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Concurrence of circles in Arbelos

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  • emmanuelantoniojosgarca
    Dear friends, Consider an Arbelos as in the figure. Let D, E, F be the highest points of semicircles AB, AC, BC, respectively. Draw a circle (D) centred at D,
    Message 1 of 9 , Mar 19
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    Dear friends,

    Consider an Arbelos as in the figure. Let D, E, F
    be the highest points of semicircles AB, AC, BC,
    respectively. Draw a circle (D) centred at D, radius CD.
    Similarly, draw  a circle (E) centred at E, radius BE and
    draw a circle (F) centred at F, radius AF.

    Then, circles (D), (E), (F) concur. 


    Thinking on a proof...


    Best regards,
    Emmanuel.


  • garciacapitan
    Nice. JC is perpendicular and equal to AB. Best regards, Francisco Javier.
    Message 2 of 9 , Mar 19
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      Nice.

      JC is perpendicular and equal to AB.

      Best regards,

      Francisco Javier.
    • Paul Yiu
      Dear Emmanuel, [EG]: Consider an Arbelos as in the figure. Let D, E, F be the highest points of semicircles AB, AC, BC, respectively. Draw a circle (D)
      Message 3 of 9 , Mar 19
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        Dear Emmanuel,

        [EG]:  Consider an Arbelos as in the figure. Let D, E, F 
        be the highest points of semicircles AB, AC, BC, 
        respectively. Draw a circle (D) centred at D, radius CD.
        Similarly, draw  a circle (E) centred at E, radius BE and
        draw a circle (F) centred at F, radius AF.

        Then, circles (D), (E), (F) concur. 

        *** Consider the square on AB, on the same side of the arbelos. 

        Construct the perpendicular to AB at C to intersect the opposite side of the square at P. 
        Since D is the center of the square, DP = DC.
        Construct the perpendicular to AB at E. 
        This gives two congruent right triangles in which EP andEB are hypotenuses. 
        Therefore, EP = EB.
        Similarly, FP= FA.
        This shows that P is a common point of the circles D(C), E(B), F(A).

        Best regards
        Sincerely
        Paul Yiu

        ___

      • Paul Yiu
        Dear Emmanuel and Francisco Javier, [EG]: Consider an Arbelos as in the figure. Let D, E, F be the highest points of semicircles AB, AC, BC, respectively.
        Message 4 of 9 , Mar 19
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          Dear Emmanuel and Francisco Javier,

          [EG]:  Consider an Arbelos as in the figure. Let D, E, F 
          be the highest points of semicircles AB, AC, BC, 
          respectively. Draw a circle (D) centred at D, radius CD.
          Similarly, draw  a circle (E) centred at E, radius BE and
          draw a circle (F) centred at F, radius AF.

          Then, circles (D), (E), (F) concur. 

          [PY]:  Consider the square on AB, on the same side of the arbelos. 

          Construct the perpendicular to AB at C to intersect the opposite side of the square at P. 
          Since D is the center of the square, DP = DC.
          Construct the perpendicular to AB at E. 
          This gives two congruent right triangles in which EP andEB are hypotenuses. 
          Therefore, EP = EB.
          Similarly, FP= FA.
          This shows that P is a common point of the circles D(C), E(B), F(A).

          *** Let the line AF intersect the semicircle BC at Y,
          the line BE intersect the semicircle AC at X, 
          and the lines AF, BE intersect at Q.
          Join CQ to intersect the semicircle AB at Z.

          Then the circle XYZ is the incircle of the arbelos.

          Best regards
          Sincerely
          Paul Yiu

          ___



        • tsihonglau
          Dear geometers, I use my notations. AC B, BA C, AB C are three semicircles clockwise. There are many properties in the attachment. Please refer to it and I
          Message 5 of 9 , Mar 19
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          Dear geometers,

          I use my notations.  AC'B, BA'C, AB'C
          are three semicircles clockwise.  There
          are many properties in the attachment.
          Please refer to it and I do not explain.

          Best regards,
          Tsihong Lau
        • emmanuelantoniojosgarca
          Dear Francisco Javier, Paul Yiu and Tsihong Lau, Thank you very much for your comments and to Paul Yiu for the proof. To Paul Yiu: I have sent a message to you
          Message 6 of 9 , Mar 20
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            Dear Francisco Javier, Paul Yiu and Tsihong Lau,

            Thank you very much for your comments and to Paul Yiu for the proof.

            To Paul Yiu:
            I have sent a message to you via gmail. Probably, you haven t seen it.

            Best regards,
            Emmanuel.
          • Angel Montesdeoca
            Application to geometry of triangle: Let ABC be a triangle and DEF the cevian triangle of a point P. Consider the three semicircles built in the opposite
            Message 7 of 9 , Mar 21
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            Application to geometry of triangle:

            Let ABC be a triangle and DEF the cevian triangle of a point P.
            Consider the three semicircles built in the opposite direction clockwise
            on the segments BC, BD and DC.
            Let Da, Db, Dc be the highest points of these semicircles, respec.

            The three circles with centers in Da, Db and Dc and passing through D,
            C and B, respectively, are concurrent at a point A' (on the
            perpendicular to BC by D and such that BC = DA' --Francisco Javier).

            Similarly, the points B' and C' are determined, proceeding cyclically on
            the sides of ABC.

            The triangles ABC and A'B'C' are perspective if and only if P is on the
            cubic circumscribed to ABC of barycentric equation:

            CyclicSum[x(SB(SC+S)y^2-SC(SB+S)z^2)]=0.

            If we consider the semicircles built in the sense
            clockwise, on BC, BD and DC segments, the points of intersection of the
            three corresponding circles are A''. B'' and C'' (reflexion of A', B'
            and C' with respect to sidelines BC, CA, AB, respec.)., the cubic is:

            CyclicSum[x(SB(SC-S)y^2-SC(SB-S)z^2)]=0.

            See more figures:
            http://amontes.webs.ull.es/otrashtm/HGT2015.htm#HG210315

            Best regards,

            Angel M.
          • Bernard Gibert
            Dear Angel, ... your cubics are pK(X1585,X4) and pK(X1586,X4) very similar to K070a,b. the loci of the perspectors are pK(X1131 x X1321, X1321) and pK(X1132 x
            Message 8 of 9 , Mar 21
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              Dear Angel,

              [AM]Let ABC be a triangle and DEF the cevian triangle of a point P.
              Consider the three semicircles built in the opposite direction clockwise 
              on the segments BC, BD and DC.
              Let Da, Db, Dc be the highest points of these semicircles, respec.

              The three circles with centers in Da, Db and Dc and passing through D, 
              C and B, respectively, are concurrent at a point A' (on the 
              perpendicular to BC by D and such that BC = DA' --Francisco Javier).

              Similarly, the points B' and C' are determined, proceeding cyclically on 
              the sides of ABC.

              The triangles ABC and A'B'C' are perspective if and only if P is on the 
              cubic circumscribed to ABC of barycentric equation:

              CyclicSum[x(SB(SC+S)y^2-SC(SB+S)z^2)]=0.

              If we consider the semicircles built in the sense 
              clockwise, on BC, BD and DC segments, the points of intersection of the 
              three corresponding circles are A''. B'' and C'' (reflexion of A', B' 
              and C' with respect to sidelines BC, CA, AB, respec.)., the cubic is:

              CyclicSum[x(SB(SC-S)y^2-SC(SB-S)z^2)]=0.

              your cubics are pK(X1585,X4) and pK(X1586,X4) very similar to K070a,b.

              the loci of the perspectors are pK(X1131 x X1321, X1321) and pK(X1132 x X1322, X1322) resp.

              Best regards

              Bernard
            • Ignacio Larrosa Cañestro
              El 19/03/2015 a las 23:44, Paul Yiu yiu@fau.edu [AdvancedPlaneGeometry] ... Let the line CD intersect the semicircle BA (not containing C) at Z . Then Z is
              Message 9 of 9 , Mar 22
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                El 19/03/2015 a las 23:44, Paul Yiu yiu@... [AdvancedPlaneGeometry]
                escribió:
                >
                >
                > Dear Emmanuel and Francisco Javier,
                >
                > [EG]: Consider an Arbelos as in the figure. Let D, E, F
                > be the highest points of semicircles AB, AC, BC,
                > respectively. Draw a circle (D) centred at D, radius CD.
                > Similarly, draw a circle (E) centred at E, radius BE and
                > draw a circle (F) centred at F, radius AF.
                >
                > Then, circles (D), (E), (F) concur.
                >
                > [PY]: Consider the square on AB, on the same side of the arbelos.
                >
                > Construct the perpendicular to AB at C to intersect the opposite side of
                > the square at P.
                > Since D is the center of the square, DP = DC.
                > Construct the perpendicular to AB at E.
                > This gives two congruent right triangles in which EP andEB are hypotenuses.
                > Therefore, EP = EB.
                > Similarly, FP= FA.
                > This shows that P is a common point of the circles D(C), E(B), F(A).
                >
                > *** Let the line AF intersect the semicircle BC at Y,
                > the line BE intersect the semicircle AC at X,
                > and the lines AF, BE intersect at Q.
                > Join CQ to intersect the semicircle AB at Z.
                >
                > Then the circle XYZ is the incircle of the arbelos.

                Let the line CD intersect the semicircle BA (not containing C) at Z'.
                Then Z is symmetric of Z' with respect to segment AB.


                --
                Best regards,

                Ignacio Larrosa Cañestro
                A Coruña (España)
                ilarrosa@...
                http://www.xente.mundo-r.com/ilarrosa/GeoGebra/
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