Let ABC be a triangle and 12, 13 the tangents to the NPC, perpendicular to BC, near to B,C, resp. Similarly the tangents 23,21 and 31,32. The 6 real...
21622
rhutson2
Feb 24, 2013 2:26 am
Let A'B'C' be the medial triangle and A"B"C" be the orthic triangle. (So A', A" are the intersections of the NPC with BC, etc.) Let Ab be the {A',A"}-Harmonic...
21623
César Lozada
cesar_e_lozada
Feb 24, 2013 2:56 am
Answers: X(1) and X(10) Regards César Lozada _____ De: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] En nombre de Antreas Hatzipolakis ...
21624
Francisco Javier
garciacapitan
Feb 24, 2013 8:14 am
Dear Randy, you can take A'B'C' the medial triangle and A"B"C" the cevian triangle of any point (p:q:r) and this gives a conic with center {(p + q) (p + r) (10...
21625
Jean-Louis Ayme
jeanlouisayme
Feb 24, 2013 1:48 pm
Dear Jean-Pierre and Mathlinkers, working on a generalization of a Nagel's result, I found in the archive of Hyacinthos the message # 19640 Dear Chris [JP] ...
21626
Barry Wolk
wolkbarry
Feb 24, 2013 10:11 pm
... There were several replies, including asking for cyclically symmetric coordinates for those intersection points. The points on the line Px+Qy+Rz=0 can...
21628
Chris Van Tienhoven
chris.vantie...
Feb 25, 2013 7:11 am
Dear friends, Thanks Barry! This brings me to a new problem. I never got into this. Given an equation f(x,y,z). How to convert it to a point in time (t):...
21632
Antreas Hatzipolakis
xpolakis
Feb 25, 2013 1:20 pm
http://anthrakitis.blogspot.gr/2013/02/feuerbach-point.html Synthetic proofs? Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal triangles of I, resp. ...
21633
jpehrmfr
Feb 25, 2013 1:49 pm
Dear Jean-Louis and other Hyacinthists this follows from two facts : if P,P' are inverse in the circumcircle of ABC, then <BPC + <BP'C = 2<BAC if P,P* are...
21634
Antreas
xpolakis
Feb 25, 2013 11:37 pm
Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal triangles of P, resp. Denote: Ab, Ac = the reflections of A' in BB', CC' Bc, Ba = the reflections of...
21635
Randy Hutson
rhutson2
Feb 26, 2013 2:47 am
Thank you, Francisco! This could be further generalized by letting A'B'C' and A"B"C" be any 2 cevian triangles. Rsndy ... [Non-text portions of this message...
21636
Antreas
xpolakis
Feb 26, 2013 9:08 am
[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...
