Spoiler for New Scientist Enigma 1769: “Crossing lines” (Follow the link to see the puzzle.)
Easier than it looks at first.
Consider at first the three lines that cross at a point. On their own they create 6 edge regions and 0 non-edge regions. Call these lines 1 through 3.
Now add new lines one at a time, each crossing all the other lines but not at a point already containing a crossing. (Can this be done? I think so; you can always make each line nearly parallel to a previous one, close enough in spacing and angle to cross the remaining lines within some epsilon of the previous line’s crossings, yet not at the same angle, so it can cross the previous line at some point. These distances and angles might need to get very small, though.) Call these lines 4 through n. The kth line, at the time it’s added, crosses (k–1) previous lines and two edges. Since the created regions can never be nonconvex, each region that is crossed is crossed only once. Therefore the kth line crosses k regions, cutting each into two. Two of the crossed regions are edge regions, and each is cut into two edge regions, so the number of edge regions increases by 2. The remaining (k–2) regions are non-edge regions, each of which is cut into two non-edge regions, so the number of non-edge regions increases by (k–2). That should remind you of triangular numbers. The number of edge regions after the nth line is 2n and the number of non-edge regions is (n–2)(n–1)/2–1. Then the solution is obtained by solving
3(2n) = (n–2)(n–1)/2–1
with the result n = 15.
Do not expect a diagram of this.