Κυριακή 5 Μαρτίου 2023
Δευτέρα 13 Ιουνίου 2022
HYACINTHOS 24185
[APH]:
Let ABC be a triangle.
Denote:The Euler lines of AaAbAc, BaBbBc, CaCbCc are concurrent.
[Angel Montesdeoca]:
The Euler lines of AaAbAc, BaBbBc, CaCbCc are concurrent. at
X(1125)= the centroid of {A,B,C,X(1)} (Darij Grinberg, 12/28/02)
A.Montesdeoca
Τετάρτη 11 Μαΐου 2022
HYACINTHOS 24287
[Tran Quang Hung]:
Let ABC be a triangle and A'B'C' the pedal triangle of H.
Denote:
Na, Nb, Nc = the NPC centers of AB'C',BC'A',CA'B', resp.
The triangles A'B'C', NaNbNc are cyclologic.
The cyclologic center (A'B'C', NaNbNc) is the orthopole of the Euler line.
Which is the cyclologic center of (NaNbNc, A'B'C') ?
[César Lozada]:
Cyclologic centers: X(125) and
Za = (2*cos(2*A)*cos(B-C)-cos(3*A)) *((cos(2*A)+cos(4*A)+1)*cos(B- C)+(-cos(A)-cos(3*A))*cos(2*( B-C))-cos(A))*sec(A) : : (trilinears)
= On lines: (4,7730), (25,8157), (571,2079), (933,3518)
= [ -2.233472799679084, -6.17738928122346, 8.948152199529426 ]
César Lozada
Τρίτη 29 Μαρτίου 2022
HYACINTHOS 6352
Floor en Lyanne van Lamoen wrote:
>
> Dear Darij and Jean-Pierre,
>
> > > [DG]:
> > > > > Prof. Clark Kimberling has updated ETC, as we have seen. But
> > there
> > > > > are still some points whose trilinears we don't know -
> > especially
> > > > I
> > > > > with my knowledge on trilinears which doesn't go further than
> > line
> > > > > equations. I am interested in the trilinears of the Schröder
> > > > point.
> > >
> > > [JPE]:
> > > > Your point is trilinear x = (b-c)^2 + a(b+c-2a)
> > > > Friendli. Jean-Pierre
> >
> > [FVL]
> > > It seems that there are more points P satisfying the
> > > condition that if A'B'C' is the pedal triangle of P, then the
> > > circumcircles of AA'P, BB'P and CC'P meet in a second point. The
> > > circumcenter seems to be one of these points. Perhaps the locus is
> > the
> > > Stammler hyperbola? I am now running out of time, later I will
> > invest
> > > more time.
>
> [JPE]:
> > This locus is the Darboux cubic. Friendly. Jean-Pierre
Now we may try to find some second intersections for given P:
We look for second points of intersection of the three circles (AA'P),
(BB'P), (CC'P) in barycentrics:
P=I gives a(2a^2-ab-ac-(b-c)^2):: see above.
P=O gives a^2(2a^2-b^2-c^2):: = X(187) = inverse of K in circumcircle.
P=L=X20 gives (-2a^2+b^2+c^2)/(-a^2+b^2+c^2):: .
Is this one in ETC?
P=X40=reflection of I through O gives a*(-2a+b+c)/(-a+b+c):: .
Is this one in ETC?
I don't think that P=H gives a well defined point (that is why I first
rejected the locus to be Darboux...).
Kind regards,
Sincerely,
Floor van Lamoen
HYACINTHOS 6321
Dear Darij and Jean-Pierre,
[DG]:
> > Prof. Clark Kimberling has updated ETC, as we have seen. But there
> > are still some points whose trilinears we don't know - especially
> I
> > with my knowledge on trilinears which doesn't go further than line
> > equations. I am interested in the trilinears of the Schröder
> point.
[JPE]:
> Your point is trilinear x = (b-c)^2 + a(b+c-2a)
> Friendli. Jean-Pierre
It seems that there are more points P satisfying the
condition that if A'B'C' is the pedal triangle of P, then the
circumcircles of AA'P, BB'P and CC'P meet in a second point. The
circumcenter seems to be one of these points. Perhaps the locus is the
Stammler hyperbola? I am now running out of time, later I will invest
more time.
Kind regards,
Sincerely,
Floor van Lamoen
Σάββατο 30 Οκτωβρίου 2021
HYACINTHOS 24180
Let ABC be a triangle.
Denote:[Angel Montesdeoca]:
The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent. at
The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent. at