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11784Re: incenters & squares

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  • jpehrmfr
    Dec 3, 2005
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      Dear Bernard
      > let P be a point on the circumcircle of triangle ABC with incenter I.
      > denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB.
      > I, Pa, Pb, Pc are the vertices of a rectangle.
      > Find P such that the rectangle becomes a square.

      Suppose that P lies on the arc BC of the circumcircle opposite to A.
      The lines PaPb and PaPc go respectively through the excenters Ib and
      The rectangle will be a square if PaI is bisector of IbPaIc.
      Hence the construction :
      Pa lies on the circle (Ca) of diameter IbIc (center Wa) such as <IbPaI
      = <IPaIc = 45°. P lies on the line WaPa.
      Do the same for P lying on the arcs CA or AB of the circumcircle.
      Friendly. Jean-Pierre
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