GC-07.05 · Tran Quang Hung’s Mixtilinear Circle in an Equilateral Triangle

Discoverer(s): Tran Quang Hung, Alexander Bogomolny, Leonard Giugiuc | Date: 2016 | Field: \(\mathbb{Q}(\sqrt{3}, \phi)\) | Loci: 3 (Lemoine \(S = 9\))

In equilateral triangle ABC with circumcenter K and circumcircle cK, inscribe the A-mixtilinear circle cF tangent to sides AB, AC and internally to cK. The line through K parallel to AB meets BC at O and cF at I and J such that \(|KO|/|OJ| = \phi\).
GC-07.05 — Tran Quang Hung's Mixtilinear Circle in an Equilateral Triangle cK: x**2 + (y - 3*sqrt(3))**2 - 108 = 0 cF: (x - 3)**2 + (y - 4*sqrt(3))**2 - 48 = 0 KO: y - 3*sqrt(3) = 0 Golden major KO = 6 Golden minor OJ = -3 + 3*sqrt(5) C = (0, 9*sqrt(3)) D = (3, 0) F = (3, 4*sqrt(3)) I = (3 - 3*sqrt(5), 3*sqrt(3)) A = (-9, 0) B = (9, 0) K = (0, 3*sqrt(3)) O = (6, 3*sqrt(3)) J = (3 + 3*sqrt(5), 3*sqrt(3)) C D F I A B K O J

Given & Constructive Steps

Given: Equilateral triangle ABC of side \(18\) with A \(= (-9,0)\), B \(= (9,0)\), C \(= (0, 9\sqrt{3})\), and circumcenter K \(= (0, 3\sqrt{3})\).

step Draw circumcircle cK of radius \(6\sqrt{3}\) around ABC.
step Inscribe A-mixtilinear circle cF centered at F \(= (3, 4\sqrt{3})\) of radius \(4\sqrt{3}\), tangent to AB at D \(= (3,0)\).
step Draw horizontal line KO (\(y = 3\sqrt{3}\)), cutting BC at O \(= (6, 3\sqrt{3})\) and cF at I and J.

Exact Symbolic Proof & Certificate

For side \(s = 18\), the A-mixtilinear inradius is \(r_A = \frac{2\sqrt{3}}{9}s = 4\sqrt{3}\) with \(|AD| = \frac{2}{3}s = 12\), placing F at \((3, 4\sqrt{3})\). Along \(y = 3\sqrt{3}\), \((x - 3)^2 + (3\sqrt{3} - 4\sqrt{3})^2 = 48\) yields \((x - 3)^2 = 45\), so \(x_J = 3 + 3\sqrt{5}\). Since \(x_K = 0\) and \(x_O = 6\), \(|KO| = 6\) and \(|OJ| = 3(\sqrt{5} - 1) = 6/\phi\), proving \(|KO|/|OJ| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

KO \(/\) OJ \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (K, O, J)

Exact Segment Lengths

\(|KO| = 6\), \(|OJ| = -3 + 3 \sqrt{5}\)

Rational Trigonometry Quadrances

\(Q(K, O) = 36\), \(Q(O, J) = 54 - 18 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 1 lines · 2 circles · 1 polygons (elementary loci \(= 3\), Lemoine \(S = 9\))

Automated Divine Scan

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

  • Section (I, K, O): long \(|KO|^2 = 36\) vs. short \(|IK|^2 = 54 - 18 \sqrt{5}\)

  • Section (K, O, J): long \(|KO|^2 = 36\) vs. short \(|OJ|^2 = 54 - 18 \sqrt{5}\)

Connected Golden Chains:

  • Chain (I — K — O — J) spanning 2 overlapping golden sections (progression < >)

Historical Commentary & Attribution

Alexander Bogomolny cataloged seven distinct golden ratios generated by a single mixtilinear circle inside an equilateral triangle.