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Re: [EMHL] Unusual concurrences and more on cubics

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  • Michael Keyton
    Problem 1 is in Honsberger with an elegant proof by John Rigby. Johnson gives it with old terminology as Corollary to 346, p. 215. In Altshiller Court it is
    Message 1 of 2 , Feb 28, 2000
      Problem 1 is in Honsberger with an elegant proof by John Rigby.
      Johnson gives it with old terminology as Corollary to 346, p. 215.
      In Altshiller Court it is Theorem 586.

      Michael Keyton

      Steve Sigur wrote:
      >
      > From: Steve Sigur <ssigur@...>
      >
      > Hello friends,
      >
      > So sorry to be absent for so long but the demands of my world have been
      > too great. I have tried to follow the discussions and, as always, have
      > found them very interesting.
      >
      > Problem 1: One of my students came across this problem in a book. Prove
      > that the midpoint to a side, the midpoint of the altitude to that side,
      > and the symmedian point are colinear. As this is a pleasant problem, I
      > will leave this as a problem for the group. One could also state this as
      > a concurrence: the 3 lines through the midpoint of a side and the
      > midpoint of the corresponding altitude concur at K, the symmedian pt.
      >
      > Problem 2: there is a weak version of the obove which can be obtained by
      > the following changes: symmedian -> incenter, orthocenter -> Gergonne
      > pt. So the new colinearity is that the incenter, the midpoint of a side,
      > and the midpoint af a Gergonne cevian are colinear.
      >
      > This started me thinking about unusual concurrences, which led (as always
      > seems to happen lately) to a cubic.
      >
      > There is another colinearity/concurrence in the strong version of the
      > problem that uses the symmedian and orthocenter. It is as follows, where
      > Ah, Bh, Ch are the midpoints of the altitudes and O is the circumcenter:
      > The lines ABh, CO, BAh concur. The point of concurrence is Ch* where the
      > * is the isogonal conjugate or conjugal.
      >
      > There are so many concurrences in this situation that there must be a
      > cubic, and there is.
      > It is one of the cubics we have been mucking around with.
      >
      > We know a cubic with all of the points used in problem 1. I call it the G
      > conCubic, but it is one of the cubics discussed in the group recently and
      > has another name I think. I prefer descriptive names. In this case the
      > name tells us that this is an isogonal cubic with pivot G. Any concubic
      > contains the 4 incenters Ix as well as the pivot, its conjugal, and ABC.
      > Other points on this cubic are the midpoints of the sides, as well as O
      > and H. These are exactly the points in the problem 1.
      >
      > A cubic organizes colinearites among points on the cubic according to its
      > group law. There were parts of this cubic's group that John and I had not
      > been able to fill in. It is now as complete as we will want it.
      >
      > First the group table and then a few comments on how to read it.
      >
      > G concubic
      >
      > equation: aa(yzz-zyy) + bb(zxx - xzz) + cc(xyy - yxx) = 0
      > pivot: G
      > constant: O, the circumcenter
      >
      > o a b c
      >
      > -II H Ah* Bh* Ch*
      > -I Mo* Ma* Mb* Mc*
      > 0 K A- B- C-
      > I Io Ia Ib Ic
      > II G A B C pivot
      > III Mo Ma Mb Mc
      > IV O Ah Bh Ch constant
      >
      > Here the Mx are the Mittenpunkts, A- the midpoint of side a, and the
      > others are as defined above.
      >
      > How to read this table. The group of a cubic is (Z cross K) where Z is
      > the integers and K is the Klein 4 group. The 4 group has the rules o+o=o,
      > o+a=a, o+b=b, o+c=c, and cyclic permutations of a+b=c.
      >
      > If you reflect a row across the Ix row, you obtain the conjugal of the
      > row.
      >
      > To be colinear, the 3 points have to satisfy the 4 group shown in the
      > columns, and the numbers in the rows (shown as Roman numerals) have to
      > add to the constant.
      >
      > Some examples: H,G,O are colinear because o+o=o and -II + II + IV =
      > IV, the constant.
      > G, A, A- are colinear since o+a=a and II + II + 0 = IV
      > Mo, Io, K are colinear since o+o=o and III + I + 0 = IV
      > K, Bh, B- are colinear since o+b=b and 0 + IV + 0 = IV
      > A-, Bh, C- are colinear since a+b=c and 0 + IV + 0 = IV
      >
      > Once you have found a colinearity, any combination of points in those
      > same rows that satisfies the 4 group will alos be colinear. This set of
      > colinearites will be a desmic triple of the three quadrangles defined by
      > the three rows.
      >
      > The most interesting desmic triples of quadrangles are those that include
      > ABC. This cubic gives two
      >
      > Mx
      > GABC
      > Mx*
      >
      > here G is the desmon and : bSB : the harmon.
      >
      > and
      >
      > H Ah* Bh* Ch*
      > G A B C
      > O Ah Bh Ch
      >
      > here G is the desmon and I think that L is the harmon point.
      >
      >
      >
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    • xpolakis@otenet.gr
      ... Yes. However, sometimes a unusual concurrence comes from a usual one. ... The usual concurrence is: Draw the circle centered on BC at A and touching
      Message 2 of 2 , Feb 29, 2000
        Steve Sigur wrote:

        >To me a concurrence is "unusual" if its elements are
        >non-symmectically defined in terms of ABC.

        Yes. However, sometimes a "unusual" concurrence comes from a "usual" one.

        Consider for example the following I posted On Thu, 17 Feb 2000:

        | Another one I discovered:
        |
        | Draw the circle centered at A' (= the foot of the bisector of the angle A)
        | and touching AB, AC at X, Y, respectively. Then the lines:
        | median from A, XY and the perpendicular to BC at A' are concurrent.

        The "usual" concurrence is:
        Draw the circle centered on BC at A' and touching BA, CA at Ba, Ca, resp.
        [Its ceneter A' is the foot of the internal bisector of the angle A]
        Draw the perpendicular to BC at A' intersecting BaCa at A*.
        Similarly define B*, C*. Then the three lines AA*, BB*, CC* are concurrent.


        A


        Ba A* Ca

        B A' A" C


        Let A" be the point that AA* intersects BC.

        We have (by a "fruitful" lemma I like to use :-)

        A"B cotB - cotw
        --- = ------------ , where w := angle(A*BC), q := angle(A*CB)
        A"C cotC - cotq

        By calculating the cots we finally get: A"B / A"C = 1

        This means that the line AA* is the median of BC, and similarly BB*, CC*
        are the medians of CA, AB, resp. [Therefore they concur]

        This concurrence is of course "usual", but if we rephrase it as above, it
        becomes "unusual" (or "unlikely").


        Antreas
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