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Four Circum Equilateral Triangles

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  • Quang Tuan Bui
    Dear All My Friends, Given triangle ABC with circumcircle (O). There is and exactly one equilateral triangle inscribed in (O) taken A as one vertex. We denote
    Message 1 of 14 , Mar 2, 2007
      Dear All My Friends,
      Given triangle ABC with circumcircle (O). There is and exactly one equilateral triangle inscribed in (O) taken A as one vertex. We denote its bottom line wrt A as La, intesection of La with sideline BC as A'. Similarly define Lb, B', Lc, C'. Three lines La, Lb, Lc bound one triangle A''B''C''. Please prove following results:
      1. ABC and A''B''C'' are perspective with perspector P on circumcircle (O).
      2. A', B', C' are collinear on one line L cutting (O) at two point Q, R.
      3. PQR are our fourth equilateral.
      Thank you and best regards,
      Bui Quang Tuan



      ---------------------------------
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    • Paul Yiu
      Dear Tuan, How wonderful. I make a quick sketch and see that your P is the isogonal conjugate of the infinite point of the Euler line. Best regards Sincerely
      Message 2 of 14 , Mar 2, 2007
        Dear Tuan,

        How wonderful. I make a quick sketch and see that your P is the isogonal
        conjugate of the infinite point of the Euler line.

        Best regards
        Sincerely
        Paul


        At 3/2/2007 08:08 AM -0800, you wrote:
        >Dear All My Friends,
        > Given triangle ABC with circumcircle (O). There is and exactly one
        > equilateral triangle inscribed in (O) taken A as one vertex. We denote
        > its bottom line wrt A as La, intesection of La with sideline BC as A'.
        > Similarly define Lb, B', Lc, C'. Three lines La, Lb, Lc bound one
        > triangle A''B''C''. Please prove following results:
        > 1. ABC and A''B''C'' are perspective with perspector P on circumcircle (O).
        > 2. A', B', C' are collinear on one line L cutting (O) at two point Q, R.
        > 3. PQR are our fourth equilateral.
        > Thank you and best regards,
        > Bui Quang Tuan
        >
        >
        >
        >---------------------------------
        >Be a PS3 game guru.
        >Get your game face on with the latest PS3 news and previews at Yahoo! Games.
        >
        >[Non-text portions of this message have been removed]
        >
        >
        >
        >
        >
        >Yahoo! Groups Links
        >
        >
        >
      • Francois Rideau
        Anh Tuan oi I think the point P is known for a long time. It is X(74). It was found by Dobbs more than 100 years ago with just your way! Friendly François ...
        Message 3 of 14 , Mar 2, 2007
          Anh Tuan oi
          I think the point P is known for a long time.
          It is X(74). It was found by Dobbs more than 100 years ago with just your
          way!
          Friendly
          François


          On 02 Mar 2007 08:53:09 -0800, Quang Tuan Bui <bqtuan1962@...> wrote:
          >
          > Dear All My Friends,
          > Given triangle ABC with circumcircle (O). There is and exactly one
          > equilateral triangle inscribed in (O) taken A as one vertex. We denote its
          > bottom line wrt A as La, intesection of La with sideline BC as A'. Similarly
          > define Lb, B', Lc, C'. Three lines La, Lb, Lc bound one triangle A''B''C''.
          > Please prove following results:
          > 1. ABC and A''B''C'' are perspective with perspector P on circumcircle
          > (O).
          > 2. A', B', C' are collinear on one line L cutting (O) at two point Q, R.
          > 3. PQR are our fourth equilateral.
          > Thank you and best regards,
          > Bui Quang Tuan
          >
          >
          > ---------------------------------
          > Be a PS3 game guru.
          > Get your game face on with the latest PS3 news and previews at Yahoo!
          > Games.
          >
          > [Non-text portions of this message have been removed]
          >
          >
          >


          [Non-text portions of this message have been removed]
        • Andreas P. Hatzipolakis
          Dear Tuan and Paul Since there are no islands in the Geometry sea , I am wondering how can we generalize it. Probably a way is this: Let ABC be a triangle
          Message 4 of 14 , Mar 2, 2007
            Dear Tuan and Paul

            Since there are no "islands" in the Geometry "sea", I am wondering
            how can we generalize it.

            Probably a way is this:

            Let ABC be a triangle and (O) its circumcircle.

            We inscribe in (O) THE isosceles triangle AB*C* with
            A = w , B* = C* = (pi-w)/2
            Denote the line B*C* as La

            Similarly the lines Lb,Lc.

            Is the Triangle bounded by the lines La,Lb,Lc and the Triangle ABC
            perspective for every given angle w (with 0 < w < pi) ?
            etc

            APH



            [Bui Quang Tuan]
            > > Given triangle ABC with circumcircle (O). There is and exactly one
            >> equilateral triangle inscribed in (O) taken A as one vertex. We denote
            >> its bottom line wrt A as La, intesection of La with sideline BC as A'.
            >> Similarly define Lb, B', Lc, C'. Three lines La, Lb, Lc bound one
            >> triangle A''B''C''. Please prove following results:
            >> 1. ABC and A''B''C'' are perspective with perspector P on
            >>circumcircle (O).
            >> 2. A', B', C' are collinear on one line L cutting (O) at two point Q, R.
            > > 3. PQR are our fourth equilateral.

            [PY]
            >How wonderful. I make a quick sketch and see that your P is the isogonal
            >conjugate of the infinite point of the Euler line.
            >


            --
          • pneagoe
            ... one equilateral triangle inscribed in (O) taken A as one vertex. We denote its bottom line wrt A as La, intesection of La with sideline BC as A . Similarly
            Message 5 of 14 , Mar 2, 2007
              --- In Hyacinthos@yahoogroups.com, Quang Tuan Bui <bqtuan1962@...>
              wrote:
              >
              > Dear All My Friends,
              > Given triangle ABC with circumcircle (O). There is and exactly
              one equilateral triangle inscribed in (O) taken A as one vertex. We
              denote its bottom line wrt A as La, intesection of La with sideline
              BC as A'. Similarly define Lb, B', Lc, C'. Three lines La, Lb, Lc
              bound one triangle A''B''C''. Please prove following results:
              > 1. ABC and A''B''C'' are perspective with perspector P on
              circumcircle (O).
              > 2. A', B', C' are collinear on one line L cutting (O) at two
              point Q, R.
              > 3. PQR are our fourth equilateral.
              > Thank you and best regards,
              > Bui Quang Tuan
              >
              >
              >
              > ---------------------------------
              > Be a PS3 game guru.
              > Get your game face on with the latest PS3 news and previews at
              Yahoo! Games.
              >
              > [Non-text portions of this message have been removed]
              >
              Dear Tuan,

              Sa=S*ctgA

              A=(1:0:0) => A=(2*(S^2):0:0)
              O=((a^2)*Sa:(b^2)*Sb:(c^2)*Sc)
              Ma is the intersection point of the lines OA and La
              AMa:MaO=3:(-1)=> Ma=(-2*(S^2)+3*(a^2)*Sa:3*(b^2)*Sb:3*(c^2)*Sc)
              let Ta be the tangent to the circumcircle at A => Ta||La
              Ta: (c^2)*y+(b^2)*z=0
              => the infinite point of La = the infinite point of Ta =
              =((c^2)-(b^2):b^2:-(c^2))
              => La: 3*(R^2)*x+(3*(R^2)-(c^2))*y+(3*(R^2)-(b^2))*z = 0
              BC: x=0
              => A'=(0:3*(R^2)-(b^2):-3*(R^2)+(c^2))
              Similarly => B'=(-3*(R^2)+(a^2):0:3*(R^2)-(c^2))
              and C'=(3*(R^2)-(a^2):-3*(R^2)+(b^2):0)
              => the points A', B' and C' are collinear.
              L:(1/(3*(R^2)-(a^2)))*x+(1/(3*(R^2)-(b^2)))*y+(1/(3*(R^2)-(c^2)))*z=0

              Best regards,
              Petrisor Neagoe
            • Francois Rideau
              Anh Tuan oi As Paul notices, your P is X(74) wrt ABC. To prove the last property of the line A B C , I think the best is to use polarity wrt the imaginary
              Message 6 of 14 , Mar 2, 2007
                Anh Tuan oi
                As Paul notices, your P is X(74) wrt ABC.
                To prove the last property of the line A'B'C', I think the best is to use
                polarity wrt the imaginary circle with equation:
                x² + y² + R²/2 = 0
                in any orthonormal frame with origin the ABC-circumcenter,
                where R is the ABC-circumradius.
                Friendly
                Francois


                [Non-text portions of this message have been removed]
              • pneagoe
                Dear Andreas, Let ABC be a triangle with circumcircle C(O) and the incircle c(I). Ca is the circle with center A and radius AI. La is the radical axis of the
                Message 7 of 14 , Mar 2, 2007
                  Dear Andreas,

                  Let ABC be a triangle with circumcircle C(O) and the incircle c(I).
                  Ca is the circle with center A and radius AI.
                  La is the radical axis of the circles Ca and C(O).
                  A' is the intersection point of the lines La and BC.
                  Similarly define Cb, Lb, B', Cc, Lc, C'. Three lines La, Lb, Lc
                  bound one triangle A"B"C".

                  1.La is tangent to the incircle c(I).
                  2.A', B', C' are collinear points.
                  3.AA", BB" and CC" are concurrent lines.

                  Best regards,
                  Petrisor Neagoe




                  --- In Hyacinthos@yahoogroups.com, "Andreas P. Hatzipolakis"
                  <xpolakis@...> wrote:
                  >
                  > Dear Tuan and Paul
                  >
                  > Since there are no "islands" in the Geometry "sea", I am wondering
                  > how can we generalize it.
                  >
                  > Probably a way is this:
                  >
                  > Let ABC be a triangle and (O) its circumcircle.
                  >
                  > We inscribe in (O) THE isosceles triangle AB*C* with
                  > A = w , B* = C* = (pi-w)/2
                  > Denote the line B*C* as La
                  >
                  > Similarly the lines Lb,Lc.
                  >
                  > Is the Triangle bounded by the lines La,Lb,Lc and the Triangle ABC
                  > perspective for every given angle w (with 0 < w < pi) ?
                  > etc
                  >
                  > APH
                  >
                • pneagoe
                  Dear Andreas, If the points A , B , C lies on the line L and the lines AA , BB , CC are concurrent at P then : 4. the point P lies on the line OI. 5. OI and
                  Message 8 of 14 , Mar 2, 2007
                    Dear Andreas,

                    If the points A', B', C' lies on the line L and
                    the lines AA", BB", CC" are concurrent at P then :

                    4. the point P lies on the line OI.
                    5. OI and L are perpendicular lines.

                    Best regards,
                    Petrisor Neagoe




                    --- In Hyacinthos@yahoogroups.com, "pneagoe" <pneagoe@...> wrote:
                    >
                    > Dear Andreas,
                    >
                    > Let ABC be a triangle with circumcircle C(O) and the incircle c(I).
                    > Ca is the circle with center A and radius AI.
                    > La is the radical axis of the circles Ca and C(O).
                    > A' is the intersection point of the lines La and BC.
                    > Similarly define Cb, Lb, B', Cc, Lc, C'. Three lines La, Lb, Lc
                    > bound one triangle A"B"C".
                    >
                    > 1.La is tangent to the incircle c(I).
                    > 2.A', B', C' are collinear points.
                    > 3.AA", BB" and CC" are concurrent lines.
                    >
                    > Best regards,
                    > Petrisor Neagoe
                    >
                    >
                    >
                  • Paul Yiu
                    Dear friends, BQT (14935) Given triangle ABC with circumcircle (O). There is and exactly one equilateral triangle inscribed in (O) taken A as one vertex. We
                    Message 9 of 14 , Mar 2, 2007
                      Dear friends,

                      BQT (14935) Given triangle ABC with circumcircle (O). There is and
                      exactly one equilateral triangle inscribed in (O) taken A as one
                      vertex. We denote its bottom line wrt A as L_a, intesection of L_a
                      with sideline BC as A'. Similarly define L_b, B', L_c, C'. Three lines
                      L_a, L_b, L_c bound one triangle A''B''C''.

                      (1) ABC and A''B''C'' are perspective at a point P on the circumcircle
                      (O).

                      (2) A', B', C' are collinear on one line L.

                      (3) If L intersects (O) at two points Q, R, then PQR is equilateral.


                      PY (14936) How wonderful. I make a quick sketch and see that your
                      P is the isogonal conjugate of the infinite point of the Euler
                      line.

                      *** The line L_a is the perpendicular to OA at the point which
                      divides OA in the ratio 3:-1.

                      FR (14939) I think the point P is known for a long time. It is X(74).
                      It was found by Dobbs more than 100 years ago with just your way!

                      APH (14940) We inscribe in (O) THE isosceles triangle AB^*C^* with
                      A = \omega , B* = C* = \frac {\pi-\omega}2. Denote the line B^*C^*
                      as L_a. Similarly the lines L_b, L_c.

                      Is the Triangle bounded by the lines L_a, L_b, L_c perspective
                      with ABC for every given angle \omega (with 0 < \omega < \pi) ?

                      *** Yes, the locus is the Jerabek hyperbola. If L_a is the
                      perpendicular to OA at the point which divides AO in the ratio t:1-t,
                      this perspector is the point

                      (\frac {a^2}{S^2-t a^2S_A}: ... : ...).


                      PN (14944, 14946):

                      Let ABC be a triangle with circumcircle (O) and the incircle (I).
                      C_a is the circle with center A and radius AI. L_a is the radical
                      axis of the circles C_a and (O). A' is the intersection point of
                      the lines L_a and BC. Similarly define C_b, L_b, B', C_c, L_c, C'.
                      Three lines L_a, L_b, L_c bound one triangle A"B"C".

                      (1) L_a is tangent to the incircle (I).

                      (2) A', B', C' are collinear points.

                      (3) AA", BB" and CC" are concurrent lines.

                      (4) The point P lies on the line OI.

                      *** This is X(65), the intersection of OI with the Jerabek hyperbola.
                      It is also the isogonal conjugate of the Schiffler point,

                      (5) OI and L are perpendicular lines.

                      *** This is the polar wrt to (O) of the exsimilicenter of (O) and (I).

                      Best regards
                      Sincerely
                      Paul
                    • Quang Tuan Bui
                      Dear All My Friends, Thank you very much for your references and remarks. I have some results: - L bound with ABC one complete quadrilateral with Miquel point
                      Message 10 of 14 , Mar 2, 2007
                        Dear All My Friends,
                        Thank you very much for your references and remarks.
                        I have some results:
                        - L bound with ABC one complete quadrilateral with Miquel point = X(110)
                        - Three lines connected A, B, C with reps Miquel circle centers are concurrent at X(477)
                        - Midpoint of QR = X(1511)
                        Best regards,
                        Bui Quang Tuan


                        Paul Yiu <yiu@...> wrote: Dear friends,

                        BQT (14935) Given triangle ABC with circumcircle (O). There is and
                        exactly one equilateral triangle inscribed in (O) taken A as one
                        vertex. We denote its bottom line wrt A as L_a, intesection of L_a
                        with sideline BC as A'. Similarly define L_b, B', L_c, C'. Three lines
                        L_a, L_b, L_c bound one triangle A''B''C''.

                        (1) ABC and A''B''C'' are perspective at a point P on the circumcircle
                        (O).

                        (2) A', B', C' are collinear on one line L.

                        (3) If L intersects (O) at two points Q, R, then PQR is equilateral.


                        PY (14936) How wonderful. I make a quick sketch and see that your
                        P is the isogonal conjugate of the infinite point of the Euler
                        line.

                        *** The line L_a is the perpendicular to OA at the point which
                        divides OA in the ratio 3:-1.

                        FR (14939) I think the point P is known for a long time. It is X(74).
                        It was found by Dobbs more than 100 years ago with just your way!

                        APH (14940) We inscribe in (O) THE isosceles triangle AB^*C^* with
                        A = \omega , B* = C* = \frac {\pi-\omega}2. Denote the line B^*C^*
                        as L_a. Similarly the lines L_b, L_c.

                        Is the Triangle bounded by the lines L_a, L_b, L_c perspective
                        with ABC for every given angle \omega (with 0 < \omega < \pi) ?

                        *** Yes, the locus is the Jerabek hyperbola. If L_a is the
                        perpendicular to OA at the point which divides AO in the ratio t:1-t,
                        this perspector is the point

                        (\frac {a^2}{S^2-t a^2S_A}: ... : ...).


                        PN (14944, 14946):

                        Let ABC be a triangle with circumcircle (O) and the incircle (I).
                        C_a is the circle with center A and radius AI. L_a is the radical
                        axis of the circles C_a and (O). A' is the intersection point of
                        the lines L_a and BC. Similarly define C_b, L_b, B', C_c, L_c, C'.
                        Three lines L_a, L_b, L_c bound one triangle A"B"C".

                        (1) L_a is tangent to the incircle (I).

                        (2) A', B', C' are collinear points.

                        (3) AA", BB" and CC" are concurrent lines.

                        (4) The point P lies on the line OI.

                        *** This is X(65), the intersection of OI with the Jerabek hyperbola.
                        It is also the isogonal conjugate of the Schiffler point,

                        (5) OI and L are perpendicular lines.

                        *** This is the polar wrt to (O) of the exsimilicenter of (O) and (I).

                        Best regards
                        Sincerely
                        Paul






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                      • jpehrmfr
                        Dear friends [QTB] ... one equilateral triangle inscribed in (O) taken A as one vertex. We denote its bottom line wrt A as La, intesection of La with sideline
                        Message 11 of 14 , Mar 3, 2007
                          Dear friends
                          [QTB]
                          > Given triangle ABC with circumcircle (O). There is and exactly
                          one equilateral triangle inscribed in (O) taken A as one vertex. We
                          denote its bottom line wrt A as La, intesection of La with sideline
                          BC as A'. Similarly define Lb, B', Lc, C'. Three lines La, Lb, Lc
                          bound one triangle A''B''C''. Please prove following results:
                          > 1. ABC and A''B''C'' are perspective with perspector P on
                          circumcircle (O).
                          > 2. A', B', C' are collinear on one line L cutting (O) at two
                          point Q, R.
                          > 3. PQR are our fourth equilateral.

                          As Antreas noticed, we have something similar if we start with an
                          isosceles triangle with vertex A
                          Here is something else, but quite similar (three directly congruent
                          triangles inscribed in (O), one with vertex A, one with vertex B, one
                          with vertex C)
                          L = homothetic of A in (O,-1/2); CL is the circle with center L going
                          through the midpoint of BC. P is any point of CL, La is perpendicular
                          at P to OP (La will enveloppe some conic with focus O and center L)
                          If La intersects (O) at Ua and Va; the rotation with center O mapping
                          A to B maps Ua and Va to Ub and Ub; similarly, the rotation with
                          center O mapping A to C maps Ua and Va to Uc and Vc. Lb = line UbVb;
                          Lc = line UcVc
                          A' = La inter BC; A'' = Lb inter Lc; ...
                          Then
                          A',B',C',O lie on a same line
                          AA'', BB'', CC'' are parallel
                          O is the incenter or an excenter of A''B''C''
                          A''B''C'' is directly similar with the pedal triangle of ABC
                          Friendly. Jean-Pierre
                        • jpehrmfr
                          Dear friends ... [JPE] ... one ... going ... perpendicular ... mapping ... UbVb; ... Here is a more symetric way to draw this figure : Ca is the circle going
                          Message 12 of 14 , Mar 3, 2007
                            Dear friends

                            > [QTB]
                            > > Given triangle ABC with circumcircle (O). There is and exactly
                            > one equilateral triangle inscribed in (O) taken A as one vertex. We
                            > denote its bottom line wrt A as La, intesection of La with sideline
                            > BC as A'. Similarly define Lb, B', Lc, C'. Three lines La, Lb, Lc
                            > bound one triangle A''B''C''. Please prove following results:
                            > > 1. ABC and A''B''C'' are perspective with perspector P on
                            > circumcircle (O).
                            > > 2. A', B', C' are collinear on one line L cutting (O) at two
                            > point Q, R.
                            > > 3. PQR are our fourth equilateral.

                            [JPE]
                            > As Antreas noticed, we have something similar if we start with an
                            > isosceles triangle with vertex A
                            > Here is something else, but quite similar (three directly congruent
                            > triangles inscribed in (O), one with vertex A, one with vertex B,
                            one
                            > with vertex C)
                            > L = homothetic of A in (O,-1/2); CL is the circle with center L
                            going
                            > through the midpoint of BC. P is any point of CL, La is
                            perpendicular
                            > at P to OP (La will enveloppe some conic with focus O and center L)
                            > If La intersects (O) at Ua and Va; the rotation with center O
                            mapping
                            > A to B maps Ua and Va to Ub and Ub; similarly, the rotation with
                            > center O mapping A to C maps Ua and Va to Uc and Vc. Lb = line
                            UbVb;
                            > Lc = line UcVc
                            > A' = La inter BC; A'' = Lb inter Lc; ...
                            > Then
                            > A',B',C',O lie on a same line
                            > AA'', BB'', CC'' are parallel
                            > O is the incenter or an excenter of A''B''C''
                            > A''B''C'' is directly similar with the pedal triangle of ABC

                            Here is a more symetric way to draw this figure :
                            Ca is the circle going through A with center the common point of the
                            perpendicular bisector of BC with the line AX[74] (The radius of this
                            circle is OH/|2.cos A|). Similarly, we get Cb, Cc (may be, there is a
                            special name for these three circles because they are closely
                            connected with the Mac Cay circles)
                            Three parallel lines going respectively through A, B, C intersect
                            again Ca, Cb, Cc respectively at A'', B'', C''; A' = BC inter
                            B''C'', ...
                            I've checked by computation, that, if we consider three directly
                            congruent non isosceles triangles inscribed in the circumcircle,
                            Ta,Tb,Tc where A is a vertex of Ta and La the opposite sideline, then
                            the points BC inter La, CA inter Lb, AB inter Lc will lie on a same
                            line only in the case above.
                            Friendly. Jean-Pierre
                          • jpehrmfr
                            Dear friends I wrote ... the ... this ... a ... then ... In fact, this is a bit tedious because the centers of the three circles Ca, Cb, Cc are homothetic in
                            Message 13 of 14 , Mar 3, 2007
                              Dear friends
                              I wrote

                              > Here is a more symetric way to draw this figure :
                              > Ca is the circle going through A with center the common point of
                              the
                              > perpendicular bisector of BC with the line AX[74] (The radius of
                              this
                              > circle is OH/|2.cos A|). Similarly, we get Cb, Cc (may be, there is
                              a
                              > special name for these three circles because they are closely
                              > connected with the Mac Cay circles)
                              > Three parallel lines going respectively through A, B, C intersect
                              > again Ca, Cb, Cc respectively at A'', B'', C''; A' = BC inter
                              > B''C'', ...
                              > I've checked by computation, that, if we consider three directly
                              > congruent non isosceles triangles inscribed in the circumcircle,
                              > Ta,Tb,Tc where A is a vertex of Ta and La the opposite sideline,
                              then
                              > the points BC inter La, CA inter Lb, AB inter Lc will lie on a same
                              > line only in the case above.

                              In fact, this is a bit tedious because the centers of the three
                              circles Ca, Cb, Cc are homothetic in (O,-1/2) of the vertices of the
                              tangential triangle.
                              That's enough for me for today.
                              Friendly. Jean-Pierre
                            • Alexey.A.Zaslavsky
                              Dear Quang! The point P is isogonally conjugated to the infinite point of Euler line, i.e the common point of the circumcircle and the Neuberg cubic distinct
                              Message 14 of 14 , Mar 5, 2007
                                Dear Quang!
                                The point P is isogonally conjugated to the infinite point of Euler line, i.e the common point of the circumcircle and the Neuberg cubic distinct from A, B, C and two circular points.

                                Sincerely Alexey

                                I
                                Dear All My Friends,
                                > Given triangle ABC with circumcircle (O). There is and exactly one
                                > equilateral triangle inscribed in (O) taken A as one vertex. We denote its
                                > bottom line wrt A as La, intesection of La with sideline BC as A'. Similarly
                                > define Lb, B', Lc, C'. Three lines La, Lb, Lc bound one triangle A''B''C''.
                                > Please prove following results:
                                > 1. ABC and A''B''C'' are perspective with perspector P on circumcircle
                                > (O).
                                > 2. A', B', C' are collinear on one line L cutting (O) at two point Q, R.
                                > 3. PQR are our fourth equilateral.
                                > Thank you and best regards,
                                >
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