The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2013volume13/FG201303index.html The editors Forum...
Dear Antreas, your conic is always homothetic to the circumconic with perspector S={3 a^4 + b^4 - 2 b^2 c^2 + c^4, a^4 + 3 b^4 - 2 a^2 c^2 + c^4, a^4 - 2 a^2...
21467
Angel
amontes1949
Jan 30, 2013 12:29 pm
A simple property of X(5020): Let ABC be an triangle, DEF is the medial triangle. Let (Ca) be a circle through E and F that is tangent to the circumcircle at...
21468
Antreas Hatzipolakis
xpolakis
Jan 30, 2013 12:44 pm
Let ABC be a triangle, L a line passing through H, intersecting BC,CA,AB at A',B',C', resp. Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp....
21469
Paul Yiu
yiuatfauedu
Jan 30, 2013 3:10 pm
Dear Antreas, [APH] Let ABC be a triangle, L a line passing through H, intersecting BC,CA,AB at A',B',C', resp. Let La, Lb, Lc be the reflections of L in...
21470
Luís Lopes
qedtexte
Jan 30, 2013 5:10 pm
Dear Hyacinthists, In the preface of a french translation of an old 1899 russian book (maybe our russian and french friends know it, Problèmes de Géométrie...
21471
Vladimir Dubrovsky
vladubr
Jan 30, 2013 6:13 pm
Dear Luis, there are no full problem statements in the preface to the 16th Russian edition of Aleksandrov's book (perhaps the last one published before he ...
21472
Antreas
xpolakis
Jan 30, 2013 7:42 pm
Dear Paul The circumcircles are coaxial, but how about the NPCs ? Are their centers collinear for lines passing through H? APH ... [PY] I think you mean...
21473
Antreas Hatzipolakis
xpolakis
Jan 30, 2013 8:56 pm
A figure here: http://anthrakitis.blogspot.gr/2013/01/collinear-npc-centers.html ... [Non-text portions of this message have been removed]...
21474
Randy Hutson
rhutson2
Jan 30, 2013 9:02 pm
Dear Angel, Here are a couple more properties of X(5020): X(5020) = homothetic center of medial triangle and 3rd antipedal triangle of X(3) X(5020) = trilinear...
21475
Randy Hutson
rhutson2
Jan 30, 2013 9:51 pm
Dear Antreas, For L = Euler line, the NPC are indeed collinear, though I have not found the line. The circumcenters lie on line X(30)X(113) (Simson line of...
21476
Antreas Hatzipolakis
xpolakis
Jan 30, 2013 10:41 pm
Dear Randy The case of the L = OK line is very interesting! So we have that for the lines L= OH,OK the circumcenters are collinear and the question is for...
21477
Paul Yiu
yiuatfauedu
Jan 30, 2013 10:54 pm
Dear Antreas and Randy, The three circumcircles are always coaxial. If L is the trilinear polar of (u:v:w), the line containing the circumcenters is ...
21478
Paul Yiu
yiuatfauedu
Jan 30, 2013 11:40 pm
Dear Antreas, [APH] Let ABC be a triangle, L a line passing through H, intersecting BC,CA,AB at A',B',C', resp. Let La, Lb, Lc be the reflections of L in...
21479
Antreas
xpolakis
Jan 31, 2013 12:05 am
Dear Paul Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is a simple...
21480
Paul Yiu
yiuatfauedu
Jan 31, 2013 12:19 am
Dear Antreas, [APH]: Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is...
21481
Paul Yiu
yiuatfauedu
Jan 31, 2013 12:25 am
Dear Antreas, [APH]: Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is...
21482
Antreas
xpolakis
Jan 31, 2013 8:38 am
Dear Paul I think it is true for the centroids as well (ie they are collinear) And, in general, how about for any point on the Euler line??? That is, let P be...
21483
Angel
amontes1949
Jan 31, 2013 1:35 pm
Dear Hyacintthists, Let ABC be a triangle, P=(u:v:w) a point and DEF the cevian triangle of P. The radical axes of the circumcircle and arbitrary circle...
21484
Randy Hutson
rhutson2
Jan 31, 2013 11:17 pm
Some final(?) observations: Let AiBiCi be the triangle formed by the radical axes of each circle (Oa, Ob, Oc) and the corresponding mixtilinear incircle. Let...
21485
rhutson2
Feb 1, 2013 12:24 am
Dear Francisco and Antreas, This perspector S is also the crosssum of X(6) and X(1350), the crosspoint of X(2) and X(3424), and also lies on line X(4)X(32). ...
21486
Antreas Hatzipolakis
xpolakis
Feb 1, 2013 8:47 am
Let ABC be a triangle, IaIbIc the excentral triangle, P a point and A'B'C' the cevian triangle of P. Denote A* = B'Ic /\ C'Ib, B* = C'Ia /\ A'Ic, C* = A'Ib /\...
21487
Antreas Hatzipolakis
xpolakis
Feb 1, 2013 10:57 am
Let ABC be a triangle, P a point on the Neuberg cubic, A'B'C' the antipedal triangle of P, La, Lb,Lc the Euler lines of PBC,PCA,PAB, resp. and Ma,Mb,Mc the...
21488
Paul Yiu
yiuatfauedu
Feb 1, 2013 9:03 pm
Dear Antreas, Let ABC be a triangle, P = (u:v:w) with cevian triangle A'B'C' and Q=(x:y:z) with anticevian triangle QaQbQc. Denote A* = B'Qc /\ C'Qb, B* =...
Let ABC be a triangle, L a line through I and La,Lb,Lc the reflections of AI, BI, CI in L. Denote: Ab,Ac = the orthogonal projections of A on Lb,Lc, resp. ...
21493
Angel
amontes1949
Feb 4, 2013 8:48 pm
... Dear Antreas According to my calculations, the Euler lines of AAbAc, BBcBa, CCaCb are concurrent on a point Q(L) on the incircle of ABC, for any line L...
21494
Francisco Javier
garciacapitan
Feb 4, 2013 8:51 pm
for L=OI, the point Q is the antipode of X3025....
21495
Francisco Javier
garciacapitan
Feb 4, 2013 9:46 pm
With respect to the question 3. For which line(s) L the point of concurrence is the Feuerbach point? I am able to see that they are two perpendicular lines, I...
