GC-06.03 · Bùi Quang Tuân’s Four Congruent Golden Circles

Discoverer(s): Bùi Quang Tuân | Date: 2015 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 5 (Lemoine \(S = 15\))

Let unit circles cA and cB pass through each other’s centers, intersecting at E and N, and let congruent unit circles cD and cC be externally tangent to both cA and cB. Then \((C, N, E, D)\) is a 4-point golden chain with \(|EN|/|ED| = |ND|/|EN| = \phi\).
GC-06.03 — Bùi Quang Tuân's Four Congruent Golden Circles cA: y**2 + (x + 1/2)**2 - 1 = 0 cB: y**2 + (x - 1/2)**2 - 1 = 0 cD: x**2 + (y - sqrt(15)/2)**2 - 1 = 0 cC: x**2 + (y + sqrt(15)/2)**2 - 1 = 0 NE: x = 0 Golden major EN = sqrt(3) Golden minor ED = (-sqrt(3) + sqrt(15))/2 C = (0, -sqrt(15)/2) A = (-1/2, 0) B = (1/2, 0) E = (0, sqrt(3)/2) N = (0, -sqrt(3)/2) D = (0, sqrt(15)/2) C A B E N D

Given & Constructive Steps

Given: Unit vesica piscis circles cA and cB centered at A \(= (-\tfrac{1}{2}, 0)\) and B \(= (\tfrac{1}{2}, 0)\).

step Mark lens vertices E \(= (0, \tfrac{\sqrt{3}}{2})\) and N \(= (0, -\tfrac{\sqrt{3}}{2})\) on symmetry axis NE.
step Locate centers D \(= (0, \tfrac{\sqrt{15}}{2})\) and C \(= (0, -\tfrac{\sqrt{15}}{2})\) at distance \(2\) from both A and B.
step Draw congruent unit circles cD and cC externally tangent to cA and cB.

Exact Symbolic Proof & Certificate

External tangency of unit circle cD to cA and cB requires \(|AD| = |BD| = 2\). Since \(|AM| = 1/2\), \(y_D = \sqrt{2^2 - (1/2)^2} = \frac{\sqrt{15}}{2}\), whereas \(y_E = \frac{\sqrt{3}}{2}\). Consequently \(|EN| = \sqrt{3}\) and \(|ED| = \frac{\sqrt{15} - \sqrt{3}}{2} = \frac{\sqrt{3}}{\phi}\), giving \(|EN|/|ED| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

EN \(/\) ED \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (N, E, D)

Exact Segment Lengths

\(|EN| = \sqrt{3}\), \(|ED| = \frac{- \sqrt{3} + \sqrt{15}}{2}\)

Rational Trigonometry Quadrances

\(Q(E, N) = 3\), \(Q(E, D) = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 1 lines · 4 circles · 0 polygons (elementary loci \(= 5\), Lemoine \(S = 15\))

Automated Divine Scan

geometor.divine.golden detected 2 exact collinear golden section(s) and 1 connected golden chain(s) in this model:

  • Section (C, N, E): long \(|NE|^2 = 3\) vs. short \(|CN|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

  • Section (N, E, D): long \(|NE|^2 = 3\) vs. short \(|ED|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

Connected Golden Chains:

  • Chain (C — N — E — D) spanning 2 overlapping golden sections (progression < >)

Historical Commentary & Attribution

Compare this four-circle packing directly with Hofstetter’s Mascheroni compass-only construction (GC-01.05 · Mascheroni Five-Circle Compass-Only Construction), which uses the same \(\sqrt{15}/2\) vs. \(\sqrt{3}/2\) radical ratio.