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

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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

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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

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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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Πέμπτη 22 Ιουλίου 2021

HYACINTHOS 21468

Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
on the circumcircle).

Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
Bc,Ba and Ca,Cb.

The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
centers are
collinear. Which is this line ? Special case: L = Euler line.

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Παρασκευή 20 Δεκεμβρίου 2019

HYACINTHOS 21959

[Antreas P. Hatzipolakis]:
> Let ABC be a triangle, P, P* two isogonala conjugate points.
>
> Denote:
>
> Ra = radical axis of (NPC_PBC), (NPC_P*BC)
>
> Rb = radical axis of (NPC_PCA), (NPC_P*CA)
>
> Rc = radical axis of (NPC_PAB), (NPC_P*AB)
>
> Which is the locus of P such that Ra,Rb,Rc are concurrent?
> The entire plane?
>
> Antreas

[Barry Wolk]:

This was already answered, and is the whole plane.
Another way of stating this result is:
 
The Poncelet points of the quads BCPP*, CAPP*, ABPP*
form a triangle which is perspective to the medial triangle.
--
Barry Wolk

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HYACINTHOS 21942

 

 [Antreas P. Hatzipolakis]:


> > Let ABC be a triangle, P, P* two isogonal conjugate points.
> >
> > Denote:
> >
> > Ra = radical axis of (NPC_PBC), (NPC_P*BC)
> >
> > Rb = radical axis of (NPC_PCA), (NPC_P*CA)
> >
> > Rc = radical axis of (NPC_PAB), (NPC_P*AB)
> >
> > Which is the locus of P such that Ra,Rb,Rc are concurrent?
> > The entire plane?
>
> I wrote this without making a figure. I had in mind
> the points P,P* = O,H: the triangles HBC,HCA, HAB
> share the same NPC, the NPC of ABC, and the NPCs of OBC, OCA, OAB
> concur at the Poncelet point of O wrt ABC, lying on the NPC of ABC.
> Therefore the radical axes in question are concurrent for
> that points.
> Now I make a figure with P,P* = G,K and it seems that
> the radical axes are again concurrent.
>
> See the figure:
>
> http://anthrakitis.blogspot.gr/2013/04/radical-axes-of-npcs.html
>
> Is it true? And if yes, which is the point of concurrence?
>
> In general???
>
> APH

[Randy Hutson]:

Antreas,

I tried various specific pairs of isogonal points, and finally for an
arbitrarily random pair, and it appears the locus is the entire plane. This
does provide an interesting mapping. Some specifics for the concurrence
point for (P,P*):

(G,K): non-ETC -0.956613251489256, which is also the Hyacinthos #16741/16782
homothetic center for line X(2)X(6), and the centroid of (degenerate) pedal
triangle of X(111).

(O,H): X(125)

(N,X54): non-ETC 4.975239945739461

(X7,X55): non-ETC 3.097837435698617

(PU(1)): non-ETC 4.098938269094193, which is the center of conic
{A,B,C,X(99),PU(37)}, complement of X(1916), and anticomplement of X(2023),
and lies on lines 2,694 3,76.

(foci of Steiner inellipse): non-ETC 0.632166489381459, which is the center
of the circumconic through the isotomic conjugates of the foci of the
Steiner inellipse; also the crosssum of X(6) and X(1380), the crosspoint of
X(2) and X(3414); also the complement of the trilinear pole of major axis of
the Steiner eliipses (line X(2)X(1341)); lies on line X(2)X(6) and on the
Steiner inellipse.

(foci of orthic inconic): non-ETC 4.706821577388139, which is the isotomic
conjugate of the polar conjugate of X(1313).

I think this is certainly worth exploring further.

Best regards,
Randy

[Cesar Lozada]:

The trilinear transformation of P( u : v : w ) is :



( f(u, v, w, a, b, c, A, B, C) : f(v, w, u, b, c, a, B, C, A) : f( w, u, v,
c, a, b, C, A, B) )



where



f(u, v, w, a, b, c, A, B, C) = u*(v^2 - w^2)*[u*( v^2 - w^2) - v*(w^2 -
u^2)*cos(C) - w*( u^2 - v^2)*cos(B) ] /a



whose inverse doesn’t seem to be easily calculable.



Regards

Cesar Lozada

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