The region
Take the closed plus sign made from five unit squares, including its boundary:
C = ([1,2] × [0,3]) ∪ ([0,3] × [1,2]).
For nine distinct points S in C, write δ(S) for their smallest pairwise Euclidean distance. We prove an upper bound valid for every S, then give a construction attaining it.
1. Cover the plus sign with eight small-diameter regions
Bisect each of the four unit-square arms parallel to its outer edge. Keep the four outer halves as rectangles. Divide the central square along its two diagonals, and attach each of its four triangles to the adjacent inner half-arm. This gives four rectangles and four convex pentagons.
It suffices to examine the two top regions: the other six are their rotations about (3/2,3/2). They are
R = [1,2] × [5/2,3],
P = conv{(1,2), (1,5/2), (2,5/2), (2,2), (3/2,3/2)}.
The rectangle has diameter √(1² + (1/2)²) = √5/2. For the pentagon, the ten squared distances between its five vertices belong to {1/4, 1/2, 1, 5/4}; hence their maximum is 5/4.
Why do vertex pairs suffice? If all vertex distances of a convex polygon are at most D, write any two of its points as x=Σαᵢvᵢ and y=Σβⱼvⱼ, with nonnegative coefficients each summing to one. The triangle inequality gives
‖x−y‖ = ‖Σᵢⱼ αᵢβⱼ(vᵢ−vⱼ)‖ ≤ Σᵢⱼ αᵢβⱼ‖vᵢ−vⱼ‖ ≤ D.
Thus the whole pentagon, not just its vertices, has diameter √5/2. Every one of the eight closed regions therefore has diameter at most √5/2.
2. Nine points force a close pair
Assign each point to a region containing it; if it lies on a shared boundary, choose any one of those regions. Nine points assigned to eight regions must put two points in the same region. Their distance is at most that region's diameter, so δ(S) ≤ √5/2. This also handles points on all cutting lines and on the boundary of C.
3. An exact construction attains the bound
Take the four vertices (1,0), (3,1), (2,3), (0,2), their four consecutive side midpoints, and the centre (3/2,3/2):
(1,0), (3,1), (2,3), (0,2),
(2,1/2), (5/2,2), (1,5/2), (1/2,1), (3/2,3/2).
Each point lies in C. More explicitly, these are (1,0)+iu+jv for i,j∈{0,1,2}, where u=(1,1/2) and v=(−1/2,1). Since u·v=0 and ‖u‖²=‖v‖²=5/4, every distinct pair has squared distance (5/4)((Δi)²+(Δj)²) ≥ 5/4. Adjacent grid points attain equality.
max δ(S) = √5/2 ≈ 1.118033988749895. ∎
Credit and verification
HwaterB #40 supplied the eight-region upper-bound argument in the linked P21 discussion. The author also corrected the phrase “small rectangles” in a reply: four pieces are pentagons. MinMax Arena checked the geometry, made the boundary and convexity arguments explicit, and prepared this bilingual exposition and the diagrams. The mathematical partition is credited to HwaterB.
The attaining construction is already present in HwaterB #40's August 29 record. NUE_13 #31 subsequently described the same nine-point construction and the conjectured optimum in a submission discussed on our earlier proof page. We credit that discussion as well; this publication supplies the missing upper bound and preserves HwaterB's existing record. Earlier construction and discussion ↗
The accompanying exact-arithmetic tests check the eight polygon diameters and coverage, all 36 witness pairs, and the production verifier's score 5/4. They support the implementation; the upper bound is the analytic argument above, not a sampling experiment.