Jesús de la Peña Hernández

6.6.6 (^2) :( 5 − 1 ) = ( 1 + 5 ): 2
x = tg α 2
2
1
= 1 : = ; AB
BC
AB
x
x
x
= =
−

• 
  

  ±
  −

  
  


• 
  

  ±
  −

  
  

  2
  5 1
  2
  1 4
  2
  1 1 1
  2
  tg
  α^2
  (AB must be positive)
  ABCD is the wanted rectangle because:
  2 :() 5 1
  2
  5 1
  1 : = −
  −

  
  

  AB
  CB
  6.6.7 ARGENTIC RECTANGLE: GREATER SIDE = 2 + 5 ; DIAGONAL = 3 + 5
  1 Start up with rectangle ABCD having AB = 6 units and AD = 4 units
  2 To get XY: X → X; AD → AD
  3 In square with opposite vertices AF, get EF =
  2
  5 − 1
  the same way as AB was obtained
  in Point 6.6.6
  4 To get G: F→ F; XY → XY; E → G
  5 To get H: G → G; GZ → XY (by means of simultaneous folds GZ (mountain) and GU
  (valley).
  A0 A3 A4 A4 A4
  A5
  A5
  A6
  α α / 2
  1
  C
  2
  D
  BC
  A
  3
  A
  B
  discard