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A circle tangent to the incircle

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  • Jean-Louis Ayme
    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
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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 tangent to the incircle
      Sincerely
      Jean-Louis




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