Posts tagged 'field-q-phi'

GC-01.01 · Euclid’s Classical Golden Section (Elements II.11 & VI.30)

Given a straight segment AB, erect the square ABCD, bisect side AD at M, and transfer MB along ray DA beyond A to E. Swinging AE onto AB at G cuts AB in extreme and mean ratio, so that \(|AB|/|AG| = |AG|/|GB| = \phi\).

GC-03.01 · Tran Quang Hung’s Square Inscribed in a Right Isosceles Semicircle

Let ABC be a right isosceles triangle inscribed in a semicircle of cO on diameter AB, and let square DEGF sit on AB with vertices D and E on legs AC and BC. Extending top side DE to meet circle cO at Q and P divides DP in the golden ratio: \(|DE|/|EP| = |DP|/|DE| = \phi\).

GC-04.01 · H. E. Huntley’s Golden Ratio in the 3-4-5 Right Triangle (1970)

In a \(3\text{-}4\text{-}5\) right triangle ABC (\(|BC|=3, |AC|=4, |AB|=5\)), let the internal bisector of \(\angle ABC\) meet leg AC at O. Circle cOC centered at O through C intersects line BO at P and Q such that \(|PQ|/|BP| = |BQ|/|PQ| = \phi\).

GC-07.01 · Tran Quang Hung’s 2:1 Rhombus & Incircle Diagonal Division

Let ABCD be a rhombus whose diagonals satisfy \(|DB| = 2|AC|\), and let cO be its inscribed circle. Circle cO intersects long diagonal DB at E and F to form a 4-point golden chain: \(|EF|/|FB| = |EB|/|EF| = \phi\).

GC-01.02 · Right-Triangle Hypotenuse Transfer (1 : 2 : √5)

On segment AB of length \(2\), erect perpendicular leg BC of length \(1 = \tfrac{1}{2}|AB|\). Circle cCB cuts hypotenuse AC at D (and ray AC at E), and circle cAD transfers AD onto AB at G, dividing AB in the golden ratio \(|AG|/|GB| = \phi\).

GC-03.02 · Tran Quang Hung’s Square Midpoint Circle Division

Let ABCD be a square with center F and side midpoints M, N, P, Q. The circle cKA on diameter AP intersects midline QN at X between F and N such that \(|FX|/|XN| = \phi\), while chord AX cuts MP at Z with \(|MZ|/|ZF| = \phi\).

GC-06.02 · Elliot McGucken & Thanos Kalogerakis’s Equal Tangent Circles

Let equal unit circles cB and cD touch externally at C, and let AF be the diameter of cB perpendicular to line of centers BD. Secant FD cuts cD at H and I such that \(|HI|/|FH| = |FI|/|HI| = |FI|/|AF| = \phi\).

GC-07.02 · John Molokach’s Golden Rhombus from Shifted 1×3 Rectangles

Stack two \(3 \times 1\) rectangles RECT1 and RECT2 shifted by one unit around center O, draw unit circle cO, and draw the two tangents to cO perpendicular to outer diagonal R1R6. Together with the horizontal outer edges, they bound a golden rhombus ABCD (\(|AC|/|BD| = \phi\)) with \(|T1K|/|KA| = \phi\).

GC-03.03 · Nguyen Thanh Dung’s 2×2 Grid Compass Arc

On a \(2 \times 2\) square grid P1P2P3P4 with horizontal midline AB, center O, and top midpoint D, extend AB by one grid unit to C. Circle cCD centered at C through D cuts AB at M in the golden ratio: \(|MB|/|AM| = |AB|/|MB| = \phi\).

GC-05.03 · Prasolov & Bùi Quang Tuân’s Equi-Area Irregular Pentagon

Let ABCDE be any convex pentagon—regular or irregular—whose five peripheral triangles \(\triangle ABE, \triangle ABC, \triangle BCD, \triangle CDE, \triangle DEA\) have equal area. Then every diagonal is parallel to the opposite side and is divided by the crossing diagonals in the golden ratio, e.g. \(|BM|/|MD| = \phi\).

GC-07.03 · Jingcheng Tong’s Area-Bisecting Chord in a 1:3 Trapezoid

Let ABCD be a trapezoid with parallel bases AD and BC satisfying \(|BC| = 3|AD|\). The chord MN parallel to the bases that bisects the area of ABCD divides both non-parallel legs AB and DC in the golden ratio: \(|AM|/|MB| = |DN|/|NC| = \phi\).

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

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-06.05 · Jerzy Kocik’s Six-Circle Apollonian Golden Window (2010)

Place two unit-diameter circles cC1, cC2 tangent at O and two flanking congruent circles cC3, cC4 inside outer circle cO. Requiring symmetric upper and lower circles cU1, cU2 to be simultaneously tangent to all five circles uniquely forces the flanking circles to have exact radius \(\phi\), outer radius \(\phi^3\), and \(|OC4|/|C4A4| = \phi\).

GC-03.06 · Ercole Suppa’s Square Midpoint Inradii Ratio (2015)

In square ABCD, let M be the midpoint of side AD. The inradii \(R\) and \(r\) of triangles \(\triangle BCM\) (incircle cEH) and \(\triangle CDM\) (incircle cFP) satisfy \(R / r = \phi\), and segment EF joining their incenters is divided by CM at K with \(|EK|/|KF| = \phi\).

GC-06.06 · John Arioni’s Golden Secant in the Yin-Yang Circle

Inscribe twin tangent circles cA and cC of radius \(\tfrac{1}{2}\) along horizontal diameter PLPR of unit circle cO, and let B be the apex of perpendicular radius OB. Secant BA cuts cA at H and I such that \(|HI|/|BH| = |BI|/|HI| = \phi\).

GC-07.06 · Exact φ-Limit Resolution of the Hooper–Curry Fibonacci Dissection

Replace integer Fibonacci lengths \((3, 5, 8, 13)\) in the \(8 \times 8 = 64\) vs. \(5 \times 13 = 65\) dissection fallacy by exact powers \((1, \phi, \phi^2, \phi^3)\). In rectangle ABCD of width \(\phi^3\) and height \(\phi\), internal vertices Q \(= (\phi, 1/\phi)\) and P \(= (\phi^2, 1)\) lie identically on diagonal AC, forming the 4-point golden chain \(|AQ|/|QP| = |AP|/|AQ| = \phi\).

GC-03.07 · Bùi Quang Tuân & Alexandre Borovik’s Greek Cross & Square

Inscribe a symmetrical five-square Greek cross CROSS and a concentric axis-aligned square S1S2S3S4 in the same circle cO. Along any inner arm line AC of the cross, the square edge S1S2 cuts AC at B such that \(|BC|/|AB| = \phi\), while \(|BD|/|CD| = \phi\).

GC-06.07 · Three Tangent Equal Circles Inscribed in a Semicircle

Three congruent unit circles cC1, cC0, cC2 stand in a tangent row on diameter AB of enclosing circle cO centered at O, touching AB at T1, O, T2 and touching cO internally. Then cO has radius \(2\phi\) and T2 divides OB in the golden ratio: \(|OT2|/|T2B| = \phi\).