HYACINTHOS 5599

Dear John,Floor,Paul and all hyacinthists,
-------------------------------------------------
[JHC]24.05.2002

>I was very intrigued indeed by Floor's problem:

> > > Show that the circumcenters of the 6 triangles, into which a
> > triangle is divided by its medians, are concyclic.

but didn't write before now because I was trying, in my usual way,
to simplify the algebra needed to find the coordinates of the center
of this circle, which I'll call "The Floor of ABC" (hoping that
Floor finds this acceptable!).

Well, I now see how to make it very simple indeed, and will
post it when I've actually done so. The argument leads to an
"location" of the Floor (F, say) as

mN + (G-K)/12tan^2(omega),
where
mN is the medial Ninecenter,
G the centroid, and
K the symmedian point.

In particular, the parallel through F to the symmedian line
hits the Euler line in the medial Ninecenter. I hope someone
will confirm this, because while fighting the algebra I actually
got several distinct answers; however, they all had the properties
that the above parallel hits the Euler line in a point with
fixed parameter p (my convention is that

p = 0 for O, 1/3 for G, 1/2 for N, 1 for H,

so that p = 1/4 for mN), and that the coefficient of G-K
is some rational multiple of cot^2(omega).

May I ask, Floor, whether you found this lovely theorem
as a consequence of some theory, or whether it was just
conjectured by "drawing the picture", so to speak?

My next message on this subject will post coordinates
and equations for everything involved.

Regards to all, John Conway >
--------------------------------------------
My message 27.05.2002

>If I didn't make mistake in calculations,the barycentrics

of "Floor" point F* of triangle ABC are

F*(t+k2-k3 : t+k3-k1 : t+k1-k2)
where t=96P^2,P is the area of triangle ABC,
k1=(bb-cc)*(2bb+2cc-aa),k2=(cc-aa)*(2cc+2aa-bb),
k3=(aa-bb)*(2aa+2bb-cc).

Also we have
F*(13a^2b^2+13a^2c^2+10b^2c^2-10a^4-4b^4-4c^4 : : ).

Second property is
(16P^2)vector(PF*)=(a^2b^2+b^2c^2+c^2a^2)vector(GX76)
where P is point on Euler line such that
PO/OH=1/3,and X76 is isotomic conjugate of
Lemoine point K.

Maybe,the following idea could be useful:
Finding the point of intersection of lines
F*X76 and Euler line we can construct the
point F*.

The problem is very nice and I have a feeling that
this point has interesting conections with the
known triangle centers.>

------------------------------------------------------
I have to apologize because of some mistakes.
We have
(96P^2)vector(PF*)=(a^2b^2+b^2c^2+c^2a^2)vector(GX76)
where P is point on Euler line such that
HP/PO=7/2,and X76 is isotomic conjugate of
Lemoine point K.

Above mentioned idea about the point of intersection
of two lines,as I could verify, was not
fruitfull.

Kind regards
Sincerely
Milorad R.Stevanovic