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A.Connes's proof of Morley's theorem

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  • Jean-Pierre.EHRMANN
    Dear Hyacinthists, may be you will be interested by the following - I try to translate a cotation from a french maths forum -: In the special book for the
    Message 1 of 1 , Oct 1, 2000
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      Dear Hyacinthists,
      may be you will be interested by the following - I try to translate a
      cotation from a french maths forum -:

      " In the special book for the 40 th birthday of IHES, we can find a nice
      proof of Morley's theorem due to Alain Connes.
      Let k be a commutative field with characteristic 0 ,
      G the group of the affine mappings from k to k , ( x -> a x + b with a<>0),
      H the morphism f -> a from G to k*,
      T = ker H (translations)
      and for f in G - T, fix(f) = b/(1-a) the fixed point of f.


      Theorem :Let f , g, h in G such as f.g, g.h, h.f and f.g.h are not in T
      and j = H(f.g.h) .
      i) and ii) are equivalent where :

      i) (f^3).(g^3).(h^3) = 1
      ii) (j^3) = 1 and fix(f.g) + j fix(g.h) + (j^2) fix(h.f) = 0

      (It can be proved with an easy computation)

      Appication : Morley's theorem :

      k is the complex field :
      angle A = 3a , angle B= 2b ,angle C=2c
      f = rotation with center A, angle 2a
      g = rotation with center B, angle 2b
      h = rotation with center C, angle 2c

      p = fix (fg) , q = fix (gh) , r = fix (hf) are the common points of the
      trisectors.
      As f^3.g^3.h^3 =1 ( because f^3, g^3, r^3 are products of reflexions w.r.t.
      two sides), we get
      p+jq+j2^r = 0 which means that p,q,r are the vertices of an equilateral
      triangle"


      Very nice; don't you think so.
      Friendly from France. Jean-Pierre
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