GC-06.01 · Bùi Quang Tuân’s Infinite Family of Golden Circle Secants

Discoverer(s): Bùi Quang Tuân | Date: 2014 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 4 (Lemoine \(S = 12\))

In unit circle cO on diameter AB, infinitely many rational-abscissa point pairs on opposite semicircles—including C1 \(= (0,1)\), D1 \(= (\tfrac{2}{3}, -\tfrac{\sqrt{5}}{3})\) and C2 \(= (-\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\), D2 \(= (\tfrac{7}{8}, -\tfrac{\sqrt{15}}{8})\)—concur on radius OB at the exact golden point M \(= (1/\phi^2, 0)\) where \(|MB|/|OM| = \phi\).
GC-06.01 — Bùi Quang Tuân's Infinite Family of Golden Circle Secants cO: x**2 + y**2 - 1 = 0 AB: y = 0 C1D1: x*(sqrt(5)/3 + 1) + 2*y/3 - 2/3 = 0 C2D2: x*(sqrt(15)/8 + sqrt(3)/2) + 11*y/8 - 7*sqrt(3)/16 + sqrt(15)/16 = 0 Golden major MB = -1/2 + sqrt(5)/2 Golden minor OM = 3/2 - sqrt(5)/2 A = (-1, 0) C1 = (0, 1) D1 = (2/3, -sqrt(5)/3) C2 = (-1/2, sqrt(3)/2) D2 = (7/8, -sqrt(15)/8) O = (0, 0) B = (1, 0) M = (3/2 - sqrt(5)/2, 0) A C1 D1 C2 D2 O B M

Given & Constructive Steps

Given: Unit circle cO centered at O \(= (0,0)\) with diameter A \(= (-1,0)\) to B \(= (1,0)\).

step Plot upper apex C1 \(= (0,1)\) and lower chord point D1 at abscissa \(x = \tfrac{2}{3}\) on cO.
step Plot a second rational-abscissa pair C2 at \(x = -\tfrac{1}{2}\) and D2 at \(x = \tfrac{7}{8}\) on opposite semicircles.
step Draw chords C1D1 and C2D2, which concur on OB at M \(= (\tfrac{3-\sqrt{5}}{2}, 0)\).

Exact Symbolic Proof & Certificate

For \(C(x, \sqrt{1-x^2})\) and \(D(y, -\sqrt{1-y^2})\), chord \(CD\) crosses \(AB\) at \(x_M = \frac{x\sqrt{1-y^2} + y\sqrt{1-x^2}}{\sqrt{1-x^2} + \sqrt{1-y^2}}\). Setting \(x_M = \frac{3 - \sqrt{5}}{2} = \frac{1}{\phi^2}\) yields Bùi Quang Tuân’s Diophantine condition \(\frac{5(1 - x^2)}{1 - y^2} = \left(\frac{3 - 2y}{1 - 2x}\right)^2\), satisfied simultaneously by \((0, \tfrac{2}{3})\), \((-\tfrac{1}{2}, \tfrac{7}{8})\), and infinitely many rational pairs.

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

MB \(/\) OM \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (O, M, B)

Exact Segment Lengths

\(|MB| = - \frac{1}{2} + \frac{\sqrt{5}}{2}\), \(|OM| = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(M, B) = \frac{3}{2} - \frac{\sqrt{5}}{2}\), \(Q(O, M) = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 3 lines · 1 circles · 0 polygons (elementary loci \(= 4\), Lemoine \(S = 12\))

Automated Divine Scan

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

  • Section (O, M, B): long \(|MB|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\) vs. short \(|OM|^2 = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\)

Historical Commentary & Attribution

At M, \(|OM| = 1/\phi^2\) and \(|MB| = 1 - 1/\phi^2 = 1/\phi\), so \(|MB|/|OM| = |OB|/|MB| = \phi\).