\section*{Problem 2 - 320922}

In the plane, there are four pairwise distinct lines given.

a) What is the maximum possible number of points that are the intersection of (at least) two of the given lines?

b) Determine which of the numbers 0, 1, 2, 3, 4, 5, 6 are possible as the number of such intersection points and which are not!
}

Let \(a, b, c, d\) be the four lines.

a) If no two of the lines are parallel and no three intersect at the same point, then \(a\) and \(b\), \(a\) and \(c\), \(a\) and \(d\), \(b\) and \(c\), \(b\) and \(d\), as well as \(c\) and \(d\) each have a unique intersection point, which is different from the others. Since two different lines can have at most one point in common, this is the maximum number of distinct intersection points.

b) We have just seen that exactly 6 intersection points can be generated if no two lines are parallel and no three intersect at the same point.

If \(a\) and \(b\) are parallel, but \(c\) and \(d\) are neither parallel to each other nor to \(a\) (and \(b\)), then \(c\) and \(d\) each have two intersection points with \(a\) and \(b\). If the intersection point of \(c\) and \(d\) is not one of these four, there are a total of 5. If in the construction above \(c\) and \(d\) are also parallel, there are only the four intersection points with the parallel lines \(a\) and \(b\).

If \(a, b\), and \(c\) are pairwise parallel to each other, but \(d\) is not, then \(a, b\), and \(c\) do not intersect pairwise, while each of them intersects with \(d\) at exactly one point, and no two of these intersection points coincide, so there are exactly 3 in total.

If all four lines are pairwise parallel, there are no, i.e., 0, intersection points.

If all four lines pass through a common point \(P\), there is exactly one intersection point, namely \(P\). There cannot be any more, as two different lines can intersect at most at one point.

It remains to show that exactly two intersection points are not possible:

Assume there were such a configuration. Then the three lines \(a, b\), and \(c\) cannot be pairwise parallel, as this would (as shown above) result in exactly 0 or exactly 3 intersection points with \(d\).

They also cannot intersect at a common point \(P\), as this would (if \(d\) passes through \(P\)) result in exactly 1 or (if \(d\) does not pass through \(P\)) exactly \(1+3=4\) intersection points. They also cannot be pairwise non-parallel, as the coincidence of two of their three intersection points would immediately imply that the third is also identical, placing us in the previous case.

Thus, of the three lines \(a, b\), and \(c\), exactly two must be parallel and the third not parallel to them. Without loss of generality, let \(a \| b\) and \(b \nVdash c\). Similarly, we can consider the three lines \(b, c\), and \(d\), of which again two must be parallel and the third not parallel to them. In the case \(b \| d\), we get the contradiction \(a\|b\| d\) analogously to above, resulting in exactly 3 intersection points, and in the remaining case \(c \| d\), we get exactly 4 intersection points (as shown above), so not exactly 2.

Therefore, every number of intersection points from 0 to 6, except for 2, can be realized.