## maybe new inequality for new radius

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• Dear Paul,Jean-Pierre and friends, ******* Happy New Year to all Hyacinthists. ... excircles of ... of the ... long been ... example, ... interesting ... lies
Message 1 of 3 , Jan 2, 2002
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Dear Paul,Jean-Pierre and friends,
*******
Happy New Year to all Hyacinthists.

Paul wrote :
> Dear Clark and friends,
>
> Happy New Year!
> Consider the circle tangent internally to each of the
excircles of
> 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,

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[Non-text portions of this message have been removed]
• ... [Jean-Pierre(?):] ... 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
Message 2 of 3 , Jan 2, 2002
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On Wed, 2 Jan 2002, Milorad Stevanovic wrote:

> Paul wrote :
> > [The] inversive image of the nine-point circle in the
> > [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 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).

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

John Conway
• Dear John, Paul, Milorad and other Hyacinthists [PY] ... OK. ... point ... [JHC] ... circle ... probably ... There is no quotation. As Paul had noticed that
Message 3 of 3 , Jan 2, 2002
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Dear John, Paul, Milorad and other Hyacinthists

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

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

There is no quotation.
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
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