GC-02.04 · Tran Quang Hung’s Reflected Equilateral Triangles

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

Reflect equilateral triangle ABC across base AB to ABC1 with circumcircle cO1. If midpoints M of BC and N of AB are joined to meet cO1 at P, then line AP divides side BC at D in the golden ratio \(|CD|/|DB| = \phi\).
GC-02.04 — Tran Quang Hung's Reflected Equilateral Triangles cO1: x**2 + (y + sqrt(3)/3)**2 - 4/3 = 0 MN: sqrt(3)*x/2 - y/2 = 0 AP: x*(-sqrt(15)/4 + sqrt(3)/4) + y*(sqrt(5)/4 + 3/4) - sqrt(15)/4 + sqrt(3)/4 = 0 BC: -sqrt(3)*x - y + sqrt(3) = 0 Golden major CD = -1 + sqrt(5) Golden minor DB = 3 - sqrt(5) C1 = (0, -sqrt(3)) O1 = (0, -sqrt(3)/3) N = (0, 0) M = (1/2, sqrt(3)/2) P = (-1/4 + sqrt(5)/4, -sqrt(3)/4 + sqrt(15)/4) A = (-1, 0) B = (1, 0) C = (0, sqrt(3)) D = (-1/2 + sqrt(5)/2, -sqrt(15)/2 + 3*sqrt(3)/2) C1 O1 N M P A B C D

Given & Constructive Steps

Given: Shared base AB from A \(= (-1, 0)\) to B \(= (1, 0)\) with reflected equilateral apices C \(= (0, \sqrt{3})\) and C1 \(= (0, -\sqrt{3})\).

step Construct equilateral triangles ABC and ABC1, and circumcircle cO1 of ABC1 centered at O1 \(= (0, -\tfrac{\sqrt{3}}{3})\).
step Join midpoint M \(= (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\) of BC and midpoint N \(= (0, 0)\) of AB, extending ray MN to meet cO1 at P.
step Draw line AP to intersect side BC at D.

Exact Symbolic Proof & Certificate

Line MN passes through the origin with slope \(\sqrt{3}\) (\(y = \sqrt{3}\,x\)). Intersecting with cO1 \((x^2 + (y + \tfrac{\sqrt{3}}{3})^2 = \tfrac{4}{3})\) gives \(4x^2 + 2x - 1 = 0\), whose positive root is \(x_P = \frac{\sqrt{5} - 1}{4} = \frac{1}{2\phi}\) and \(y_P = \frac{\sqrt{3}}{2\phi}\), so \(|NP| = 1/\phi\) along NM (of length \(1\)). Because MN is the midline parallel to AC and N bisects AB, projecting P from A onto BC yields \(|CD|/|DB| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

CD \(/\) DB \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (C, D, B)

Exact Segment Lengths

\(|CD| = -1 + \sqrt{5}\), \(|DB| = 3 - \sqrt{5}\)

Rational Trigonometry Quadrances

\(Q(C, D) = 6 - 2 \sqrt{5}\), \(Q(D, B) = 14 - 6 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

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

Automated Divine Scan

geometor.divine.golden detected 2 exact collinear golden section(s) and 0 connected golden chain(s) in this model:

  • Section (N, P, M): long \(|NP|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\) vs. short \(|PM|^2 = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\)

  • Section (C, D, B): long \(|CD|^2 = 6 - 2 \sqrt{5}\) vs. short \(|DB|^2 = 14 - 6 \sqrt{5}\)

Historical Commentary & Attribution

Contributed by Tran Quang Hung (Cut-the-Knot), this construction produces golden divisions simultaneously along midline MN (\(|NP|/|MN| = \phi\)) and along side BC (\(|CD|/|DB| = \phi\)).