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Dual triangles

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  • César Lozada
    Dear Geometers: According to ETC (just before X(2979)): Suppose DEF is a triangle in the plane of triangle ABC. Let D be the isogonal conjugate of the point
    Message 1 of 2 , Apr 3, 2012
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      Dear Geometers:

      According to ETC (just before X(2979)):

      Suppose DEF is a triangle in the plane of triangle ABC. Let D' be the
      isogonal conjugate of the point of intersection of line EF and the line at
      infinity. Define E' and F' cyclically. The triangle D'E'F' is here named the
      dual of DEF. The vertices of D'E'F' lie on the circumcircle, and D'E'F' is
      similar to DEF. The duality is between the sidelines EF, FD, DE and the
      points D', E', F', respectively. For example, if E and F remain fixed and D
      varies, then D' remains fixed, while E'F' varies. Actually D'E'F' is the
      dual of any triangle homothetic to DEF. Indeed, DEF need not be a triangle
      but can be the union of three concurrent lines. (Proof of similarity follows
      from Theorem 6E in TCCT, as the "gamma triangle" there is the dual of a
      triangle whose sidelines are respectively perpendicular to those of DEF.)

      Suppose U is a point having cevian triangle DEF and dual triangle D'E'F'.
      Then there exists a point DC(U) whose circum-anticevian triangle (TCCT, p.
      201) is D'E'F'. The mapping DC is given for U = u : v : w (trilinears) by

      DC(U) = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv).



      To construct DC(U) from U, let A' = AD'∩BC, and let A" be the harmonic
      conjugate of A' with respect to B and C. Define B" and C" cyclically. The
      lines AA", BB", CC" concur in DC(U). Also, DC(U) = U-isoconjugate of the
      crosssum of U and X(6).

      Inversely, the circum-anticevian triangle of a point P is the dual of a
      point CD(P), given for P = p : q : r by the inverse of the DC-mapping; that
      is:

      CD(P) = bc/(-a/p + b/q + c/r) : ca/(a/p - b/q + c/r) : ab/(a/p + b/q - c/r)
      = isogonal conjugate of X(6)-Ceva conjugate of P.



      -----------------

      I can get this from another construction:

      Let DA’B’C’ be the cevian triangle of U in DABC. The circle passing
      through {U, B’, C’} cuts (AC) and (AB) again in A_B and A_C and A” is
      the midpoint of A_B and A_C. B” and C” are built cyclically.

      Then DA”B”C” and DABC are perspective and, for U=u : v : w (trilinears),
      the perspector is P = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv) = DC(U)

      QUESTIONS: Are both constructions equivalent? Why?



      Note: The vertex of my DA”B”C” are not on the circumcircle of DABC

      ----------------

      I note that equation leading to the inverse mapping has a square root, i.e.:

      CD(P)=sqrt( b-2 c2- / (-a/p + b/q + c/r)2 ) : …..

      which should be written as:

      CD(P)= b c- / abs(-a/p + b/q + c/r ) : …..



      QUESTION: Why is the absolute value function not relevant?



      Regards

      Cesar Lozada







      [Non-text portions of this message have been removed]
    • CESAR
      Better writing: Dear Geometers: According to ETC (just before X(2979)): Suppose DEF is a triangle in the plane of triangle ABC. Let D be the isogonal
      Message 2 of 2 , Apr 3, 2012
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        Better writing:

        Dear Geometers:
        According to ETC (just before X(2979)):
        Suppose DEF is a triangle in the plane of triangle ABC. Let D' be the isogonal conjugate of the point of intersection of line EF and the line at infinity. Define E' and F' cyclically. The triangle D'E'F' is here named the dual of DEF. The vertices of D'E'F' lie on the circumcircle, and D'E'F' is similar to DEF. The duality is between the sidelines EF, FD, DE and the points D', E', F', respectively. For example, if E and F remain fixed and D varies, then D' remains fixed, while E'F' varies. Actually D'E'F' is the dual of any triangle homothetic to DEF. Indeed, DEF need not be a triangle but can be the union of three concurrent lines. (Proof of similarity follows from Theorem 6E in TCCT, as the "gamma triangle" there is the dual of a triangle whose sidelines are respectively perpendicular to those of DEF.)
        Suppose U is a point having cevian triangle DEF and dual triangle D'E'F'. Then there exists a point DC(U) whose circum-anticevian triangle (TCCT, p. 201) is D'E'F'. The mapping DC is given for U = u : v : w (trilinears) by
        DC(U) = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv).

        To construct DC(U) from U, let A' = AD'∩BC, and let A" be the harmonic conjugate of A' with respect to B and C. Define B" and C" cyclically. The lines AA", BB", CC" concur in DC(U). Also, DC(U) = U-isoconjugate of the crosssum of U and X(6).
        Inversely, the circum-anticevian triangle of a point P is the dual of a point CD(P), given for P = p : q : r by the inverse of the DC-mapping; that is:
        CD(P) = bc/(-a/p + b/q + c/r) : ca/(a/p - b/q + c/r) : ab/(a/p + b/q - c/r) = isogonal conjugate of X(6)-Ceva conjugate of P.

        -----------------
        I can get this from another construction:
        Let A'B'C' be the cevian triangle of U in triangle ABC. The circle passing through {U, B', C'} cuts (AC) and (AB) again in A_B and A_C and A" is the midpoint of A_B and A_C. B" and C" are built cyclically.
        Then triangles A"B"C" and ABC are perspective and, for U=u : v : w (trilinears), the perspector is P = vw/(bv + cw) : wu/(cw + au) : uv/(au + bv) = DC(U)
        QUESTIONS: Are both constructions equivalent? Why?

        Note: The vertex of my triangle A"B"C" are not on the circumcircle of ABC
        ----------------
        I note that equation leading to the inverse mapping has a square root, i.e.:
        CD(P)=sqrt( b^2 c^2 / (-a/p + b/q + c/r)^2 ) : …..
        which should be written as:
        CD(P)= bc/abs(-a/p + b/q + c/r ): …..

        QUESTION: Why is the absolute value function not relevant?

        Regards
        Cesar Lozada
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