1. Three points per strip
Suppose six points are pairwise more than 1 apart. Split the rectangle into the lower strip y ≤ 1/2 and the upper strip y ≥ 1/2. Two points of one strip differ vertically by at most 1/2, so horizontally by more than √3/2; four of them would span more than 3√3/2 > 2. Each strip therefore holds exactly three points, and none lies on y = 1/2.
2. The band width η
Let η be the largest of the lower points’ y and the upper points’ 1 − y, and put s = √(1 − η²), α = 2 − 2s. Points of one strip are horizontally more than s apart, so the sorted lower abscissae satisfy
x₁ < α, s < x₂ < 2 − s, x₃ > 2s
and the upper ones u₁, u₂, u₃ likewise. All three intervals have length α, so |xᵢ − uᵢ| < α and each matched pair differs vertically by more than √(1 − α²). Every lower y and every upper 1 − y is then below η′ = 1 − √(1 − α²), so η < η′.
3. The contradiction
With t = 1 − s we have 0 < t ≤ 1 − √3/2 < 1/5 when η > 0, and
η′ = 1 − √(1 − 4t²) < 4t² < √t < √(2t − t²) = η
The three steps are equivalent to 0 < 4t² < 1, 16t³ < 1 and t < 1, so η′ < η, contradicting η < η′. If η = 0 the three lower points lie on y = 0, pairwise more than 1 apart, spanning more than 2: again impossible. So six points always have a pair within distance 1.
4. Attaining construction
(0,0), (1,0), (2,0), (0,1), (1,1), (2,1)
The minimum distance of this grid is exactly 1. ∎
Credit and precision
NUE_13 #31 emailed this proof on 3 October, stating that DeepSeek-V4-Pro generated it. MinMax Arena checked every step and prepared this bilingual text. This adopted contribution earns one permanent +2 proof award.
Unlike n = 4 and 5, the optimum is exactly attainable on the nine-decimal grid: the grid’s coordinates are integers. The instance closes through the existing proven-optimal mechanism, with history retained.