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14954Re: Four Circum Equilateral Triangles

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  • jpehrmfr
    Mar 3, 2007
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      Dear friends
      I wrote

      > Here is a more symetric way to draw this figure :
      > Ca is the circle going through A with center the common point of
      > perpendicular bisector of BC with the line AX[74] (The radius of
      > circle is OH/|2.cos A|). Similarly, we get Cb, Cc (may be, there is
      > special name for these three circles because they are closely
      > connected with the Mac Cay circles)
      > Three parallel lines going respectively through A, B, C intersect
      > again Ca, Cb, Cc respectively at A'', B'', C''; A' = BC inter
      > B''C'', ...
      > I've checked by computation, that, if we consider three directly
      > congruent non isosceles triangles inscribed in the circumcircle,
      > Ta,Tb,Tc where A is a vertex of Ta and La the opposite sideline,
      > the points BC inter La, CA inter Lb, AB inter Lc will lie on a same
      > line only in the case above.

      In fact, this is a bit tedious because the centers of the three
      circles Ca, Cb, Cc are homothetic in (O,-1/2) of the vertices of the
      tangential triangle.
      That's enough for me for today.
      Friendly. Jean-Pierre
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