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Re : [EMHL] Fermat point problem

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  • Milorad Stevanovic
    Dear Barry and your student, You wrote ... I have the following proof. AA =BB =CC =d. 2d^2=aa+bb+cc+4Dsqrt3,where D=area(ABC).
    Message 1 of 1 , Oct 1, 2003
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      Dear Barry and your student,
      You wrote
      >The first Fermat point F is the intersection of the lines AA', BB',
      >CC', where the triangles A'BC, B'CA, and C'AB are equilateral.
      >Let d be the circle through A'B'C'. Line AFA' meets d at A' and A*,
      >and B* and C* on d are defined cyclically. Show that
      > |AA'| = |AA*| + |BB*| + |CC*| .
      I have the following proof.
      2d^2=aa+bb+cc+4Dsqrt3,where D=area(ABC).
      analogously BB*==(R1^2-O1B^2)/d,
      CC*==(R1^2-O1C^2)/d and we have
      Thus we need to prove that
      3R1^2-(O1A^2+O1B^2+O1C^2)]=d^2 (1).
      From Leibnitz theorem we have
      O1A^2+O1B^2+O1C^2=3O1G^2+(1/3)(aa+bb+cc) (2).
      From (1) and (2) it follows that we need to prove that
      Since centroid G1 of triangle A'B'C' is equal to centroid G of
      triangle ABC then we have that O1G^2=R1^2-(1/9)(a1^2+b1^2+c1^2) (3).
      Thus we have to prove that
      a1^2+b1^2+c1^2=3d^2+a^2+b^2+c^2 =m(5).
      From triangles FB'C',FC'A',FA'B' we get
      Now we have to prove that
      2(A'F^2+B'F^2+C'F^2)+B'F*C'F+C'F*A'F+A'F*B'F=m (6).
      Since A'F=d-AF,B'F=d-BF,C'F=d-CF we have to prove that
      2(AF^2+BF^2+CF^2)+BF*CF+CF*AF+AF*BF=aa+bb+cc (7).
      Since AF=(1/2dsqrt3)[4D+(bb+cc-aa)sqrt3] and analogously
      BF,CF and 2d^2=a^2+b^2+c^2+4Dsqrt3 after some
      calculation we confirm that (7) is true.
      Best regards
      Milorad R.Stevanovic

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