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Re: [EMHL] Re: excircle and circumcircle

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  • Alexey.A.Zaslavsky
    ... Dear Darij! I think that I is in HaHb. In this case the equivalence of 4 and 5 may be proved by trygonomethric transformations. The triangles ABC and HaHbC
    Message 1 of 3 , Jan 27, 2004
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      >
      >>> Let AX and BY are the bisectrix of triangle ABC,
      >>> O, R - the center and radius of its circumcircle,
      >>> I1, r1 - the center and radius of excircle
      >>> touching the sideline AB, A', B', C' - the
      >>> touching points of excircle with BC, CA, AB.
      >>> Then next conditions are equivalent:
      >>> 1. R=r1.
      >>> 2. O is in line XY.
      >>> 3. O is the orthocenter of A'B'C' (I suppose
      >>> also that A'B'C' is autopolar with respect to
      >>> circumcicle of ABC).
      >>> 4. cosA+cosB=cosC.
      >
      >You can add the following equivalent condition:
      >5. If I is the incenter of triangle ABC, and Hb
      >and Hc are the feet of the altitudes from B and C,
      >then I lies on HbHc.
      >
      Dear Darij!
      I think that I is in HaHb. In this case the equivalence of 4 and 5 may be
      proved by trygonomethric transformations. The triangles ABC and HaHbC are
      similar with coefficient cosC, so CL=CC1cosC, where CL is the bisectrix of
      HaHbC, CC1 - the bisectrix of ABC. But CI/CC1=(sinA+sinB)/(sinA+sinB+sinC)
      and by 4 I=L.

      Sincerely Alexey
    • Alexey.A.Zaslavsky
      Dear Darij! ... The equivalence of 2 and 5 is a particular case of next fact. Let given a triangle ABC and a point P. A1, B1 are the common point of AP and BC,
      Message 2 of 3 , Jan 28, 2004
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        Dear Darij!
        >
        >The equivalence of the conditions 2. and 5. was the
        >subject of Problem 4 in the Bundeswettbewerb
        >Mathematik (German National Mathematics Olympiad)
        >2002, 2 round. In fact, it was a very hard problem;
        >I was lucky enough to know the method of trilinear
        >coordinates to solve it!
        >

        The equivalence of 2 and 5 is a particular case of next fact. Let given a
        triangle ABC and a point P. A1, B1 are the common point of AP and BC, BP and
        AC. Q - a point in A1B1, Q' is isogonally conjugated to Q, A2, B2 are the
        common points of AQ' and BC, BQ' and AC. Then the point P' isogonally
        conjugated to P is in A2B2. This can be easily proved by trilinear
        coordinates.

        Sincerely Alexey
      • Darij Grinberg
        Dear Alexey, ... Exactly. My apologies for the typo. ... Yes, and this was exactly the same generalization I found in 2002 when I solved the problem!
        Message 3 of 3 , Jan 28, 2004
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          Dear Alexey,

          In Hyacinthos message #9144, you wrote:

          >> I think that I is in HaHb.

          Exactly. My apologies for the typo.

          In Hyacinthos message #9145, you wrote:

          >> The equivalence of 2 and 5 is a particular case
          >> of next fact. Let given a triangle ABC and a
          >> point P. A1, B1 are the common point of AP and
          >> BC, BP and AC. Q - a point in A1B1, Q' is
          >> isogonally conjugated to Q, A2, B2 are the
          >> common points of AQ' and BC, BQ' and AC. Then
          >> the point P' isogonally conjugated to P is in
          >> A2B2. This can be easily proved by trilinear
          >> coordinates.

          Yes, and this was exactly the same generalization
          I found in 2002 when I solved the problem!

          Sincerely,
          Darij Grinberg
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