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10195Re: [EMHL] Re: IMO 2004 Problem 1 revisited

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  • Antreas P. Hatzipolakis
    Aug 2, 2004
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      Dear Darij

      >[APH] Let Ra, Rb, Rc be the orthogonal
      >projections of H on the cevians of I [ie
      >on the angle bisectors of A,B,C, resp.]
      >In my acute-angled triangle figure, the Euler lines
      >of BCRa, CARb, ABRc are concurrent.

      [DG]:
      >Good idea... but unfortunately, zooming in on my
      >dynamic geometry sketch shows that they are not
      >exactly concurrent...


      Hmmmm...... Apo idees .... allo tipota!!
      [free translation: I am full of ideas!!]

      Here is another one (idea, I mean):

      What point is (where is lying on?) the Radical Center of the
      circumcircles of the triangles bounded by the lines:
      (AB,AC, parallel - to - BC - through - Ra)
      (BC,BA, parallel - to - CA - through - Rb)
      (CA,CB, parallel - to - AB - through - Rc) ?

      Now, let Ta, Tb, Tc be three points on the angle bisectors
      AI, BI, CI such that:

      ITa / IRa = ITb / IRb = ITc / IRc = t

      As t varies, which is the locus of the Radical Center of the
      circumcircles of the triangles bounded by the lines:

      (AB,AC, parallel - to - BC - through - Ta)
      (BC,BA, parallel - to - CA - through - Tb)
      (CA,CB, parallel - to - AB - through - TC) ?

      Greetings

      Antreas



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