GC-04.04 · Bùi Quang Tuân’s Van Obel Division in a Right Isosceles Triangle

Discoverer(s): Bùi Quang Tuân | Date: 2014 | Field: \(\mathbb{Q}(\phi)\) | Loci: 6 (Lemoine \(S = 18\))

In right isosceles triangle ABC with hypotenuse midpoint M and midpoint O of altitude AM, circle cOM cuts OC and OB at D and F. Cevians BD and CF concur on OM at E, dividing OM in the golden ratio: \(|OE|/|EM| = \phi\).
GC-04.04 — Bùi Quang Tuân's Van Obel Division in a Right Isosceles Triangle cOM: x**2 + (y - 1)**2 - 1 = 0 AM: x = 0 OC: x + 2*y - 2 = 0 OB: x - 2*y + 2 = 0 BD: x*(-1 + sqrt(5)/5) + y*(2*sqrt(5)/5 + 2) - 2 + 2*sqrt(5)/5 = 0 CF: x*(-1 + sqrt(5)/5) + y*(-2 - 2*sqrt(5)/5) - 2*sqrt(5)/5 + 2 = 0 Golden major OE = -1/2 + sqrt(5)/2 Golden minor EM = 3/2 - sqrt(5)/2 A = (0, 2) D = (2*sqrt(5)/5, 1 - sqrt(5)/5) F = (-2*sqrt(5)/5, 1 - sqrt(5)/5) B = (-2, 0) C = (2, 0) M = (0, 0) O = (0, 1) E = (0, 3/2 - sqrt(5)/2) A D F B C M O E

Given & Constructive Steps

Given: Right isosceles triangle ABC with B \(= (-2,0)\), C \(= (2,0)\), A \(= (0,2)\), midpoint M \(= (0,0)\), and O \(= (0,1)\).

step Draw circle cOM centered at O of radius \(1\) (passing through A and M).
step Join OC and OB to meet cOM internally at D and F.
step Draw cevians BD and CF meeting on symmetry axis AM at E.

Exact Symbolic Proof & Certificate

In triangle \(\triangle OBC\), \(|OB| = |OC| = \sqrt{5}\) and \(|OD| = |OF| = 1\), so D divides OC in ratio \(|OD|/|DC| = 1/(\sqrt{5}-1) = \phi/2\), and symmetrically \(|OF|/|FB| = \phi/2\). By Van Obel’s theorem on \(\triangle OBC\) with cevians OM, BD, CF concurring at E, \(|OE|/|EM| = |OD|/|DC| + |OF|/|FB| = \phi/2 + \phi/2 = \phi\).

Machine-Verified Symbolic Certificate

Invariant

Exact Symbolic Value

Primary Golden Ratio

OE \(/\) EM \(= \frac{1}{2} + \frac{\sqrt{5}}{2}\) on collinear triple (O, E, M)

Exact Segment Lengths

\(|OE| = - \frac{1}{2} + \frac{\sqrt{5}}{2}\), \(|EM| = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

Rational Trigonometry Quadrances

\(Q(O, E) = \frac{3}{2} - \frac{\sqrt{5}}{2}\), \(Q(E, M) = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\), \(Q_{\text{long}}/Q_{\text{short}} = \frac{\sqrt{5}}{2} + \frac{3}{2} = \phi^2\)

Algebraic Number Field

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

Géométrographie Loci

2 given points · 5 lines · 1 circles · 1 polygons (elementary loci \(= 6\), Lemoine \(S = 18\))

Automated Divine Scan

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

  • Section (M, E, O): long \(|EO|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\) vs. short \(|ME|^2 = \frac{7}{2} - \frac{3 \sqrt{5}}{2}\)

  • Section (E, O, A): long \(|OA|^2 = 1\) vs. short \(|EO|^2 = \frac{3}{2} - \frac{\sqrt{5}}{2}\)

Connected Golden Chains:

  • Chain (M — E — O — A) spanning 2 overlapping golden sections (progression < <)

Historical Commentary & Attribution

Notice the elegance of the Van Obel proof: each symmetric side contributes half of \(\phi\), summing immediately to \(\phi\) at the concurrence point E.