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The Brocard points of a quadrilateral (Geoff Millin)

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  • Darij Grinberg
    An interesting thread was started in the geometry-college group: http://mathforum.org/epigone/geometry-college/pringkheldeh ... Subject: Brocard points of
    Message 1 of 15 , Jun 1, 2004
      An interesting thread was started in the
      geometry-college group:

      http://mathforum.org/epigone/geometry-college/pringkheldeh

      ----------------------------------------------------

      Subject: Brocard points of quadrilaterals
      Author: geoff millin <cgmillin*tiscali.co.uk>
      replace the sign * by @

      Correction:

      ABCD is a cyclic quadrilateral. I have proved
      that a necessary and sufficient condition for
      it to have Brocard points is that the products
      of the opposite sides are equal. Ie
      AB.CD = AD.BC.

      I expect this is a well-known fact, but I can
      find no reference to it on the web.

      ----------------------------------------------------

      Subject: Re: Brocard points of quadrilaterals
      Author: Darij Grinberg

      Dear Geoff Millin,

      You wrote:

      >> ABCD is a cyclic quadrilateral. I have proved
      >> that a necessary and sufficient condition for
      >> it to have Brocard points is that the products
      >> of the opposite sides are equal. Ie
      >> AB.CD = AD.BC.

      Congratulations for having discovered a nice new
      fact in Euclidean Geometry! A cyclic
      quadrilateral ABCD satisfying AB * CD = AD * BC
      is called a "harmonic quadrilateral"; such
      quadrilaterals were studied in the 19th century,
      and many properties of them were uncovered. Just
      two examples: The line AC is a symmedian in the
      triangles DAB and BCD simultaneously, and the
      line BD is a symmedian in the triangles ABC and
      CDA simultaneously. And: There exists an
      inversion mapping the points A, B, C, D to the
      vertices of a square. But I have never seen your
      result about Brocard points mentioned anywhere,
      it is most likely new.

      I have tried to explore your Brocard points by
      dynamic geometry software. One of the nice
      properties (I have no proof yet) of these points
      is that the two "Brocard angles" are equal, much
      like in a triangle. In other words, if you take
      a harmonic quadrilateral ABCD and its two
      Brocard points R and R', with

      < RAB = < RBC = < RCD = < RDA and
      < R'BA = < R'CB = < R'DC = < R'AD,

      then

      < RAB = < RBC = < RCD = < RDA
      = < R'BA = < R'CB = < R'DC = < R'AD.

      Equivalently, the Brocard points R and R' are
      "isogonal conjugates" with respect to the
      quadrilateral ABCD, i. e. the lines AR and AR'
      are symmetric with respect to the angle
      bisector of the angle DAB, and similarly for
      the other pairs of lines. Generally, not every
      point in the plane of our quadrilateral ABCD
      has an isogonal conjugate, but these Brocard
      points have isogonal conjugates, namely the
      isogonal conjugate of one Brocard point is the
      other one.

      I don't see, however, any formula of the kind
      cot w = cot A + cot B + cot C for the Brocard
      angle of a harmonic quadrilateral.

      May I ask you how you proved your original
      result? Did you perform a calculation using
      coordinates or do you have a synthetic proof?
      Do you also have an idea about my above
      conjectures?

      Sincerely,
      Darij Grinberg

      ----------------------------------------------------
    • dick tahta
      Darij raised some interesting questions about harmonic quadrilaterals. ... Casey, A sequel to Euclid , has various theorems on harmonic n-gons, including a
      Message 2 of 15 , Jun 1, 2004
        Darij raised some interesting questions about harmonic quadrilaterals.

        > I don't see, however, any formula of the kind
        > cot w = cot A + cot B + cot C for the Brocard
        > angle of a harmonic quadrilateral.

        Casey, 'A sequel to Euclid', has various theorems on harmonic n-gons,
        including a formula for the Bocard angle ascribed to Tucker, namely:
        cosec^w = cosec^A + cosec^B + ....

        Casey also gives the result (for n=4) about equal products of opposite
        sides. It is also in Johnson's Advanced Euclidean Geometry' (Th 133) which
        also quotes (Th504) further Brocard properties ascribed to Tucker.

        Dick Tahta
      • Darij Grinberg
        Dear Dick, ... Thank you very much for the information. My dynamic sketch confirms csc^2 w = csc^2 A + csc^2 B for a harmonic quadrilateral. But what is the
        Message 3 of 15 , Jun 1, 2004
          Dear Dick,

          In Hyacinthos message #9830, you wrote:

          >> Casey, 'A sequel to Euclid', has various
          >> theorems on harmonic n-gons, including a
          >> formula for the Bocard angle ascribed to
          >> Tucker, namely:
          >> cosec^w = cosec^A + cosec^B + ....

          Thank you very much for the information.
          My dynamic sketch confirms

          csc^2 w = csc^2 A + csc^2 B

          for a harmonic quadrilateral. But what is
          the definition of an harmonic n-gon?

          Sincerely,
          Darij Grinberg
        • Alexey.A.Zaslavsky
          Dear Geoff Millin and Darij! ... This results are very interesting. I have one question. It is known that the projections of two isogonally conjugated points
          Message 4 of 15 , Jun 2, 2004
            Dear Geoff Millin and Darij!
            >
            >>> ABCD is a cyclic quadrilateral. I have proved
            >>> that a necessary and sufficient condition for
            >>> it to have Brocard points is that the products
            >>> of the opposite sides are equal. Ie
            >>> AB.CD = AD.BC.
            >
            >
            >I have tried to explore your Brocard points by
            >dynamic geometry software. One of the nice
            >properties (I have no proof yet) of these points
            >is that the two "Brocard angles" are equal, much
            >like in a triangle. In other words, if you take
            >a harmonic quadrilateral ABCD and its two
            >Brocard points R and R', with
            >
            > < RAB = < RBC = < RCD = < RDA and
            > < R'BA = < R'CB = < R'DC = < R'AD,
            >
            >then
            >
            > < RAB = < RBC = < RCD = < RDA
            > = < R'BA = < R'CB = < R'DC = < R'AD.
            >
            This results are very interesting. I have one question. It is known that the
            projections of two isogonally conjugated points to the sidelines of ABCD are
            in the circle. The center of this circle is the midpoint of segment RR' and
            it is in Gauss line of ABCD. What is this center if R and R' are the Brocard
            points?

            Sincerely Alexey
          • Darij Grinberg
            Dear Alexey, ... This is very interesting! At first, consider the Brocard points R and R of the harmonic quadrilateral ABCD, and the symmedian point K (this
            Message 5 of 15 , Jun 2, 2004
              Dear Alexey,

              In Hyacinthos message #9842, you wrote:

              >> It is known that the projections of two
              >> isogonally conjugated points to the
              >> sidelines of ABCD are in the circle. The
              >> center of this circle is the midpoint of
              >> segment RR' and it is in Gauss line of
              >> ABCD. What is this center if R and R'
              >> are the Brocard points?

              This is very interesting! At first, consider
              the Brocard points R and R' of the harmonic
              quadrilateral ABCD, and the symmedian point
              K (this is the point whose distances to the
              sides of the quadrilateral are proportional
              to the lengths of the sides). Also consider
              the circumcenter O of our quadrilateral.
              Then, the points O, K, R and R' lie on one
              circle, which is analogous to the "Brocard
              circle" of a triangle. The points O and K
              are the endpoints of a diameter of this
              circle. The points R and R' lie
              symmetrically with respect to this diameter.
              The circle through the points O, K, R and R'
              also passes through the midpoints of the
              diagonals AC and BD. The line joining these
              midpoints meets the diameter OK at the
              midpoint of the segment RR'. The angles
              < ROK and < R'OK are equal to the Brocard
              angle of the quadrilateral ABCD.

              Would be very nice if anybody had a proof to
              all of this...

              Sincerely,
              Darij Grinberg
            • Alexey.A.Zaslavsky
              Dear Darij! ... I can prove some of this facts. Let P is the first Brocard point and the lines AP, BP, CP, DP intersect the circumcircle in B , C , D , A .
              Message 6 of 15 , Jun 4, 2004
                Dear Darij!
                >
                >This is very interesting! At first, consider
                >the Brocard points R and R' of the harmonic
                >quadrilateral ABCD, and the symmedian point
                >K (this is the point whose distances to the
                >sides of the quadrilateral are proportional
                >to the lengths of the sides). Also consider
                >the circumcenter O of our quadrilateral.
                >Then, the points O, K, R and R' lie on one
                >circle, which is analogous to the "Brocard
                >circle" of a triangle. The points O and K
                >are the endpoints of a diameter of this
                >circle. The points R and R' lie
                >symmetrically with respect to this diameter.
                >The circle through the points O, K, R and R'
                >also passes through the midpoints of the
                >diagonals AC and BD. The line joining these
                >midpoints meets the diameter OK at the
                >midpoint of the segment RR'. The angles
                >< ROK and < R'OK are equal to the Brocard
                >angle of the quadrilateral ABCD.
                >
                >Would be very nice if anybody had a proof to
                >all of this...
                >
                I can prove some of this facts. Let P is the first Brocard point and the
                lines AP, BP, CP, DP intersect the circumcircle in B', C', D', A'. Then the
                quadrilateral A'B'C'D' is the rotation of ABCD with center O and angle 2x,
                where x is the Brocard angle. So P is the second Brocard point Q' of
                A'B'C'D', OP=OQ and angle POQ=2x. Now let K is the common point of AC and
                BD, K' - the common point of A'C' and B'D'. Then P is in KK' (see my paper
                in "Kvant" N.1, 1996). Similarly Q is in KK'', where K'' is the reflection
                of K' in OK. As K''Q=KP and POQ=K'OK=KOK''=2x, P and Q are the midpoints of
                KK' and KK" respectively. So the angles OPK and OQK are direct and P, Q are
                in the circle with diameter OK. Also it is easy to find x from equation
                sin^4(x)=sin(A-x)sin(B-x)sin(C-x)sin(D-x), but I don't understand why ABCD
                must be harmonic.

                Sincerely Alexey
              • Alexey.A.Zaslavsky
                Dear Darij! ... There is one correction. P is the midpoint of KK because OK=OK . This follows from next theorem. Let four chords A_iB_i concur in point C. D
                Message 7 of 15 , Jun 4, 2004
                  Dear Darij!
                  >>
                  >>
                  >I can prove some of this facts. Let P is the first Brocard point and the
                  >lines AP, BP, CP, DP intersect the circumcircle in B', C', D', A'. Then the
                  >quadrilateral A'B'C'D' is the rotation of ABCD with center O and angle 2x,
                  >where x is the Brocard angle. So P is the second Brocard point Q' of
                  >A'B'C'D', OP=OQ and angle POQ=2x. Now let K is the common point of AC and
                  >BD, K' - the common point of A'C' and B'D'. Then P is in KK' (see my paper
                  >in "Kvant" N.1, 1996). Similarly Q is in KK'', where K'' is the reflection
                  >of K' in OK. As K''Q=KP and POQ=K'OK=KOK''=2x, P and Q are the midpoints of
                  >KK' and KK" respectively. So the angles OPK and OQK are direct and P, Q are
                  >in the circle with diameter OK.
                  There is one correction. P is the midpoint of KK' because OK=OK'. This
                  follows from next theorem. Let four chords A_iB_i concur in point C. D is
                  the common point of A_1A_2 and A_3A_4, E - the common point of B_1B_2 and
                  B_3B_4. Then C, D, E are collinear and in Klein's model of Lobachevsky
                  geometry C is the midpoint of DE.
                  >
                  Sincerely Alexey
                • Alexey.A.Zaslavsky
                  Dear colleagues! I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then 1) The Brocard points exist iff the symmedians of triangles
                  Message 8 of 15 , Jun 7, 2004
                    Dear colleagues!
                    I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then
                    1) The Brocard points exist iff the symmedians of triangles
                    A_{i-1}A_iA_{i+1} from A_i concur in point L.
                    2) Two Brocard points P, Q are in the circle with diameter OL and the angles
                    POL and QOL are equal to brocard angle.
                    If 1) is correct then I can prove 2)

                    Sincerely Alexey
                  • jpehrmfr
                    Dear Alexey ... the angles ... And what do you think of the following conjecture : A cyclic polygon has Brocard points if and only if the polygon is the
                    Message 9 of 15 , Jun 8, 2004
                      Dear Alexey
                      > I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then
                      > 1) The Brocard points exist iff the symmedians of triangles
                      > A_{i-1}A_iA_{i+1} from A_i concur in point L.
                      > 2) Two Brocard points P, Q are in the circle with diameter OL and
                      the angles
                      > POL and QOL are equal to brocard angle.
                      > If 1) is correct then I can prove 2)

                      And what do you think of the following conjecture :
                      A cyclic polygon has Brocard points if and only if the polygon is
                      the inverse of a regular polygon?
                      Friendly. Jean-Pierre
                    • Alexey.A.Zaslavsky
                      Dear Jean-Pierre! ... I think that this two conditions are equivalent. If the point L exists then the projective transformation T conserving the circumcircle
                      Message 10 of 15 , Jun 8, 2004
                        Dear Jean-Pierre!
                        >> I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then
                        >> 1) The Brocard points exist iff the symmedians of triangles
                        >> A_{i-1}A_iA_{i+1} from A_i concur in point L.
                        >
                        >And what do you think of the following conjecture :
                        >A cyclic polygon has Brocard points if and only if the polygon is
                        >the inverse of a regular polygon?
                        >
                        I think that this two conditions are equivalent. If the point L exists then
                        the projective transformation T conserving the circumcircle and such that
                        T(L)=O transforms our polygon to regular. In the circle this transformation
                        coincide with some inversion.

                        My condition 1) follows from next fact. Let A, B, C, D are in the circle
                        with center O. P is the point such that the angles PAB, PBC and PCD are
                        equal, Q is the point such that the angles QDC, QCB and QBA are equal (P is
                        the common point distinct from B of two circles: one pass through A and B
                        and touches BC, and another pass through B and C and touches CD), L is the
                        common point of symmedians BX and CY of triangles ABC and BCD. Then O, P, Q,
                        L are in the circle.
                        It seems that this fact can be proved by calculation, but it is interesting
                        to find the synthetic proof.

                        Sincerely Alexey
                      • jpehrmfr
                        Dear Alexey [AZ] ... Then ... [JPE] ... [AZ] ... exists then ... such that ... transformation ... Suppose that A_1,...,A_n has Brocard points P,Q; let M be a
                        Message 11 of 15 , Jun 8, 2004
                          Dear Alexey
                          [AZ]
                          > >> I have a next hypothesis. Let A_1...A_n is the cyclic polygon.
                          Then
                          > >> 1) The Brocard points exist iff the symmedians of triangles
                          > >> A_{i-1}A_iA_{i+1} from A_i concur in point L.
                          [JPE]
                          > >And what do you think of the following conjecture :
                          > >A cyclic polygon has Brocard points if and only if the polygon is
                          > >the inverse of a regular polygon?
                          [AZ]
                          > I think that this two conditions are equivalent. If the point L
                          exists then
                          > the projective transformation T conserving the circumcircle and
                          such that
                          > T(L)=O transforms our polygon to regular. In the circle this
                          transformation
                          > coincide with some inversion.

                          Suppose that A_1,...,A_n has Brocard points P,Q; let M be a Poncelet
                          point of the pencil generated by the circumcircle (O) of the polygon
                          and the circle OPQ. the line MA_k intersects again (O) at B_k.
                          Then B_1,...,B_n is regular.
                          Of course, it is necessary to prove that such a point M is real, ie
                          that (O) and the circle OPQ cannot intersect.
                          It is necessary too to prove the reciprocal.
                          I think that it could be possible to name these two points M the
                          isodynamic points of the Brocardian polygon because
                          MA_(i+k)/M_A(i-k) = A_iA_(i+k)/A_iA_(i-k) for all i and k.
                          Friendly. Jean-Pierre
                        • Alexey.A.Zaslavsky
                          ... Yes, this is the same transformation. Let we considere a sphere with diameter circle (O) and the pole N. The perpendicular to the plan (O) from my point L
                          Message 12 of 15 , Jun 9, 2004
                            >[AZ]
                            >> I think that this two conditions are equivalent. If the point L
                            >exists then
                            >> the projective transformation T conserving the circumcircle and
                            >such that
                            >> T(L)=O transforms our polygon to regular. In the circle this
                            >transformation
                            >> coincide with some inversion.
                            >
                            >Suppose that A_1,...,A_n has Brocard points P,Q; let M be a Poncelet
                            >point of the pencil generated by the circumcircle (O) of the polygon
                            >and the circle OPQ. the line MA_k intersects again (O) at B_k.
                            >Then B_1,...,B_n is regular.
                            >Of course, it is necessary to prove that such a point M is real, ie
                            >that (O) and the circle OPQ cannot intersect.
                            >It is necessary too to prove the reciprocal.
                            >I think that it could be possible to name these two points M the
                            >isodynamic points of the Brocardian polygon because
                            >MA_(i+k)/M_A(i-k) = A_iA_(i+k)/A_iA_(i-k) for all i and k.
                            >
                            Yes, this is the same transformation. Let we considere a sphere with
                            diameter circle (O) and the pole N. The perpendicular to the plan (O) from
                            my point L intersect the sphere in point L' (the plan (O) divide L' and N).
                            Then the line NL' intersect (O) in your point M. (That is the corresponding
                            between Klein and Poincare models of Lobachevsky geometry). So the inversion
                            with center M is equivalent to projective transformation with center L.

                            Sincerely Alexey
                          • jpehrmfr
                            Dear Alexey, suppose that ABC is a variable isosceles triangle (in B) inscribed in a fixed circle with center o; let M be a fixed point and A B C the cevian
                            Message 13 of 15 , Jun 9, 2004
                              Dear Alexey,
                              suppose that ABC is a variable isosceles triangle (in B) inscribed
                              in a fixed circle with center o; let M be a fixed point and A'B'C'
                              the cevian triangle of M wrt ABC. It is very easy to check that the
                              B'-symedian of A'B'C' intersects the line OM at a point L
                              independant of the choice of ABC.
                              If S is a point of the circle such as OM and OS are perpendicular, L
                              is the projection upon OM of the reflection of O wrt SM. More over L
                              lies on the half-line OM and inside the circle.
                              From this it follows that if B_1,...,B_n is a regular polygon
                              inscribed in the circle, if A_k is the second intersection of MB_k
                              with the circle, every A_k symedian of A_(k-1),A_k,A_(k+) goes
                              through L.
                              That's probably a part of the proof you're waiting for.
                              Friendly. Jean-Pierre
                            • jpehrmfr
                              Dear Alexey, I wrote ... Of course, I was meaning the circumcevian triangle of M wrt ABC. Sorry for the typo. ... L ... L
                              Message 14 of 15 , Jun 9, 2004
                                Dear Alexey,
                                I wrote
                                > suppose that ABC is a variable isosceles triangle (in B) inscribed
                                > in a fixed circle with center o; let M be a fixed point and A'B'C'
                                > the cevian triangle of M wrt ABC.

                                Of course, I was meaning the circumcevian triangle of M wrt ABC.
                                Sorry for the typo.

                                > It is very easy to check that the
                                > B'-symedian of A'B'C' intersects the line OM at a point L
                                > independant of the choice of ABC.
                                > If S is a point of the circle such as OM and OS are perpendicular,
                                L
                                > is the projection upon OM of the reflection of O wrt SM. More over
                                L
                                > lies on the half-line OM and inside the circle.
                                > From this it follows that if B_1,...,B_n is a regular polygon
                                > inscribed in the circle, if A_k is the second intersection of MB_k
                                > with the circle, every A_k symedian of A_(k-1),A_k,A_(k+) goes
                                > through L.
                                > That's probably a part of the proof you're waiting for.
                                > Friendly. Jean-Pierre
                              • Darij Grinberg
                                ... These remarks started an interesting thread, which I could not follow since one half of the mathematics involved was new to me, but let me just make one
                                Message 15 of 15 , Jun 9, 2004
                                  In Hyacinthos message #9843, I wrote:

                                  >> This is very interesting! At first, consider
                                  >> the Brocard points R and R' of the harmonic
                                  >> quadrilateral ABCD, and the symmedian point
                                  >> K (this is the point whose distances to the
                                  >> sides of the quadrilateral are proportional
                                  >> to the lengths of the sides). Also consider
                                  >> the circumcenter O of our quadrilateral.
                                  >> Then, the points O, K, R and R' lie on one
                                  >> circle, which is analogous to the "Brocard
                                  >> circle" of a triangle. The points O and K
                                  >> are the endpoints of a diameter of this
                                  >> circle. The points R and R' lie
                                  >> symmetrically with respect to this diameter.
                                  >> The circle through the points O, K, R and R'
                                  >> also passes through the midpoints of the
                                  >> diagonals AC and BD. The line joining these
                                  >> midpoints meets the diameter OK at the
                                  >> midpoint of the segment RR'. The angles
                                  >> < ROK and < R'OK are equal to the Brocard
                                  >> angle of the quadrilateral ABCD.

                                  These remarks started an interesting thread,
                                  which I could not follow since one half of the
                                  mathematics involved was new to me, but let me
                                  just make one remark: In a harmonic
                                  quadrilateral ABCD, the point K is the point
                                  of intersection of the diagonals AC and BD.
                                  [This is quite easy to prove.] Now, since the
                                  midpoints M and N of the diagonals AC and BD
                                  are the orthogonal projections of the
                                  circumcenter O on these diagonals, the angles
                                  < OMK and < ONK are right angles. Hence, M and
                                  N lie on the circle with diameter OK.

                                  Darij Grinberg
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