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3 nine point circles and the Euler line

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  • KHOA LU
    Dear friends, I think this one is right but I haven t thought about it at all. So I think I ll call this one my conjecture : Let ABC be a triangle and I_a,
    Message 1 of 4 , Oct 1, 2004
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      Dear friends,
      I think this one is right but I haven't thought about it at all.
      So I think I 'll call this one my conjecture :
      Let ABC be a triangle and I_a, I_b, I_c are excenters,
      Let (W_a), (W_b), (W_c) are respectively the nine point
      circles of I_aBC, I_bCA, I_cAB . Prove that the 3 radical axises of (W_a)
      and (W_b), (W_b) and (W_c), (W_c) and (W_a) and the Euler line of ABC are concurent. What is its barycentric coordinate ?
      Thanks.
      Best regards,
      Khoa Lu Nguyen.


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    • jpehrmfr
      Dear Hyacithists ... all. ... axises of (W_a) ... of ABC are concurent. What is its barycentric coordinate ? ... Your radical center is the complement of
      Message 2 of 4 , Oct 1, 2004
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        Dear Hyacithists
        --- In Hyacinthos@yahoogroups.com, KHOA LU <treegoner@y...> wrote:
        > Dear friends,
        > I think this one is right but I haven't thought about it at
        all.
        > So I think I 'll call this one my conjecture :
        > Let ABC be a triangle and I_a, I_b, I_c are excenters,
        > Let (W_a), (W_b), (W_c) are respectively the nine point
        > circles of I_aBC, I_bCA, I_cAB . Prove that the 3 radical
        axises of (W_a)
        > and (W_b), (W_b) and (W_c), (W_c) and (W_a) and the Euler line
        of ABC are concurent. What is its barycentric coordinate ?
        > Thanks.
        > Best regards,
        > Khoa Lu Nguyen.

        Your radical center is the complement of the Schiffler point and
        appears as X(442) in ETC.
        Friendly. Jean-Pierre
      • Paul Yiu
        Dear Khoa Lu, [KLN]: ... Let ABC be a triangle and I_a, I_b, I_c are excenters, ... *** You are right. Let A_b and A_c be the pedal (orthogonal
        Message 3 of 4 , Oct 1, 2004
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          Dear Khoa Lu,

          [KLN]: ... Let ABC be a triangle and I_a, I_b, I_c are excenters,
          > Let (W_a), (W_b), (W_c) are respectively the nine point
          > circles of I_aBC, I_bCA, I_cAB . Prove that the 3 radical axises of
          > (W_a)
          > and (W_b), (W_b) and (W_c), (W_c) and (W_a) and the Euler line of ABC
          > are concurent. What is its barycentric coordinate ?

          *** You are right. Let A_b and A_c be the pedal (orthogonal projections) of
          A on I_aI_b and I_aI_c respectively, then the radical axis of (W_b) and
          (W_c) is the same as the Euler line
          of AA_bA_c. From ETC, this line and the two analogous line interesect at
          X(442), which has barycentric coordinates

          ((b+c)(a^2(b+c)+2abc-(b+c)(b-c)^2) : ... : ...).

          This same point also appears in a recent message of Juan Carlos Salazar
          (Hyacinthos 10323). See Darij's 10342.

          Best regards
          Sincerely
          Paul


          > Thanks.
          > Best regards,
          > Khoa Lu Nguyen.
          >
          >
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        • Darij Grinberg
          Dear Khoa Lu, Jean-Pierre and Paul, ... I don t have a synthetic proof, but I have some remarks. After Theorem F2 of my Hyacinthos message #10584 (with
          Message 4 of 4 , Oct 3, 2004
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            Dear Khoa Lu, Jean-Pierre and Paul,

            In Hyacinthos message #10563, Khoa Lu Nguyen wrote:

            >> Let ABC be a triangle and I_a, I_b, I_c are
            >> excenters,
            >> Let (W_a), (W_b), (W_c) are respectively the
            >> nine point circles of I_aBC, I_bCA, I_cAB .
            >> Prove that the 3 radical axises of (W_a) and
            >> (W_b), (W_b) and (W_c), (W_c) and (W_a) and
            >> the Euler line of ABC are concurent.

            In Hyacinthos message #10565, Paul Yiu wrote:

            >> Let A_b and A_c be the pedal (orthogonal
            >> projections) of A on I_aI_b and I_aI_c
            >> respectively, then the radical axis of (W_b)
            >> and (W_c) is the same as the Euler line of
            >> AA_bA_c. From ETC, this line and the two
            >> analogous line interesect at X(442),

            I don't have a synthetic proof, but I have some
            remarks.

            After Theorem F2 of my Hyacinthos message #10584
            (with modified notations), if I is the incenter of
            triangle ABC, and A'b and A'c are the orthogonal
            projections of the point A on the lines CI and BI,
            respectively, then the Euler line of triangle
            AA'bA'c is the radical axis of the nine-point
            circles of triangles CIA and AIB.

            After application of A-extraversion, this result
            takes the following form: With Ab and Ac being the
            orthogonal projections of the point A on the lines
            CIa and BIa, respectively, the Euler line of
            triangle AAbAc is the radical axis of the
            nine-point circles of triangles CIaA and AIaB.

            Meantime Paul Yiu's Hyacinthos message #10565
            cited above shows that the Euler line of triangle
            AAbAc is the radical axis of the nine-point
            circles of triangles CIbA and AIcB.

            Hence, the radical axis of the nine-point circles
            of triangles CIaA and AIaB coincides with the
            radical axis of the nine-point circles of
            triangles CIbA and AIcB. On the other hand, one
            can easily see synthetically that the radical
            axis of the nine-point circles of triangles CIaA
            and CIbA coincides with the radical axis of the
            nine-point circles of triangles AIaB and AIcB
            (in fact, both times, the radical axis is the
            line joining the midpoints of the segments CA
            and AB). Hence, the nine-point circles x, y, z,
            w of the triangles CIaA, AIaB, AIcB and CIbA
            form a quadruple of circles with the property
            that the radical axis of the circles x and y
            coincides with the radical axis of the circles
            z and w, and that the radical axis of the
            circles y and z coincides with the radical axis
            of the circles w and x.

            Finally, note that the Euler line of triangle
            AAbAc is not only the radical axis of two
            pairs of nine-point circles, but it has also,
            at least, three interesting points lying on
            it:

            - the midpoint of the segment AIa;
            - the A-Feuerbach point of triangle ABC;
            - the complement of the Schiffler point of
            triangle ABC (this is X(442)).

            Thanks to all of you for this wonderful
            problem. Note that I did not consider the
            relations of this with Juan Carlos' problem in
            Hyacinthos message #10342, since this would be
            too much work (the problem is like a hydra:
            each time one proofs one part of it, three new
            conjectures turn up).

            Sincerely,
            Darij Grinberg
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