Posts tagged 'tran-quang-hung'

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-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-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-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-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-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-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-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-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 \(|OF|/|FW1| = \phi\).