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Re: [EMHL] Re: Perspector

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  • Darij Grinberg
    Hi Treegoner, Nice to see you at Hyacinthos! ... Actually, they are not. X, Y, Z are defined as the touch points of (I_a), (I_b), (I_c) with the nine-point
    Message 1 of 6 , Sep 4, 2004
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      Hi Treegoner,

      Nice to see you at Hyacinthos!

      In Hyacinthos message #10379, you wrote:

      >> If I 'm not mistaken, X, Y, Z are the touch
      >> points of (I_a) and BC, (I_b) and CA, (I_c)
      >> and AB.

      Actually, they are not. X, Y, Z are defined as
      the touch points of (I_a), (I_b), (I_c) with
      the nine-point circle of triangle ABC.

      >> Then I think P can't lie on the Euler line
      >> of ABC.
      >>
      >> Because I did prove some properties :
      >>
      >> 1) UVW and I_aI_bI_c are homothetic. Then
      >> their two Euler lines must be parallel
      >>
      >> 2) UVW and ABC are perspective at a point
      >> H which is the circumcenter of UVW.
      >>
      >> 3) UVW and XYZ are perspective at P which
      >> is the orthocenter of UVW.

      In fact, what you are speaking about are nice
      facts, however not directly related to Juan
      Carlos' perspector, since your points X, Y, Z
      are not his points X, Y, Z. You can find some
      other properties of your configuration in
      Hyacinthos message #6729.

      Sincerely,
      Darij Grinberg
    • KHOA LU
      Dear friends, You re right Darij. I misunderstood the problem because Carlos wrote NPC which disconcerted me. But anyway, I have an idea. 1) If you call A ,
      Message 2 of 6 , Sep 4, 2004
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        Dear friends,

        You 're right Darij. I misunderstood the problem because Carlos wrote NPC which disconcerted me.

        But anyway, I have an idea.

        1) If you call A', B' be respectively the projections of I_a, and I_b on AB, then U, V, A', B', X, Y lie on the same circle.

        2) The orthocenter of ABC is the circumcenter of UVW (my previous post)

        From 1) you 'll obtain XYZ and UVW are perspective at P which lies on the line joining the two circumcenters of ABC and UVW which is the Euler line of ABC (from 2)).

        In fact, I get stuck with the first one. I spent 3 hours proving that but I didn't make it out. You can just only prove UVXY is cyclic, but I just elaborate two more points A', B'. I 'm quite sure of it (90%) although I do not have a solution to it. I always believe in "my sixth sense of geometry" though sometimes they are wrong. From this proof (if it 's right) you 'll get the equality I think interesting :

        Call A_1, A_2 be the projections of I_a on AB, AC, B_1, B_2 be the projections of I_b on BC, AB, C_1, C_2 be the projections of I_a on AC, BC. Then

        XA_1 . YB_1 . ZC_1 = XA_2 . YB_2 . ZC_2.

        I 'm tired now. I 'll try it again tomorrow. I hope someone will find it out.

        Treegoner

        PS Hey Darij, how can you remember the number of the point X(?) in ETC ? See you.

        Darij Grinberg <darij_grinberg@...> wrote: Hi Treegoner,

        Nice to see you at Hyacinthos!

        In Hyacinthos message #10379, you wrote:

        >> If I 'm not mistaken, X, Y, Z are the touch
        >> points of (I_a) and BC, (I_b) and CA, (I_c)
        >> and AB.

        Actually, they are not. X, Y, Z are defined as
        the touch points of (I_a), (I_b), (I_c) with
        the nine-point circle of triangle ABC.

        >> Then I think P can't lie on the Euler line
        >> of ABC.
        >>
        >> Because I did prove some properties :
        >>
        >> 1) UVW and I_aI_bI_c are homothetic. Then
        >> their two Euler lines must be parallel
        >>
        >> 2) UVW and ABC are perspective at a point
        >> H which is the circumcenter of UVW.
        >>
        >> 3) UVW and XYZ are perspective at P which
        >> is the orthocenter of UVW.

        In fact, what you are speaking about are nice
        facts, however not directly related to Juan
        Carlos' perspector, since your points X, Y, Z
        are not his points X, Y, Z. You can find some
        other properties of your configuration in
        Hyacinthos message #6729.

        Sincerely,
        Darij Grinberg


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      • Darij Grinberg
        Dear Treegoner, ... That s right. Juan Carlos has communicated this result to me. However, I have no synthetic proof yet. ... Indeed, this follows from a
        Message 3 of 6 , Sep 6, 2004
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          Dear Treegoner,

          In Hyacinthos message #10384, you wrote:

          >> 1) If you call A', B' be respectively the
          >> projections of I_a, and I_b on AB, then U,
          >> V, A', B', X, Y lie on the same circle.

          That's right. Juan Carlos has communicated
          this result to me. However, I have no
          synthetic proof yet.

          >> 2) The orthocenter of ABC is the
          >> circumcenter of UVW

          Indeed, this follows from a simple angle
          chase once it is shown that the orthocenter
          of triangle ABC lies on the lines AU, BV and
          CW.

          >> From 1) you 'll obtain XYZ and UVW are
          >> perspective at P which lies on the line
          >> joining the two circumcenters of ABC and
          >> UVW which is the Euler line of ABC
          >> (from 2)).

          Oh, that's nice! In fact, I see what you do:
          From 1), the points V, W, Y, Z lie on one
          circle; similarly, the points W, U, Z, X lie
          on one circle, and similarly, the points U,
          V, X, Y lie on one circle. The pairwise
          radical axes of these circles are the lines
          XU, YV and ZW, and hence these lines concur
          at one point, the radical center P of these
          three circles. And thus, XP * PU = YP * PV
          = ZP * PW is the common power of this point
          P with respect to every of these three
          circles. But the equation XP * PU = YP * PV
          = ZP * PW can be also rewritten as
          PX * PU = PY * PV = PZ * PW, and this
          yields that there exists an inversion (with
          real radius or with imaginary radius) which
          maps the points X, Y, Z to the points U, V,
          W. And now one readily sees that the
          circumcenters of triangles XYZ and UVW lie
          on one line with the point P. But the
          circumcenter of triangle XYZ is the
          nine-point center of triangle ABC, and the
          circumcenter of triangle UVW is the
          orthocenter of triangle ABC. Hence, the
          point P lies on the Euler line of triangle
          ABC.

          >> Call A_1, A_2 be the projections of I_a
          >> on AB, AC, B_1, B_2 be the projections of
          >> I_b on BC, AB, C_1, C_2 be the
          >> projections of I_a on AC, BC. Then
          >>
          >> XA_1 . YB_1 . ZC_1 = XA_2 . YB_2 . ZC_2.

          Thanks, a nice formula about the
          ubiquituous Feuerbach points...

          >> PS Hey Darij, how can you remember the
          >> number of the point X(?) in ETC ?

          Well, I don't remember anything :-)

          I just draw a triangle ABC with sides
          6, 9, 13, then draw the point of this
          triangle whose number (in the ETC) I want
          to find, then I measure its distance to
          the side BC of this triangle, and then I
          look up in the "Search" list of
          Kimberling's ETC:

          http://faculty.evansville.edu/ck6/encyclopedia/search.html

          For more details, see the MathLinks thread

          http://www.mathlinks.ro/Forum/viewtopic.php?t=4437

          where I explained the method to Grobber.

          Friendly,
          Darij Grinberg
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