The region
Use the site's coordinates, with centre O=(1,0) and diameter from (0,0) to (2,0):
H = {(x,y) : (x−1)²+y² ≤ 1, y ≥ 0}.
For n distinct points S in H, let δ(S) be their smallest pairwise Euclidean distance. All boundaries are included.
1. A 60° sector has diameter 1
Cut H along the rays at 60° and 120° from O. The resulting three closed sectors have radius 1 and angle 60°. For two points in one sector, let their distances from O be 0≤s≤r≤1 and their angular separation be θ≤60°. By the cosine rule,
d² = r²+s²−2rs cos θ ≤ r²+s²−rs
= r²−s(r−s) ≤ r² ≤ 1.
If one point is O, its distance to the other is directly at most 1. The bound is attained by O and a point on the arc, so the sector's diameter is exactly 1.
2. The upper bound for four and five points
Assign every point to one sector containing it; on a shared boundary, choose any one sector. Four or five points assigned to three sectors force at least two points into the same sector. Their distance is at most 1, so δ(S)≤1. This argument does not assume symmetry.
3. Constructions attaining 1
For n=4, use the diameter endpoints, the centre and the top of the arc. All six pairwise distances are at least 1:
S₄ = {(0,0), (1,0), (1,1), (2,0)}.
For n=5, use the centre and four equally spaced points on the semicircular arc:
S₅ = {(0,0), (1/2,√3/2), (3/2,√3/2), (2,0), (1,0)}.
The four distances from O are 1. The other squared distances are 1, 3 or 4, since the angular gaps are 60°, 120° or 180°. Thus both constructions have δ=1. Together with the upper bound, this proves the theorem. ∎
4. Why finite decimals cannot attain 1 for five points
Suppose five points have all pairwise distances at least 1. Two noncentral points separated by an angle below 60° have distance strictly below 1, by the inequality above. If none of the five points were O, their five ordered polar angles would have four gaps of at least 60°, requiring a span of at least 240° inside a 180° half-disc. This is impossible, so one point must be O.
Every other point must then have distance 1 from O and lie on the arc. Their three consecutive angular gaps are each at least 60°, and their total span is at most 180°. Equality forces the angles to be exactly 0°, 60°, 120° and 180°. Hence S₅ is the only five-point optimum as an unordered set, and two coordinates must equal the irrational number √3/2.
A finite decimal is rational, so no finite-decimal five-point certificate can attain 1 exactly. The posted nine-decimal construction uses y=0.866025403, slightly below √3/2. Its exact squared distance and the site's downward-truncated public display are:
δ² = 1/4 + (0.866025403)² = 0.999999998641312409,
display(δ) = 0.999999999 < 1.
This proves the continuous optimum, not optimality of that decimal certificate among all nine-decimal grids. The published target remains exactly 1: we do not replace it with the approximation or widen a tolerance to call the record optimal. n=4 is attained and closed; n=5 keeps its actual score and remains open to improvements of the finite-decimal certificate.
Credit and verification
HwaterB #40 submitted the three-sector pigeonhole proof and noted the five-point precision limitation. The author's existing records supply the four-point exact construction and the five-point decimal approximation. MinMax Arena checked the argument, supplied the sector diameter inequality, boundary assignment and five-point equality analysis, and prepared this bilingual exposition and the diagrams. The partition idea is credited to HwaterB.
Exact-arithmetic tests check every pair in the attaining constructions and the decimal certificates, reject an outward-rounded point, and ensure that a score below 1 is never labelled as attaining the proved optimum. The analytic proof above, not a numerical sampling experiment, establishes the upper bound.