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Some remarks on a NPC concurrence

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  • Darij Grinberg
    In Hyacinthos message #9979, Antreas considered the configuration arising from a triangle ABC, an arbitrary point P in its plane, and the circumcevian triangle
    Message 1 of 40 , Sep 30, 2004
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      In Hyacinthos message #9979, Antreas considered the
      configuration arising from a triangle ABC, an arbitrary
      point P in its plane, and the circumcevian triangle
      A'B'C' of the point P. The reflections of the points A',
      B', C' in the lines BC, CA, AB were denoted by A*, B*,
      C*.

      In Hyacinthos message #9983, I wrote about this
      configuration:

      >> The lines AA*, BB*, CC* concur if and only if
      >> the point P lies on the line at infinity or on
      >> the Jerabek hyperbola. Notably, if P is an
      >> infinite point, the point of concurrence of
      >> the lines AA*, BB*, CC* seems to be the
      >> isogonal conjugate of the point P with respect
      >> to the orthic triangle of triangle ABC, and
      >> hence should lie on the nine-point circle of
      >> triangle ABC.

      There is actually much more to say about this
      concurrence. Here is a listing of properties:

      (1) Let g be an arbitrary line in the plane of triangle
      ABC, and let Ha, Hb, Hc be the feet of the altitudes of
      triangle ABC. Let ga, gb, gc be the parallels to the
      line g through the points Ha, Hb, Hc, and let a', b', c'
      be the reflections of the lines ga, gb, gc in the lines
      BC, CA, AB, respectively. Then, the lines a', b', c'
      concur at one point V on the nine-point circle of
      triangle ABC.

      (This is a result by Floor van Lamoen.)

      (2) If P is the infinite point of the line g, then the
      point V is the isogonal conjugate of P with respect to
      triangle HaHbHc.

      (And now our configuration enters, with slightly
      modified definition:)

      (3) If the parallels to the line g through the points A,
      B, C intersect the circumcircle of triangle ABC at the
      points A', B', C' (apart from the points A, B, C), and
      if A*, B*, C* are the reflections of the points A', B',
      C' in the lines BC, CA, AB, then the points A*, B*, C*
      are the reflections of the points A, B, C in the point
      V.

      (And this makes clear that the lines AA*, BB*, CC*
      concur at V.)

      (Now a few off-topic properties, meaning that these
      properties do not really involve A*, B*, C*:)

      (4) If h is the perpendicular to the line g from the
      circumcenter O of triangle ABC, then the triangle A'B'C'
      is the reflection of the triangle ABC in the line h.

      (5) The lines A'A*, B'B*, C'C*, i. e. the perpendiculars
      to the lines BC, CA, AB through the points A', B', C',
      concur at a point Q on the circumcircle of triangle ABC.

      (6) The Simson line of this point Q with respect to
      triangle ABC is parallel to g.

      PROOFS.

      In the following, we work with directed angles modulo
      180; all angles are measured in degrees.

      There are two proofs for assertion (1). The first proof
      is an angle chase:

      Let the line a' meet the nine-point circle of triangle
      ABC at a point V (apart from the point Ha); then, if we
      succeed to prove that this point V also lies on the
      lines b' and c', then the proof of (1) is complete.
      Hence, it remains to prove that this point V also lies
      on the lines b' and c'.

      Since the point V lies on the nine-point circle, we have
      < HaVHb = < HaHcHb. In other words,
      < (a'; HbV) = < HaHcHb. It is well-known that
      < HaHcHb = 2 < BCA, and thus < (a'; HbV) = 2 < BCA.

      On the other hand, since the line a' is the reflection
      of the line ga in BC, we have < (a'; BC) = < (BC; ga).
      Since ga || g, this becomes < (a'; BC) = < (BC; g).
      Similarly, < (b'; CA) = < (CA; g), and thus

      < (a'; b') = < (a'; BC) + < (BC; CA) - < (b'; CA)
      = < (BC; g) + < (BC; CA) - < (CA; g)
      = (< (BC; g) - < (CA; g)) + < (BC; CA)
      = < (BC; CA) + < (BC; CA)
      = 2 < (BC; CA) = 2 < BCA = < (a'; HbV).

      Thus, b' || HbV, and since the lines b' and HbV have the
      point Hb in common, they must coincide, i. e. the point
      V lies on the line b'. Similarly, V also lies on c'.
      This completes the first proof of (1).

      The second proof of (1) is merely philosophical:
      Consider the infinite point P common to the parallel
      lines g, ga, gb, gc. The lines ga, gb, gc are the
      cevians of this point P with respect to the triangle
      HaHbHc. Now, it is well-known that the lines BC, CA, AB
      are the angle bisectors (internal or external, this
      doesn't matter) of this triangle HaHbHc. Hence, the
      lines a', b', c' are the reflections of the cevians of
      the point P with respect to the triangle HaHbHc in the
      respective angle bisectors; thus, they concur at the
      isogonal conjugate of P with respect to triangle HaHbHc.
      And since the point P is an infinite point, its isogonal
      conjugate lies on the circumcircle of triangle HaHbHc,
      i. e. on the nine-point circle of triangle ABC. This
      proves (1) again.

      The proof of (2) is contained in the second proof of (1)
      given above.

      The proof of (3) is a kind of angle chase again: We will
      show that the point B* is the reflection of B in V. (The
      similar assertions about C* and A* will then follow by
      analogy.)

      In order to prove that the point B* is the reflection of
      B in V, we follow the "working backwards" strategy: We
      *define* the point B* as the reflection of B in V, then
      we *define* the point B' as the reflection of B* in CA,
      and then we try to show that the point B' is the point
      of intersection of the parallel to g through B with the
      circumcircle of triangle ABC.

      Consider the midpoints Ma, Mb, Mc of the sides BC, CA,
      AB of triangle ABC. The points Ha, Hb, Hc, Ma, Mb, Mc
      and V all lie on the nine-point circle of triangle ABC.
      Thus, < MaVMc = < MaMbMc; but clearly < MaMbMc = < ABC,
      and thus < MaVMc = < ABC.

      Since the point B* is the reflection of B in V, the
      point V is the midpoint of BB*. On the other hand, the
      point Ma is the midpoint of BC. Hence, VMa || B*C.
      Similarly, VMc || B*A. Thus,
      < (B*C; B*A) = < (VMa; VMc), or, in other words,
      < CB*A = < MaVMc. Together with < MaVMc = < ABC, this
      becomes < CB*A = < ABC.

      Since the point B' is the reflection of B* in CA, we
      have < CB'A = - < CB*A = - < ABC = < CBA, and thus, the
      point B' lies on the circumcircle of triangle ABC. So
      the only thing left to show is that the point B' lies on
      the parallel to g through B; once this will be shown,
      the proof of (3) will be complete.

      Let S be the reflection of B in Hb. Then, Hb is the
      midpoint of BS. On the other hand, V is the midpoint of
      BB*. Thus, HbV || SB*; in other words, b' || SB*.

      Now, since S is the reflection of B in Hb, we can also
      say that S is the reflection of B in CA. On the other
      hand, B* is the reflection of B' in CA. Thus, the lines
      BB' and SB* are symmetric to each other with respect to
      the line CA. Also, per definitionem, the lines gb and b'
      are symmetric to each other with respect to the line CA.
      Since b' || SB*, it follows that gb || BB'. Together
      with gb || g, this implies g || BB', and thus, the point
      B' lies on the parallel to g through B. This completes
      the proof of (3).

      (The last part of the proof of (3) could be also done
      without introducing S, but doing an angle chase
      instead.)

      The assertion (4) is trivial: Since AA' is a chord in
      the circumcircle of triangle ABC, whose center is O, the
      perpendicular bisector of the segment AA' passes through
      O. This perpendicular bisector is also perpendicular to
      g (since AA' || g); thus, this perpendicular bisector is
      the line h, and the points A and A' are symmetric to
      each other with respect to h. Similarly, the points B
      and B' are symmetric to each other with respect to h,
      and the same holds for C and C'. This proves (4).

      As for (5), let's call Q the point where the
      perpendicular to BC through A' meets the circumcircle of
      triangle ABC (apart from A', of course). We will show
      that this point Q lies on the perpendiculars to CA and
      AB through B' and C', too.

      Since Q lies on the circumcircle, we have
      < A'QB' = < A'C'B'. Because of (4), the triangles ABC
      and A'B'C' are indirectly congruent, so
      < A'C'B' = - < ACB. Consequently,
      < A'QB' = - < ACB = < BCA. In other words,
      < (A'Q; B'Q) = < (BC; CA). Since the point Q lies on the
      perpendicular to BC through A', we have
      < (BC; A'Q) = 90. Thus,

      < (CA; B'Q) = < (CA; BC) + < (BC; A'Q) + < (A'Q; B'Q)
      = < (CA; BC) + 90 + < (BC; CA) = 90,

      and thus, the point Q also lies on the perpendicular to
      CA through B'. Similarly, Q lies on the perpendicular to
      AB through C'. This completes the proof of (5).

      Last but not least, a proof of (6): The Simson line of
      the point Q with respect to triangle ABC passes through
      the orthogonal projections X, Y, Z of Q on the sides BC,
      CA, AB. We have to prove that YZ || g.

      Well, since < AYQ = 90 and < AZQ = 90, the points Y and
      Z lie on the circle with diameter AQ, so that
      < YZA = < YQA, and consequently,

      < (YZ; AB) = < YZA = < YQA = < (QY; AQ)
      = < (QY; CA) + < (CA; AQ)
      = 90 + < (CA; AQ)
      (since QY is perpendicular to CA)
      = 90 + < CAQ
      = 90 + < CC'Q (cyclic)
      = 90 + < (CC'; C'Q)
      = 90 + < (g; C'Q) (since CC' || g)
      = 90 + < (g; AB) + < (AB; C'Q)
      = 90 + < (g; AB) + 90
      (since C'Q is perpendicular to AB)
      = < (g; AB),

      so that YZ || g, and we are done.

      Darij Grinberg
    • Darij Grinberg
      Dear Jean-Pierre, ... Indeed, that s clear (it s all about the angles). [...] ... Yes, of course, it doesn t remain true. Maybe this could give a good locus...
      Message 40 of 40 , Oct 2, 2004
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        Dear Jean-Pierre,

        In Hyacinthos message #10581, you wrote:

        >> Note that the result remains true if, instead
        >> of the midpoints of the sides, you take for
        >> MaMbMc a triangle inscribed in ABC and
        >> directly similar with ABC : the lines a',b',c'
        >> concur at a point of the circle MaMbMc.

        Indeed, that's clear (it's all about the angles).

        [...]

        >> Unfortunately, if instead of three parallel
        >> lines ga,gb,gc, we consider three concurrent
        >> lines ga,gb,gc, it dosn't seem that the
        >> result remains true.

        Yes, of course, it doesn't remain true. Maybe
        this could give a good locus...

        Sincerely,
        Darij Grinberg
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