Loading ...
Sorry, an error occurred while loading the content.

Goormaghtigh and isogonals

Expand Messages
  • brozolo
    Dear Hyacinthists, I m writing to send you an elementary proof - not using Menelaos theorem nor projective methods - of the theorem you call Goormaghtigh
    Message 1 of 1 , Jan 16, 2007
    View Source
    • 0 Attachment
      Dear Hyacinthists, I'm writing to send you an elementary proof - not
      using Menelaos theorem nor projective methods - of the theorem you
      call "Goormaghtigh theorem" (see Hyacinthos messages #8664, #8745, #9302).
      I don't know which was the original proof of Goormaghtigh.

      The theorem says:
      If A', B', C' are the points where the reflections of lines PA, PB, PC
      in a
      given line through P meet the sides BC, CA, AB then A', B', C' are
      collinear.

      If P is on the circumcircle the result is well known since it can be
      easily proved tha P is the Miquel point of points A', B' C' with respect
      triangle ABC. In fact, using directed angles, we have:
      from ang. A'BC' = ang. CBA = ang. CPA = ang. A'PC' follows that P,
      A', B, C'
      lie on a circle,
      from ang. B'CA' = ang. ACB = ang. APB = ang. B'PA' follows that P,
      A', B',
      C lie on a circle,
      from ang. CA'B' = ang. CPB' = ang. BPC' = ang. BA'C' follows that A',
      B', C'
      lie on a line.
      So in this case A'B'C' is the "generalized" Simson line of P wrt ABC.

      If P is not on the circumcircle then call Q its isogonal conjugate wrt
      triangle ABC and Q' its isogonal conjugate wrt triangle A B'C'.
      We want to prove that Q=Q':
      ang. BQC = ang. BAC + ang. CPB (since P and Q are isogonal
      conjugates in triangle ABC)
      = ang. C'AB'+ ang .B'PC' =
      = ang. C'Q'B' (since P and Q are isogonal conjugates in
      triangle AB'C')

      Now I will use the result that if P and Q are isogonal conjugates wrt
      to a triangle ABC and X, Y, Z are the reflections of P in the sides of
      ABC then Q is the circumcenter of the XYZ triangle (for a proof see
      Darij Grinberg's paper on isogonals in his site:
      de.geocities.com/darij_grinberg/).

      If X, Y, Z, are the reflections of P in the sides BC, CA, AB of
      triangle ABC
      and X' its reflection in the side B'C' of triangle AB'C' then:
      ang. ZXY = ang. BQC (because QC is orthogonal to XY, QB is
      orthogonal to XZ)
      ang. ZX'Y = ang. C'Q'B' (because Q'B' is orthogonal to X'Y, Q'C' is
      orthogonal X'Z)
      so, since ang. C'Q'B' = ang. BQC, we get ang. ZXY = ang. ZX'Y.
      From this follows that X, Y, Z, X' are concyclic,
      but the center of the XYZ-circle is Q while the center of the
      X'YZ-circle is
      Q' so Q and Q' are the same point.

      In a similar way it is found that Q is the isogonal point of P also wrt
      triangles A'BC' and A'B'C.
      Now, since Q is the isogonal conjugate of P with respect to triangle
      A'BC'
      and triangle AB'C' we have:
      ang. BC'A' = ang. AC'A' = ang. AC'P + ang. PC'Q + ang. QC'A' = ang.
      QC'B' +
      ang.PC'Q + ang.BC'P = ang.BC'B' = ang. AC'B'; this means that A', B',
      C' are
      on a line. QED.

      It can also be easily seen just from elementary reasoning that A'B'C'
      envelopes an inconic whose foci are P and Q as the "reflection line"
      through
      P varies, this conic is an
      ellipse if P is inside the XYZ-circle, an hyperbola if it is outside (a
      parabola if P is an infinite point or on the circumcircle).
      A sketch of proof is: if T is the point where QX' meets B'C' then PT=TX'
      because B'C' is the perp. bisector of PX',
      so, if P is inside the XYZ-circle, QT+PT=QT+TX'=QX'=cost. -> the locus
      of T
      is an ellipse and B'C' is the tangent in T (B'T' is the bisector of angle
      PTX'); if P is outside...

      The idea of using the fact that triangles ABC AB'C', A'BC', A'B'C
      share the
      same isogonal conj. point of P to prove this theorem is due to Gerhard
      Hessenberg.
      I've found it in his book "Grundlagen der Geometrie", published in 1930.
      His proof is not elementary because it uses the "Calculus of
      Reflections" (an approach, similar to analytic geometry, based on
      group-theoretic calculations), here I've just tried to "translate" his
      idea into the language of elementary geometry and directed angles.

      Ciao!
      Marcello Tarquini
    Your message has been successfully submitted and would be delivered to recipients shortly.