[APH] ... See the construction of the circles here: http://anthrakitis.blogspot.gr/2013/02/circles-of-quadrilateral.html Note that we get a quadrilateral...
21637
Ignacio Larrosa Ca...
ilarrosa
Feb 26, 2013 9:35 am
... From: "Antreas" <anopolis72@...> To: <Hyacinthos@yahoogroups.com> Sent: Tuesday, February 26, 2013 10:07 AM Subject: [EMHL] Re: Quadrilateral Problem...
21638
Angel
amontes1949
Feb 26, 2013 12:57 pm
Dear Antreas, If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle. If P=H the lines La,Lb,Lc concur in X(1986)=...
21639
Hyacinthos@yahoogroup...
Feb 26, 2013 7:42 pm
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the Hyacinthos group. File :...
21640
Angel
amontes1949
Feb 26, 2013 7:43 pm
More information on the algebraic curve of degree nine (Gamma): - Passes through the vertices of the triangle ABC. - The vertices are multiple points of order...
21641
Antreas Hatzipolakis
xpolakis
Feb 26, 2013 8:15 pm
Dear Angel Thank you. I came to this configuration trying to find three homocentric (concentric) circles but not by construction (ie not by taking a point as...
21642
Antreas
xpolakis
Feb 26, 2013 10:22 pm
Let ABC be a triangle and A'B'C' the cevian triangle of I. Denote: B'a, C'a = the reflections of B',C' in AA', resp. A'a = (perpendicular to BB' from B'a) /\...
21643
Antreas Hatzipolakis
xpolakis
Feb 26, 2013 11:34 pm
Let ABC be a triangle, P a point and (Ia), (Ib), (Ic) the excircles. Denote: Pa = inversive of P in (Ia), Pab = Inversive of Pa in (Ib) and Pac = Inversive of...
21644
Antreas Hatzipolakis
xpolakis
Feb 27, 2013 6:24 am
[APH] ... More for P = H: The NPCs of A'AbAc, B'BcBa, C'CaCb are concurrent on the NPC of A'B'C' (On the Poncelet point of H with respect A'B'C' ie the point...
21645
Francisco Javier
garciacapitan
Feb 27, 2013 6:42 am
The locus has a 12th degree equation, 52596 terms long....
21646
Antreas Hatzipolakis
xpolakis
Feb 27, 2013 7:15 am
No simple point (O,H,I, etc) is lying on it? aph On Wed, Feb 27, 2013 at 8:42 AM, Francisco Javier ... [Non-text portions of this message have been removed]...
21647
Antreas Hatzipolakis
xpolakis
Feb 27, 2013 11:36 am
Let ABC be a point, L a line and A',B',C' the orth. projections of A,B,C in L, resp. Let P be a point on L and Pa,Pb,Pc its reflections in AA',BB',CC', resp. ...
21648
Angel
amontes1949
Feb 28, 2013 12:42 am
Dear Antreas If the equation (barycentric) of L is px + qy + rz = 0, when P moves on the line L the lines PdP through the point L' of first coordinate: L'=(...
21649
Antreas Hatzipolakis
xpolakis
Feb 28, 2013 7:03 pm
Let ABC be a triangle, P a point and (a),(Ob),(Oc) the circles with diameters PA,PB,PC, resp. The circles (Ob), (Oc) intersect in point A1, other than P, on BC...
21650
Francois Rideau
francoisride...
Feb 28, 2013 7:03 pm
I thought it was the orthopole of L. So I was wrong Francois ... [Non-text portions of this message have been removed]...
21651
rhutson2
Feb 28, 2013 10:44 pm
Antreas, The circumcircles are concurrent at non-ETC 0.812149174855220, which is, equivalently: X(1)-Ceva conjugate of X(36) Antigonal conjugate, wrt incentral...
21652
Antreas Hatzipolakis
xpolakis
Feb 28, 2013 11:26 pm
Dear Randy Thanks. I guess that the point for the orthic version (ie ABC is orthic triangle of reference triangle) is not in ETC as well. Now, as for...
21653
Sung Hyun Lim
bbblow
Mar 1, 2013 2:55 pm
Dear Everyone, It's been a long time since I've written. I'd like to ask a question: Does anyone know of mathematical journals that treat classical/Euclidean ...
21654
rhutson2
Mar 1, 2013 5:24 pm
Dear Antreas, For the orthic version (P=X(4), case 1.), the circumcircles do not concur, but their radical center is non-ETC -10.266788994809312, which lies on...
21655
Antreas
xpolakis
Mar 1, 2013 10:02 pm
Dear Randy Sorry for not being clear. Let Q be the point of concurrence of the circles wrt ABC. Now, let Q' be the point of concurrence wrt orthic triangle. ...
21658
Antreas
xpolakis
Mar 3, 2013 11:50 pm
Let ABC be a triangle and PaPbPc the pedal triangle of point P. Denote: A'B'C' the triangle bounded by the radical axes: ...
21659
Francisco Javier
garciacapitan
Mar 4, 2013 7:24 am
The locus for a quartic through O and a quintic trhough H and nine point center N. For H the perspector is not in ETC: it is the isotomic conjugate of the...
21660
Antreas Hatzipolakis
xpolakis
Mar 4, 2013 8:33 am
Dear Francisco, Thanks! We can replace O with any other fixed point Q, but I had chosen O because it is the one with the simplest tripolar coordinates...
21661
Jean-Louis Ayme
jeanlouisayme
Mar 5, 2013 11:03 am
Dear Hyacinthists, an article concerning “La droite de Simson de pôle Fe’’ has been put on my website. http://perso.orange.fr/jl.ayme   vol. 7 ...
21662
Antreas Hatzipolakis
xpolakis
Mar 5, 2013 7:51 pm
Dear Jean-Louis “La droite de Simson de pôle Fe" is not clear. In fact you mean, the Simson line with respect the pedal triangle of I [intouch triangle] of...
21663
Antreas Hatzipolakis
xpolakis
Mar 5, 2013 10:01 pm
Let ABC be a triangle and IaIbIc the excentral triangle (ie Ia,Ib,Ic are the excenters). Denote: IaaIabIac = the pedal triangle of Ia IbbIbcIba = the pedal...
21664
Randy Hutson
rhutson2
Mar 5, 2013 10:58 pm
Antreas, It sounded promising, but the lines are not concurrent. Regarding Jean-Louis' original article, I was unable to translate the French, so this may have...
21665
Randy Hutson
rhutson2
Mar 5, 2013 11:40 pm
Antreas, A'B'C', A^B^C^ are not orthologic. However, A^B^C^ is perspective to ABC at X(262), and to A"B"C" at non-ETC 43.220195027499022, which is also the...
21666
Jean-Louis Ayme
jeanlouisayme
Mar 6, 2013 6:10 am
Dear Hyacinthists, yes , I agree with your remark... I was too speed , but the title of my article written in french is OK. Sincerely Jean-Louis ...
21667
Antreas Hatzipolakis
xpolakis
Mar 6, 2013 10:48 am
Dear Randy I guess that only "sounds promising" is also the case of the Apollonius circle instead of NPC! That is, we replace the NPC (which touches the...
