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Some theorems on Miquel points

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  • Darij Grinberg <darij_grinberg@web.de>
    Dear Floor van Lamoen, and Hyacinthos, Since we have started to investigate Miquel points, here are two forgotten theorems on them. The Miquel point
    Message 1 of 8 , Jan 1, 2003
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      Dear Floor van Lamoen, and Hyacinthos,

      Since we have started to investigate Miquel points, here are
      two "forgotten" theorems on them.

      The Miquel point configuration is defined as follows:

      On the sidelines BC, CA, AB of a triangle ABC, points A', B', C' are
      chosen and the circles AB'C', BC'A', CA'B' are drawn. Then, as we
      know, these circles have a common point P (the Miquel point).

      Now the two theorems:

      1. If X is an arbitrary point in the plane, then the second points of
      intersection of the lines XA, XB, XC with the circles AB'C', BC'A',
      CA'B' (the first intersections being A, B, C) are concyclic with P
      and X.

      [Jacques Hadamard: Lecons de Geometrie Elementaire, exercise 344. I
      don't know of a proof.]

      2. Let Ma, Mb, Mc be the centers of the circles AB'C', BC'A', CA'B'
      [it is well-known that triangles MaMbMc and ABC are similar], let M'
      be the circumcenter of triangle MaMbMc, and M the circumcenter of
      triangle ABC. Then M'M = M'P.

      [This is a theorem by Peter Baum; proposed in the little German
      periodical "Die Wurzel" (see the website wurzel.org), in the 12/98
      issue. Solved by Sefket Arslanagic in the 5/99 issue; the solution
      was quite involved.]

      Darij Grinberg
    • jpehrmfr <jean-pierre.ehrmann@wanadoo.fr>
      Dear Darij ... CA B ... M ... It is well known (Steiner 1827) that Ma,Mb,Mc,M,P are concyclic (Miquel cirle of the quadrilateral). Hence, this theorem is not
      Message 2 of 8 , Jan 1, 2003
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        Dear Darij

        > 2. Let Ma, Mb, Mc be the centers of the circles AB'C', BC'A',
        CA'B'
        > [it is well-known that triangles MaMbMc and ABC are similar], let
        M'
        > be the circumcenter of triangle MaMbMc, and M the circumcenter of
        > triangle ABC. Then M'M = M'P.
        >
        > [This is a theorem by Peter Baum; proposed in the little German
        > periodical "Die Wurzel" (see the website wurzel.org), in the 12/98
        > issue. Solved by Sefket Arslanagic in the 5/99 issue; the solution
        > was quite involved.]

        It is well known (Steiner 1827) that Ma,Mb,Mc,M,P are concyclic
        (Miquel cirle of the quadrilateral). Hence, this theorem is not a
        recent one.
        Happy new year to all Hyacinthists. Jean-Pierre
      • jpehrmfr <jean-pierre.ehrmann@wanadoo.fr>
        Dear Darij ... are ... of ... BC A , ... I ... Using oriented angles of lines (modulo Pi)
        Message 3 of 8 , Jan 1, 2003
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          Dear Darij

          > Since we have started to investigate Miquel points, here are
          > two "forgotten" theorems on them.
          >
          > The Miquel point configuration is defined as follows:
          >
          > On the sidelines BC, CA, AB of a triangle ABC, points A', B', C'
          are
          > chosen and the circles AB'C', BC'A', CA'B' are drawn. Then, as we
          > know, these circles have a common point P (the Miquel point).
          >
          > Now the two theorems:
          >
          > 1. If X is an arbitrary point in the plane, then the second points
          of
          > intersection of the lines XA, XB, XC with the circles AB'C',
          BC'A',
          > CA'B' (the first intersections being A, B, C) are concyclic with P
          > and X.
          >
          > [Jacques Hadamard: Lecons de Geometrie Elementaire, exercise 344.
          I
          > don't know of a proof.]

          Using oriented angles of lines (modulo Pi)
          <PXaX = <PXaA = <PC'A = <PC'B = <PXbB = <PXbX and Xa,Xb,P,X are
          concyclic...
          Happy New Year to all of you. Jean-Pierre
        • Darij Grinberg <darij_grinberg@web.de>
          (Answer to Hyacinthos message #6240) ... If I understand correctly, you assume that A , B , C are collinear. But they need not be! Therefore, the Baum theorem
          Message 4 of 8 , Jan 3, 2003
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            (Answer to Hyacinthos message #6240)

            In message #6240, Jean-Pierre Ehrmann wrote:

            >> > [DG] 2. Let Ma, Mb, Mc be the centers of the circles AB'C',
            >> > BC'A', CA'B' [it is well-known that triangles MaMbMc and
            >> > ABC are similar], let M' be the circumcenter of triangle
            >> > MaMbMc, and M the circumcenter of triangle ABC. Then
            >> > M'M = M'P.
            >> >
            >> > [This is a theorem by Peter Baum; proposed in the little
            >> > German periodical "Die Wurzel" (see the website
            >> > wurzel.org), in the 12/98 issue. Solved by Sefket
            >> > Arslanagic in the 5/99 issue; the solution was quite
            >> > involved.]
            >>
            >> [JPE] It is well known (Steiner 1827) that Ma,Mb,Mc,M,P are
            >> concyclic (Miquel cirle of the quadrilateral). Hence, this
            >> theorem is not a recent one.

            If I understand correctly, you assume that A', B', C' are collinear.
            But they need not be! Therefore, the Baum theorem is a partial
            generalization of the Miquel circle in a quadrilateral, and not a
            corollary.

            Sorry for the possible-to-misunderstand use of the term "Miquel
            point".

            Sincerely,
            Darij Grinberg
          • jpehrmfr <jean-pierre.ehrmann@wanadoo.fr>
            Dear Darij, ... collinear. ... Yes, you are perfectly right; I thought that you were talking about the Miquel point of a complete quadrilateral. With my
            Message 5 of 8 , Jan 3, 2003
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              Dear Darij,

              > >> > [DG] 2. Let Ma, Mb, Mc be the centers of the circles AB'C',
              > >> > BC'A', CA'B' [it is well-known that triangles MaMbMc and
              > >> > ABC are similar], let M' be the circumcenter of triangle
              > >> > MaMbMc, and M the circumcenter of triangle ABC. Then
              > >> > M'M = M'P.
              > >> >
              > >> > [This is a theorem by Peter Baum; proposed in the little
              > >> > German periodical "Die Wurzel" (see the website
              > >> > wurzel.org), in the 12/98 issue. Solved by Sefket
              > >> > Arslanagic in the 5/99 issue; the solution was quite
              > >> > involved.]
              > >>
              > >> [JPE] It is well known (Steiner 1827) that Ma,Mb,Mc,M,P are
              > >> concyclic (Miquel cirle of the quadrilateral). Hence, this
              > >> theorem is not a recent one.
              >
              > If I understand correctly, you assume that A', B', C' are
              collinear.
              > But they need not be! Therefore, the Baum theorem is a partial
              > generalization of the Miquel circle in a quadrilateral, and not a
              > corollary.
              >
              > Sorry for the possible-to-misunderstand use of the term "Miquel
              > point".

              Yes, you are perfectly right; I thought that you were talking about
              the Miquel point of a complete quadrilateral.
              With my apologizes for the misunderstanding. Jean-Pierre
            • jpehrmfr <jean-pierre.ehrmann@wanadoo.fr>
              Dear Darij, I ve tried to find an easy synthetic proof of Braun theorem ... are ... CA B ... M ... PaPbPc is the pedal triangle of P. A direct similitude
              Message 6 of 8 , Jan 4, 2003
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                Dear Darij,
                I've tried to find an easy synthetic proof of Braun theorem
                > The Miquel point configuration is defined as follows:
                >
                > On the sidelines BC, CA, AB of a triangle ABC, points A', B', C'
                are
                > chosen and the circles AB'C', BC'A', CA'B' are drawn. Then, as we
                > know, these circles have a common point P (the Miquel point).
                >
                > Now the two theorems:
                >
                > 2. Let Ma, Mb, Mc be the centers of the circles AB'C', BC'A',
                CA'B'
                > [it is well-known that triangles MaMbMc and ABC are similar], let
                M'
                > be the circumcenter of triangle MaMbMc, and M the circumcenter of
                > triangle ABC. Then M'M = M'P.
                >
                > [This is a theorem by Peter Baum; proposed in the little German
                > periodical "Die Wurzel" (see the website wurzel.org), in the 12/98
                > issue. Solved by Sefket Arslanagic in the 5/99 issue; the solution
                > was quite involved.]

                PaPbPc is the pedal triangle of P.
                A direct similitude with center P, angle T, ratio 1/cos(T) maps
                PaPbPc to A'B'C'.
                As the circumcenter of PPbPc is the midpoint A'' of AP, the
                similitude maps this midpoint to Ma; hence the similitude maps the
                midpoint I of PM (circumcenter of A''B''C'') to M'(circumcenter of
                MaMbMc).
                As PM'/PI = cos(T) and <IPM' = T, it follows that M' lies on the
                perpendicular bisector of PM.
                Friendly. Jean-Pierre
              • jpehrmfr <jean-pierre.ehrmann@wanadoo.fr>
                Dear Darij, I wrote ... Of course, I was meaning Baum theorem. Friendly. Jean-Pierre
                Message 7 of 8 , Jan 4, 2003
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                  Dear Darij, I wrote

                  > I've tried to find an easy synthetic proof of Braun theorem
                  Of course, I was meaning Baum theorem.
                  Friendly. Jean-Pierre
                • jpehrmfr <jean-pierre.ehrmann@wanadoo.fr>
                  We have PM /PI = 1/cos(T) and not PM /PI. With my apologizes. Jean-Pierre
                  Message 8 of 8 , Jan 4, 2003
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                    We have PM'/PI = 1/cos(T) and not PM'/PI.
                    With my apologizes. Jean-Pierre
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