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More Orth. Proj. of H (was: IMO 2004 Problem 1 revisited)

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  • Antreas P. Hatzipolakis
    Let ABC be a triangle, P a point, A , B , C the orth. projections of H on AP, BP, CP, resp. and Ea, Eb, Ec the midpoints of AH, BH, CH (ie EaEbEc = the Euler
    Message 1 of 11 , Jul 31, 2004
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      Let ABC be a triangle, P a point, A', B', C' the orth.
      projections of H on AP, BP, CP, resp. and Ea, Eb, Ec
      the midpoints of AH, BH, CH (ie EaEbEc = the Euler
      triangle of ABC)

      Which is the locus of P such that A'B'C', EaEbEc
      are perspective? (Locus of the perspectors?)

      I think that I, N are points of the locus.

      APH
      --
    • Darij Grinberg
      Dear Antreas and Bernard, Thanks for another surprising perspector! In Hyacinthos message #10184, Bernard Gibert ... I think that the perspector of the
      Message 2 of 11 , Aug 1, 2004
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        Dear Antreas and Bernard,

        Thanks for another surprising perspector!

        In Hyacinthos message #10184, Bernard Gibert
        wrote:

        >> > [APH] Let Ra, Rb, Rc be the orthogonal
        >> > projections of H on the cevians of I, [ie
        >> > on the angle bisectors of A,B,C, resp.]
        >> > and SaSbSc the pedal triangle of I (aka
        >> > intouch triangle of ABC).
        >> >
        >> > Are the triangles RaRbRc, SaSbSc perspective?
        >>
        >> They are indeed at : [...]

        I think that the perspector of the triangles
        RaRbRc and SaSbSc is the point X(847) of the
        triangle SaSbSc (i. e. of the intouch triangle).

        In fact, if we rewrite the problem from the
        viewpoint of triangle SaSbSc, we get the
        following:

        Let ABC be a triangle. Further, let H' be the
        orthocenter of the tangential triangle of
        triangle ABC, and let D, E, F be the
        orthogonal projections of the point H' on the
        perpendicular bisectors of the segments BC,
        CA, AB. Then show that the lines AD, BE, CF
        are concurrent at the point X(847) of
        triangle ABC.

        In fact, by the definition of X(847), the
        line AX(847) passes through the point
        BAb /\ CAc, where the point Ab is the point
        of intersection of the A-altitude of
        triangle ABC with the perpendicular
        bisector of the side CA, and the point Ac
        is is the point of intersection of the
        A-altitude with the perpendicular bisector
        of the side AB. Thus, the lines
        AD = AX(847), BAb and CAc are concurrent.

        In fact, it turns out (conjecture!) that the
        following four lines are concurrent:

        AD, BAb, CAc and X(5)X(6)X(68)X(155).

        I would be glad to see a proof of this (BTW,
        I know the synthetic proof of the
        collinearity of the points X(5), X(6), X(68),
        X(155), so you can define the line
        X(5)X(6)X(68)X(155) by any two of them).

        Sincerely,
        Darij Grinberg
      • Antreas P. Hatzipolakis
        ... In my acute-angled triangle figure, the Euler lines of BCRa, CARb, ABRc are concurrent. Is it true ? APH --
        Message 3 of 11 , Aug 1, 2004
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          >[APH] Let Ra, Rb, Rc be the orthogonal
          >projections of H on the cevians of I [ie
          >on the angle bisectors of A,B,C, resp.]

          In my acute-angled triangle figure, the Euler lines
          of BCRa, CARb, ABRc are concurrent.

          Is it true ?

          APH
          --
        • Darij Grinberg
          Dear Antreas, ... Good idea... but unfortunately, zooming in on my dynamic geometry sketch shows that they are not exactly concurrent... Friendly, Darij
          Message 4 of 11 , Aug 1, 2004
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            Dear Antreas,

            In Hyacinthos message #10188, you wrote:

            >> In my acute-angled triangle figure, the
            >> Euler lines of BCRa, CARb, ABRc are concurrent.

            Good idea... but unfortunately, zooming in on my
            dynamic geometry sketch shows that they are not
            exactly concurrent...

            Friendly,
            Darij Grinberg
          • Paul Yiu
            Dear Antreas, [APH]: Let ABC be a triangle, P a point, A , B , C the orth. projections of H on AP, BP, CP, resp. and Ea, Eb, Ec the midpoints of AH, BH, CH
            Message 5 of 11 , Aug 2, 2004
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              Dear Antreas,

              [APH]: Let ABC be a triangle, P a point, A', B', C' the orth.
              projections of H on AP, BP, CP, resp. and Ea, Eb, Ec
              the midpoints of AH, BH, CH (ie EaEbEc = the Euler
              triangle of ABC)

              Which is the locus of P such that A'B'C', EaEbEc
              are perspective? (Locus of the perspectors?)

              I think that I, N are points of the locus.


              *** Yes, the locus the quintic

              cyclic sum a^2(S_{CA}+S_{AB}+2S_{BC})v^2w^2(S_Bv - S_Cw)
              + cyclic uvw S_A(S_B-S_C)(b^2c^2u^2- a^2S_Avw) = 0.

              It also contains H (trivially), X_{80}, and X_{1263}. Note that X(80)
              and X(1263) are respectively the reflection conjugates of I and N
              respectively.

              I have verified that the quintic is indeed invariant reflection
              conjugation. [The reflection conjugate of a point P is the common
              point of the reflections of the circles PBC, PCA, PAB in the sidelines
              BC, CA, AB respectively].

              Best regards
              Sincerely
              Paul
            • Antreas P. Hatzipolakis
              Dear Darij ... Hmmmm...... Apo idees .... allo tipota!! [free translation: I am full of ideas!!] Here is another one (idea, I mean): What point is (where is
              Message 6 of 11 , Aug 2, 2004
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                Dear Darij

                >[APH] Let Ra, Rb, Rc be the orthogonal
                >projections of H on the cevians of I [ie
                >on the angle bisectors of A,B,C, resp.]
                >In my acute-angled triangle figure, the Euler lines
                >of BCRa, CARb, ABRc are concurrent.

                [DG]:
                >Good idea... but unfortunately, zooming in on my
                >dynamic geometry sketch shows that they are not
                >exactly concurrent...


                Hmmmm...... Apo idees .... allo tipota!!
                [free translation: I am full of ideas!!]

                Here is another one (idea, I mean):

                What point is (where is lying on?) the Radical Center of the
                circumcircles of the triangles bounded by the lines:
                (AB,AC, parallel - to - BC - through - Ra)
                (BC,BA, parallel - to - CA - through - Rb)
                (CA,CB, parallel - to - AB - through - Rc) ?

                Now, let Ta, Tb, Tc be three points on the angle bisectors
                AI, BI, CI such that:

                ITa / IRa = ITb / IRb = ITc / IRc = t

                As t varies, which is the locus of the Radical Center of the
                circumcircles of the triangles bounded by the lines:

                (AB,AC, parallel - to - BC - through - Ta)
                (BC,BA, parallel - to - CA - through - Tb)
                (CA,CB, parallel - to - AB - through - TC) ?

                Greetings

                Antreas



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