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Re: Simson lines

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  • jpehrmfr
    Dear Alexei ... prove next interesting fact. ... P_A the point on the line PA such that IA is the bisector (internal or external) of angle PIP_A. The points
    Message 1 of 15 , Jul 1, 2008
      Dear Alexei
      > The polar transformation and the properties of Simson lines allow to
      prove next interesting fact.
      > Let I be the incenter of ABC. For an arbitrary point P we denote as
      P_A the point on the line PA such that IA is the bisector (internal or
      external) of angle PIP_A. The points P_B, P_C are defined similarly.
      > 1. If P_A, P_B, P_C are collinear then this line touches the incircle
      of ABC.
      > 2. The locus of points P with this property is the line.
      > Can anybody prove this without using the polar transformation?

      Here is a better presentation of a generalization :
      Starting with a point J in the plane of ABC, the perpendicular lines
      through J to JA, JB, JC intersect BC, CA, AB respectively at A', B', C'.
      It is well known that A',B',C' are on a same line L(J).
      For any point M,
      Ma is the point of the line AM such as <MJA = <AJMa
      Mb is the point of the line BM such as <MJB = <BJMb
      Mc is the point of the line CM such as <MJC = <CJMc
      Then Ma,Mb,Mc are collinear if and only if M lies on L(J)
      In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the
      inscribed conic with focus J
      This can be proved with projective tools using the fact that M->Ma is
      the harmonic homology with center A and axis the line JA'
      * The harmonic homology (I don't know if this English name is correct)
      with center O, axis L maps M to his harmonic conjugate wrt O and OM
      inter L


      I think that, if we consider three harmonic homologies with centers
      A,B,C, the locus of M with three colinear images is generally a cubic,
      even when the three axis concur. In the particular case of three axis
      La,Lb,Lc going through a point J and such as there exists an involution
      swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in three
      lines :
      the fixed lines of the involution and the line through La inter BC, Lb
      inter CA, Lc inter AB
      Friendly. Jean-Pierre
    • jpehrmfr
      Dear Alxei ... lines ... B , C . ... is ... correct) ... cubic, ... axis ... involution ... three ... Lb ... Now, if we consider a line L with trilinear pole P
      Message 2 of 15 , Jul 1, 2008
        Dear Alxei
        I wrote :
        > Here is a better presentation of a generalization :
        > Starting with a point J in the plane of ABC, the perpendicular
        lines
        > through J to JA, JB, JC intersect BC, CA, AB respectively at A',
        B', C'.
        > It is well known that A',B',C' are on a same line L(J).
        > For any point M,
        > Ma is the point of the line AM such as <MJA = <AJMa
        > Mb is the point of the line BM such as <MJB = <BJMb
        > Mc is the point of the line CM such as <MJC = <CJMc
        > Then Ma,Mb,Mc are collinear if and only if M lies on L(J)
        > In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the
        > inscribed conic with focus J
        > This can be proved with projective tools using the fact that M->Ma
        is
        > the harmonic homology with center A and axis the line JA'
        > * The harmonic homology (I don't know if this English name is
        correct)
        > with center O, axis L maps M to his harmonic conjugate wrt O and OM
        > inter L
        >
        >
        > I think that, if we consider three harmonic homologies with centers
        > A,B,C, the locus of M with three colinear images is generally a
        cubic,
        > even when the three axis concur. In the particular case of three
        axis
        > La,Lb,Lc going through a point J and such as there exists an
        involution
        > swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in
        three
        > lines :
        > the fixed lines of the involution and the line through La inter BC,
        Lb
        > inter CA, Lc inter AB

        Now, if we consider a line L with trilinear pole P and intersecting
        the sidelines of ABC at A',B',C', we get :
        Suppose that M is a point of L; the harmonic homology with center A,
        axis JA' maps M to Ma,.., similarly Mb,Mc
        Then
        Ma, Mb, Mc are collinear
        the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the inscribed conic
        with perspector the cevian product J*P.
        Looking at that, I've realized something which should be interesting
        for Bernard :
        For any point J, the cevian product J*o(J), where o(J) is the
        orthocorrespondant of J, is the perspector of the inscribed conic
        with focus J.
        Friendly. Jean-Pierre
      • Alexey Zaslavsky
        Dear colleagues! Is next fact known? Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is
        Message 3 of 15 , Jun 6, 2012
          Dear colleagues!
          Is next fact known?
          Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.

          Sincerely Alexey

          [Non-text portions of this message have been removed]
        • Daniel
          Yes, it is a particular case of S-triangles. Two triangles ABC and A B C inscribed in the same circle are S-triangles iff arc(AA )+arc(BB )+arc(CC )=0 (mod
          Message 4 of 15 , Jun 6, 2012
            Yes, it is a particular case of S-triangles.

            Two triangles ABC and A'B'C' inscribed in the same circle are S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
            The Simson lines of A',B',C', with respect to ABC and the Simson lines of A,B,C, with respect to A'B'C' are concurrent in the midpoint of [HH'] where H and H' are the orthocenters of ABC and A'B'C'.

            The triangles CAB and CPQ are S-triangles iff PQ is parallel to AB.
            So, the Simson lines of P,Q, and C are concurrent. But the Simson line of C is the altitude from C.

            See
            [1] Traian Lalescu, A Class of Remarkable Triangles,
            Gazeta Matematica , vol XX, feb. 1915 ,p 213
            (in Romanian)

            [2] Trajan Lalesco, La geometrie du triangle ,Bucharest,
            1937, Paris, Libraire Vuibert

            Sincerely
            Daniel Vacaretu


            --- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:
            >
            > Dear colleagues!
            > Is next fact known?
            > Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.
            >
            > Sincerely Alexey
            >
            > [Non-text portions of this message have been removed]
            >
          • Nikolaos Dergiades
            Dear Daniel, Very good observation and proof. So we can generalize Alexey s problem as follows: If L is a line and S is the isogonal conjugate of its infinite
            Message 5 of 15 , Jun 6, 2012
              Dear Daniel,
              Very good observation and proof.
              So we can generalize Alexey's problem
              as follows:
              If L is a line and S is the isogonal conjugate
              of its infinite point and the chord PQ of the
              circumcircle of ABC is parallel to L then the
              triangles ABC, SPQ are S-triangles.
              Hence the Simson lines of P, Q relative to ABC
              meet at M' midpoint of HH'. The locus of H' orthocenter
              of SPQ is the perpendicular line from S to L.
              Hence the locus of M' is the perpendicular line
              from the midpoint M of HS to L.

              If L is parallel to BC then S=A, M=H.
              Best regards
              Nikos Dergiades


              > Από: Daniel <dvacaretu@...>
              > Yes, it is a particular case of
              > S-triangles.
              >
              > Two triangles ABC and A'B'C' inscribed in the same circle
              > are S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
              > The Simson lines of A',B',C', with respect to ABC and 
              > the Simson lines of A,B,C, with respect to A'B'C' are
              > concurrent in the midpoint of [HH'] where H and H' are the
              > orthocenters of ABC and A'B'C'.
              >
              > The triangles CAB and CPQ are S-triangles iff PQ is parallel
              > to AB.
              > So, the Simson lines of P,Q, and C are concurrent. But the
              > Simson line of C is the altitude from C.

              > "Alexey Zaslavsky" <zasl@...> wrote:
              > >
              > > Dear colleagues!
              > > Is next fact known?
              > > Let P, Q be two point on the circumcircle of triangle
              > ABC. Then their Simson lines meet on the altitude from C iff
              > PQ is parallel to AB.
            • Cosmin
              An easier proof would be to just recall that the Simson line bisects the segment determined by the orthocenter of the triangle and the point, thus if we let X,
              Message 6 of 15 , Jun 7, 2012
                An easier proof would be to just recall that the Simson line bisects the segment determined by the orthocenter of the triangle and the point, thus if we let X, Y be the midpoints of HP and HQ (where H is the orthocenter of ABC), the result follows at once by Desargues' theorem applied to triangles PXM, QYN, where M, N are the projections of P, Q on AB.

                Best,
                Cosmin

                --- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:
                >
                > Dear colleagues!
                > Is next fact known?
                > Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.
                >
                > Sincerely Alexey
                >
                > [Non-text portions of this message have been removed]
                >
              • Barry Wolk
                ... A routine calculation using complex numbers as coordinates gives the following:   If A,B,C,D,E are on the unit circle, then the Simson lines of D and of
                Message 7 of 15 , Jun 7, 2012
                  > From: Daniel <dvacaretu@...>
                  >
                  > Yes, it is a particular case of S-triangles.
                  >
                  > Two triangles ABC and A'B'C' inscribed in the same circle are
                  > S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
                  > The Simson lines of A',B',C', with respect to ABC and  the Simson
                  > lines of A,B,C, with respect to A'B'C' are concurrent in the
                  > midpoint of [HH'] where H and H' are the orthocenters of ABC and
                  > A'B'C'.

                  A routine calculation using complex numbers as coordinates gives
                  the following:   If A,B,C,D,E are on the unit circle, then the Simson
                  lines of D and of E (wrt ABC) meet at (A+B+C+D+E+ABC/(DE))/2
                   
                  This formula is simpler than expected, even though it was known
                  in advance that it had to have some symmetry and scaling properties.
                  And the S-triangle result follows easily from this.
                  --
                  Barry Wolk
                • Alexey Zaslavsky
                  Dear Daniel, Nikos, Cosmin and Barry! Yes, I also found these general facts. Thank you for the references. Also the envelop of Simson lines is the astroid and
                  Message 8 of 15 , Jun 7, 2012
                    Dear Daniel, Nikos, Cosmin and Barry!
                    Yes, I also found these general facts. Thank you for the references. Also the envelop of Simson lines is the astroid and its center coincides with the center of NPC. And what is the locus of point P such that the circumcevian triangle of P is S-triangle? It is clear that this is some cubic, but which?

                    Sincerely Alexey


                    ----- Original Message -----
                    From: Barry Wolk
                    To: Hyacinthos@yahoogroups.com
                    Sent: Thursday, June 07, 2012 11:33 PM
                    Subject: [EMHL] Re: Simson lines



                    > From: Daniel <dvacaretu@...>
                    >
                    > Yes, it is a particular case of S-triangles.
                    >
                    > Two triangles ABC and A'B'C' inscribed in the same circle are
                    > S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
                    > The Simson lines of A',B',C', with respect to ABC and the Simson
                    > lines of A,B,C, with respect to A'B'C' are concurrent in the
                    > midpoint of [HH'] where H and H' are the orthocenters of ABC and
                    > A'B'C'.

                    A routine calculation using complex numbers as coordinates gives
                    the following: If A,B,C,D,E are on the unit circle, then the Simson
                    lines of D and of E (wrt ABC) meet at (A+B+C+D+E+ABC/(DE))/2

                    This formula is simpler than expected, even though it was known
                    in advance that it had to have some symmetry and scaling properties.
                    And the S-triangle result follows easily from this.
                    --
                    Barry Wolk





                    [Non-text portions of this message have been removed]
                  • Bernard Gibert
                    Dear Alexey, ... see K024 in http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html Best regards Bernard [Non-text portions of this message have been
                    Message 9 of 15 , Jun 8, 2012
                      Dear Alexey,

                      > Yes, I also found these general facts. Thank you for the references. Also the envelop of Simson lines is the astroid and its center coincides with the center of NPC. And what is the locus of point P such that the circumcevian triangle of P is S-triangle? It is clear that this is some cubic, but which?

                      see K024 in

                      http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html

                      Best regards

                      Bernard

                      [Non-text portions of this message have been removed]
                    • Alexey Zaslavsky
                      Dear Bernard! see K024 in http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html Thank you. Sincerely Alexey [Non-text
                      Message 10 of 15 , Jun 8, 2012
                        Dear Bernard!

                        see K024 in

                        http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html

                        Thank you.

                        Sincerely Alexey

                        [Non-text portions of this message have been removed]
                      • Antreas Hatzipolakis
                        Let ABC be a triangle P a point and A B C , A B C the pedal, circumcevian triangles of P, resp. Denote: A*, B*, C* = the other (than A , B , C )
                        Message 11 of 15 , Jun 29
                          Let ABC be a triangle P a point and A'B'C', A"B"C" the pedal, circumcevian triangles of P, resp.

                          Denote:

                          A*, B*, C* = the other (than A", B", C")  intersections of A"A', B"B', C"C' and the circumcircle, resp.

                          S1, S2, S3 = the Simson lines of A*, B*, C*, resp.

                          AsBsCs = the triangle bounded by S1, S2, S3.

                          Which is the locus of P such that:

                          1. ABC, AsBsCs are parallelogic?
                          The entire plane?

                          2. S1, S2, S3 are concurrent?

                          APH
                        • Antreas Hatzipolakis
                          [APH]: Let ABC be a triangle P a point and A B C , A B C the pedal, circumcevian triangles of P, resp. Denote: A*, B*, C* = the other (than A , B , C )
                          Message 12 of 15 , Jun 30
                            [APH]:

                            Let ABC be a triangle P a point and A'B'C', A"B"C" the pedal, circumcevian triangles of P, resp.

                            Denote:

                            A*, B*, C* = the other (than A", B", C")  intersections of A"A', B"B', C"C' and the circumcircle, resp.

                            S1, S2, S3 = the Simson lines of A*, B*, C*, resp.

                            AsBsCs = the triangle bounded by S1, S2, S3.

                            Which is the locus of P such that:

                            1. ABC, AsBsCs are parallelogic?
                            The entire plane?

                            2. S1, S2, S3 are concurrent?


                            [César Lozada]:

                             

                            2) Simson lines concur for P on: {sidelines} \/ { Line at infinity} \/ {circum-sixtic q6 through ETC’s:  4, 64, 84})

                             

                            q6:  Σ[ v*w*((SB-SC)*SB*SC*a*v^2*w^2+( ((SB-SA)*S^2+3*SA^2*SB)*b*v-(( SC-SA)*S^2+3*SA^2*SC)*c*w)*u^ 3)] = 0 (trilinears)

                             

                            ETC-pairs (P,Z2(P)=point of concurrence): (4, 4), (64, 5562), (84,10)

                             

                            1)      Locus for parallelogic ABC and AsBsCs = the entire plane

                             

                            The parallelogic center A->As is Za(P) = Isogonal(Reflection(Isogonal( P), O)),

                            i.e., the Isogonal-conjugate of the-reflection-in-O of the isogonal-conjugate-of-P.

                             

                            The parallelogic center Zs(P)=As->A has a complicated expression. There are no ETC pairs (P,Zs(P)) for non-concurrent Simson lines.

                             

                            Points Zs(P) seem not to be of interest: long coordinates and very few relations with existing 13600+ ETC centers.

                             

                            César Lozada

                             

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