GC-07.04 · Tran Quang Hung’s 60° Isosceles Trapezoid Diagonal Division

Discoverer(s): Tran Quang Hung | Date: 2015 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 3 (Lemoine \(S = 9\))

In half-hexagon isosceles trapezoid ABCD (\(|DA| = |AB| = |BC|\) and \(\angle C = \angle D = 60^\circ\)), reflect B across C to K. Circle cKB centered at K through B cuts diagonal AC internally at M such that \(|CM|/|MA| = |CA|/|CM| = \phi\).
GC-07.04 — Tran Quang Hung's 60° Isosceles Trapezoid Diagonal Division cKB: (x - 3)**2 + (y + sqrt(3))**2 - 16 = 0 BK: 2*sqrt(3)*x + 2*y - 4*sqrt(3) = 0 AC: sqrt(3)*x + 3*y - 2*sqrt(3) = 0 Golden major CM = -sqrt(3) + sqrt(15) Golden minor MA = -sqrt(15) + 3*sqrt(3) B = (1, sqrt(3)) K = (3, -sqrt(3)) D = (-2, 0) C = (2, 0) A = (-1, sqrt(3)) M = (7/2 - 3*sqrt(5)/2, -sqrt(3)/2 + sqrt(15)/2) B K D C A M

Given & Constructive Steps

Given: Half-hexagon trapezoid ABCD with base D \(= (-2,0)\) to C \(= (2,0)\) and top A \(= (-1,\sqrt{3})\) to B \(= (1,\sqrt{3})\).

step Extend leg BC past C by \(|CK| = |BC| = 2\) to K \(= (3, -\sqrt{3})\).
step Draw diagonal AC (of length \(2\sqrt{3}\), perpendicular to DA).
step Draw circle cKB of radius \(4\) centered at K, cutting AC at M.

Exact Symbolic Proof & Certificate

Vectors \(\vec{CA} = (-3, \sqrt{3})\) (\(|CA| = 2\sqrt{3}\)) and \(\vec{CK} = (1, -\sqrt{3})\) (\(|CK| = 2\)) satisfy \(\vec{CA}\cdot\vec{CK} = -6\), so \(\angle ACK = 120^\circ\). Parameterizing M on CA by \(x = |CM|/|CA|\), the Law of Cosines on \(\triangle KCM\) with \(|KM| = 4\) gives \(12x^2 + 12x + 4 = 16\), i.e. \(x^2 + x - 1 = 0\), whence \(x = 1/\phi\) and \(|CM|/|MA| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

CM \(/\) MA \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (A, M, C)

Exact Segment Lengths

\(|CM| = - \sqrt{3} + \sqrt{15}\), \(|MA| = - \sqrt{15} + 3 \sqrt{3}\)

Rational Trigonometry Quadrances

\(Q(C, M) = 18 - 6 \sqrt{5}\), \(Q(M, A) = 42 - 18 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\sqrt{3}, \phi)\)

Géométrographie Loci

2 given points · 2 lines · 1 circles · 1 polygons (elementary loci \(= 3\), Lemoine \(S = 9\))

Automated Divine Scan

geometor.divine.golden detected 1 exact collinear golden section(s) and 0 connected golden chain(s) in this model:

  • Section (A, M, C): long \(|MC|^2 = 18 - 6 \sqrt{5}\) vs. short \(|AM|^2 = 42 - 18 \sqrt{5}\)

Historical Commentary & Attribution

Parallel lines through M transfer this golden section simultaneously onto top base AB and leg BC, inscribing a golden parallelogram inside the half-hexagon.