (iv) one point incident with three lines, and (v) no point incident with $>2$ lines.Makes $5$ configurations.

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  • $\begingroup$ Thank you! This gives me a clearer understanding. $\endgroup$ – mroch241 Feb 15 '18 at 21:26

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