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14944Re: [EMHL] Four Circum Equilateral Triangles

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  • pneagoe
    Mar 2, 2007
      Dear Andreas,

      Let ABC be a triangle with circumcircle C(O) and the incircle c(I).
      Ca is the circle with center A and radius AI.
      La is the radical axis of the circles Ca and C(O).
      A' is the intersection point of the lines La and BC.
      Similarly define Cb, Lb, B', Cc, Lc, C'. Three lines La, Lb, Lc
      bound one triangle A"B"C".

      1.La is tangent to the incircle c(I).
      2.A', B', C' are collinear points.
      3.AA", BB" and CC" are concurrent lines.

      Best regards,
      Petrisor Neagoe

      --- In Hyacinthos@yahoogroups.com, "Andreas P. Hatzipolakis"
      <xpolakis@...> wrote:
      > Dear Tuan and Paul
      > Since there are no "islands" in the Geometry "sea", I am wondering
      > how can we generalize it.
      > Probably a way is this:
      > Let ABC be a triangle and (O) its circumcircle.
      > We inscribe in (O) THE isosceles triangle AB*C* with
      > A = w , B* = C* = (pi-w)/2
      > Denote the line B*C* as La
      > Similarly the lines Lb,Lc.
      > Is the Triangle bounded by the lines La,Lb,Lc and the Triangle ABC
      > perspective for every given angle w (with 0 < w < pi) ?
      > etc
      > APH
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