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• The golden number geometry ;

• Squaring the circle ;

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Squaring the circle

The purpose is to draw a square whose surface is equal to a circle.
If one chooses a circle whose ray is the unit, it is necessary to trace a square whose side is equal to since the surface of the circle is equal to .

First Part : Geometrical construction of the number (Kochansky method)

• create a circle centered at B, of radius BC = 1 (unit), trace the two diameters AC and OJ, OJ is perpendicular to OJ at A.

• On the AC perpendicular, at C, trace three equals segments
  
  CD = DE = EF = 1 (unit)
  

• Trace an arc of circle centered at J, of radius JB = 1 (unit), this intersect the main circle at G and K, GK being one of the inscribed equilateral triangle sides
  
  an angle (A B G) = 30°
  

• Draw BG that intersect the perpendicular in A at H.
  
  The segment FH is equal to
  

Demonstration :

• Draw IH, perpendicular at CF

• ABH triangle : tg 30° = AH/AB= 0,57735 (AB = 1 unit)

• HIF right-angled triangle triangle HIF, HI = AC = 2 units

• IF = CF – CI and CI = AH =  0.57735

• HI² +IF² = HF² (IF = 3 – 0.57735)

• HF² = 4 + (3 – 0.57735)²

• So : HF = 9,86923

• HF = 3,141533

Second part : geometrical construction of

• Either FH = , one prolongs FH in M such as HM = 1 (unit)

• Let Q be the medium of MF

• Draw the arc of circle MLF (centered on Q)

• Draw HF, perpendicular in H, it intersect the arc of circle at L,
  
  LH =
  

Demonstration :

Triangles MFL and LHM are right-angled and similar : LH/HM = FH/LH

Consequently  :

• LH² = HM *FH, HM = 1 (unit)

• LH²= FH =

• LH  = 

Conclusion :

The surface of the square LPNH whose side is equal tois close to surface equal toof the circle centered on H and of radius HM = 1.

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