Posts tagged 'field-pentagonal'

GC-05.05 — Tran Quang Hung's Six-Pentagon Reflection Star AB: y = 0 S1: x*sqrt(5 - sqrt(5))*(-sqrt(10)/2 - sqrt(2)) + y*(3/2 + 3*sqrt(5)/2) + sqrt(10 - 2*sqrt(5))/2 = 0 S2: x*sqrt(5 - sqrt(5))*(-3*sqrt(2) - sqrt(10))/2 + y*(-sqrt(5) - 1) + sqrt(5 - sqrt(5))*(sqrt(10) + 3*sqrt(2)) = 0 Golden major AG = -1 + sqrt(5) Golden minor GB = 3 - sqrt(5) C = (1/2 + sqrt(5)/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) D = (0, sqrt(10)*sqrt(5 - sqrt(5))/4 + 3*sqrt(2)*sqrt(5 - sqrt(5))/4) E = (-sqrt(5)/2 - 1/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) R11 = (-sqrt(5)/2 - 1/2, -sqrt(10)*sqrt(5 - sqrt(5))/4 - sqrt(2)*sqrt(5 - sqrt(5))/4) R13 = (1/2 + sqrt(5)/2, -sqrt(10)*sqrt(5 - sqrt(5))/4 - sqrt(2)*sqrt(5 - sqrt(5))/4) R21 = (sqrt(5)/2 + 3/2, -sqrt(2)*sqrt(5 - sqrt(5))/2) R23 = (sqrt(5)/2 + 5/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) R31 = (1 + sqrt(5), sqrt(10)*sqrt(5 - sqrt(5))/4 + 3*sqrt(2)*sqrt(5 - sqrt(5))/4) R33 = (-1/2 + sqrt(5)/2, sqrt(2)*sqrt(5 - sqrt(5)) + sqrt(10)*sqrt(5 - sqrt(5))/2) R41 = (1/2 - sqrt(5)/2, sqrt(2)*sqrt(5 - sqrt(5)) + sqrt(10)*sqrt(5 - sqrt(5))/2) R43 = (-sqrt(5) - 1, sqrt(10)*sqrt(5 - sqrt(5))/4 + 3*sqrt(2)*sqrt(5 - sqrt(5))/4) R51 = (-5/2 - sqrt(5)/2, sqrt(2)*sqrt(5 - sqrt(5))/4 + sqrt(10)*sqrt(5 - sqrt(5))/4) R53 = (-3/2 - sqrt(5)/2, -sqrt(2)*sqrt(5 - sqrt(5))/2) H = (2, 0) A = (-1, 0) B = (1, 0) G = (-2 + sqrt(5), 0) C D E R11 R13 R21 R23 R31 R33 R41 R43 R51 R53 H A B G

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.