GC-03.05 · Tran Quang Hung’s 3×3 Grid Incenter Division

Discoverer(s): Tran Quang Hung | Date: 2016 | Field: \(\mathbb{Q}(\sqrt{2}, \phi)\) | Loci: 2 (Lemoine \(S = 6\))

In a \(3 \times 3\) grid AQ1Q2Q3 on \([0,3]^2\), let S \(= (1,1)\) and T \(= (2,2)\) be opposite corners of the central unit cell, and let U \(= (3,1)\) and V \(= (1,3)\). The incenter P of isosceles triangle AUV divides diagonal ST of the central cell in the golden ratio: \(|PT|/|SP| = \phi\).
GC-03.05 — Tran Quang Hung's 3×3 Grid Incenter Division cPT: (x - 5/2 + sqrt(5)/2)**2 + (y - 5/2 + sqrt(5)/2)**2 - 2*(-1/2 + sqrt(5)/2)**2 = 0 AT: -3*x + 3*y = 0 Golden major PT = (-sqrt(2) + sqrt(10))/2 Golden minor SP = (-sqrt(10) + 3*sqrt(2))/2 Q2 = (3, 3) Q3 = (0, 3) K1 = (2, 1) K2 = (1, 2) U = (3, 1) V = (1, 3) A = (0, 0) Q1 = (3, 0) S = (1, 1) T = (2, 2) P = (5/2 - sqrt(5)/2, 5/2 - sqrt(5)/2) Q2 Q3 K1 K2 U V A Q1 S T P

Given & Constructive Steps

Given: Outer \(3 \times 3\) square AQ1Q2Q3 with A \(= (0,0)\) and Q1 \(= (3,0)\).

step Mark central cell diagonal vertices S \(= (1,1)\) and T \(= (2,2)\) along symmetry axis AT.
step Form lattice isosceles triangle AUV with U \(= (3,1)\) and V \(= (1,3)\) (whose base UV has midpoint T).
step Construct incenter P and incircle cPT tangent to UV at T.

Exact Symbolic Proof & Certificate

Triangle AUV has side lengths \(|AU| = |AV| = \sqrt{10}\) and \(|UV| = 2\sqrt{2}\). By the angle-bisector theorem on altitude AT (\(|AT| = 2\sqrt{2}\)), \(|PT|/|AP| = (|UV|/2)/|AU| = \sqrt{2}/\sqrt{10} = 1/\sqrt{5}\), so \(x_P = 2 \cdot \frac{\sqrt{5}}{\sqrt{5} + 1} = \frac{5 - \sqrt{5}}{2}\). Hence \(2 - x_P = \frac{\sqrt{5}-1}{2} = 1/\phi\) and \(x_P - 1 = \frac{3-\sqrt{5}}{2} = 1/\phi^2\), proving \(|PT|/|SP| = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

PT \(/\) SP \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (S, P, T)

Exact Segment Lengths

\(|PT| = \frac{- \sqrt{2} + \sqrt{10}}{2}\), \(|SP| = \frac{- \sqrt{10} + 3 \sqrt{2}}{2}\)

Rational Trigonometry Quadrances

\(Q(P, T) = 3 - \sqrt{5}\), \(Q(S, P) = 7 - 3 \sqrt{5}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

\(\mathbb{Q}(\sqrt{2}, \phi)\)

Géométrographie Loci

2 given points · 1 lines · 1 circles · 3 polygons (elementary loci \(= 2\), Lemoine \(S = 6\))

Automated Divine Scan

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

  • Section (S, P, T): long \(|PT|^2 = 3 - \sqrt{5}\) vs. short \(|SP|^2 = 7 - 3 \sqrt{5}\)

  • Section (P, T, Q2): long \(|TQ2|^2 = 2\) vs. short \(|PT|^2 = 3 - \sqrt{5}\)

Connected Golden Chains:

  • Chain (S — P — T — Q2) spanning 2 overlapping golden sections (progression < <)

Historical Commentary & Attribution

Because T is simultaneously a grid vertex and the tangency point of incircle cPT on base UV, the inradius of \(\triangle AUV\) equals \(|PT| = \sqrt{2}/\phi\).