Re: [EMHL] Unusual concurrences and more on cubics
- 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.
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
> 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
> here G is the desmon and : bSB : the harmon.
> 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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- Steve Sigur wrote:
>To me a concurrence is "unusual" if its elements areYes. However, sometimes a "unusual" concurrence comes from a "usual" one.
>non-symmectically defined in terms of ABC.
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.
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").