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related to the Lucas circles

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  • pohoata_cosmin2000
    Greetings, I see that in Paul Yiu, Introduction to the geometry of the triangle, it is mentioned that the perspector of triangle formed by the points of
    Message 1 of 5 , Aug 5, 2007
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      Greetings,
      I see that in Paul Yiu, Introduction to the geometry of the triangle,
      it is mentioned that the perspector of triangle formed by the points
      of tangency of the Lucas circles and ABC is X371, the isogonal of the
      Vecten point.

      Isn't it true that this point is the radical center of the circles
      circumscribed the three squares inscribed in ABC ?

      Sorry if it's a dumb question, I didn't see it mentioned. In my sketch
      the collinearity of the circumcenter of ABC, radical center of the
      Lucas circles and X371 followed from the above remark, so that's why I
      ask.


      Thank you and best regards,
      Cosmin Pohoata
    • garciacapitan
      Hello, For now, what I have seen is that the centers of these squares are perspective with ABC, having as perspector X486, the inner Vecten point. I don t know
      Message 2 of 5 , Aug 5, 2007
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        Hello,

        For now, what I have seen is that the centers of these squares are
        perspective with ABC, having as perspector X486, the inner Vecten
        point.

        I don't know also if this is new.


        --- In Hyacinthos@yahoogroups.com, "pohoata_cosmin2000"
        <pohoata_cosmin2000@...> wrote:
        >
        > Greetings,
        > I see that in Paul Yiu, Introduction to the geometry of the
        triangle,
        > it is mentioned that the perspector of triangle formed by the points
        > of tangency of the Lucas circles and ABC is X371, the isogonal of
        the
        > Vecten point.
        >
        > Isn't it true that this point is the radical center of the circles
        > circumscribed the three squares inscribed in ABC ?
        >
        > Sorry if it's a dumb question, I didn't see it mentioned. In my
        sketch
        > the collinearity of the circumcenter of ABC, radical center of the
        > Lucas circles and X371 followed from the above remark, so that's
        why I
        > ask.
        >
        >
        > Thank you and best regards,
        > Cosmin Pohoata
        >
      • Nikolaos Dergiades
        ... No it not correct. I think you conclude this by a sketch. If we call these circes C1, C2, C3 then the radical axis of C2, C3 seems to pass through X371 but
        Message 3 of 5 , Aug 6, 2007
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          > I see that in Paul Yiu, Introduction to the geometry
          > of the triangle,
          > it is mentioned that the perspector of triangle
          > formed by the points
          > of tangency of the Lucas circles and ABC is X371,
          > the isogonal of the
          > Vecten point.
          >
          > Isn't it true that this point is the radical center
          > of the circles
          > circumscribed the three squares inscribed in ABC ?

          No it not correct.
          I think you conclude this by a sketch.
          If we call these circes C1, C2, C3
          then the radical axis of C2, C3 seems to pass
          through X371 but if you calculate the distance
          of X371 from the above axis you'll see it is not zero.
          The barycentrics of the radical center are
          sinAsinAcosA /(sinA+cosA), . . , . .
          while the barycentrics of X371 are
          sinA(sinA+cosA), . . , . .

          Best regards
          Nikos Dergiades



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        • garciacapitan
          Dear, Cosmin Pohoata, Nikos, and all, I was wrong, the perspector is X485, the outer Vecten point Best regards, ... points
          Message 4 of 5 , Aug 7, 2007
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            Dear, Cosmin Pohoata, Nikos, and all,

            I was wrong, the perspector is X485, the outer Vecten point

            Best regards,



            --- In Hyacinthos@yahoogroups.com, "garciacapitan"
            <garciacapitan@...> wrote:
            >
            > Hello,
            >
            > For now, what I have seen is that the centers of these squares are
            > perspective with ABC, having as perspector X486, the inner Vecten
            > point.
            >
            > I don't know also if this is new.
            >
            >
            > --- In Hyacinthos@yahoogroups.com, "pohoata_cosmin2000"
            > <pohoata_cosmin2000@> wrote:
            > >
            > > Greetings,
            > > I see that in Paul Yiu, Introduction to the geometry of the
            > triangle,
            > > it is mentioned that the perspector of triangle formed by the
            points
            > > of tangency of the Lucas circles and ABC is X371, the isogonal of
            > the
            > > Vecten point.
            > >
            > > Isn't it true that this point is the radical center of the circles
            > > circumscribed the three squares inscribed in ABC ?
            > >
            > > Sorry if it's a dumb question, I didn't see it mentioned. In my
            > sketch
            > > the collinearity of the circumcenter of ABC, radical center of the
            > > Lucas circles and X371 followed from the above remark, so that's
            > why I
            > > ask.
            > >
            > >
            > > Thank you and best regards,
            > > Cosmin Pohoata
            > >
            >
          • pohoata_cosmin2000
            ... Dear Nikos, you are right, I misscalculated something. Thank you for pointing out the mistake.
            Message 5 of 5 , Aug 7, 2007
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              --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...>
              wrote:

              > No it not correct.
              > I think you conclude this by a sketch.
              > If we call these circes C1, C2, C3
              > then the radical axis of C2, C3 seems to pass
              > through X371 but if you calculate the distance
              > of X371 from the above axis you'll see it is not zero.
              > The barycentrics of the radical center are
              > sinAsinAcosA /(sinA+cosA), . . , . .
              > while the barycentrics of X371 are
              > sinA(sinA+cosA), . . , . .
              >
              > Best regards
              > Nikos Dergiades


              Dear Nikos, you are right, I misscalculated something. Thank you for
              pointing out the mistake.
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