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Three Triangle Circles

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  • Antreas
    Let ABC be a triangle and (Kab), (Kac) the two congruent and tangent circles such that (Kab) is tangent to the sides of the internal angle B and (Kac) to the
    Message 1 of 3 , Feb 22, 2013
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      Let ABC be a triangle and (Kab), (Kac) the two congruent
      and tangent circles such that (Kab) is tangent to the
      sides of the internal angle B and (Kac) to the sides of the
      internal angle C.
      Let (Kabb), (Kacc) be the two congruent tangent circles such that
      (Kabb) is tangent externally to (Kab) and to AB, and (Kacc) to
      (Kac) and to AC.

      The quadrilateral KabKabbKaccKac is inscriptible. Let (K1) be its incircle. Similarly define (K2), (K3).

      (For external angles we get another three circles.)

      K1K2K3 is perspective with triangles????
      Is it with ABC?

      APH
    • Nikolaos Dergiades
      Dear Andreas, Do you have a construction of circles (Kabb), (Kacc)? If A1 is the mid point of (Kab), (Kac) then the triangles ABC, A1B1C1 are perspective at
      Message 2 of 3 , Feb 23, 2013
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        Dear Andreas,

        Do you have a construction of circles
        (Kabb), (Kacc)?

        If A1 is the mid point of (Kab), (Kac)
        then the triangles ABC, A1B1C1 are perspective
        at X(1123) isotomic conjugate of (bc+S:ca+S:ab+S)

        If the circles
        (Kabb) is tangent externally to (Kab) and to BC, and (Kacc)to
        (Kac) and to BC then the triangles ABC, K1K2K3
        are perspective at the isotomic conjugate of
        (SA + 2S + 2bc : SB + 2S +2ca : . . .)

        But your problem seems to be difficult.

        Best regards
        Nikos Dergiades


        > Let ABC be a triangle and (Kab),
        > (Kac) the two congruent
        > and tangent circles such that (Kab) is tangent to the
        > sides of the internal angle B and (Kac) to the sides of the
        >
        > internal angle C.
        > Let (Kabb), (Kacc) be the two congruent tangent circles such
        > that
        > (Kabb) is tangent externally to (Kab) and to AB, and (Kacc)
        > to
        > (Kac) and to AC.
        >
        > The quadrilateral KabKabbKaccKac is inscriptible. Let (K1)
        > be its incircle. Similarly define (K2), (K3).
        >
        > (For external angles we get another three circles.)
        >
        > K1K2K3 is perspective with triangles????
        > Is it with ABC?
        >
        > APH
        >
        >
        >
        > ------------------------------------
        >
        > Yahoo! Groups Links
        >
        >
        >     Hyacinthos-fullfeatured@yahoogroups.com
        >
        >
      • Antreas Hatzipolakis
        Dear Nikos, A pure geometric construction of (Kabb), (Kacc) would be quite difficult, since even the construction for lines instead of circles (Kab), (Kac) is
        Message 3 of 3 , Feb 23, 2013
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          Dear Nikos,

          A pure geometric construction of (Kabb), (Kacc) would be quite difficult,
          since even the construction for lines instead of circles (Kab), (Kac)
          is difficult !!
          (it is the problem of constucting two mutually tangent congruent
          circlles, each one of whose is tangent to the sides of opposite angles
          of quadrilateral)

          But algebraically/trigonometrically we can compute the radius of the first
          circles (Kab), (Kac), and after that the radius of (Kabb), (Kacc), more or less
          easily I guess!!

          APH

          On Sat, Feb 23, 2013 at 4:18 PM, Nikolaos Dergiades <ndergiades@...> wrote:
          > Dear Andreas,
          >
          > Do you have a construction of circles
          > (Kabb), (Kacc)?
          >
          > If A1 is the mid point of (Kab), (Kac)
          > then the triangles ABC, A1B1C1 are perspective
          > at X(1123) isotomic conjugate of (bc+S:ca+S:ab+S)
          >
          > If the circles
          > (Kabb) is tangent externally to (Kab) and to BC, and (Kacc)to
          > (Kac) and to BC then the triangles ABC, K1K2K3
          > are perspective at the isotomic conjugate of
          > (SA + 2S + 2bc : SB + 2S +2ca : . . .)
          >
          > But your problem seems to be difficult.
          >
          > Best regards
          > Nikos Dergiades
          >
          >
          >> Let ABC be a triangle and (Kab),
          >> (Kac) the two congruent
          >> and tangent circles such that (Kab) is tangent to the
          >> sides of the internal angle B and (Kac) to the sides of the
          >>
          >> internal angle C.
          >> Let (Kabb), (Kacc) be the two congruent tangent circles such
          >> that
          >> (Kabb) is tangent externally to (Kab) and to AB, and (Kacc)
          >> to
          >> (Kac) and to AC.
          >>
          >> The quadrilateral KabKabbKaccKac is inscriptible. Let (K1)
          >> be its incircle. Similarly define (K2), (K3).
          >>
          >> (For external angles we get another three circles.)
          >>
          >> K1K2K3 is perspective with triangles????
          >> Is it with ABC?
          >>
          >> APH
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