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Re: A locus group

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  • xpolakis
    ... I think that the lines O aHa, O bHb, O cHc are parallels to AI,BI,CI, resp. through the vertices A ,B ,C of the cevian triangle of the isotomic conjugate
    Message 1 of 3 , Sep 7, 2008
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      --- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:

      > Let ABC be a triangle, Oa,Ob,Oc and Ha,Hb,Hc the
      > circumcenters, orthocenters, resp. Of IBC,ICA,IAB, resp.
      > and O'a,O'b,O'c the reflections of Oa,Ob,Oc
      > in BC,CA,AB, resp.
      >
      > Are the lines O'aHa, O'bHb, O'cHc concurrent?

      I think that the lines O'aHa, O'bHb, O'cHc
      are parallels to AI,BI,CI, resp.
      through the vertices A',B',C' of the
      cevian triangle of the isotomic conjugate of I.

      Anyway, we have a locus problem:

      Let ABC be a triangle, P a point, P*
      its isotomic conjugate, and A1B1C1, A2B2C2
      their cevian triangles.

      Let La,Lb,Lc be the parallels to AA1, BB1, CC1
      through A2,B2,C2, resp.

      Which is the locus of P such that La,Lb,Lc are
      concurrent? The whole plane??

      APH
    • jan.vonk
      Dear Antreas and other Hyacinthists, this configuration is very interesting, but huge amounts of results are really popping out when I slightly modify your
      Message 2 of 3 , Sep 7, 2008
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        Dear Antreas and other Hyacinthists,

        this configuration is very interesting, but huge amounts of results
        are really popping out when I slightly modify your idea. See below.
        First, your conjectures:

        > Let ABC be a triangle, Oa,Ob,Oc and Ha,Hb,Hc the
        > circumcenters, orthocenters, resp. Of IBC,ICA,IAB, resp.
        > and O'a,O'b,O'c the reflections of Oa,Ob,Oc
        > in BC,CA,AB, resp.
        >
        > Are the lines O'aHa, O'bHb, O'cHc concurrent?

        Yes, they are. As you already pointed out, O'aHa and OaI are parallel.

        To see why this is so, one could for example observe that IHa =
        2R.cos(BIC), where R is the circumradius of BIC.
        On the other hand, we can easiley see that Oa is the midpoint of arc
        BC in the circumcircle of ABC. It is well known that OaB = OaC = OaI =
        R, so we can now quickly see that indeed OaO'a = 2R.cos(BIC) = IHa.
        Because OaO'a and IHa are parallel, the conclusion follows.

        The point of concurrency in the above problem is the Nagel point of ABC.

        > Let ABC be a triangle, P a point, P*
        > its isotomic conjugate, and A1B1C1, A2B2C2
        > their cevian triangles.
        >
        > Let La,Lb,Lc be the parallels to AA1, BB1, CC1
        > through A2,B2,C2, resp.
        >
        > Which is the locus of P such that La,Lb,Lc are
        > concurrent? The whole plane??

        Indeed! The lines are concurrent at the anticomplement of P, which is
        obvious if one considers the anticomplementary triangle A'B'C'.
        La,Lb,Lc clearly pass through A',B',C' and the conclusion follows.
        This applies in the above case, the Nagel point is the anticomplement
        of I.

        After checking all your cases, only the following are whole plane
        concurrent:

        > 1. O'aHa, O'bHb, O'cHc

        Concurrent at the Nagel point

        > 2. O'aGa, O'bGb, O'cGc

        Concurrent! I have no idea at which point though.

        > 3. O'aNa, O'bNb, O'cNc

        Concurrent at I.
        It's easy to see that O'a is the circumcenter of BCHa. This triangle
        however has Na as it's nine point center, so O'aNa is the Euler line
        of BCHa, which passes through it's orthocenter I.

        > 10. N'aOa, N'bOb, N'cOc

        Yes! Very nice, these three lines are concurrent at the inverse of I
        in the circumcircle.

        _______________________________________________________

        So 4 of your 12 triples were concurrent, that's a nice score!
        But I think I can do better ;-)

        Why, instead of considering the reflections of these centers in AB,BC
        and AC, not considering the reflections of these centers in the
        MIDPOINTS of AB,BC,AC.

        So for example H'a,H'b,H'c are the reflections of Ha,Hb,Hc in the
        midpoints of BC,AC,AB.

        I think if we do this, ALL the mentioned lines are concurrent.
        Some special interesting cases:

        4) H'aOa, H'bOb, H'cOc

        H'a,H'b,H'c are the excenters of ABC in this case, so the point of
        concurrence is I.

        10) N'aOa, N'bOb, N'cOc

        The point of concurrency is the anticomplement of the Feuerbach point

        _______________________________________________________

        Further questions:

        - In general, what can be said about the 12 points of concurrency we
        met just now?

        - I strongly conjecture the following, is it true?

        Let ABC be a triangle and P a point on the circumcircle or Neuberg
        cubic of ABC. Call Oa,Ob,Oc and Ha,Hb,Hc the circumcenters and
        orthocenters of BPC,APC,APB.
        Define Sa(k1),Sb(k1),Sc(k1) and Ta(k2),Tb(k2),Tc(k2) as the points on
        OaHa, ObHb, OcHc which divide (OiHi) (i=a,b,c) in the ratio k1 and k2
        respectively.
        Furthermore, call S'a(k1),S'b(k1),S'c(k1) and T'a(k2),T'b(k2),T'c(k2)
        their reflections in the midpoints of BC,AC,AB.

        S'a(k1)Ta(k2), S'b(k1)Tb(k2), S'c(k1)Tb(k2) are concurrent.
        Sa(k1)T'a(k2), Sb(k1)T'b(k2), Sc(k1)T'b(k2) are concurrent.

        - What can be said about the point of concurrency in the above theorem
        (if true)?

        Best regards,
        Jan
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