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Conjectures (was: F.G. - M.)

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  • xpolakis@xxxxxx.xxxxxxxxxxxxx.xxxxxxxxxx
    ... The conjecture was: Let ABC be a triangle. From A we draw parallels to b- and c- altitudes intersecting the BC at Ba, Ca respectively. Similarly, we define
    Message 1 of 5 , Jan 2, 2000
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      John Conway wrote:

      > I'm afraid I did not return to your conjectures yesterday. Instead,
      >I've filled all the available space in my copy of today's New York Times
      >with calculations in an attempt to verify Steve's assertion that the
      >ex- and extra- Gergonne triangles are in perspective.

      The conjecture was:

      Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
      intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
      and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
      are concurrent.

      Synthetically, we can prove it by finding a point which belongs to
      each one of the lines. But this is not always easy! Sometimes it is, as
      in the following example:
      Let A'B'C' be the orthic of a given triangle ABC. From B', C' draw
      oaralles to the lines AC'B, AB'C, respectively, intersecting at A".
      Similarly define B", C". Then the Euler lines of the triangles ABC,
      A"B'C', B"C'A', C"A'B' are concurrent.
      They are concurrent because the four triangles share the same orthocenter.
      But which is the analogous common point in the conjecture?

      Analytically, using trilinears, we can prove it this way:

      Let H, H_a, H_b, H_c be the orthocenters, and O, O_a, O_b, O_c the
      circumcenters of the triangles ABC, ABaCa, BCbAb, CBcAc, respectively.

      Trilinears of Ha, Hb, Hc (in respect to triangle ABC):

      (-sinBsinC : cosAcosC : cosAcosB)

      (-sinCsinA : cosBcosA : cosBcosC)

      (-sinAsinB : cosCcosB : cosCcosA)


      Trilinears of Oa, Ob, Oc (in respect to triangle ABC):

      (-cosA : cosB : cosC)

      (cosA : -cosB : cosC)

      (cosA : cosB : -cosC)


      Notation:
      sinA := D, cosA := d; sinB := E, cosB := e; sinC := F, cosC := f

      Trilinears of Oa, Ha:

      (-EF : df : de)

      (-d : e : f)


      Equation of Line OaHa (Euler line of ABaCa):

      |df de| |-EF de| |-EF df|
      x*| | - y*| | + z*| | = 0
      |e f | |-d f| |-d f |


      Similarly, Euler lines of BCbAb, CBcAc:

      |ed ef| |-FD ef| |-FD ed|
      x*| | - y*| | + z*| | = 0
      |f d| |-e d| |-e d |


      |fe fd| |-DE fd| |-DE fe|
      x*| | - y*| | + z*| | = 0
      |d e| |-f e| |-f e |


      The three Euler lines concur if:


      | |df de| |-EF de| |-EF df| |
      | | | -| | | | |
      | |e f | |-d f| |-d f | |
      | |
      | |ed ef| |-FD ef| |-FD ed| |
      | | | -| | | | | = 0
      | |f d| |-e d| |-e d | |
      | |
      | |fe fd| |-DE fd| |-DE fe| |
      | | | -| | | | |
      | |d e| |-f e| |-f e | |


      Hmmm..... We need a whole month for all algebraic-trigonometrical
      calculations!

      > It occurred to me that the triangle got by reflecting ABC in its
      >altitudes might be as interesting as the ones your conjectures were
      >about.

      The three triangles got by reflecting ABC about its altitudes have
      the same orthocenter as the reference triangle.
      So the four Euler lines of these triangles concur.


      Antreas
    • wolkb@xx.xxxxxxxxx.xx
      ... According to my calculations, no three of those four lines are concurrent. I used Maple to do the messy algebra. Incidentally, onelist.com suppresses email
      Message 2 of 5 , Jan 4, 2000
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        Antreas wrote:

        > The conjecture was:
        >
        > Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
        > intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
        > and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
        > are concurrent.

        According to my calculations, no three of those four lines are concurrent.
        I used Maple to do the messy algebra.

        Incidentally, onelist.com suppresses email addresses in the headers of posted
        articles. I suggest that posters include their address in their sig-file.
        --
        Barry Wolk <wolkb@...>
      • xpolakis@xxxxxx.xxxxxxxxxxxxx.xxxxxxxxxx
        ... Thanks, Barry. Hope that the other conjecture (about the perpendicular bisectors of BaCa, CbAb, AcBc) is true! Antreas
        Message 3 of 5 , Jan 4, 2000
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          Barry Wolk wrote:

          >Antreas wrote:
          >
          >> The conjecture was:
          >>
          >> Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
          >> intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
          >> and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
          >> are concurrent.
          >
          >According to my calculations, no three of those four lines are concurrent.
          >I used Maple to do the messy algebra.


          Thanks, Barry.
          Hope that the other conjecture (about the perpendicular bisectors of
          BaCa, CbAb, AcBc) is true!

          Antreas
        • Paul Yiu
          Dear Antreas, As Barry just pointed out, these Euler lines do not concur. The same is true for the Brocard axes. However, your original conjecture [Message 3
          Message 4 of 5 , Jan 4, 2000
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            Dear Antreas,

            As Barry just pointed out, these Euler lines do not concur.
            The same is true for the Brocard axes. However, your
            original conjecture [Message 3 in Digest 3, Saturday,
            26/12/99] is true.

            The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.

            The homogeneous barycentric coordinates of the intersection
            is the point

            a^2(a^8-4a^6(b^2+c^2)
            +2a^4(3b^4-b^2c^2+3c^4)
            -4a^2(b^2-c^2)^2(b^2+c^2)
            +(b^2-c^2)^2(b^4+6b^2c^2+c^4))
            : ... : ...

            In P-perpendicularity, the coordinates are
            much more elegant, simply

            f(g+h)(g^2+h^2-f^2) : ... : ...

            Best regards.
            Sincerely,
            Paul

            ----------
            From: Antreas P. Hatzipolakis[SMTP:xpolakis@...]
            Reply To: Hyacinthos@onelist.com
            Sent: Sunday, January 02, 2000 8:43 AM
            To: Hyacinthos@onelist.com
            Subject: [EMHL] Conjectures (was: F.G. - M.)

            From: xpolakis@... (Antreas P. Hatzipolakis)

            John Conway wrote:

            > I'm afraid I did not return to your conjectures yesterday. Instead,
            >I've filled all the available space in my copy of today's New York Times
            >with calculations in an attempt to verify Steve's assertion that the
            >ex- and extra- Gergonne triangles are in perspective.

            The conjecture was:

            Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
            intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
            and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
            are concurrent.

            Synthetically, we can prove it by finding a point which belongs to
            each one of the lines. But this is not always easy! Sometimes it is, as
            in the following example:
            Let A'B'C' be the orthic of a given triangle ABC. From B', C' draw
            oaralles to the lines AC'B, AB'C, respectively, intersecting at A".
            Similarly define B", C". Then the Euler lines of the triangles ABC,
            A"B'C', B"C'A', C"A'B' are concurrent.
            They are concurrent because the four triangles share the same orthocenter.
            But which is the analogous common point in the conjecture?

            Analytically, using trilinears, we can prove it this way:

            Let H, H_a, H_b, H_c be the orthocenters, and O, O_a, O_b, O_c the
            circumcenters of the triangles ABC, ABaCa, BCbAb, CBcAc, respectively.

            Trilinears of Ha, Hb, Hc (in respect to triangle ABC):

            (-sinBsinC : cosAcosC : cosAcosB)

            (-sinCsinA : cosBcosA : cosBcosC)

            (-sinAsinB : cosCcosB : cosCcosA)


            Trilinears of Oa, Ob, Oc (in respect to triangle ABC):

            (-cosA : cosB : cosC)

            (cosA : -cosB : cosC)

            (cosA : cosB : -cosC)


            Notation:
            sinA := D, cosA := d; sinB := E, cosB := e; sinC := F, cosC := f

            Trilinears of Oa, Ha:

            (-EF : df : de)

            (-d : e : f)


            Equation of Line OaHa (Euler line of ABaCa):

            |df de| |-EF de| |-EF df|
            x*| | - y*| | + z*| | = 0
            |e f | |-d f| |-d f |


            Similarly, Euler lines of BCbAb, CBcAc:

            |ed ef| |-FD ef| |-FD ed|
            x*| | - y*| | + z*| | = 0
            |f d| |-e d| |-e d |


            |fe fd| |-DE fd| |-DE fe|
            x*| | - y*| | + z*| | = 0
            |d e| |-f e| |-f e |


            The three Euler lines concur if:


            | |df de| |-EF de| |-EF df| |
            | | | -| | | | |
            | |e f | |-d f| |-d f | |
            | |
            | |ed ef| |-FD ef| |-FD ed| |
            | | | -| | | | | = 0
            | |f d| |-e d| |-e d | |
            | |
            | |fe fd| |-DE fd| |-DE fe| |
            | | | -| | | | |
            | |d e| |-f e| |-f e | |


            Hmmm..... We need a whole month for all algebraic-trigonometrical
            calculations!

            > It occurred to me that the triangle got by reflecting ABC in its
            >altitudes might be as interesting as the ones your conjectures were
            >about.

            The three triangles got by reflecting ABC about its altitudes have
            the same orthocenter as the reference triangle.
            So the four Euler lines of these triangles concur.


            Antreas

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          • Paul Yiu
            Dear Antreas, I should have remarked a simple geometric property of this intersection of the ``first perpendicular bisectors of the orthial triangles. Its
            Message 5 of 5 , Jan 4, 2000
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              Dear Antreas,

              I should have remarked a simple geometric
              property of this intersection of the ``first''
              perpendicular bisectors of the orthial triangles.
              Its distances from the vertices A, B, C are in
              the proportions of

              cot A : cot B : cot C.

              This follows from a remark in your second message
              [12/26/99] that

              The "bases" BaCa, CbAb, BcAc of these triangles are equal to:

              atanBtanC, btanCtanA, ctanAtanB.

              The circumradii of these triangles are therefore

              (a tan B tan C)/(sin (pi-A)) = 2R tan B tan C

              etc,

              This result has generalizations. I shall write about this
              later.

              Best regards.
              Sincerely,
              Paul
              ----------
              From: Paul Yiu[SMTP:yiu@...]
              Reply To: Hyacinthos@onelist.com
              Sent: Tuesday, January 04, 2000 3:18 PM
              To: 'Hyacinthos@onelist.com'
              Subject: RE: [EMHL] Conjectures (was: F.G. - M.)

              Dear Antreas,

              As Barry just pointed out, these Euler lines do not concur.
              The same is true for the Brocard axes. However, your
              original conjecture [Message 3 in Digest 3, Saturday,
              26/12/99] is true.

              The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.

              The homogeneous barycentric coordinates of the intersection
              is the point

              a^2(a^8-4a^6(b^2+c^2)
              +2a^4(3b^4-b^2c^2+3c^4)
              -4a^2(b^2-c^2)^2(b^2+c^2)
              +(b^2-c^2)^2(b^4+6b^2c^2+c^4))
              : ... : ...

              In P-perpendicularity, the coordinates are
              much more elegant, simply

              f(g+h)(g^2+h^2-f^2) : ... : ...

              Best regards.
              Sincerely,
              Paul

              ----------
              From: Antreas P. Hatzipolakis[SMTP:xpolakis@...]
              Reply To: Hyacinthos@onelist.com
              Sent: Sunday, January 02, 2000 8:43 AM
              To: Hyacinthos@onelist.com
              Subject: [EMHL] Conjectures (was: F.G. - M.)

              From: xpolakis@... (Antreas P. Hatzipolakis)

              John Conway wrote:

              > I'm afraid I did not return to your conjectures yesterday. Instead,
              >I've filled all the available space in my copy of today's New York Times
              >with calculations in an attempt to verify Steve's assertion that the
              >ex- and extra- Gergonne triangles are in perspective.

              The conjecture was:

              Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
              intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
              and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
              are concurrent.

              Synthetically, we can prove it by finding a point which belongs to
              each one of the lines. But this is not always easy! Sometimes it is, as
              in the following example:
              Let A'B'C' be the orthic of a given triangle ABC. From B', C' draw
              oaralles to the lines AC'B, AB'C, respectively, intersecting at A".
              Similarly define B", C". Then the Euler lines of the triangles ABC,
              A"B'C', B"C'A', C"A'B' are concurrent.
              They are concurrent because the four triangles share the same orthocenter.
              But which is the analogous common point in the conjecture?

              Analytically, using trilinears, we can prove it this way:

              Let H, H_a, H_b, H_c be the orthocenters, and O, O_a, O_b, O_c the
              circumcenters of the triangles ABC, ABaCa, BCbAb, CBcAc, respectively.

              Trilinears of Ha, Hb, Hc (in respect to triangle ABC):

              (-sinBsinC : cosAcosC : cosAcosB)

              (-sinCsinA : cosBcosA : cosBcosC)

              (-sinAsinB : cosCcosB : cosCcosA)


              Trilinears of Oa, Ob, Oc (in respect to triangle ABC):

              (-cosA : cosB : cosC)

              (cosA : -cosB : cosC)

              (cosA : cosB : -cosC)


              Notation:
              sinA := D, cosA := d; sinB := E, cosB := e; sinC := F, cosC := f

              Trilinears of Oa, Ha:

              (-EF : df : de)

              (-d : e : f)


              Equation of Line OaHa (Euler line of ABaCa):

              |df de| |-EF de| |-EF df|
              x*| | - y*| | + z*| | = 0
              |e f | |-d f| |-d f |


              Similarly, Euler lines of BCbAb, CBcAc:

              |ed ef| |-FD ef| |-FD ed|
              x*| | - y*| | + z*| | = 0
              |f d| |-e d| |-e d |


              |fe fd| |-DE fd| |-DE fe|
              x*| | - y*| | + z*| | = 0
              |d e| |-f e| |-f e |


              The three Euler lines concur if:


              | |df de| |-EF de| |-EF df| |
              | | | -| | | | |
              | |e f | |-d f| |-d f | |
              | |
              | |ed ef| |-FD ef| |-FD ed| |
              | | | -| | | | | = 0
              | |f d| |-e d| |-e d | |
              | |
              | |fe fd| |-DE fd| |-DE fe| |
              | | | -| | | | |
              | |d e| |-f e| |-f e | |


              Hmmm..... We need a whole month for all algebraic-trigonometrical
              calculations!

              > It occurred to me that the triangle got by reflecting ABC in its
              >altitudes might be as interesting as the ones your conjectures were
              >about.

              The three triangles got by reflecting ABC about its altitudes have
              the same orthocenter as the reference triangle.
              So the four Euler lines of these triangles concur.


              Antreas




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