- Dear Paul,Jean-Pierre and friends,

*******

Happy New Year to all Hyacinthists.

Paul wrote :> Dear Clark and friends,

excircles of

>

> Happy New Year!

> Consider the circle tangent internally to each of the

> triangle ABC. This circle can be constructed as the inversive image

of the

> nine-point circle in the radical circle of the excircles. It has

long been

> known that the radius of this circle is (r^2+s^2)/(4r). See, for

example,

> Problem 1864 and solution, Crux Math. 20 (1994) 174--175. What is

interesting

> is that the center is the point X(970) in ETC. As such, this center

lies on

> the Brocard axis. Is there a simple explanation of this fact?

Your circle intersects BC at (in barycentrics)

(0,s-cc,cc) and (0,bb,s-bb) where s = -2abc/(a+b+c).

Thus your circle is a Tucker circle and he is centered on OK.

In fact, as the radical center of the excircles is the Spieker point

S, the center of your circle is the common point of OK and SN

(N=NPcenter), which is, as you've noticed X(970).

Friendly. Jean-Pierre

*******

Let r* be the radius of this circle.Then we have the following inequality

r* >= (R1 + 5r1)/2,where R1 and r1 are respectively circumradius and

inradius of triangle with vertices in excircle-centers of tringle ABC.

In proof we use the formulas 4r*r = s^2 + r^2, s = (r1xs1)/R1,

2rxR1 = (s1)^2 - (2R1 + r1)^2, (x denotes multiplication),and two

Gerretsen's inequalities s^2 <= 4(R^2) + 4Rr + 3(r^2), s^2 >= 16Rr - 5(r^2),

applied on excircle-centers triangle,and following equality

(8R1)r* = 4(r1^2) + {(4r1^2)(2R1 + r1)^2}/{s1^2 - (2R1 +r1)^2} + s1^2 - (2R1 + r1)^2.

Weaker inequalities are,obviously, r* >= (7r1)/2 > = 7r as consequences of

R >= 2r.Maybe,this is a new inequality.

Kind regards,

Sincerely,

Milorad R.Stevanovic

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

[Non-text portions of this message have been removed] - On Wed, 2 Jan 2002, Milorad Stevanovic wrote:

> Paul wrote :

[Jean-Pierre(?):]

> > [The] inversive image of the nine-point circle in the

> >radical circle of the excircles [has] radius (r^2+s^2)/(4r).

> > [The] center[,] X(970)[,] lies on the Brocard axis.

> > Is there a simple explanation of this fact?

> [...] Thus your circle is a Tucker circle and he is centered on OK.

I suspect there must also be some connection with "the Conway circle"

> In fact, as the radical center of the excircles is the Spieker point

> S, the center of your circle is the common point of OK and SN

> (N=NPcenter), which is, as you've noticed X(970).

(center Io), since the radius of that is root(rr+ss). There's probably

a vast generalization that describes all the Tucker circles in some

such way (which maybe Jean-Pierre is quoting).

John Conway - Dear John, Paul, Milorad and other Hyacinthists

[PY]> > > [The] inversive image of the nine-point circle in the

OK.

> > >radical circle of the excircles [has] radius (r^2+s^2)/(4r).

> > > [The] center[,] X(970)[,] lies on the Brocard axis.

> > > Is there a simple explanation of this fact?

>

> [Jean-Pierre(?):]

>

> > [...] Thus your circle is a Tucker circle and he is centered on

> > In fact, as the radical center of the excircles is the Spieker

point

> > S, the center of your circle is the common point of OK and SN

[JHC]

> > (N=NPcenter), which is, as you've noticed X(970).

> I suspect there must also be some connection with "the Conway

circle"

> (center Io), since the radius of that is root(rr+ss). There's

probably

> a vast generalization that describes all the Tucker circles in some

There is no quotation.

> such way (which maybe Jean-Pierre is quoting).

As Paul had noticed that the circle was centered on OK, I've just

looked whether he could be a Tucker circle (it's enough to check that

bb BL + cc CL' = 0 where L,L' are the common points of BC and the

circle) and the answer was yes but I'm unable to understand why.

More over, I think that I'm not the first one to notice that this

circle is a Tucker circle but I've never heard of that.

Happy 2002 to all Hyacinthists.

Friendly. Jean-Pierre