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Fuhrmann triangle

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  • Darij Grinberg <darij_grinberg@web.de>
    Let ABC be a triangle, and A , B , C be the midpoints of the arcs BC, CA, AB on the circumcircle of ABC. Denote by A , B , C the reflections of A , B , C in
    Message 1 of 7 , Jan 9, 2003
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      Let ABC be a triangle, and A', B', C' be the midpoints of the arcs
      BC, CA, AB on the circumcircle of ABC. Denote by A", B", C" the
      reflections of A', B', C' in BC, CA, AB. Then the triangle A"B"C" is
      called FUHRMANN TRIANGLE of ABC.

      It is well-known that the circumcenter of A"B"C" is the midpoint M of
      segment HN, where H is the orthocenter and N the Nagel point of
      triangle ABC. One can easily prove that this midpoint M is also the
      reflection of the incenter I of ABC in the nine-point center O', and
      also the reflection of the circumcenter O in the Spieker center I'
      (what an analogy between incenter and circumcenter!).

      Something very interesting was stated by Milorad Stevanovich in
      Hyacinthos message #5897: The orthocenter of the Fuhrmann triangle
      A"B"C" is the incenter of I of ABC.

      This is, as you can easily see, equivalent to the following fact: The
      nine-point centers of ABC and the Fuhrmann triangle A"B"C" coincide.

      However, I do not know a proof of this theorem either in Milorad's or
      in my form.

      Another conjecture which I have found:

      The perpendiculars from A to B"C", from B to C"A", and from C to A"B"
      concur at the point X(80), namely the reflection of the incenter I in
      the Feuerbach point of ABC.

      (The fact that they concur follows from the orthologicity of
      triangles ABC and A"B"C", but I suppose it should be not so easy to
      prove that the concurrence is X(80).)

      -- I could keep on finding new and new interesting properties of the
      Fuhrmann triangle, but I know that nobody will read a long message;
      so I conclude with only one little note: The Fuhrmann triangle has
      some properties similar to those of the two Brocard triangles,
      although it is not perspective to ABC. Altogether, the Fuhrmann
      triangle should be much better known as it is.

      Darij Grinberg
    • ben_goss_ro
      ... is ... of ... and ... This isn t strictly about the Fuhrmann triangle, but it s related to it. I was playing around in Euklides trying to show that H is on
      Message 2 of 7 , Apr 18, 2004
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        --- In Hyacinthos@yahoogroups.com, "Darij Grinberg
        <darij_grinberg@w...>" <darij_grinberg@w...> wrote:
        > Let ABC be a triangle, and A', B', C' be the midpoints of the arcs
        > BC, CA, AB on the circumcircle of ABC. Denote by A", B", C" the
        > reflections of A', B', C' in BC, CA, AB. Then the triangle A"B"C"
        is
        > called FUHRMANN TRIANGLE of ABC.
        >
        > It is well-known that the circumcenter of A"B"C" is the midpoint M
        of
        > segment HN, where H is the orthocenter and N the Nagel point of
        > triangle ABC. One can easily prove that this midpoint M is also the
        > reflection of the incenter I of ABC in the nine-point center O',
        and
        > also the reflection of the circumcenter O in the Spieker center I'
        > (what an analogy between incenter and circumcenter!).

        This isn't strictly about the Fuhrmann triangle, but it's related to
        it. I was playing around in Euklides trying to show that H is on the
        Fhurmann circle, and I thought this property could be generalized:

        ABC is a triangle and P is a point in its plane. AP, BP, CP cut the
        circumcicle of ABC in A', B', C' respectively. A", B", C" are the
        reflections of A', B', C" in BC, CA, AB respectively. Show that
        A"B"C" is similar to A'B'C' (a) and H lies on the circumcircle of
        A"B"C" (b).

        I it stupid for me to mention this? :-) (I mean, is it really well-
        known?). Anyway, I don't have a proof. Does ne1 have a nice proof (no
        monstrous computations)?

        It's easy to calculate the angles between HA", HB", HC", so it's
        enough to give a proof for either (a) or (b) and the other one will
        follow.

        Thanks!
      • Darij Grinberg
        Dear Grobber, ... This was mentioned before in Hyacinthos message #9493 by Jean-Pierre Ehrmann and proven (synthetically!) in the messages #9500 and #9503. A
        Message 3 of 7 , Apr 18, 2004
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          Dear Grobber,

          In Hyacinthos message #9713, you wrote:

          >> ABC is a triangle and P is a point in its plane.
          >> AP, BP, CP cut the circumcicle of ABC in A', B',
          >> C' respectively. A", B", C" are the reflections
          >> of A', B', C" in BC, CA, AB respectively. Show
          >> that A"B"C" is similar to A'B'C' (a) and H lies
          >> on the circumcircle of A"B"C" (b).

          This was mentioned before in Hyacinthos message
          #9493 by Jean-Pierre Ehrmann and proven
          (synthetically!) in the messages #9500 and #9503.
          A reference to Hagge 1908 was given in message
          #9506, and I gave some additional results in
          #9511. By the way, I am still searching for
          synthetic proofs of the conjectures (3) and (5)
          from my message #9511. Actually, these two
          conjectures can be united under

          (6) There exists an indirect similitude mapping
          the points A, B, C, A', B', C' and P to the
          points X, Y, Z, A2, B2, C2 and P

          (where the notations are taken from Hyacinthos
          message #9511). I would be grateful for any
          help regarding this conjecture.

          Sincerely,
          Darij Grinberg
        • ben_goss_ro
          ... is ... of ... and ... The ... or ... A B ... in ... I ll make use of the fact (proven by Milorad Stevanovich) that I is the orthocenter of A B C and of
          Message 4 of 7 , Apr 19, 2004
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            --- In Hyacinthos@yahoogroups.com, "Darij Grinberg
            <darij_grinberg@w...>" <darij_grinberg@w...> wrote:
            > Let ABC be a triangle, and A', B', C' be the midpoints of the arcs
            > BC, CA, AB on the circumcircle of ABC. Denote by A", B", C" the
            > reflections of A', B', C' in BC, CA, AB. Then the triangle A"B"C"
            is
            > called FUHRMANN TRIANGLE of ABC.
            >
            > It is well-known that the circumcenter of A"B"C" is the midpoint M
            of
            > segment HN, where H is the orthocenter and N the Nagel point of
            > triangle ABC. One can easily prove that this midpoint M is also the
            > reflection of the incenter I of ABC in the nine-point center O',
            and
            > also the reflection of the circumcenter O in the Spieker center I'
            > (what an analogy between incenter and circumcenter!).
            >
            > Something very interesting was stated by Milorad Stevanovich in
            > Hyacinthos message #5897: The orthocenter of the Fuhrmann triangle
            > A"B"C" is the incenter of I of ABC.
            >
            > This is, as you can easily see, equivalent to the following fact:
            The
            > nine-point centers of ABC and the Fuhrmann triangle A"B"C" coincide.
            >
            > However, I do not know a proof of this theorem either in Milorad's
            or
            > in my form.
            >
            > Another conjecture which I have found:
            >
            > The perpendiculars from A to B"C", from B to C"A", and from C to
            A"B"
            > concur at the point X(80), namely the reflection of the incenter I
            in
            > the Feuerbach point of ABC.

            I'll make use of the fact (proven by Milorad Stevanovich) that I is
            the orthocenter of A"B"C" and of the known fact that IO'=R/2-r, where
            O' is the Euler center of ABC.

            First of all, the centers of Fuhrmann's circle, of Euler's circle and
            of the cirumcircle are collinear and I lies on this line, so X(80)
            also lies on this line. Further on we can see that MI*MX(80)=(R-2r)(R-
            2r+2r)=R^2-2Rr=OI^2=MH^2, so X(80) is the inverse of I with respect
            to the Fuhrmann circle.

            The last remark shows that triangles MIA" and MA"X(80) are similar,
            so MI/MA"=IA"/A"X(80), but MI=R-2r and MA"=OI=sqrt(R^2-2Rr), so A"X
            (80)=IA"*sqrt(R/(R-2r)) (*).

            A"C=A'C=IA' and the system I, A", B", C" is similar to I, A', B', C'
            (because A"B"C" is similar to A'B'C' as proven in some messages
            around 9500-9511 for the general case and I is the orthocenter of
            both A'B'C' and A"B"C"), so IA"/IA'=(circumradius of A"B"C")/R=sqrt
            (R^2-2Rr)/R=>IA'=IA"*sqrt(R/(R-2r)) (**).

            From (*) and (**) we find A"C=A"X(80). In the same way we show B"C=B"X
            (80), so A"B" is the perpendicular bisector of CX(80), and from all
            the analogous relations we find what we want: the perpendiculars from
            A, B, C to B"C", C"A", A"B" concur at X(80).

            (Here A', B', C' are the midpoints of the smaller arcs BC, CA, AB)
          • Darij Grinberg
            Dear Grobber, ... Thanks - this remark of you about perpendicular bisectors has opened my eyes to see a much simpler proof: The main property of X(80) is the
            Message 5 of 7 , Apr 19, 2004
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              Dear Grobber,

              In Hyacinthos message #9724, you wrote:

              >> A"B" is the perpendicular bisector of CX(80),
              >> and from all the analogous relations we find
              >> what we want: the perpendiculars from A, B,
              >> C to B"C", C"A", A"B" concur at X(80).

              Thanks - this remark of you about perpendicular
              bisectors has opened my eyes to see a much simpler
              proof:

              The main property of X(80) is the following: If
              the points Ja, Jb, Jc are the reflections of the
              excenters Ia, Ib, Ic of triangle ABC in the
              sidelines BC, CA, AB, then the lines AJa, BJb, CJc
              and the circles BCJa, CAJb, ABJc all pass through
              X(80). Now, it follows that the circle BCX(80)
              coincides with the circle BCJa. Hence, the
              circumcenter of triangle BCX(80) is the
              circumcenter of triangle BCJa. But since the
              triangle BCJa is the reflection of the triangle
              BCIa in the line BC, the circumcenter of triangle
              BCJa is the reflection of the circumcenter of
              triangle BCIa in BC, i. e. the reflection of A' in
              BC (since A' is the circumcenter of triangle BCIa,
              what is very easy to prove), i. e. the point A".
              In other words, the point A" is the circumcenter
              of triangle BCX(80). Similarly, the points B" and
              C" are the circumcenters of triangles CAX(80) and
              ABX(80). Consequently, the lines B"C", C"A", A"B"
              are the perpendicular bisectors of the segments
              AX(80), BX(80), CX(80). Qed..

              Now you may ask: How to prove that the lines AJa,
              BJb, CJc and the circles BCJa, CAJb, ABJc all pass
              through the point X(80) which is initially defined
              as the reflection of the incenter I in the
              Feuerbach point of triangle ABC ?

              Well, this is quite tricky. At first, you consider
              four arbitrary points A, B, C, D in the plane, and
              show by some angle chasing that the nine-point
              circles of the triangles ABC, BCD, CDA, DAB concur
              at one point P. We call this point P the "Poncelet
              point" of the points A, B, C, D. This Poncelet
              point P also lies on the pedal circles of the
              points A, B, C, D with respect to the triangles
              BCD, CDA, DAB, ABC. [The "pedal circle" of a point
              with respect to a triangle is defined as the
              circle passing through the orthogonal projections
              of the point on the sidelines of the triangle.]
              Prove this by angle chase again (this time, you
              must use the feet of the altitudes of the
              triangles BCD, CDA, DAB, ABC; these feet lie on
              the nine-point circles and on the appropriate
              pedal circles as well!). Now let D' be the
              reflection of D in the Poncelet point P. Then you
              will easily show (angle chase again) that
              < BD'C = - < BDC, < CD'A = - < CDA and
              < AD'B = - < ADB. This point D' is called the
              "antigonal conjugate" of the point D with respect
              to triangle ABC; the word "antigonal" means
              nothing but the fact that the angles at which we
              see the triangle's sides from this point are
              oppositely equal to the corresponding angles for
              D (i. e. the fact that < BD'C = - < BDC, etc.).
              Antigonal conjugate is NOT the same as isogonal
              conjugate. (In fact, these two kinds of
              conjugation are interrelated: One can prove that
              the antigonal conjugate D' of D is the inverse of
              the isogonal conjugate of the inverse of D;
              hereby, the word "inverse" of a point always means
              the image of the point in the inversion with
              respect to the circumcircle of triangle ABC. But
              we won't need this result in our proof.)

              Okay, well, we have introduced the notions of the
              Poncelet point and the antigonal point. Now, let D
              be the incenter I of triangle ABC. What is the
              Poncelet point P of the points A, B, C, D = I ?
              Well, the Poncelet point P must lie on the
              nine-point circles of triangles ABC, BCD, CDA, DAB
              and on the pedal circles of the points A, B, C, D
              with respect to the triangles BCD, CDA, DAB, ABC;
              however, if you consider only two of these eight
              circles, namely the nine-point circle of triangle
              ABC and the pedal circle of the point D = I with
              respect to triangle ABC, then already these two
              circles have only one common point, namely the
              Feuerbach point of triangle ABC (this is just the
              great Feuerbach theorem). Hence, the Poncelet
              point P is the Feuerbach point of triangle ABC.
              Now, you consider the antigonal conjugate D' of D
              with respect to triangle ABC. By definition, D' is
              the reflection of the incenter I in the Feuerbach
              point P; hence, D' is X(80). But from the equatins
              < BD'C = - < BDC, < CD'A = - < CDA and
              < AD'B = - < ADB, we can derive by a simple angle
              chase that the point D' lies on the circles BCJa,
              CAJb, ABJc, and in the next step we can easily
              conclude by another bit of angle calculation that
              this point D' also lies on the lines AJa, BJb,
              CJc. Proof complete!

              This was quite much theory now; feel free to ask
              me if you don't understand something, I have
              omitted many details. Also, what I have written is
              only a little part of the theory around Poncelet
              points and antigonal conjugates. If you are
              interested, I can tell you more of the story.

              Sincerely,
              Darij Grinberg
            • Eric Danneels
              Dear Darij, in message 9725 you wrote about the antigonal conjugate: ... ... I think it is: The antigonal conjugate D of D is the isogonal conjugate of the
              Message 6 of 7 , May 4 11:15 AM
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                Dear Darij,

                in message 9725 you wrote about the antigonal conjugate:

                ....
                > One can prove that
                > the antigonal conjugate D' of D is the inverse of
                > the isogonal conjugate of the inverse of D;
                > hereby, the word "inverse" of a point always means
                > the image of the point in the inversion with
                > respect to the circumcircle of triangle ABC.
                ...

                I think it is:

                The antigonal conjugate D' of D is the isogonal conjugate of
                the inverse of the isogonal conjugate of D

                I suppose you've interchanged the terms inverse and isogonal
                conjugate ?


                Greetings from Bruges


                Eric Danneels
              • Darij Grinberg
                Dear Eric, ... Oh yes, sorry. Thanks for the remark. By the way, the proof is easy: If D1 is the isogonal conjugate of D, then we can readily see that
                Message 7 of 7 , May 5 12:16 PM
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                  Dear Eric,

                  In Hyacinthos message #9761, you wrote:

                  >> I think it is:
                  >>
                  >> The antigonal conjugate D' of D is the
                  >> isogonal conjugate of the inverse of the
                  >> isogonal conjugate of D
                  >>
                  >> I suppose you've interchanged the terms
                  >> inverse and isogonal conjugate ?

                  Oh yes, sorry. Thanks for the remark. By
                  the way, the proof is easy: If D1 is the
                  isogonal conjugate of D, then we can
                  readily see that < BD1C = < BAC - < BDC.
                  If D2 is the inverse of D1, then another
                  angle chase shows < BD2C = 2 < BAC - < BDC,
                  so that < BD2C = < BAC + < BDC. Now, if
                  D3 is the isogonal conjugate of D2, then
                  < BD3C = < BAC - < BD2C, so that
                  < BD3C = - < BDC = < BD'C, and similarly
                  < CD3A = < CD'A and < AD3B = < AD'B, and
                  consequently, D3 coincides with the
                  antigonal conjugate D' of D.

                  Sincerely,
                  Darij Grinberg
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