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An equilateral triangle in plane geometry

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  • yeuemtrondoitb85
    Dear Geometers, Let ABC be a triangle in plane geometry, A0B0C0 be the excentral triangle of ABC, A1B1C1 be the excentral triangle of A0B0C0, Define: AiBiCi be
    Message 1 of 3 , May 31, 2017

      Dear Geometers,


      Let ABC be a triangle in plane geometry, A0B0C0 be the excentral triangle of ABC, A1B1C1 be the excentral triangle of A0B0C0, Define: AiBiCi be the excentral triangle of A{i-1}B{i-1}C{i-1} with i=1,2,3,...Then show that AnBnCn be an equilateral triangle when n \to  \infty


      Best regards

      Sincerely

      Dao Thanh Oai


    • yeuemtrondoitb85
      Dear Geometers, I give a proof of the conjecture: An=[(-1)^(n-1)+2^n]/2^nA/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3+2^(n+1)]/2^(n+1)C/3
      Message 2 of 3 , May 31, 2017
        Dear Geometers,

        I give a proof of the conjecture:

        An=[(-1)^(n-1)+2^n]/2^nA/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3+2^(n+1)]/2^(n+1)C/3

        Bn=[(-1)^(n-1)+2^n]/2^nB/3+[(-1)^n+2^(n+1)]/2^(n+1)C/3+2^(n+1)]/2^(n+1)A/3

        Cn=[(-1)^(n-1)+2^n]/2^nC/3+[(-1)^n+2^(n+1)]/2^(n+1)A/3+2^(n+1)]/2^(n+1)B/3

        So when n--> \infity then An=Bn=Cn=180^0/3=60^0

        The perspector of AnBnC with ABC?

        Best regards
        Sincerely
        Dao Thanh Oai
      • yeuemtrondoitb85
        Dear Geometers, I am sorry, I eidted a typo I give a proof of the conjecture:
        Message 3 of 3 , May 31, 2017
          Dear Geometers,

          I am sorry, I eidted a typo

          I give a proof of the conjecture:

          An=[(-1)^(n-1)+2^n]/2^nA/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3+[(-1)^n+2^(n+1)]/2^(n+1)C/3

          Bn=[(-1)^(n-1)+2^n]/2^nB/3+[(-1)^n+2^(n+1)]/2^(n+1)C/3+[(-1)^n+2^(n+1)]/2^(n+1)A/3

          Cn=[(-1)^(n-1)+2^n]/2^nC/3+[(-1)^n+2^(n+1)]/2^(n+1)A/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3

          So when n--> \infity then An=Bn=Cn=180^0/3=60^0

          The perspector of AnBnCn with ABC?

          Best regards
          Sincerely
          Dao Thanh Oai
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