## 2 new points on the Soddy-line

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• 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
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
• 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
> 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.
• ... 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
> > 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
• ... 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
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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