GC-02.06 · Regular Hexagon & External Square Compass Sweep

Discoverer(s): Tran Quang Hung | Date: 2014 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 3 (Lemoine \(S = 9\))

Given regular hexagon ABCDEF and external square ABHG on side AB, circle cAH centered at A through opposite square corner H cuts hexagon sides EF at I and BC at L in the golden ratio: \(|FI|/|IE| = |BL|/|LC| = \phi\).
GC-02.06 — Regular Hexagon & External Square Compass Sweep cAH: x**2 + y**2 - 8 = 0 EF: sqrt(3)*x - y + 2*sqrt(3) = 0 BC: -sqrt(3)*x + y + 2*sqrt(3) = 0 Golden major FI = -1 + sqrt(5) Golden minor IE = 3 - sqrt(5) C = (3, sqrt(3)) D = (2, 2*sqrt(3)) G = (0, -2) H = (2, -2) L = (sqrt(5)/2 + 3/2, -sqrt(3)/2 + sqrt(15)/2) A = (0, 0) B = (2, 0) E = (0, 2*sqrt(3)) F = (-1, sqrt(3)) I = (-3/2 + sqrt(5)/2, sqrt(3)/2 + sqrt(15)/2) C D G H L A B E F I

Given & Constructive Steps

Given: Regular hexagon ABCDEF on side AB from A \(= (0, 0)\) to B \(= (2, 0)\).

step Construct regular hexagon ABCDEF of side \(2\) above AB.
step Construct external square ABHG of side \(2\) below AB and diagonal AH of length \(2\sqrt{2}\).
step Draw circle cAH centered at A through H, intersecting side EF at I and side BC at L.

Exact Symbolic Proof & Certificate

In \(\triangle AFI\), \(|AF| = 2\), \(\angle AFI = 120^\circ\), and radius \(|AI| = |AH| = 2\sqrt{2}\). Letting \(t = |FI|\), the Law of Cosines gives \(8 = 4 + t^2 - 2(2)t\cos 120^\circ = t^2 + 2t + 4\), so \(t^2 + 2t - 4 = 0\) and \(t = \sqrt{5} - 1 = 2/\phi\). Since side \(|FE| = 2\), the remaining segment is \(|IE| = 2 - (\sqrt{5} - 1) = 3 - \sqrt{5} = 2/\phi^2\), proving \(|FI|/|IE| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

FI \(/\) IE \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (F, I, E)

Exact Segment Lengths

\(|FI| = -1 + \sqrt{5}\), \(|IE| = 3 - \sqrt{5}\)

Rational Trigonometry Quadrances

\(Q(F, I) = 6 - 2 \sqrt{5}\), \(Q(I, E) = 14 - 6 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

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

Automated Divine Scan

geometor.divine.golden detected 2 exact collinear golden section(s) and 0 connected golden chain(s) in this model:

  • Section (F, I, E): long \(|FI|^2 = 6 - 2 \sqrt{5}\) vs. short \(|IE|^2 = 14 - 6 \sqrt{5}\)

  • Section (B, L, C): long \(|BL|^2 = 6 - 2 \sqrt{5}\) vs. short \(|LC|^2 = 14 - 6 \sqrt{5}\)

Historical Commentary & Attribution

Notice the structural duality between Tran Quang Hung’s hexagon construction (Cut-the-Knot) and Michel Bataille’s equilateral triangle construction: replacing the \(60^\circ\) interior angle with the hexagon’s \(120^\circ\) angle flips \(t^2 - 2t - 4 = 0\) to \(t^2 + 2t - 4 = 0\), bringing the intersection inside side EF.