GC-06.02 · Elliot McGucken & Thanos Kalogerakis’s Equal Tangent Circles

Discoverer(s): Elliot McGucken, Thanos Kalogerakis | Date: 2016 | Field: \(\mathbb{Q}(\phi)\) | Loci: 4 (Lemoine \(S = 12\))

Let equal unit circles cB and cD touch externally at C, and let AF be the diameter of cB perpendicular to line of centers BD. Secant FD cuts cD at H and I such that \(|HI|/|FH| = |FI|/|HI| = |FI|/|AF| = \phi\).
GC-06.02 — Elliot McGucken & Thanos Kalogerakis's Equal Tangent Circles cB: x**2 + y**2 - 1 = 0 cD: y**2 + (x - 2)**2 - 1 = 0 BD: y = 0 FD: x + 2*y - 2 = 0 Golden major HI = 2 Golden minor FH = -1 + sqrt(5) D = (2, 0) A = (0, -1) B = (0, 0) C = (1, 0) F = (0, 1) H = (2 - 2*sqrt(5)/5, sqrt(5)/5) I = (2*sqrt(5)/5 + 2, -sqrt(5)/5) D A B C F H I

Given & Constructive Steps

Given: Externally tangent unit circles cB at B \(= (0,0)\) and cD at D \(= (2,0)\), touching at C \(= (1,0)\).

step Erect perpendicular diameter AF of cB with F \(= (0,1)\) and A \(= (0,-1)\).
step Draw secant line FD through center D of circle cD.
step Mark intersections H (near) and I (far) of FD with cD.

Exact Symbolic Proof & Certificate

Right triangle \(\triangle FBD\) has legs \(|BF| = 1\) and \(|BD| = 2\), giving \(|FD| = \sqrt{5}\). Subtracting and adding radius \(1\) of cD yields \(|FH| = \sqrt{5} - 1 = 2/\phi\), diameter \(|HI| = 2\), and \(|FI| = \sqrt{5} + 1 = 2\phi\). Hence \(|HI|/|FH| = \phi\) (McGucken) and \(|FI|/|AF| = \phi\) (Kalogerakis).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

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

Exact Segment Lengths

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

Rational Trigonometry Quadrances

\(Q(H, I) = 4\), \(Q(F, H) = 6 - 2 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 2 lines · 2 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 (F, H, I): long \(|HI|^2 = 4\) vs. short \(|FH|^2 = 6 - 2 \sqrt{5}\)

Historical Commentary & Attribution

This construction geometricizes the Heron-Ptolemy \(1\text{-}2\text{-}\sqrt{5}\) right triangle purely through two tangent unit circles.