- 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 - Dear Eric,

Thanks for the nice idea.

> Consider a triangle ABC with intouch triangle A'B'C'.

They are the common points of the pencil circles orthogonal to the pencil of

> 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.

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. - --- In Hyacinthos@yahoogroups.com, Floor en Lyanne van Lamoen

<fvanlamoen@p...> wrote:>

and C-

> 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-

> > Soddy-circle into congruent circles.

pencil of

>

> They are the common points of the pencil circles orthogonal to the

> 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 - Edit:
> It's natural to ask for the "special" points of this

Just for fun, replace the Soddy circles radii with a,b,c

> 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.

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