GC-06.06 · John Arioni’s Golden Secant in the Yin-Yang Circle

Discoverer(s): John Arioni | Date: 2016 | Field: \(\mathbb{Q}(\phi)\) | Loci: 4 (Lemoine \(S = 12\))

Inscribe twin tangent circles cA and cC of radius \(\tfrac{1}{2}\) along horizontal diameter PLPR of unit circle cO, and let B be the apex of perpendicular radius OB. Secant BA cuts cA at H and I such that \(|HI|/|BH| = |BI|/|HI| = \phi\).
GC-06.06 — John Arioni's Golden Secant in the Yin-Yang Circle cO: x**2 + y**2 - 1 = 0 cA: y**2 + (x + 1/2)**2 - 1/4 = 0 cC: y**2 + (x - 1/2)**2 - 1/4 = 0 BA: x - y/2 + 1/2 = 0 Golden major HI = 1 Golden minor BH = -1/2 + sqrt(5)/2 PL = (-1, 0) PR = (1, 0) A = (-1/2, 0) C = (1/2, 0) O = (0, 0) B = (0, 1) H = (-1/2 + sqrt(5)/10, sqrt(5)/5) I = (-1/2 - sqrt(5)/10, -sqrt(5)/5) PL PR A C O B H I

Given & Constructive Steps

Given: Unit circle cO centered at O \(= (0,0)\) with apex B \(= (0,1)\).

step Bisect radii OPL and OPR at A \(= (-\tfrac{1}{2}, 0)\) and C \(= (\tfrac{1}{2}, 0)\) and draw inner circles cA and cC through O.
step Join apex B to center A of circle cA.
step Mark intersections H and I of secant BA with cA.

Exact Symbolic Proof & Certificate

Right triangle \(\triangle AOB\) has legs \(|OA| = 1/2\) and \(|OB| = 1\), so \(|BA| = \sqrt{5}/2\). Since circle cA has radius \(1/2\) and diameter \(|HI| = 1\), we obtain \(|BH| = \frac{\sqrt{5}-1}{2} = 1/\phi\) and \(|BI| = \frac{\sqrt{5}+1}{2} = \phi\), proving \(|HI|/|BH| = |BI|/|HI| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

HI \(/\) BH \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (B, H, I)

Exact Segment Lengths

\(|HI| = 1\), \(|BH| = - \frac{1}{2} + \frac{\sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(H, I) = 1\), \(Q(B, H) = \frac{3}{2} - \frac{\sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\phi)\)

Géométrographie Loci

2 given points · 1 lines · 3 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 (I, H, B): long \(|IH|^2 = 1\) vs. short \(|HB|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

Historical Commentary & Attribution

Notice that BO is tangent to both inner circles cA and cC at O, making \(|BO|^2 = |BH|\cdot|BI| = 1\) an immediate instance of Euclid III.36.