Posts tagged 'golden-chain'

GC-02.01 · George Odom’s Equilateral Midline & Circumcircle Chord (1983)

Let ABC be an equilateral triangle inscribed in circle cO, and let L and M be the midpoints of sides AB and AC. Extending midline LM to meet circumcircle cO at X and Y divides the chord in the golden ratio: \(|LM|/|MY| = |LY|/|LM| = \phi\).

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-05.01 · Claudius Ptolemy’s Regular Pentagon & Pentagram (Almagest I.10)

In unit circle cO with perpendicular radii OB and OC, let M bisect OB. Circle cMC cuts AB at E with \(|OB|/|EO| = \phi\) and chord \(|CE|\) equal to the side of inscribed regular pentagon CP1P2P3P4, whose pentagram diagonals P1P4, CP2, CP3 cut one another in golden chains \(|P1K2|/|K2P4| = \phi\).

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-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-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-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-06.03 · Bùi Quang Tuân’s Four Congruent Golden Circles

Let unit circles cA and cB pass through each other’s centers, intersecting at E and N, and let congruent unit circles cD and cC be externally tangent to both cA and cB. Then \((C, N, E, D)\) is a 4-point golden chain with \(|EN|/|ED| = |ND|/|EN| = \phi\).

GC-01.04 · Hofstetter’s 2004 Five-Step Secant Division

Given points A and B, draw line AB and unit circles cAB, cBA meeting at C and D, with cAB meeting ray BA beyond A at E. Circle cEB cuts cBA above AB at F, and secant DF divides AB at G in the golden ratio \(|AG|/|GB| = \phi\) in exactly five Euclidean loci.

GC-03.04 · Tran Quang Hung & Elliot McGucken’s 3×3 Grid Circle

In a \(3 \times 3\) square grid S1AS3S4 centered at O, let C be the upper-right vertex of the central grid cell, A the outer upper-right corner, and T the adjacent trisection vertex on the top edge. Circle cOT cuts diagonal segment CA at B such that \(|CB|/|BA| = \phi\).

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-06.04 · Tran Quang Hung & Elliot McGucken’s Three Tangent Circles

Let congruent unit circles cA, cB, cC be mutually externally tangent at D, E, F, and let circle cO pass through their centers A, B, C. Line DF intersects cO at J and G to form a 4-point golden chain \(|DF|/|FG| = |DG|/|DF| = \phi\).

GC-01.05 · Mascheroni Five-Circle Compass-Only Construction

Using a compass alone and zero straightedge lines, five circles constructed from given points A and B generate four collinear points Y, C, D, X such that \(|CX|/|CD| = |CD|/|DX| = \phi\).

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

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-05.05 · Tran Quang Hung’s Six-Pentagon Reflection Star

Reflect regular pentagon ABCDE across each of its five sides to obtain five surrounding congruent pentagons. Joining the outer lateral vertices across two intervening pentagons (such as chord S1) cuts every side of ABCDE in the golden ratio, dividing AB at G and H in a 4-point golden chain.

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-07.05 · Tran Quang Hung’s Mixtilinear Circle in an Equilateral Triangle

In equilateral triangle ABC with circumcenter K and circumcircle cK, inscribe the A-mixtilinear circle cF tangent to sides AB, AC and internally to cK. The line through K parallel to AB meets BC at O and cF at I and J such that \(|KO|/|OJ| = \phi\).

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-02.07 · Butterfly Chords Astride an Equilateral Triangle

Let I be the midpoint of radius OJ of circle cO, XT the chord perpendicular to OJ through I, and ICD an equilateral triangle symmetric about OJ with C and D on cO. If rays CI and DI meet cO again at A and B, then chords BC and AD cut XT at Y and Z with \(|YZ|/|ZT| = |YT|/|YZ| = |YZ|/|XY| = \phi\).

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: \(|OT2|/|T2B| = \phi\).

GC-03.08 · John Arioni’s 30°-Rotated Unit Square & Diagonal Circle

Rotate unit square DGHJ clockwise by \(30^\circ\) around vertex D to obtain congruent square DCBA. Circle cAC centered at A through opposite vertex C intersects line DG at E and F such that \(|DG|/|GE| = |DE|/|DG| = \phi\).