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10187Re: [EMHL] Re: IMO 2004 Problem 1 revisited

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  • Darij Grinberg
    Aug 1, 2004
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      Dear Antreas and Bernard,

      Thanks for another surprising perspector!

      In Hyacinthos message #10184, Bernard Gibert

      >> > [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.]
      >> > and SaSbSc the pedal triangle of I (aka
      >> > intouch triangle of ABC).
      >> >
      >> > Are the triangles RaRbRc, SaSbSc perspective?
      >> They are indeed at : [...]

      I think that the perspector of the triangles
      RaRbRc and SaSbSc is the point X(847) of the
      triangle SaSbSc (i. e. of the intouch triangle).

      In fact, if we rewrite the problem from the
      viewpoint of triangle SaSbSc, we get the

      Let ABC be a triangle. Further, let H' be the
      orthocenter of the tangential triangle of
      triangle ABC, and let D, E, F be the
      orthogonal projections of the point H' on the
      perpendicular bisectors of the segments BC,
      CA, AB. Then show that the lines AD, BE, CF
      are concurrent at the point X(847) of
      triangle ABC.

      In fact, by the definition of X(847), the
      line AX(847) passes through the point
      BAb /\ CAc, where the point Ab is the point
      of intersection of the A-altitude of
      triangle ABC with the perpendicular
      bisector of the side CA, and the point Ac
      is is the point of intersection of the
      A-altitude with the perpendicular bisector
      of the side AB. Thus, the lines
      AD = AX(847), BAb and CAc are concurrent.

      In fact, it turns out (conjecture!) that the
      following four lines are concurrent:

      AD, BAb, CAc and X(5)X(6)X(68)X(155).

      I would be glad to see a proof of this (BTW,
      I know the synthetic proof of the
      collinearity of the points X(5), X(6), X(68),
      X(155), so you can define the line
      X(5)X(6)X(68)X(155) by any two of them).

      Darij Grinberg
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