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Re: [EMHL] Re: A MALFATTI - LIKE PROBLEM

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  • Ignacio Larrosa Cañestro
    ... The perspector seems to depend only on the two tangents circles to (Ka), (Kb) and (K,c), an is aligned with its centres, as also the perspector defined
    Message 1 of 6 , Jan 19, 2013
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      El 19/01/2013 17:33, Vladimir Dubrovsky escribió:
      > Dear Antreas,
      >
      > the perspector is called the Eppstein point (one of the two possible) of
      > the triange formed by the centers of the three initial circles. See, for
      > example, Eppstein, David. “Tangent Spheres and Triangle Centers.” The
      > American Mathematical Monthly 108 (2001), pp. 63-66.
      >
      >

      The perspector seems to depend only on the two tangents circles to (Ka),
      (Kb) and (K,c), an is aligned with its centres, as also the perspector
      defined similarly by
      (Ka), (Kb) and (K,c) and the inner tangent circle:

      http://www.xente.mundo-r.com/ilarrosa/GeoGebra/Malfatti_like_problem.html

      >> **
      >>
      >>
      >> The perspecivity problem is independent from the triangle!
      >>
      >> Let (Ka), (Kb), (Kc) be three mutually tangent circles (externally)
      >> : (Kb), (Kc) are tangent at A* and (Kc), (Ka) at B* and (Ka), (Kb)
      >> at C*. The circle tangent internally to three circles touches (Ka)
      >> at A', (Kb) at B', and (Kc) at C'.
      >>
      >> The lines A'A*, B'B*, C'C* are concurrent (??)
      >>
      >> Probably it is a well-known problem !!!
      >>
      >> APH
      >>
      >>
      >> --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
      >>>
      >>>
      >>> --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
      >>>> Let ABC be a triangle. To draw three circles, each of which
      >>>> is tangent to the other two and to one side of ABC and to
      >>>> the circumcircle of ABC.
      >>>> Note: We can replace circumcircle with an other circle
      >>>> (NPC for example)
      >>>>
      >>>> Figure here
      >>>> http://anthrakitis.blogspot.gr/2013/01/a-malfatti-like-problem.html
      >>>>
      >>>> APH
      >>>
      >>> Perspective Triangles.
      >>>
      >>> Let (Ka), (Kb), (Kc) be the three circles.
      >>>
      >>> (Ka) is tangent to the circumcircle at A', to the BC at A" and to the
      >> other two circles (Kb), (Kc) at C*, B*, resp.
      >>> (Kb) is tangent to the circumcircle at B', to the CA at B" and to the
      >> other two circles (Kc), (Ka) at A*, C*, resp.
      >>> (Kc) is tangent to the circumcircle at C', to the AB at C" and to the
      >> other two circles (Ka), (Kb) at B*, A*, resp.
      >>> The triangles A'B'C', A*B*C* are perspective (??)
      >>>
      >>> APH

      --
      Best regards,

      Ignacio Larrosa Cañestro
      A Coruña (España)
      ilarrosa@...
      http://www.xente.mundo-r.com/ilarrosa/GeoGebra/
    • Antreas
      Dear David Thanks! In my configuration Eppstein point is wrt KaKbKc. The question now is which are the h. coordinates of the point wrt ABC. aph ps Eppstein
      Message 2 of 6 , Jan 19, 2013
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        Dear David

        Thanks!

        In my configuration Eppstein point is wrt KaKbKc. The question now
        is which are the h. coordinates of the point wrt ABC.

        aph

        ps
        Eppstein points
        http://mathworld.wolfram.com/FirstEppsteinPoint.html
        http://mathworld.wolfram.com/SecondEppsteinPoint.html


        --- In Hyacinthos@yahoogroups.com, Vladimir Dubrovsky wrote:
        >
        > Dear Antreas,
        >
        > the perspector is called the Eppstein point (one of the two possible) of
        > the triange formed by the centers of the three initial circles. See, for
        > example, Eppstein, David. “Tangent Spheres and Triangle Centers.” The
        > American Mathematical Monthly 108 (2001), pp. 63-66.
        >
        > Best,
        > Vladimir
        >
        >
        > 2013/1/19 Antreas
        >
        > > **
        > >
        > >
        > > The perspecivity problem is independent from the triangle!
        > >
        > > Let (Ka), (Kb), (Kc) be three mutually tangent circles (externally)
        > > : (Kb), (Kc) are tangent at A* and (Kc), (Ka) at B* and (Ka), (Kb)
        > > at C*. The circle tangent internally to three circles touches (Ka)
        > > at A', (Kb) at B', and (Kc) at C'.
        > >
        > > The lines A'A*, B'B*, C'C* are concurrent (??)
        > >
        > > Probably it is a well-known problem !!!
        > >
        > > APH
        > >
        > >
        > > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
        > > >
        > > >
        > > >
        > > > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
        > > > >
        > > > > Let ABC be a triangle. To draw three circles, each of which
        > > > > is tangent to the other two and to one side of ABC and to
        > > > > the circumcircle of ABC.
        > > > > Note: We can replace circumcircle with an other circle
        > > > > (NPC for example)
        > > > >
        > > > > Figure here
        > > > > http://anthrakitis.blogspot.gr/2013/01/a-malfatti-like-problem.html
        > > > >
        > > > > APH
        > > >
        > > >
        > > > Perspective Triangles.
        > > >
        > > > Let (Ka), (Kb), (Kc) be the three circles.
        > > >
        > > > (Ka) is tangent to the circumcircle at A', to the BC at A" and to the
        > > other two circles (Kb), (Kc) at C*, B*, resp.
        > > >
        > > > (Kb) is tangent to the circumcircle at B', to the CA at B" and to the
        > > other two circles (Kc), (Ka) at A*, C*, resp.
        > > >
        > > > (Kc) is tangent to the circumcircle at C', to the AB at C" and to the
        > > other two circles (Ka), (Kb) at B*, A*, resp.
        > > >
        > > > The triangles A'B'C', A*B*C* are perspective (??)
        > > >
        > > > APH
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