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Re: [EMHL] Area Sum And Distance Are Constant Values

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  • Quang Tuan Bui
    Dear Francois and Petrisor, Thank you very much for your valuable advices and interesting remarks. All my results bellow are for orthogonal projection case: -
    Message 1 of 7 , Feb 7, 2007
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      Dear Francois and Petrisor,
      Thank you very much for your valuable advices and interesting remarks. All my results bellow are for orthogonal projection case:
      - Constant A1A2 = Abs(SA)*S^2*a*/(a^2*b^2*c^2)
      - Constant Area(BB1B2) + Area(CC1C2) = S^3/(2*b^2*c^2)
      (Abs(X) = Absolute value of X)
      Based on your remarks I make some notes as following:
      1. There are two points on sideline BC such that A1 = X and A2 = X. In this case, six points A1B1C1, A2B2C2 are on (as Francois remark) Second Lemoine Circle.
      2. There is one point on sideline BC such that Area(A1B1C1) = Area(A2B2C2) and another interesting facts as Petrisor remark.
      3. In Petrisor's case: if A3, A4 are two A-points on CA (when X taken on CA) and A5, A6 are two A-points on AB (when X taken on AB) and H is orthocenter of ABC then:
      - Area(HA1A2) = Area(HA3A4) = Area(HA5A6)
      - A1, A2, A3, A4, A5, A6 are on one conic.
      Thank you again and best regards,
      Bui Quang Tuan


      Neagoe Petrisor <pneagoe@...> wrote: Dear Quang Tuan Bui,

      In your problem, if X is orthogonal projection of A on BC
      then the perpendicular bisectors of A1A2, B1B2, C1C2 are
      concurrent in a point P. The point P lies on the A-symedian.

      Friendly,
      Petrisor Neagoe


      Quang Tuan Bui wrote:
      Dear All My Friends,
      Given triangle ABC and any point X moving on sideline BC.
      C1 = orthogonal projection of X on AB (projection f1)
      B1 = orthogonal projection of C1 on AC (projection f2)
      A1 = orthogonal projection of B1 on BC (projection f3)
      B2 = orthogonal projection of X on AC (projection f2)
      C2 = orthogonal projection of B2 on AB (projection f1)
      A2 = orthogonal projection of C2 on BC (projection f3)
      Please prove following facts:
      1. Area(BB1B2) + Area(CC1C2) = constant (area are calculated with direction)
      2. Distance A1A2 = constant
      If we use any other three affine projections (f1, f2, f3) instead of orthogonal then (1) is still true but distance in (2) is not constant.
      Dear All My Friends,
      Today there are exactly 500 members in our group.
      Best wishes to all our members, My Friends!
      Many thanks to Antreas Hatzipolakis, Our Group Founder!
      Best regards,
      Bui Quang Tuan


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    • Quang Tuan Bui
      Dear Francois and Petrisor, I am sorry for missing one value: 3. In Petrisor s case: - Area(HA1A2) = Area(HA3A4) = Area(HA5A6) = = S*SA*SB*SC/(2*a^2*b^2*c^2) =
      Message 2 of 7 , Feb 7, 2007
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        Dear Francois and Petrisor,
        I am sorry for missing one value:
        3. In Petrisor's case:
        - Area(HA1A2) = Area(HA3A4) = Area(HA5A6) =
        = S*SA*SB*SC/(2*a^2*b^2*c^2) = Area(ABC)*cos(A)*cos(B)*cos(C)
        Thank you and best regards,
        Bui Quang Tuan


        Quang Tuan Bui <bqtuan1962@...> wrote: Dear Francois and Petrisor,
        Thank you very much for your valuable advices and interesting remarks. All my results bellow are for orthogonal projection case:
        - Constant A1A2 = Abs(SA)*S^2*a*/(a^2*b^2*c^2)
        - Constant Area(BB1B2) + Area(CC1C2) = S^3/(2*b^2*c^2)
        (Abs(X) = Absolute value of X)
        Based on your remarks I make some notes as following:
        1. There are two points on sideline BC such that A1 = X and A2 = X. In this case, six points A1B1C1, A2B2C2 are on (as Francois remark) Second Lemoine Circle.
        2. There is one point on sideline BC such that Area(A1B1C1) = Area(A2B2C2) and another interesting facts as Petrisor remark.
        3. In Petrisor's case: if A3, A4 are two A-points on CA (when X taken on CA) and A5, A6 are two A-points on AB (when X taken on AB) and H is orthocenter of ABC then:
        - Area(HA1A2) = Area(HA3A4) = Area(HA5A6)
        - A1, A2, A3, A4, A5, A6 are on one conic.
        Thank you again and best regards,
        Bui Quang Tuan


        Neagoe Petrisor
        wrote: Dear Quang Tuan Bui,

        In your problem, if X is orthogonal projection of A on BC
        then the perpendicular bisectors of A1A2, B1B2, C1C2 are
        concurrent in a point P. The point P lies on the A-symedian.

        Friendly,
        Petrisor Neagoe


        Quang Tuan Bui wrote:
        Dear All My Friends,
        Given triangle ABC and any point X moving on sideline BC.
        C1 = orthogonal projection of X on AB (projection f1)
        B1 = orthogonal projection of C1 on AC (projection f2)
        A1 = orthogonal projection of B1 on BC (projection f3)
        B2 = orthogonal projection of X on AC (projection f2)
        C2 = orthogonal projection of B2 on AB (projection f1)
        A2 = orthogonal projection of C2 on BC (projection f3)
        Please prove following facts:
        1. Area(BB1B2) + Area(CC1C2) = constant (area are calculated with direction)
        2. Distance A1A2 = constant
        If we use any other three affine projections (f1, f2, f3) instead of orthogonal then (1) is still true but distance in (2) is not constant.
        Dear All My Friends,
        Today there are exactly 500 members in our group.
        Best wishes to all our members, My Friends!
        Many thanks to Antreas Hatzipolakis, Our Group Founder!
        Best regards,
        Bui Quang Tuan




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