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Re: A nice little problem

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  • Bernard Gibert
    ... Dear John and all, I checked this again and I did not see any flaw in my proof, but I want to stay humble just in case... Let s see : I will use Clark s
    Message 1 of 14 , Feb 3, 2000
      >> on 1/27/00 6:13 PM Bernard Gibert wrote
      >
      >>> I think the one with diameter OH* [H* = isogonal of isotomic of H] is
      >>> another one.
      > [John wrote]
      > Did you check this again? I did, and don't think it works.
      >
      Dear John and all,

      I checked this again and I did not see any flaw in my proof, but I want to
      stay humble just in case...

      Let's see :

      I will use Clark's wonderful ETC
      [congratulations Clark, it is a master piece !]

      H = X(4)
      H' = isotomic of H = X(69)
      H* = isogonal of H' = X(25)

      H* is the pole of the orthic axis with respect to the circumcircle,
      therefore, if I call I the foot of the orthic axis on Euler line [it was the
      central point of my involution], the polar line of I with respect to the
      circumcircle goes through H* and is the perpendicular at H* to Euler line.

      Then, the circle with diameter OI [inverse of the perpendicular in H*] is
      orthogonal to the circumcircle and the inversion with respect to the
      circumcircle maps the orthic axis to the circle with diameter OH*.

      So, I and H* are inverse with respect to the circumcircle.
      That should do the trick.

      regards

      bernard
    • Quang Tuan
      Dear All My Friends, I would like to add some remarks to this topic: 1). Clark Kimberling already put this property in his ETC, at X(4) Orthocenter item:
      Message 2 of 14 , Nov 21, 2011
        Dear All My Friends,

        I would like to add some remarks to this topic:

        1). Clark Kimberling already put this property in his ETC, at X(4) Orthocenter item:

        http://faculty.evansville.edu/ck6/encyclopedia/ETC.html#X4

        2). I. F. Sharygin, Problems on Geometry. Plane geometry, Nauka, Moscow, 1982, page 46, problem 188:

        On the lines AB and AC take points M and N respectively. To prove that common chord of two circles with diameters CM an BN pass through intersection point of altitudes of triangle ABC.

        3). V. V. Prasolov, Problems in Plane Geometry, Part 1, Nauka, Moscow, 1986, page 64, problem 3.23 (b):

        On sides BC, CA, AB of acute triangle ABC take any points A1, B1, C1. To prove that three common chords of pairs of circles with diameters AA1, BB1, CC1 pass through intersection point of altitudes of triangle ABC.

        4). Recently we can read the problem in book Mathematical Olympiad Challenges of T. Andreescu, R. Gelca (Birkhäuser, 2004, 5th printing, 1.3.8 (p. 12)) as in:

        http://www.cut-the-knot.org/Curriculum/Geometry/PHQCollinearity.shtml

        Best regards,
        Bui Quang Tuan

        --- In Hyacinthos@yahoogroups.com, Floor van Lamoen <f.v.lamoen@...> wrote:
        >
        > Dear all,
        >
        > At the moment it seems that I only have time to do tiny problems. And
        > probably this one is already known, but I was surprised by it:
        >
        > Let A'B'C' be a triangle inscribed in ABC (so A' lies on BC, etc.).
        > Determine the radical center of the circles with diameters AA', BB' and
        > CC' respectively.
        >
        > Kind regards,
        > Floor.
        >
      • Quang Tuan
        Dear All My Friends, There is another interesting reference for this fact: H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967, Theorems 2.45, 2.46
        Message 3 of 14 , Nov 23, 2011
          Dear All My Friends,

          There is another interesting reference for this fact:

          H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967, Theorems 2.45, 2.46

          We can read detail at:

          http://www.cut-the-knot.org/Curriculum/Geometry/CircleOnCevian.shtml

          Best regards,
          Bui Quang Tuan

          --- In Hyacinthos@yahoogroups.com, "Quang Tuan" <bqtuan1962@...> wrote:
          >
          > Dear All My Friends,
          >
          > I would like to add some remarks to this topic:
          >
          > 1). Clark Kimberling already put this property in his ETC, at X(4) Orthocenter item:
          >
          > http://faculty.evansville.edu/ck6/encyclopedia/ETC.html#X4
          >
          > 2). I. F. Sharygin, Problems on Geometry. Plane geometry, Nauka, Moscow, 1982, page 46, problem 188:
          >
          > On the lines AB and AC take points M and N respectively. To prove that common chord of two circles with diameters CM an BN pass through intersection point of altitudes of triangle ABC.
          >
          > 3). V. V. Prasolov, Problems in Plane Geometry, Part 1, Nauka, Moscow, 1986, page 64, problem 3.23 (b):
          >
          > On sides BC, CA, AB of acute triangle ABC take any points A1, B1, C1. To prove that three common chords of pairs of circles with diameters AA1, BB1, CC1 pass through intersection point of altitudes of triangle ABC.
          >
          > 4). Recently we can read the problem in book Mathematical Olympiad Challenges of T. Andreescu, R. Gelca (Birkhäuser, 2004, 5th printing, 1.3.8 (p. 12)) as in:
          >
          > http://www.cut-the-knot.org/Curriculum/Geometry/PHQCollinearity.shtml
          >
          > Best regards,
          > Bui Quang Tuan
          >
          > --- In Hyacinthos@yahoogroups.com, Floor van Lamoen <f.v.lamoen@> wrote:
          > >
          > > Dear all,
          > >
          > > At the moment it seems that I only have time to do tiny problems. And
          > > probably this one is already known, but I was surprised by it:
          > >
          > > Let A'B'C' be a triangle inscribed in ABC (so A' lies on BC, etc.).
          > > Determine the radical center of the circles with diameters AA', BB' and
          > > CC' respectively.
          > >
          > > Kind regards,
          > > Floor.
          > >
          >
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