Posts tagged 'field-q-sqrt3-phi'

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-06.01 · Bùi Quang Tuân’s Infinite Family of Golden Circle Secants

In unit circle cO on diameter AB, infinitely many rational-abscissa point pairs on opposite semicircles—including C1 \(= (0,1)\), D1 \(= (\tfrac{2}{3}, -\tfrac{\sqrt{5}}{3})\) and C2 \(= (-\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})\), D2 \(= (\tfrac{7}{8}, -\tfrac{\sqrt{15}}{8})\)—concur on radius OB at the exact golden point M \(= (1/\phi^2, 0)\) where \(|MB|/|OM| = \phi\).

GC-02.02 · Michel Bataille’s Equilateral Triangle & External Square

Erect equilateral triangle ABC and external square BCDE on opposite sides of shared base BC. Circle cCE centered at C through opposite square vertex E cuts ray AB beyond B at F so that \(|AF|/|AB| = |AB|/|BF| = \phi\).

GC-01.03 · Lemoine–Reusch–Hofstetter 5-Step Division

Given points A and B on line AB, let unit vesica circles cAB and cBA meet at C and D. Circle cCA cuts cAB at E and chord CD at F; then circle cEF cuts AB internally at G and externally at G₁ so that \(|AG|/|GB| = |G_1A|/|AB| = \phi\).

GC-02.03 · Bùi Quang Tuân’s Side-and-Altitude Equilateral Triangles

Let M be the midpoint of base BC of equilateral triangle ABC, and erect a second equilateral triangle AMN on altitude AM. Then circle cBA centered at B through A cuts side MN at P in the golden ratio \(|MN|/|MP| = |MP|/|PN| = \phi\).

GC-04.03 · Jo Niemeyer’s Three Equal Leaning Segments

Place segment A1B1 perpendicular to line L at A1, lean equal segment B2A2 from midpoint B2 of A1B1 onto L, and lean a third equal segment B3A3 from midpoint B3 of B2A2 onto L. Then A2 divides A1A3 in the golden ratio: \(|A1A2|/|A2A3| = \phi\).

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-02.04 · Tran Quang Hung’s Reflected Equilateral Triangles

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-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-07.04 · Tran Quang Hung’s 60° Isosceles Trapezoid Diagonal Division

In half-hexagon isosceles trapezoid ABCD (\(|DA| = |AB| = |BC|\) and \(\angle C = \angle D = 60^\circ\)), reflect B across C to K. Circle cKB centered at K through B cuts diagonal AC internally at M such that \(|CM|/|MA| = |CA|/|CM| = \phi\).

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-02.05 · Nested Equilateral Triangles on the Incircle

Inscribe equilateral triangle FGH in the incircle cIn of equilateral triangle ABC so that extended sides HF, FG, GH pass through vertices A, C, B respectively. Then \(|FH|/|AF| = |FG|/|CF_1|\) divides each secant in the golden ratio \(|FH|/|AF| = \phi\).

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-01.06 · Hofstetter’s Fixed-Aperture Rusty-Compass Division

With a rusty compass locked permanently to aperture \(r = |AB|\), circles cAB and cBA determine midpoint M via CD; a third unit circle cM centered at M meets cBA above AB at F, and line DF cuts AB at G in the golden ratio \(|AG|/|GB| = \phi\).

GC-02.06 · Regular Hexagon & External Square Compass Sweep

Given regular hexagon ABCDEF and external square ABHG on side AB, circle cAH centered at A through opposite square corner H cuts hexagon sides EF at I and BC at L in the golden ratio: \(|FI|/|IE| = |BL|/|LC| = \phi\).

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-03.08 · 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 \(|DG|/|GE| = |DE|/|DG| = \phi\).