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A MALFATTI - LIKE PROBLEM

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  • Antreas
    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
    Message 1 of 6 , Jan 19, 2013
      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
    • Antreas
      ... 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
      Message 2 of 6 , Jan 19, 2013
        --- 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
      • Antreas
        The perspecivity problem is independent from the triangle! Let (Ka), (Kb), (Kc) be three mutually tangent circles (externally) ... at C*. The circle tangent
        Message 3 of 6 , Jan 19, 2013
          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
          >
        • Vladimir Dubrovsky
          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
          Message 4 of 6 , Jan 19, 2013
            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 <anopolis72@...>

            > **
            >
            >
            > 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
            > >
            >
            >
            >


            [Non-text portions of this message have been removed]
          • 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 5 of 6 , Jan 19, 2013
              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 6 of 6 , Jan 19, 2013
                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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