On-Sphere Chord-Midpoint Inner-Product Identity
ProvedProblem97.inner_chord_eq_two_mul_inner_midpointconvex-geometryerdos-97-96-countinginner-productinscribed-anglelean-formalization
Let O,c,x,y be points in the Euclidean plane with x, y, and c at the same distance from O. Then the inner product of x−c and y−c equals twice the inner product of midpoint(x,y)−O and midpoint(x,y)−c.
Preamble
/- Generated theorem stub from Erdos9796Proof.P97.Cap.ArcInscribedAngle by Stage 2 proof cut; source SHA-256 229631a65f8abd10a13cc09e09bfa34327c4b03427c96d3053e9f4b1fef392da -/ import Definitions.Def_Erdos9796Counting_Adapter import Definitions.Def_Erdos9796Counting_Foundation import Mathlib.Analysis.InnerProductSpace.Basic open Problem97 /-! # On-sphere chord-arc inner-product nonnegativity For three points `c, x, y` on a 2D Euclidean sphere of center `O` and radius `r`, if `c` lies on the closed half-plane of the chord `xy` containing `O`, then the inscribed angle at `c` subtending `xy` is at most `π/2`; equivalently `⟪x - c, y - c⟫_ℝ ≥ 0`. This is the inscribed-angle / Thales-style step of the MEC arc-angle chain: once `c` is pinned to the major-arc side (the half-plane of `xy` containing the center `O`), the inner-product nonnegativity feeds the algebraic step `Problem97.dist_midpoint_ge_half_of_inner_nn` to close the on-sphere version of the cap-arc midpoint inequality. The proof reduces the inscribed-angle theorem to a clean algebraic identity: under the sphere hypothesis `‖x - O‖ = ‖y - O‖ = ‖c - O‖`, one has `⟪x - c, y - c⟫_ℝ = 2 · ⟪m - O, m - c⟫_ℝ`, where `m := midpoint ℝ x y`. The right-hand side is `≥ 0` exactly when `c` lies in the closed half-plane of the chord `xy` containing the center — the chord-side characterization of the major arc. -/ open scoped EuclideanGeometry InnerProductSpace
Formal statement
theorem Problem97.inner_chord_eq_two_mul_inner_midpoint
{O c x y : ℝ²}
(hxO : ‖x - O‖ = ‖c - O‖) (hyO : ‖y - O‖ = ‖c - O‖) :
⟪x - c, y - c⟫_ℝ =
2 * ⟪midpoint ℝ x y - O, midpoint ℝ x y - c⟫_ℝ := by sorrySource