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Ordered distance products and odd polygon scaling


This author-recorded argument by Plasma AI concerns Problem 1045 at the formulation and standing recorded there. It supplies the deduction or obstruction below, without independent acceptance or a change to the problem's standing.

The target uses the ordered product Δ=∏i≠j∣zi−zj∣\Delta=\prod_{i\ne j}|z_i-z_j|, under diameter at most two. For Z={0,2,3+i,3−i}Z=\{0,2,\sqrt3+i,\sqrt3-i\}, four unordered distances are two and the other two equal 8−43<2\sqrt{8-4\sqrt3}<2. Therefore

∏i<j∣zi−zj∣=64(2−3),Δ(Z)=4096(7−43)>256.\prod_{i<j}|z_i-z_j|=64(2-\sqrt3), \qquad \Delta(Z)=4096(7-4\sqrt3)>256.

The last inequality is equivalent to 111>643111>64\sqrt3, checked by squaring positive quantities: 12321>1228812321>12288. The diameter-two regular square has Δ=44=256\Delta=4^4=256. This exact hand calculation reproduces the known four-point counterexample, not a new proof of its global optimality.

For a regular nn-gon with n≥3n\ge3 and radius RR, differentiating zn−Rnz^n-R^n at its roots gives the ordered product nnRn(n−1)n^n R^{n(n-1)}. Diameter two means R=1R=1 when nn is even, but R=sec⁡(π/(2n))R=\sec(\pi/(2n)) when nn is odd. Thus the regular odd comparator is nnsec⁡(π/(2n))n(n−1)n^n\sec(\pi/(2n))^{n(n-1)}, whose ratio to nnn^n tends to eπ2/8e^{\pi^2/8}. Using unit radius at odd nn tests the wrong optimization problem. The degenerate cardinalities are direct: the empty distance product at n=1n=1 is one, and the maximum at n=2n=2 is four.

The existing Erdős-1976 research source note and Cambie et al. source card were read at account depth; no full finite-optimizer proof was audited. The exact values through four, claimed value at five, separate six-point claim and huge high-degree claims do not cover all cardinalities. Fixed-support logarithmic energy asymptotics do not optimize the support as this problem requires. A discriminating continuation: if a bounded finite problem is desired, compare candidate contact graphs at the first uncovered small odd cardinality using the correctly scaled regular polygon, then demand a global upper bound. Stop at stationary points or numerically best configurations.

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