GC-06.07 · Three Tangent Equal Circles Inscribed in a Semicircle

Discoverer(s): Alexander Bogomolny | Date: 2010 | Field: \(\mathbb{Q}(\phi)\) | Loci: 6 (Lemoine \(S = 18\))

Three congruent unit circles cC1, cC0, cC2 stand in a tangent row on diameter AB of enclosing circle cO centered at O, touching AB at T1, O, T2 and touching cO internally. Then cO has radius \(2\phi\) and T2 divides OB in the golden ratio: \(|OT2|/|T2B| = \phi\).
GC-06.07 — Three Tangent Equal Circles Inscribed in a Semicircle cO: x**2 + y**2 - (1 + sqrt(5))**2 = 0 cC1: (x + 2)**2 + (y - 1)**2 - 1 = 0 cC0: x**2 + (y - 1)**2 - 1 = 0 cC2: (x - 2)**2 + (y - 1)**2 - 1 = 0 AB: y = 0 OC2: -x + 2*y = 0 Golden major OT2 = 2 Golden minor T2B = -1 + sqrt(5) T1 = (-2, 0) C1 = (-2, 1) C0 = (0, 1) C2 = (2, 1) A = (-sqrt(5) - 1, 0) S = (2 - 2*sqrt(5)/5, 1 - sqrt(5)/5) K = (-2 - 2*sqrt(5)/5, -1 - sqrt(5)/5) O = (0, 0) T2 = (2, 0) B = (1 + sqrt(5), 0) T1 C1 C0 C2 A S K O T2 B

Given & Constructive Steps

Given: Origin O \(= (0,0)\) and outer circle base tangency T2 \(= (2,0)\).

step Erect three tangent unit circles cC1, cC0, cC2 centered at \((-2,1), (0,1), (2,1)\) resting on AB.
step Join O to outer center C2 \(= (2,1)\) (\(|OC2| = \sqrt{5}\)) and extend by unit radius \(1\) to outer tangency K.
step Draw enclosing circle cO of radius \(|OK| = \sqrt{5} + 1 = 2\phi\), cutting base line AB at A and B.

Exact Symbolic Proof & Certificate

Center C2 \(= (2,1)\) lies at distance \(\sqrt{2^2 + 1^2} = \sqrt{5}\) from O, so internal tangency of unit circle cC2 to cO gives enclosing radius \(R = \sqrt{5} + 1 = 2\phi\). Along radius OB, \(|OT2| = 2\) and \(|T2B| = 2\phi - 2 = 2/\phi\), hence \(|OT2|/|T2B| = \phi\) (and simultaneously along OK, \(|SK|/|OS| = 2 / (\sqrt{5}-1) = \phi\)).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

OT2 \(/\) T2B \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (O, T2, B)

Exact Segment Lengths

\(|OT2| = 2\), \(|T2B| = -1 + \sqrt{5}\)

Rational Trigonometry Quadrances

\(Q(O, T2) = 4\), \(Q(T2, B) = 6 - 2 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 2 lines · 4 circles · 0 polygons (elementary loci \(= 6\), Lemoine \(S = 18\))

Automated Divine Scan

geometor.divine.golden detected 4 exact collinear golden section(s) and 2 connected golden chain(s) in this model:

  • Section (A, T1, O): long \(|T1O|^2 = 4\) vs. short \(|AT1|^2 = 6 - 2 \sqrt{5}\)

  • Section (A, O, T2): long \(|AO|^2 = 2 \sqrt{5} + 6\) vs. short \(|OT2|^2 = 4\)

  • Section (T1, O, B): long \(|OB|^2 = 2 \sqrt{5} + 6\) vs. short \(|T1O|^2 = 4\)

  • Section (O, T2, B): long \(|OT2|^2 = 4\) vs. short \(|T2B|^2 = 6 - 2 \sqrt{5}\)

Connected Golden Chains:

  • Chain (A — T1 — O — B) spanning 2 overlapping golden sections (progression < <)

  • Chain (A — O — T2 — B) spanning 2 overlapping golden sections (progression > >)

Historical Commentary & Attribution

In the Japanese Sangaku temple-tablet tradition, packing three equal circles inside a semicircle yields \(R / r = 2\phi\) as the fundamental enclosing ratio.