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2 new points on the Soddy-line

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  • Eric Danneels
    Dear Hyacinthians, thanks to an idea from Floor in a private mail, I found 2 new interesting points on the Soddy-line. Consider a triangle ABC with intouch
    Message 1 of 4 , Dec 28 10:53 AM
      Dear Hyacinthians,

      thanks to an idea from Floor in a private mail, I found 2 new
      interesting points on the Soddy-line.

      Consider a triangle ABC with intouch triangle A'B'C'.
      Let A* be the intersection of BC and B'C'
      and define B* and C* similarly.
      The circles with centers A*, B* and C* through
      respectively A', B' and C' intersect in 2 points
      on the Soddy line with barycentric coordinates:

      (Ra +- sin(60).a : Rb +- sin(60).b : Rc +- sin(60).c)

      where Ra, Rb and Rc are the radii of the excircles.

      If we represent the incenter by I and the Gergonne point by G
      these points can be written as

      (Ri + 4.Ro).G + 2.sin(60).s.I
      where
      RI = inradius
      Ro= circumradius
      s = semiperimeter

      Other properties of these points

      - they are harmonically conjugated wrt G and I;
      - the Fletcher-point X(1323) is their midpoint;
      - an inversion centered at these points transforms the A-, B- and C-
      Soddy-circle into congruent circles.


      ---------

      To all a happy and a healthy New Year

      Eric











      Consoid
    • Floor en Lyanne van Lamoen
      Dear Eric, Thanks for the nice idea. ... They are the common points of the pencil circles orthogonal to the pencil of circles generated by the incircle and the
      Message 2 of 4 , Dec 28 1:48 PM
        Dear Eric,

        Thanks for the nice idea.

        > Consider a triangle ABC with intouch triangle A'B'C'.
        > Let A* be the intersection of BC and B'C'
        > and define B* and C* similarly.
        > The circles with centers A*, B* and C* through
        > respectively A', B' and C' intersect in 2 points
        > on the Soddy line with barycentric coordinates:
        >
        > (Ra +- sin(60).a : Rb +- sin(60).b : Rc +- sin(60).c)
        >
        > where Ra, Rb and Rc are the radii of the excircles.
        >
        > If we represent the incenter by I and the Gergonne point by G
        > these points can be written as
        >
        > (Ri + 4.Ro).G + 2.sin(60).s.I
        > where
        > RI = inradius
        > Ro= circumradius
        > s = semiperimeter
        >
        > Other properties of these points
        >
        > - they are harmonically conjugated wrt G and I;
        > - the Fletcher-point X(1323) is their midpoint;
        > - an inversion centered at these points transforms the A-, B- and C-
        > Soddy-circle into congruent circles.

        They are the common points of the pencil circles orthogonal to the pencil of
        circles generated by the incircle and the inner and outer Soddy-circles
        (containing the circles making a given angle to the A-, B- and C-Soddy
        circles, or "cosines" as Hauke recently described to us).

        Happy, Healthy and Fruitful 2006 to all Hyacinthos members!

        Kind regards,
        Floor.
      • Hauke Reddmann
        ... and C- ... pencil of ... circles ... Soddy ... Here is my :-) way to define the Fletcher point: In the framework of my power line approach, if the
        Message 3 of 4 , Jan 2, 2006
          --- In Hyacinthos@yahoogroups.com, Floor en Lyanne van Lamoen
          <fvanlamoen@p...> wrote:
          >
          > Dear Eric,
          >
          > Thanks for the nice idea.
          >
          > > Consider a triangle ABC with intouch triangle A'B'C'.
          > > Let A* be the intersection of BC and B'C'
          > > and define B* and C* similarly.
          > > The circles with centers A*, B* and C* through
          > > respectively A', B' and C' intersect in 2 points
          > > on the Soddy line with barycentric coordinates:
          > >
          > > (Ra +- sin(60).a : Rb +- sin(60).b : Rc +- sin(60).c)
          > >
          > > where Ra, Rb and Rc are the radii of the excircles.
          > >
          > > If we represent the incenter by I and the Gergonne point by G
          > > these points can be written as
          > >
          > > (Ri + 4.Ro).G + 2.sin(60).s.I
          > > where
          > > RI = inradius
          > > Ro= circumradius
          > > s = semiperimeter
          > >
          > > Other properties of these points
          > >
          > > - they are harmonically conjugated wrt G and I;
          > > - the Fletcher-point X(1323) is their midpoint;
          > > - an inversion centered at these points transforms the A-, B-
          and C-
          > > Soddy-circle into congruent circles.
          >
          > They are the common points of the pencil circles orthogonal to the
          pencil of
          > circles generated by the incircle and the inner and outer Soddy-
          circles
          > (containing the circles making a given angle to the A-, B- and C-
          Soddy
          > circles, or "cosines" as Hauke recently described to us).
          >
          Here is "my" :-) way to define the Fletcher point:
          In the framework of my "power line" approach, if the cosine
          of the intersection angle is=q and the radius of the
          intersecting circle=r, it's center being (x,y),
          then for the Kimberling triangle
          x=(2qr+3)/3,y=(89qr+120)/12/sqrt(35), and
          1129q^2r^2+4240qr-560r^2+1600.
          (For the general case, likewise A1q^2r^2+A2qr+A3r^2+A4=0)

          It's natural to ask for the "special" points of this
          equation. Besides q=0 (radical center/incenter) and q=+-1
          (Apollonius/Soddy circles) these are [limit of r=0&q=oo]
          (not named yet), r=oo (point of infinity of Soddy line),
          and the values for q and r where the discriminant of
          the above equation vanishes.
          These are q=+-i/sqrt(3), r=i*sqrt(3)*40/53
          (Gergonne point, I mentioned that - the other sign of q
          gives #390, since they have to lie symmetrically to the
          incenter) and r=+-i*40*sqrt(3/1129),o=i*53/sqrt(3387)
          (Fletcher point and its reflection in #1 which has no
          number yet).
          Note that q and r are imaginary so you can't interpret
          this easily geometrically but this crosscancels in (x,y).

          Hauke
        • Hauke Reddmann
          ... Just for fun, replace the Soddy circles radii with a,b,c instead of s-a,s-b,s-c. The power line and radical center stays the same (neither negating all
          Message 4 of 4 , Jan 2, 2006
            Edit:
            > It's natural to ask for the "special" points of this
            > equation. Besides q=0 (radical center/incenter) and q=+-1
            > (Apollonius/Soddy circles) these are [limit of r=0&q=oo]
            > (not named yet), r=oo (point of infinity of Soddy line),
            > and the values for q and r where the discriminant of
            > the above equation vanishes.

            Just for fun, replace the Soddy circles radii with a,b,c
            instead of s-a,s-b,s-c. The "power line" and radical center
            stays the same (neither negating all radii nor shifting
            all radii by a constant does this), the points of r1=r2 are
            #347&unknown [reflection of #347 in #1].
            I could redefine known points in this context all the new
            year long without coming to an end :-)

            Hauke
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