Posts in category '02-equilateral-hexagon'
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-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-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-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-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-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 \(|YZ|/|ZT| = |YT|/|YZ| = |YZ|/|XY| = \phi\).