Σάββατο 9 Νοεμβρίου 2024

HYACINTHOS 1139

5246

Re: [EMHL] A 3D triangular problem



 

Antreas P. Hatzipolakis



 

Dec 31, 2002

 

«»

On Sunday, April 14, 2002, at 08:26 PM, Steve Sigur wrote:


 

> I have a serious reason for wanting this answer. In barycentric
> coordinates,
> most of the triangle centers are rational functions of the sides a, b,
> c. In
> the field extension sense, we also know the points that are made from
> adding
> root2 and root3 to the formulas (these are the Vecten-like and
> Fermat-like
> points). I was wondering if there were any root5 extensions that had any
> relevance?
>

 

I am not sure what you mean exactly, but
here is a first thought:

Construct regular pentagons on the sides of ABC.
Let A',B',C' be their "apexes".

The triangles ABC, A'B'C' are perspective
with barycentric perspector:

(1/(cotA +- cot72) ::) =

(1/(cotA +- sqrt(5 + 2sqrt5)) ::)

[+,- for outwardly,inwardly constructed 5gons)

APH


 

Δευτέρα 13 Ιουνίου 2022

HYACINTHOS 24185

[APH]:

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

Aa, Ab, Ac = the orthogonal projections of Na on IA, IB, IC, resp.

Ba, Bb, Bc = the orthogonal projections of Nb on IA, IB, IC, resp.

Ca, Cb, Cc = the orthogonal projections of Nc on IA,IB, IC, resp.

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:

Ab, Ac = the orthogonal projections of A on NB, NC, resp.

(Oab), (Oac) = the circles with diameters AAb,AAc, resp.

Similarly (Obc), (Oba) and (Oca), (Ocb)

R1 = the radical axis of (Oab), (Oac)
R2 = the radical axis of (Obc), (Oba)
R3 = the radical axis of (Oca), (Ocb)

The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent.

[Angel Montesdeoca]:

The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent. at

X(195) = X(5)-CEVA CONJUGATE OF X(3)

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