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Isogonal conjugate of nine points circle wrt its triangle

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  • mbabelian
    Dear all (1).$ABC$ a triangle and $I$ is its Incenter. (2)Incircle of $ABC$ meet $BC,CA,AB$ at $D,E,F$ respectively. (3)$D$ is the midpoint of $IP$. (4)$N $ is
    Message 1 of 3 , Jul 9, 2012
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      Dear all

      (1).$ABC$ a triangle and $I$ is its Incenter.
      (2)Incircle of $ABC$ meet $BC,CA,AB$ at $D,E,F$ respectively.
      (3)$D$ is the midpoint of $IP$.
      (4)$N'$ is the Isogonal conjugate of $N$,the center of nine points circle of $DEF$ wrt triangle $DEF$.

      prove that $N',A,P$ lie on a line.

      -------------------------------------------
      are there any article or paper about N's Isogonal conjugate?
      -------------------------------------------

      With Regards.
    • Francisco Javier
      Your point N is X(3649) = KS(INTOUCH TRIANGLE) However, N , A and P are never collinear, except when ABC is isosceles with AB=AC. Best regards, Francisco
      Message 2 of 3 , Jul 9, 2012
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        Your point N' is X(3649) = KS(INTOUCH TRIANGLE)

        However, N', A and P are never collinear, except when ABC is isosceles with AB=AC.

        Best regards,

        Francisco Javier.


        --- In Hyacinthos@yahoogroups.com, "mbabelian" <mbabelian@...> wrote:
        >
        > Dear all
        >
        > (1).$ABC$ a triangle and $I$ is its Incenter.
        > (2)Incircle of $ABC$ meet $BC,CA,AB$ at $D,E,F$ respectively.
        > (3)$D$ is the midpoint of $IP$.
        > (4)$N'$ is the Isogonal conjugate of $N$,the center of nine points circle of $DEF$ wrt triangle $DEF$.
        >
        > prove that $N',A,P$ lie on a line.
        >
        > -------------------------------------------
        > are there any article or paper about N's Isogonal conjugate?
        > -------------------------------------------
        >
        > With Regards.
        >
      • Francisco Javier
        As Chandan Banerjee have just pointed out from Argentina, this is true: reflection of I on N is collinear with A,P. Good luck at IMO!
        Message 3 of 3 , Jul 9, 2012
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          As Chandan Banerjee have just pointed out from Argentina, this is true: reflection of I on N' is collinear with A,P.

          Good luck at IMO!

          --- In Hyacinthos@yahoogroups.com, "mbabelian" <mbabelian@...> wrote:
          >
          > Dear all
          >
          > (1).$ABC$ a triangle and $I$ is its Incenter.
          > (2)Incircle of $ABC$ meet $BC,CA,AB$ at $D,E,F$ respectively.
          > (3)$D$ is the midpoint of $IP$.
          > (4)$N'$ is the Isogonal conjugate of $N$,the center of nine points circle of $DEF$ wrt triangle $DEF$.
          >
          > prove that $N',A,P$ lie on a line.
          >
          > -------------------------------------------
          > are there any article or paper about N's Isogonal conjugate?
          > -------------------------------------------
          >
          > With Regards.
          >
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