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two Lemuan's points

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  • Alexey.A.Zaslavsky
    Dear colleagues! Does anybody know the non projective proof of next fact. Let A B C is the circumcevian triangle of ABC with respect to point P, L, L are the
    Message 1 of 9 , Jun 4, 2004
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      Dear colleagues!
      Does anybody know the non projective proof of next fact.
      Let A'B'C' is the circumcevian triangle of ABC with respect to point P, L,
      L' are the Lemuan's point of ABC and A'B'C'. Then P is in LL'.
      In fact if we concern projective transformations saving the circumcircle
      then the Lemuan point is invariant. So it is sufficient to concern an
      evident case when P=O.

      Sincerely Alexey
    • jpehrmfr
      Dear Alexey ... point P, L, ... The harmonic homology with center P and axis the polar line of P wrt the circumcircle fixes the circumcircle and maps A,B,C,L
      Message 2 of 9 , Jun 4, 2004
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        Dear Alexey
        > Does anybody know the non projective proof of next fact.
        > Let A'B'C' is the circumcevian triangle of ABC with respect to
        point P, L,
        > L' are the Lemuan's point of ABC and A'B'C'. Then P is in LL'.
        The harmonic homology with center P and axis the polar line of P wrt
        the circumcircle fixes the circumcircle and maps A,B,C,L to
        A',B',C',L'.
        Hence, P lies on LL'.
        Friendly. Jean-Pierre
      • Darij Grinberg
        Dear Alexey, ... (It is actually Lemoine , but better say symmedian points .) ... I have found an elementary proof of this fact, hoping that you will like
        Message 3 of 9 , Jun 20, 2004
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          Dear Alexey,

          In Hyacinthos message #9856, you wrote:

          >> Let A'B'C' is the circumcevian triangle of
          >> ABC with respect to point P, L, L' are the
          >> Lemuan's point

          (It is actually "Lemoine", but better say
          "symmedian points".)

          >> of ABC and A'B'C'. Then P is in LL'.

          I have found an elementary proof of this fact,
          hoping that you will like it.

          The following lemma is important:

          LEMMA 1. Let BCD and B'C'D' be two triangles such
          that the lines BB', CC', DD' are concurent at a
          point A. Let B1, C1, D1 be three points on the
          lines CD, DB, BC such that the lines BB1, CC1,
          DD1 concur at a point M. Finally, let the lines
          AB1, AC1 and AD1 meet the lines C'D', D'B' and
          B'C' at three points B1', C1', D1'. Then, the
          lines B'B1', C'C1', D'D1' and AM concur.

          Here are two proofs of this lemma:

          PROJECTIVE PROOF.

          Using projective transformations, Lemma 1 is
          trivial: Undertake such a projective
          transformation which maps the axis of perspective
          of triangles BCD and B'C'D' to the line at
          infinity. Then, the (new) triangles BCD and
          B'C'D' are homothetic (since opposite sidelines
          are parallel). The center of homothety is A.
          Then, the homothety with center A mapping
          triangle BCD to triangle B'C'D' must, of course,
          map the points B1, C1, D1 to the points B1', C1',
          D1', respectively; thus, the point M lying on the
          lines BB1, CC1, DD1 is mapped to a point M1 lying
          on the lines B'B1', C'C1', D'D1' and on the line
          AM, of course (what follows from the properties
          of a homothety). Hence we are done.

          ELEMENTARY PROOF.

          Since the triangles BCD and B'C'D' are
          perspective, i. e. the lines BB', CC' and DD'
          concur (at A), the Desargues theorem yields that
          the points X = CD /\ C'D', Y = DB /\ D'B' and
          Z = BC /\ B'C' are collinear, where the sign /\
          means "intersection" (i. e. the intersection of
          two lines g and h is denoted by g /\ h).

          Now consider the triangles BB'Y and B1B1'X. Since
          the points BB' /\ B1B1' = A, B'Y /\ B1'X = D' and
          YB /\ XB1 = D are collinear, the Desargues
          theorem shows that the lines BB1, B'B1' and YX
          concur. In other words, the point of intersection
          S of the lines BB1 and B'B1' lies on the line
          XYZ. Similarly, the point of intersection T of
          the lines DD1 and D'D1' lies on the line XYZ,
          too.

          Finally we apply the Desargues theorem to the
          triangles DD'T and BB'S: Since the lines DB, D'B'
          and TS concur (at Y), the points DD' /\ BB',
          D'T /\ B'S and TD /\ SB are collinear. But
          DD' /\ BB' = A, D'T /\ B'S = D'D1' /\ B'B1' and
          TD /\ SB = DD1 /\ BB1 = M. Hence, the points A,
          D'D1' /\ B'B1' and M are collinear. In other
          words, the lines D'D1', B'B1' and AM concur.
          Similarly, the lines D'D1', C'C1' and AM concur.
          Hence, all four lines B'B1', C'C1', D'D1' and AM
          concur at one point. Proof complete.

          (See also the MathLinks thread

          http://mathlinks.ro/viewtopic.php?t=5117 )

          Okay, now consider our two triangles ABC and
          A'B'C', and their respective tangential triangles
          XYZ and X'Y'Z'. The symmedian point of a triangle
          is the perspector of this triangle with its
          tangential triangle; hence, the lines AX, BY, CZ
          meet at the symmedian point L of triangle ABC,
          and the lines A'X', B'Y', C'Z' meet at the
          symmedian point L' of triangle A'B'C'.

          It is easy to see that not only the lines AA',
          BB', CC', but also the lines XX', YY', ZZ' pass
          through the point P. In fact, the line XX' passes
          through the poles X and X' of the lines BC and
          B'C' with respect to the circumcircle of triangle
          ABC, and hence it is the polar of the point of
          intersection of the lines BC and B'C'. But from
          the cyclic quadrilateral BCC'B', it follows that
          the polar of the point of intersection of the
          lines BC and B'C' passes through the point of
          intersection of the lines BB' and CC', i. e.
          through P. Hence, the line XX' passes through P,
          and similarly for the lines YY' and ZZ'.

          Now, consider the triangles XYZ and X'Y'Z'. The
          lines XX', YY', ZZ' are concurrent at the point
          P. The points A, B, C are three points on the
          lines YZ, ZX, XY, and we know that the lines XA,
          YB, ZC concur at the point L. Furthermore, the
          lines PA, PB and PC meet the lines Y'Z', Z'X',
          X'Y' at the points A', B', C' (since the points
          A', B', C' lie on the lines PA, PB, PC and on the
          lines Y'Z', Z'X', X'Y'). Hence, an application of
          Lemma 1 yields that the lines X'A', Y'B', Z'C'
          and PL concur. In other words, the point L' where
          the lines X'A', Y'B', Z'C' meet must lie on the
          line PL. Hence, the points P, L and L' are
          collinear. Proof complete.

          Your theorem has a very interesting special case:
          Namely, let P be the orthocenter H of triangle
          ABC. Then, the points A', B', C' are the
          intersections of the altitudes of triangle ABC
          with the circumcircle (different from A, B, C,
          respectively), and, equivalently, the points A',
          B', C' are the reflections of the orthocenter H
          in the sides BC, CA, AB. Thus, if DEF is the
          orthic triangle of the triangle ABC, then the
          triangle A'B'C' is the image of the triangle DEF
          in the homothety with center H and factor 2.
          Hence, the symmedian point L' of triangle A'B'C'
          is the image of the symmedian point L1 of
          triangle DEF in this homothety. It follows that
          the point L' lies on the line HL1. Now, since by
          your theorem the points P ( = H ), L and L' are
          collinear, it follows that all four points H, L,
          L' and L1 are collinear. Now forget L' and
          paraphrase the collinearity of the points H, L
          and L1 as follows:

          The orthocenter of a triangle, the symmedian
          point of the triangle, and the symmedian point of
          the orthic triangle are collinear.

          Finally, if Ia, Ib, Ic are the excenters and I is
          the incenter of triangle ABC, then the triangle
          ABC is the orthic triangle of the triangle
          IaIbIc. Also, the orthocenter of triangle IaIbIc
          is the incenter I of triangle ABC, and the
          symmedian point of triangle IaIbIc is the Mitten
          point of triangle ABC (this is X(9) in
          Kimberling's ETC). Hence, we summarize:

          The incenter, the Mitten point and the symmedian
          point of any triangle are collinear.

          This fact is rather trivial if you use trilinear
          coordinates, but I have not seen any synthetic
          proof before. Thanks for your general result.

          Friendly,
          Darij Grinberg
        • Antreas P. Hatzipolakis
          On 20-06-04, Darij Grinberg wrote ... In other words: Let ABC be a triangle, P a point, A B C the circumcevian triangle of P, and
          Message 4 of 9 , Jun 20, 2004
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            On 20-06-04, "Darij Grinberg" <darij_grinberg@...> wrote
            (partly):

            >Dear Alexey,
            >
            >In Hyacinthos message #9856, you wrote:
            >
            >>> Let A'B'C' is the circumcevian triangle of
            >>> ABC with respect to point P, L, L' are the
            >>> Lemuan's point
            >
            >(It is actually "Lemoine", but better say
            >"symmedian points".)
            >
            >>> of ABC and A'B'C'. Then P is in LL'.

            In other words:

            Let ABC be a triangle, P a point, A'B'C'
            the circumcevian triangle of P, and K,K'
            the Lemoine points of ABC, A'B'C', resp.
            The locus of P such that P,K,K' are collinear
            is the entire plane.

            Now, how about if we replace Ks with other
            triangle centers (O[: trivial case], H,I,G,N etc)

            Which are the corresponding loci of P ?

            APH



            --
          • Darij Grinberg
            Dear Antreas, ... I have tried P = X(i) with 1
            Message 5 of 9 , Jun 20, 2004
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              Dear Antreas,

              In Hyacinthos message #9924, you wrote:

              >> Let ABC be a triangle, P a point, A'B'C'
              >> the circumcevian triangle of P, and K,K'
              >> the Lemoine points of ABC, A'B'C', resp.
              >> The locus of P such that P,K,K' are collinear
              >> is the entire plane.
              >>
              >> Now, how about if we replace Ks with other
              >> triangle centers

              I have tried P = X(i) with 1 <= i <= 10, and
              the only i's such that the locus is the
              entire plane were X(3) - trivial indeed - and
              X(6).

              What is rather interesting is the locus of all
              P such that the orthocenters of triangles ABC
              and A'B'C' are collinear with P. This locus
              passes through X(1), X(3) and X(4), through
              the vertices of the triangle ABC and through
              the feet of its altitudes. The locus doesn't
              pass through X(2), X(i) for 5 <= i <= 10 and
              i = 13 and i = 15. However, note that the
              circumcircle of triangle ABC is obviously part
              of the locus, hence, after dividing away the
              circumcircle, I don't know whether the
              vertices of triangle ABC remain. Another part
              of the locus is the line at infinity, what is
              very easy to see, too. Maybe some specialist
              - Bernard? - could provide more facts?

              Sincerely,
              Darij Grinberg
            • Bernard Gibert
              Dear Antreas and Darij, ... your locus with Hs is the line at infinity and Q021. http://perso.wanadoo.fr/bernard.gibert/curves/q021.html I will add this to my
              Message 6 of 9 , Jun 20, 2004
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                Dear Antreas and Darij,

                > [APH] Let ABC be a triangle, P a point, A'B'C'
                > >> the circumcevian triangle of P, and K,K'
                > >> the Lemoine points of ABC, A'B'C', resp.
                > >> The locus of P such that P,K,K' are collinear
                > >> is the entire plane.
                > >> Now, how about if we replace Ks with other
                > >> triangle centers
                >
                > [DG] What is rather interesting is the locus of all
                > P such that the orthocenters of triangles ABC
                > and A'B'C' are collinear with P. This locus
                > passes through X(1), X(3) and X(4), through
                > the vertices of the triangle ABC and through
                > the feet of its altitudes. The locus doesn't
                > pass through X(2), X(i) for 5 <= i <= 10 and
                > i = 13 and i = 15. However, note that the
                > circumcircle of triangle ABC is obviously part
                > of the locus, hence, after dividing away the
                > circumcircle, I don't know whether the
                > vertices of triangle ABC remain. Another part
                > of the locus is the line at infinity, what is
                > very easy to see, too. Maybe some specialist
                > - Bernard? - could provide more facts?

                your locus with Hs is the line at infinity and Q021.

                http://perso.wanadoo.fr/bernard.gibert/curves/q021.html

                I will add this to my web site.

                Best regards

                Bernard

                [Non-text portions of this message have been removed]
              • Bernard Gibert
                Dear Antreas, ... Hs : Q021 (my previous message) Gs : Q019 + Linf + C(O,R) Ns : Linf + a bicircular sextic which I plan to add to my web site... Best regards
                Message 7 of 9 , Jun 20, 2004
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                  Dear Antreas,

                  > [APH] Let ABC be a triangle, P a point, A'B'C'
                  > the circumcevian triangle of P, and K,K'
                  > the Lemoine points of ABC, A'B'C', resp.
                  > The locus of P such that P,K,K' are collinear
                  > is the entire plane.
                  >
                  > Now, how about if we replace Ks with other
                  > triangle centers (O[: trivial case], H,I,G,N etc)
                  >
                  > Which are the corresponding loci of P ?

                  Hs : Q021 (my previous message)

                  Gs : Q019 + Linf + C(O,R)

                  Ns : Linf + a bicircular sextic which I plan to add to my web site...

                  Best regards

                  Bernard

                  [Non-text portions of this message have been removed]
                • Antreas P. Hatzipolakis
                  Dear Bernard, ... Note that all three points (H,G,N) are collinear. Probably we get interesting curves (as loci) for other points on the same line (Euler line)
                  Message 8 of 9 , Jun 20, 2004
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                    Dear Bernard,

                    >> [APH] Let ABC be a triangle, P a point, A'B'C'
                    >> the circumcevian triangle of P, and K,K'
                    >> the Lemoine points of ABC, A'B'C', resp.
                    >> The locus of P such that P,K,K' are collinear
                    >> is the entire plane.
                    >>
                    >> Now, how about if we replace Ks with other
                    >> triangle centers (O[: trivial case], H,I,G,N etc)
                    >>
                    >> Which are the corresponding loci of P ?

                    [BG]:

                    >Hs : Q021 (my previous message)
                    >
                    >Gs : Q019 + Linf + C(O,R)
                    >
                    >Ns : Linf + a bicircular sextic which I plan to add to my web site...

                    Note that all three points (H,G,N) are collinear.

                    Probably we get interesting curves (as loci) for other points
                    on the same line (Euler line)

                    How about the Ls [de Longchamps], for example?


                    Greetings

                    Antreas

                    --
                  • Bernard Gibert
                    Dear Antreas, ... Let X be a center on the Euler line such that OX = k OH (vectors, k¡Á0) and X the same center in A B C . P, X, X are collinear iff P lies
                    Message 9 of 9 , Jun 21, 2004
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                      Dear Antreas,

                      > >> [APH] Let ABC be a triangle, P a point, A'B'C'
                      > >> the circumcevian triangle of P, and K,K'
                      > >> the Lemoine points of ABC, A'B'C', resp.
                      > >> The locus of P such that P,K,K' are collinear
                      > >> is the entire plane.
                      > >>
                      > >> Now, how about if we replace Ks with other
                      > >> triangle centers (O[: trivial case], H,I,G,N etc)
                      > >>
                      > >> Which are the corresponding loci of P ?
                      >
                      > [BG]:
                      >
                      > >Hs : Q021 (my previous message)
                      > >
                      > >Gs : Q019 + Linf + C(O,R)
                      > >
                      > >Ns : Linf + a bicircular sextic which I plan to add to my web site...
                      >
                      > Note that all three points (H,G,N) are collinear.
                      >
                      > Probably we get interesting curves (as loci) for other points
                      > on the same line (Euler line)
                      >
                      > How about the Ls [de Longchamps], for example?

                      Let X be a center on the Euler line such that OX = k OH (vectors, k≠0)
                      and X' the same center in A'B'C'.
                      P, X, X' are collinear iff P lies on a bicircular sextic passing
                      through A, B, C (double points), O, X1113, X1114, the centers of the
                      Apollonian circles, the cevians of X in ABC, two other points E1, E2 on
                      the Euler line and on the circle centered at X468 orthogonal to the
                      circumcircle. Its has two real asymptotes parallel to those of the
                      rectangular circum-hyperbola passing through Y such that OY = (1-k)/2
                      OH.
                      This sextic decomposes into the circumcircle and a circular quartic iff
                      X = G. This is Q019.

                      Best regards

                      Bernard

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