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Re: Re [EMHL] Limit of sequence of the same position points

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  • ndergiades
    Dear Milorad, [MRS] ... If this property of the lines D1E1,D2E2,D3E3 is a method of construction of the point P then a second method is by drawing the lines
    Message 1 of 2 , Jun 29, 2002
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      Dear Milorad,

      [MRS]
      >>The following problem could be of interest:
      >>Let Po be a point in triangle AoBoCo,and A1B1C1 be
      >>cevian triangle of Po,P1 be the point in triangle A1B1C1
      >>with same position wrt to A1B1C1 as Po wrt to AoBoCo,
      >>A2B2C2 be cevian triangle of P1 in triangle A1B1C1,
      >>P2 be the point in triangle A2B2C2 with the
      >>same position wrt A2B2C2 as Po wrt to AoBoCo and so on.
      >>If Po(x:y:z)(in barycentrics),then the limit point P' of
      >>sequence of points Po,P1,P2,...,Pn,...
      >>is given by(if my calculations are good)
      >> P'(x(y+z):y(z+x):z(x+y)).
      >>From a given point Po it is posible to construct
      >>point P' by ruler and compass.
      >>Sorry,if this problem and solutions are known.
      >-------------------------------------
      >I think it would be better to say that point Pj
      >have the same Cevian ratios wrt to triangle AjBjCj
      >for all j.
      >If D1,D2,D3 are the midpoints of B0C0,C0A0,A0B0 and
      >if E1,E2,E3 are the midpoints of cevians A0A1,B0B1,C0C1
      >then the lines D1E1,D2E2,D3E3 intersect in point P'.
      >Special cases are
      >1.P=H,point P' is Lemoine point K.
      >2.P=Ge (Gergonne point),point P' is incenter I.
      >(all wrt to A0B0C0).


      If this property of the lines D1E1,D2E2,D3E3
      is a method of construction of the point P' then
      a second method is by drawing the lines A0M1, B0M2, C0M3
      where M1 is the mid point of B1C1 etc
      because P=[x : y : z] B1 = [x : 0 : z] = [x(x+y) : 0 :
      z(x+y)]
      C1 = [x : y : 0] = [x(x+z) : y(x+z) : 0] and hence the mid
      point
      M1 = [x(x+y)+x(x+z) : y(x+z) : z(x+y)] or that the line A0M1
      passes through the point P' etc and a third method is by
      drawing the complement of the isotomic conjugate of P
      since P' = [x(y+z) : y(z+x) : z(x+y)] = [1/y+1/z :1/z+1/x
      :1/x+1/y]

      Best regards
      Nikos Dergiades
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