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Re: [EMHL] Darboux and a line with 5 points GENERALIZED

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  • Quim Castellsaguer
    It is known. I recalled it in an earlier message to Hyacinthos. You can find it in A Generalization of a Theorem of Euler , by D.Mitrea and M.Mitrea, American
    Message 1 of 2 , Mar 5, 2003
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      It is known. I recalled it in an earlier message to Hyacinthos. You can find
      it in "A Generalization of a Theorem of Euler", by D.Mitrea and M.Mitrea,
      American Mathematical Monthly January 1994.

      I would like to find a construction of something stated in this article: how
      to construct a triangle A"B"C" congruent to a given one ABC and perspective
      and orthologic to another given triangle A'B'C'. I have managed to construct
      the perspective part, but not yet the orthologic. If somebody can help ...

      Quim Castellsaguer
      Barcelona

      En/Na "Darij Grinberg " ha escrit:

      > In Hyacinthos #6609, I wrote:
      >
      > >> Dear friends,
      > >>
      > >> Most of us know the following theorem (which
      > >> can be proven with trilinears):
      > >>
      > >> DARBOUX THEOREM
      > >>
      > >> If ABC is a triangle, P is a point with pedal
      > >> triangle XYZ. Then the lines AX, BY, CZ
      > >> concur if and only if the point P, the
      > >> isogonal conjugate P' of P and the Longchamps
      > >> point L of triangle ABC are collinear.
      > >>
      > >> But here is MY CONJECTURE:
      > >>
      > >> If the lines AX, BY, CZ concur, then the
      > >> line through P, P' and L also contains the
      > >> point of concurrence of the lines AX, BY, CZ.
      >
      > GENERALIZATION (conjecture):
      >
      > If two orthologic triangles are perspective,
      > then the perspector is collinear with the
      > two orthologic centers!
      >
      > NOTE: Two triangles are called orthologic if
      > the perpendiculars from the vertices of the
      > first triangle to the sides of the second
      > triangle concur, and if the perpendiculars
      > from the vertices of the first triangle to
      > the sides of the second triangle concur. In
      > fact, these two conditions are equivalent,
      > and the two points of concurrences are
      > called orthologic centers of the triangles.
      >
      > I am quite sure that this generalization is
      > known (Johnson?). If not, it will be a great
      > discovery:-)
      >
      > More is true:
      >
      > If two orthologic triangles are perspective,
      > then the perspector is collinear with the
      > two orthologic centers, and the join of the
      > three points is orthogonal to their
      > perspectrix!
      >
      > [BTW, this theorem reminds of III) in
      > message #6228, but I don't think it is
      > equivalent.]
      >
      > Darij Grinberg
      >
      > >>
      > >> Evident COROLLARY of Darboux Theorem (evident
      > >> if we know the basic theorems on pedal
      > >> circles):
      > >>
      > >> The line through P, P' and L passes through
      > >> the circumcenter of XYZ.
      > >>
      > >> Nontrivial EXAMPLE: If P is the Bevan point
      > >> X(40), then we get the collinear points
      > >> P = Bevan point X(40);
      > >> P' = Hexyl point X(84);
      > >> L = Longchamps point X(20);
      > >> AX /\ BY /\ CZ = Nagel point X(8);
      > >> circumcenter of XYZ = X(1158).
      > >> True!
      > >>
      > >> Can somebody prove my conjecture?
      > >>
      > >> Darij Grinberg
      >
      >
      >
      > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
    • Darij Grinberg
      I am currently overrun by new theorems and conjectures around pedal- cevian points. I would admire any (synthetic) proof of one of them. (1) If a triangle XYZ
      Message 2 of 2 , Mar 6, 2003
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        I am currently overrun by new theorems and conjectures around pedal-
        cevian points. I would admire any (synthetic) proof of one of them.

        (1) If a triangle XYZ is a pedal-cevian triangle inscribed in a
        triangle ABC, and the perpendiculars from X to BC, etc. concur at P,
        and the cevians AX, BY, CZ concur at Q, then P, Q and the isogonal
        conjugate of P are collinear.

        (2) In the same configuration, P, the isogonal conjugate of Q and the
        circumcenter of ABC are collinear.

        (3) In particular: If Q is the Longchamps point of ABC, then the
        circumcenter of ABC is the midpoint between P and the isogonal
        conjugate of Q.

        (4) Back to the general case: If the perpendicular to BC at X meets
        AB at K, and the perpendicular to AB at Z meets BC at K', and Tb is
        the intersection of AK' and CK - i. e., K is some kind of
        generalization of the b-Triplex point of triangle ABC -, then we can
        easily prove (by Pappos) that K lies on the line PQ. What is much
        more difficult: The intersection of BK with CA lies on the line
        through P, the isogonal conjugate of Q and the circumcenter of ABC.

        Darij Grinberg
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