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Re: [EMHL] Problem

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  • Francois Rideau
    Dear Alexei I think it is only a matter about product of central symmetries along each sides of ABC. For example, look at the side BC. Let A be the middle of
    Message 1 of 18 , May 15, 2007
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      Dear Alexei
      I think it is only a matter about product of central symmetries along each
      sides of ABC.
      For example, look at the side BC. Let A' be the middle of segment BC and L
      be the orthogonal projection on side BC of the orthocenter H.
      Let s(A') and s(L) be the central symmetry respectively wrt A' and L .
      Then A2 = s(L).s(A').s(L) (A1) = s(U) (A1)
      where S(U) is the central symmetry wrt U = s(L)(A') and we are done.
      Friendly
      François


      On 5/14/07, Алексей Мякишев <alex_geom@...> wrote:
      >
      > Dear friends!
      > One construction:
      > Given tiangle ABC. On BC line we mark two points: A1 and A2, so that
      > AA1=AB and AA2=AC. And perpendicylar bissector to that segment.
      > So we do with two others sides.
      > Point of intersection is symmetry with center O of H.
      > Is it obvious?
      > Best regards,
      > Alexei
      >
      > [Non-text portions of this message have been removed]
      >
      >
      >


      [Non-text portions of this message have been removed]
    • Nikolaos Dergiades
      Dear Alexei ... Perhaps you have a typo. A1 is the symmetric of B wrt the altitude AH A2 is the symmetric of C wrt the altitude AH The segment A1A2 is the
      Message 2 of 18 , May 15, 2007
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        Dear Alexei

        > One construction:
        > Given tiangle ABC. On BC line we mark two points:
        > A1 and A2, so that AA1=AB and AA2=AC. And
        > perpendicylar bissector to that segment.
        > So we do with two others sides.
        > Point of intersection is symmetry with center O of
        > H.
        > Is it obvious?

        Perhaps you have a typo.
        A1 is the symmetric of B wrt the altitude AH
        A2 is the symmetric of C wrt the altitude AH
        The segment A1A2 is the symmetric of BC
        wrt the altitude AH and hence the perpendicular
        bisector of BC that passes through O
        is symmetric to the perpendicular bisector of A1A2
        that passes through the symmetric of O with center H.

        Best regards
        Nikos Dergiades






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      • Huub van Kempen
        Dear Hyacinthos, Given a circle (M,r) and two lines l and m outside the circle. The points A and B lie on the circumference of the circle. Question: To
        Message 3 of 18 , May 2, 2008
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          Dear Hyacinthos,

          Given a circle (M,r) and two lines l and m outside the circle.
          The points A and B lie on the circumference of the circle.
          Question: To construct the points C and D on the circumference so that
          AC and BD intersect on line l and CD is parallel to line m.

          Regards,

          Huub van Kempen
          _________________________________________________________________
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        • jpehrmfr
          Dear Huub van Kempen ... [UV] is the diameter of the circle perpendicular to m. If CD is a variable chord parallel to m, the locus of the common point of AC
          Message 4 of 18 , May 2, 2008
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            Dear Huub van Kempen

            > Given a circle (M,r) and two lines l and m outside the circle.
            > The points A and B lie on the circumference of the circle.
            > Question: To construct the points C and D on the circumference so that
            > AC and BD intersect on line l and CD is parallel to line m.

            [UV] is the diameter of the circle perpendicular to m.
            If CD is a variable chord parallel to m, the locus of the common point
            of AC and BD is the rectangular hyperbola through A,B,U,V (the center
            of the hyperbola is the midpoint of AB)
            So you have just to find the common points of l with the hyperbola
            above.
            Friendly. Jean-Pierre
          • maedu@hotmail.com
            If you put C on circle, and a parallel to m by C, then the locus of the intersection of AC and BD is a hyperbole passing by A, B, and the points of contact of
            Message 5 of 18 , May 2, 2008
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              If you put C on circle, and a parallel to m by C, then the locus of the intersection of AC and BD is a hyperbole passing by A, B, and the points of contact of two parallels to m tangents to the circle.
              The points of intersection of this hyperbola with line l are the solutions.

              Martin


              From: Huub van Kempen
              Sent: Friday, May 02, 2008 11:04 AM
              To: hyacinthos@yahoogroups.com
              Subject: [EMHL] Problem


              Dear Hyacinthos,

              Given a circle (M,r) and two lines l and m outside the circle.
              The points A and B lie on the circumference of the circle.
              Question: To construct the points C and D on the circumference so that
              AC and BD intersect on line l and CD is parallel to line m.

              Regards,

              Huub van Kempen
              __________________________________________________________
              Probeer Live Search: de zoekmachine van de makers van MSN!
              http://www.live.com/?searchOnly=true

              [Non-text portions of this message have been removed]





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