[For the case where the points are distributed uniformly. Otherwise, it could go as high as 31. But never 32.]
Just checking in… Thanks all for your replies. My daughter and I will have to brush up on this!
[Following up on my last post: Just to be completely explicit, this is assuming the connections along which we cut between pairs of points are such that two separate such connections only intersect at at most one point (e.g., as with straight lines).
Let me give a more detailed explanation:
We start with one piece. Each cut between two points which we then make starts off in some piece (splitting it from one into two), and transitions into further pieces a number of times (again, splitting each piece it encounters). Thus, each cut increases the number of pieces by 1 + the number of times it intersects a previous cut (not counting endpoints of such cuts).
And thus, the total number of pieces we end up with is equal to our 1 original piece + the number of cuts + the number of times a cut intersects previous cuts.
In our case, the number of cuts is, of course, n choose 2.
As for the number of times a cut intersects previous cuts, it depends. But if we assume that no three cuts intersect at the same point, then this is the same as the number of points of intersection between cuts (more generally, without this assumption, a point of intersection between m cuts must be counted m - 1 times).
Note that the cut from A to B will have to intersect the cut from C to D if, in order around the circle, these points are arranged A, C, B, D. If we now furthermore assume, as at the top, that no pair of cuts intersects at more than one point, then this exhausts the intersections between cuts; of any four points, there will be one pair of cuts between them which intersect, and indeed, their intersection will be in just in one intersection point.
And thus we may conclude that, overall, the number of pieces will be 1 + (n choose 2) + (n choose 4).
If we maintain the condition that no pair of cuts intersects at more than one point, but allow more than two cuts to simultaneously intersect at a point, we can only change from this to even fewer pieces [for, as noted before, a point of intersection between m cuts contributes m - 1 to this total, whereas if it was jiggled a little, we would instead have a greater contribution of (m choose 2)]. This is what happens when we consider 6 points evenly spaced around our circle, connected by straight lines; we have a simultaneous intersection of 3 diameters in the center, contributing (3 - 1) = 2 rather than (3 choose 2) = 3 to the total number of pieces, dropping it from the generic 31 to a mere 30 in this case. [Note that 6 is the minimal number of points at which a particular case can diverge from the generic in this fashion, as we will need at least three disjoint pairs to cause a simultaneous inner intersection of three or more lines]
We ought also note that if we drop the condition that no pair of cuts intersects at more than one point, we can have arbitrarily many pieces (for two cuts could zig-zag across each other over and over, creating as many pieces as you like).
Finally, returning to our formula 1 + (n choose 2) + (n choose 4), why does this behave like 2[sup]n - 1[/sup] up through n = 5, and then suddenly diverge at n = 6?
Well, note that 1 can be rewritten as ((n - 1) choose 0), while (n choose 2) can be rewritten as ((n - 1) choose 1) + ((n - 1) choose 2), and (n choose 4) can be rewritten as ((n - 1) choose 3) + ((n - 1) choose 4). Thus, our formula is the same as the binomial theorem expansion of (1 + x)[sup]n - 1[/sup] up through degree 4, evaluated at x = 1. So it is guaranteed to match 2[sup]n - 1[/sup] so long as n - 1 <= 4, falling short once we move beyond that.]
[Perhaps a better word than “intersect” would’ve been “cross”. Alas, nothing I write is ever perfect.]
Indistinguishable: The J. R. R. Tolkien of mathematics writing.