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5246
→ Re: [EMHL] A 3D triangular problem
Antreas P. Hatzipolakis
Dec 31, 2002
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On Sunday, April 14, 2002, at 08:26 PM, Steve Sigur wrote:
I am not sure what you mean exactly, but |
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
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
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
Dear friends,
I found the following:
Let ABC be triangle, A'B'C' its anticomplementary triangle, and A''B''C'' the circuncevian triangle of the infinite point of the Euler line of ABC, that is A''B''C'' are the second intersection points of the circumcircle and the parallel lines to Euler line through A,
B, C, respectively.
The conic ABCA'A'', with center the midpoint of BC is the locus of points P such the Euler lines of PBC and ABC are parallel.
The three conics ABCA'A'', ABCB'B'', and aBCC'C'' go through X1138.
Francisco Javier Garcia Capitan
Antreas P. Hatzipolakis
Variations and Points
http://anthrakitis.blogspot.gr/2013/04/concurrent-radical-axes.html
APH
--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle, A'B'C' the orthic triangle and
> A"B"C" the Euler triangle (ie A",B",C" are the second
> intersections of NPC with AA',BB',CC', resp. = midpoints
> of AH,BH,CH, resp.)
>
> Denote:
>
> Ra = the radical axis of ((B', B'C"),(C',C'B"))
>
> Rb = the radical axis of ((C', C'A"),(A',A'C"))
>
> Rc = the radical axis of ((A', A'B"),(B',B'A"))
>
> The Ra,Rb,Rc are concurrent.
>
> Point?
>
> Antreas
>
Antreas P. Hatzipolakis
Let ABC be a triangle, A'B'C' the orthic triangle and
A"B"C" the Euler triangle (ie A",B",C" are the second
intersections of NPC with AA',BB',CC', resp. = midpoints
of AH,BH,CH, resp.)
Denote:
Ra = the radical axis of ((B', B'C"),(C',C'B"))
Rb = the radical axis of ((C', C'A"),(A',A'C"))
Rc = the radical axis of ((A', A'B"),(B',B'A"))
The Ra,Rb,Rc are concurrent.
Point?
APH
[Antreas P. Hatzipolakis]:
Let ABC be a triangle and A'B'C' the antipedal triangle of I (excentral triangle).
A*, B*, C* = X(110) of A'BC, B'CA, C'AB.
The circles with diameters AA*, BB*, CC* and the circumcircle of ABC are concurrent.
Which is the point of concurrence?
APH
*** X(2222)
Angel Montesdeoca
Dear friends:
A concurrency at X1659.
Given a triangle ABC, construct the circles with BC, CA, AB as diameters, then construct the three circles touching internally two of them and externally the third one. The lines joining the centers of these triangles and the corresponding vertices concur at X1659 (Yiu-Paasche point).
See:
http://garciacapitan.blogspot.com.es/2013/02/a-concurrency-at-x1659.html
Love for everybody,
Francisco Javier García Capitán
Dear Antreas,
The triangles ABC, A'B'C' are perspective.
Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)
Best regards.
Angel Montesdeoca
--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
>
> 3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)
> the reflections of the NPC (N) in the perp. bisectors
> OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical
> axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
> The triangles ABC, A'B'C' are perspective.
>
> Perspector?
>
> APH
--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
>
> X(5390) = EULER-MORLEY-ZHAO POINT
>
> Barycentrics (unknown)
> Let DEF be the classical Morley triangle. The Euler lines of the three
> triangles AEF, BFD, CDE
> appear to concur in a point for which barycentric coordinates remain
> to be discovered.
> Construction by Zhao Yong of Anhui, China, October 2, 2012.
>
> http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
>
>
--- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
>
> Dear Hyacinthists,
>
> What is the point P for which P of the incentral triangle = X(11)?
>
> What is the point Q for which Q of the incentral triangle = X(115)?
>
> What is the point R for which R of the incentral triangle = X(3024)?
>
> Since X(11), X(115) and X(3024) lie on the incentral circle, P, Q and R lie on the circumcircle of ABC.
>
> I have checked the usual suspects, X(74), X(98)-X(112), X(476), and their antipodes, but no luck.
>
> Thanks in advance,
> Randy Hutson
>
--- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
>
> Friends,
>
> I would think this has been covered before, but I am unable to find anything on it:
>
> The locus of the centers of all circumhyperbolas passing through X(1) and a point on the Euler line is an ellipse with center X(1125). It passes through X(11) (center of Feuerbach Hyperbola) and X(1015) (Exsimilicenter of Moses Circle and Incircle). Are there any other Kimberling centers on this ellipse? I have checked the centers of circumhyperbolas passing through X(1) and Euler line points up to X(297), and have not found any. X(1015) is the center of hyperbola {A,B,C,X(1),X(2)}. The hyperbola {A,B,C,X(1),X(3)} also passes through X(29), and its center is not a Kimberling center.
>
> Is anything else known about this ellipse?
>
> Regards,
> B. Randy Hutson
>
> Let us celebrate the beginning of the cycle of this year
> by ETC cyclic points exploration.
> Nikos Dergiades
>I am wondering what geometrical properties [locus, envelope] hasMy limited understanding of a dual comes from here
>the "dual" of the MacBeath inconic ie the MacBeath circumconic
>[= the conic centered at N, and passing through A,B,C.]
