GC-02.05 · Nested Equilateral Triangles on the Incircle

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

Inscribe equilateral triangle FGH in the incircle cIn of equilateral triangle ABC so that extended sides HF, FG, GH pass through vertices A, C, B respectively. Then \(|FH|/|AF| = |FG|/|CF_1|\) divides each secant in the golden ratio \(|FH|/|AF| = \phi\).
GC-02.05 — Nested Equilateral Triangles on the Incircle cIn: x**2 + y**2 - 1 = 0 AH: x*(3*sqrt(5)/8 + 15/8) + y*(-sqrt(15)/8 - sqrt(3)/8) + sqrt(3)/4 + sqrt(15)/4 = 0 BG: -3*x/4 + y*(sqrt(15)/4 + sqrt(3)) + sqrt(3)/4 + sqrt(15)/4 = 0 CF: x*(-9/8 - 3*sqrt(5)/8) + y*(-7*sqrt(3)/8 - sqrt(15)/8) + sqrt(3)/4 + sqrt(15)/4 = 0 Golden major FH = sqrt(3) Golden minor AF = (-sqrt(3) + sqrt(15))/2 O = (0, 0) G = (sqrt(15)/4, -1/4) B = (-sqrt(3), -1) C = (sqrt(3), -1) A = (0, 2) F = (-sqrt(15)/8 + sqrt(3)/8, 1/8 + 3*sqrt(5)/8) H = (-sqrt(15)/8 - sqrt(3)/8, 1/8 - 3*sqrt(5)/8) O G B C A F H

Given & Constructive Steps

Given: Equilateral triangle ABC of circumradius \(2\) centered at O \(= (0, 0)\) with unit incircle cIn.

step Construct equilateral triangle ABC and its unit incircle cIn centered at O.
step Inscribe equilateral triangle FGH of side \(\sqrt{3}\) in cIn such that F lies on segment AH, G on CF, and H on BG.
step Draw the three three-fold symmetric secants AH, BG, and CF.

Exact Symbolic Proof & Certificate

For vertex A \(= (0, 2)\) and unit circle cIn, the power of A with respect to cIn is \(|OA|^2 - 1^2 = 4 - 1 = 3\). Along secant AFH, any equilateral triangle FGH inscribed in cIn has side \(|FH| = \sqrt{3}\). By the Tangent-Secant / Power-of-a-Point theorem, \(|AF| \cdot |AH| = |AF|(|AF| + \sqrt{3}) = 3\). Dividing by \(3\) gives \((|AF|/\sqrt{3})^2 + (|AF|/\sqrt{3}) - 1 = 0\), so \(|AF| = \sqrt{3}/\phi\) and \(|FH|/|AF| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

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

Exact Segment Lengths

\(|FH| = \sqrt{3}\), \(|AF| = \frac{- \sqrt{3} + \sqrt{15}}{2}\)

Rational Trigonometry Quadrances

\(Q(F, H) = 3\), \(Q(A, F) = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

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

Automated Divine Scan

geometor.divine.golden detected 3 exact collinear golden section(s) and 0 connected golden chain(s) in this model:

  • Section (H, F, A): long \(|HF|^2 = 3\) vs. short \(|FA|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

  • Section (B, H, G): long \(|HG|^2 = 3\) vs. short \(|BH|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

  • Section (F, G, C): long \(|FG|^2 = 3\) vs. short \(|GC|^2 = \frac{9}{2} - \frac{3 \sqrt{5}}{2}\)

Historical Commentary & Attribution

Discovered by Tran Quang Hung (Cut-the-Knot), this rotational-pinwheel configuration embeds three simultaneous golden sections with \(C_3\) symmetry inside \(\mathbb{Q}(\sqrt{3}, \phi)\).