## A circle tangent to the incircle

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• Dear Hyacinthists, an article IN ENGLISH concerning a circle tangent to the incircle has been put on my website http://perso.orange.fr/jl.ayme vol. 6 A circle
Message 1 of 9 , Sep 24, 2010
Dear Hyacinthists,
an article IN ENGLISH concerning a circle tangent to the incircle has been put on my website
http://perso.orange.fr/jl.ayme vol. 6 A circle tangent to the incircle
Sincerely
Jean-Louis

[Non-text portions of this message have been removed]
• There are three circles touching the incircle at three points, say, Ta,Tb,Tc. A natural question is: Is the triangle ABC and the triangle TaTbTc perspective?
Message 2 of 9 , Sep 25, 2010
There are three circles touching the incircle at three points,
say, Ta,Tb,Tc.
A natural question is:
Is the triangle ABC and the triangle TaTbTc perspective?

That is:

Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
of H,I, resp. The circle (A,AB"=AC") intersects AA' at A* between
A,A'. Denote A1 := BC /\ IA*. The circle with diameter AA1 touches
the incircle (I) at Ta. Similarly Tb, Tc.

Question:

Are the triangles ABC, TaTbTc perspective?

APH

--- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
>
> Dear Hyacinthists,
> an article IN ENGLISH concerning a circle tangent to the incircle has been put on my website
> http://perso.orange.fr/jl.ayme vol. 6 A circle tangent to the incircle
> Sincerely
> Jean-Louis
• Dear Antreas and Hyacinthists, I have consider this question when I was working on this subject. I have tried to use the Rabinowitz s result
Message 3 of 9 , Sep 25, 2010
Dear Antreas and Hyacinthists,
I have consider this question when I was working on this subject. I have tried to use the Rabinowitz's result (http://perso.orange fr/jl.ayme vol. 3) to prove this concurrence but I haven't found a way to prove it synthetically.
Any isea?
Sincerely
Jean-Louis

--- En date de : Sam 25.9.10, Antreas <anopolis72@...> a écrit :

De: Antreas <anopolis72@...>
Objet: [EMHL] Re: A circle tangent to the incircle
À: Hyacinthos@yahoogroups.com
Date: Samedi 25 septembre 2010, 10h21

There are three circles touching the incircle at three points,
say, Ta,Tb,Tc.
A natural question is:
Is the triangle ABC and the triangle TaTbTc perspective?

That is:

Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
of H,I, resp. The circle (A,AB"=AC") intersects AA' at A* between
A,A'. Denote A1 := BC /\ IA*. The circle with diameter AA1 touches
the incircle (I) at Ta. Similarly Tb, Tc.

Question:

Are the triangles ABC, TaTbTc perspective?

APH

--- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
>
> Dear Hyacinthists,
> an article IN ENGLISH concerning a circle tangent to the incircle has been put on my website
> http://perso.orange.fr/jl.ayme vol. 6 A circle tangent to the incircle
> Sincerely
> Jean-Louis

[Non-text portions of this message have been removed]
• Dear Antreas, the triangles TaTbTc is perspective with ABC at X1336.
Message 4 of 9 , Sep 25, 2010
Dear Antreas,

the triangles TaTbTc is perspective with ABC at X1336.

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> There are three circles touching the incircle at three points,
> say, Ta,Tb,Tc.
> A natural question is:
> Is the triangle ABC and the triangle TaTbTc perspective?
>
> That is:
>
> Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
> of H,I, resp. The circle (A,AB"=AC") intersects AA' at A* between
> A,A'. Denote A1 := BC /\ IA*. The circle with diameter AA1 touches
> the incircle (I) at Ta. Similarly Tb, Tc.
>
> Question:
>
> Are the triangles ABC, TaTbTc perspective?
>
> APH
>
> --- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@> wrote:
> >
> > Dear Hyacinthists,
> > an article IN ENGLISH concerning a circle tangent to the incircle has been put on my website
> > http://perso.orange.fr/jl.ayme vol. 6 A circle tangent to the incircle
> > Sincerely
> > Jean-Louis
>
• Dear friends, 1. X(1336) is a point on the circumscribed hyperbola through X(1) and X(2). 2. The triangle A1B1C1 is perspective with ABC at X(175)=
Message 5 of 9 , Sep 25, 2010
Dear friends,

1. X(1336) is a point on the circumscribed hyperbola through X(1) and X(2).
2. The triangle A1B1C1 is perspective with ABC at X(175)= Isoperimetric Point. Interesting about X(175):
X(175) and X(176) are common points of hyperbola's (A,(foci=B,C)),(B,(foci=C,A)),(C,(foci=A,B)).

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> Dear Antreas,
>
> the triangles TaTbTc is perspective with ABC at X1336.
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> >
> > There are three circles touching the incircle at three points,
> > say, Ta,Tb,Tc.
> > A natural question is:
> > Is the triangle ABC and the triangle TaTbTc perspective?
> >
> > That is:
> >
> > Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
> > of H,I, resp. The circle (A,AB"=AC") intersects AA' at A* between
> > A,A'. Denote A1 := BC /\ IA*. The circle with diameter AA1 touches
> > the incircle (I) at Ta. Similarly Tb, Tc.
> >
> > Question:
> >
> > Are the triangles ABC, TaTbTc perspective?
> >
> > APH
> >
> > --- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@> wrote:
> > >
> > > Dear Hyacinthists,
> > > an article IN ENGLISH concerning a circle tangent to the incircle has been put on my website
> > > http://perso.orange.fr/jl.ayme vol. 6 A circle tangent to the incircle
> > > Sincerely
> > > Jean-Louis
> >
>
• Dear Friends [Jean-Louis Ayme] ... [APH] ... [Francisco Javier] ... I think that the extraversion is: Let ABC be a triangle, A B C , A aA bA c the pedal
Message 6 of 9 , Sep 25, 2010
Dear Friends

[Jean-Louis Ayme]
> > > an article IN ENGLISH concerning a circle tangent to
> > > the incircle has been put on my website
> > > http://perso.orange.fr/jl.ayme vol. 6 A circle tangent
> > > to the incircle

[APH]
> > There are three circles touching the incircle at three points,
> > say, Ta,Tb,Tc.
> > A natural question is:
> > Is the triangle ABC and the triangle TaTbTc perspective?
> >
> > That is:
> >
> > Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
> > of H,I, resp. The circle (A,AB"=AC") intersects AA' at A* between
> > A,A'. Denote A1 := BC /\ IA*. The circle with diameter AA1
> > touches the incircle (I) at Ta. Similarly Tb, Tc.
> >
> > Question:
> >
> > Are the triangles ABC, TaTbTc perspective?

[Francisco Javier]
> the triangles TaTbTc is perspective with ABC at X1336.

I think that the "extraversion" is:

Let ABC be a triangle, A'B'C', A"aA"bA"c the pedal
triangles of H, Ia, resp.
The circle (A, A"b = A"c) intersects AA' at A**, with A' between
A and A**.

Denote A2 := BC /\ IaA**. The circle with diameter AA2
touches externally the excircle (Ia) at Sa.
Similarly Sb, Sc.

ABC, SaSbSc should be perspective as well.

APH
• If XYZ is the cevian triangle of X1336, then Sa and Ta are harmonic conjugates with respect A,X; thus we have that SaSbSc is perspective with ABC at X1336
Message 7 of 9 , Sep 25, 2010
If XYZ is the cevian triangle of X1336, then Sa and Ta are harmonic conjugates with respect A,X; thus we have that SaSbSc is perspective with ABC at X1336 also.

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Dear Friends
>
> [Jean-Louis Ayme]
> > > > an article IN ENGLISH concerning a circle tangent to
> > > > the incircle has been put on my website
> > > > http://perso.orange.fr/jl.ayme vol. 6 A circle tangent
> > > > to the incircle
>
> [APH]
> > > There are three circles touching the incircle at three points,
> > > say, Ta,Tb,Tc.
> > > A natural question is:
> > > Is the triangle ABC and the triangle TaTbTc perspective?
> > >
> > > That is:
> > >
> > > Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
> > > of H,I, resp. The circle (A,AB"=AC") intersects AA' at A* between
> > > A,A'. Denote A1 := BC /\ IA*. The circle with diameter AA1
> > > touches the incircle (I) at Ta. Similarly Tb, Tc.
> > >
> > > Question:
> > >
> > > Are the triangles ABC, TaTbTc perspective?
>
> [Francisco Javier]
> > the triangles TaTbTc is perspective with ABC at X1336.
>
> I think that the "extraversion" is:
>
> Let ABC be a triangle, A'B'C', A"aA"bA"c the pedal
> triangles of H, Ia, resp.
> The circle (A, A"b = A"c) intersects AA' at A**, with A' between
> A and A**.
>
> Denote A2 := BC /\ IaA**. The circle with diameter AA2
> touches externally the excircle (Ia) at Sa.
> Similarly Sb, Sc.
>
> ABC, SaSbSc should be perspective as well.
>
> APH
>
• [APH] ... [Francisco Javier] ... So we have one theorem instead of two! TaTbTc, SaSbSc are perspective at X(1336) APH
Message 8 of 9 , Sep 26, 2010
[APH]
>> Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
>> of H,I, resp. The circle (A,AB"=AC") intersects AA' at A* between
>> A,A'. Denote A1 := BC /\ IA*. The circle with diameter AA1
>> touches the incircle (I) at Ta. Similarly Tb, Tc.
>> [extraversion]:
>> Let ABC be a triangle, A'B'C', A"aA"bA"c the pedal
>> triangles of H, Ia, resp.
>> The circle (A, A"b = A"c) intersects AA' at A**, with A' between
>> A and A**.
>> Denote A2 := BC /\ IaA**. The circle with diameter AA2
>> touches externally the excircle (Ia) at Sa.
>> Similarly Sb, Sc.

[Francisco Javier]
> The triangles TaTbTc is perspective with ABC at X1336.
>
> If XYZ is the cevian triangle of X1336, then Sa and Ta are
> harmonic conjugates with respect A,X; thus we have that SaSbSc
> is perspective with ABC at X1336 also.

So we have one theorem instead of two!

TaTbTc, SaSbSc are perspective at X(1336)

APH
• [APH] ... [Francisco Javier] ... [APH] ... Now, let T1T2T3 be the triangle bounded by the tangents to Incircle (I) at Ta,Tb,Tc, and S1S2S3 the triangle bounded
Message 9 of 9 , Sep 30, 2010
[APH]
> >> Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles
> >> of H,I, resp. The circle (A,AB"=AC") intersects AA' at A*
> >> between A,A'. Denote A1 := BC /\ IA*. The circle with
> >> diameter AA1 touches the incircle (I) at Ta. Similarly Tb, Tc.
> >> [extraversion]:
> >> Let ABC be a triangle, A'B'C', A"aA"bA"c the pedal
> >> triangles of H, Ia, resp.
> >> The circle (A, A"b = A"c) intersects AA' at A**, with A'
> >> between A and A**.
> >> Denote A2 := BC /\ IaA**. The circle with diameter AA2
> >> touches externally the excircle (Ia) at Sa.
> >> Similarly Sb, Sc.

[Francisco Javier]
> > The triangles TaTbTc is perspective with ABC at X1336.
> >
> > If XYZ is the cevian triangle of X1336, then Sa and Ta are
> > harmonic conjugates with respect A,X; thus we have that SaSbSc
> > is perspective with ABC at X1336 also.

[APH]
> So we have one theorem instead of two!
>
> TaTbTc, SaSbSc are perspective at X(1336)

Now, let T1T2T3 be the triangle bounded by the tangents to Incircle
(I) at Ta,Tb,Tc, and S1S2S3 the triangle bounded by the tangents
to excircles (Ia),(Ib),(Ic) at Sa, Sb, Sc, resp.

Is the Orthic triangle A'B'C' perspective with T1T2T3 and
S1S2S3 ??

APH
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