The Kepler Conjecture 2025, 49 - Contravening Graphs

Let R be a standard region in a contravening decomposition star D, the boundary R is a simple polygon with at most 8 edges or a nonpolygonal standard regions (n(R) = 7, 7, 8, 8, 8). If τ_R(D) ≥ t_n, n = n(R), and σ_R(D) ≤ s_n for 5 ≤ n ≤ 8. A standard cluster (R, D) with triangular region R, then σ_R(D) ≤ 1 pt. If R is not triangular, then σ_R(D) ≤ 0. For standard clusters, τ_R(D) ≥ 0. Triangular standard regions don't contain any enclosed vertices. A quadrilateral region doesn't enclose vertices of height at most 2t₀. A union of standard regions F, and the boundary of F consists of four edges. If the area of F is at most 2π, then there is at most one enclosed vertex over F. If F is the union of 2 standard regions, one triangular, one pentagonal, meeting at a vertex of type (1, 0, 1), then τ_F(D) ≥ 11.16 pt.

An exceptional standard region R with r different interior angles, pairwise nonadjacent and s.t. each is at most 1.32, then τ_R(D) ≥ t_n + r(1.47)pt. Every interior angle of standard regions is at least 0.8638. If the standard region is not a triangle, it's at least 1.153. The central vertex of a flat quarter is defined to be the one not on the triangle formed by the origin and the diagonal. An interior angle at a corner v of a non-triangular standard region is at most 1.32, then there's a flat quarter over R with central vertex is v. A contravening decomposition star D has a tame plane graph G(D).