Content of type 'construction'

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-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-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-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-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-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-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-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-04.02 · Johannes Kepler’s Geometric-Progression Right Triangle (1597)

Divide hypotenuse AB at H in the golden ratio (\(|AH|/|HB| = \phi\)), and erect perpendicular HC to semicircle cOB on diameter AB. Right triangle ABC has side lengths in geometric progression \(1 : \sqrt{\phi} : \phi\) and side quadrances in progression \(1 : \phi : \phi^2\).

GC-05.02 · Dao Thanh Oai’s Regular Pentagon Edge Perpendicular

In regular pentagon ABCDE, erect line EF perpendicular to side DE at vertex E. Line EF intersects opposite non-adjacent side AB internally at F such that \(|FB|/|AF| = |AB|/|FB| = \phi\).

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-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-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-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-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-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-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-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-05.04 · Tran Quang Hung’s Shared-Base Regular Pentagon & Hexagon

Erect regular hexagon ABCDEF and regular pentagon AMNPF on the same side AF. The perpendiculars MQ and PR to diagonal MP at M and P intersect upper hexagon sides BC and DE at Q and R such that \(|CQ|/|QB| = |DR|/|RE| = \phi\).

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-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-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-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-05.06 · Tran Quang Hung’s Fivefold Wu Xing Equilateral Star

Arrange five congruent equilateral triangles OV1W1 … OV5W5 of unit side with fivefold rotational symmetry around shared vertex O. The pentagram star joining outer vertices V1 … V5 cuts the radial sides of the equilateral triangles in the golden ratio, e.g. chord V2V5 cuts OW1 at F with \(|OF|/|FW1| = \phi\).

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-01.07 · Bùi Quang Tuân’s Five-Step Three-Circle Secant

On line L through O, draw circle cO meeting L at A and B, circle cBA meeting L at C, and circle cCO meeting cO above L at M and cBA below L at N. Then chord MN cuts cO at P with \(|MN|/|MP| = |MP|/|PN| = \phi\).

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 \(|DG|/|GE| = |DE|/|DG| = \phi\).