Let ABC be a triangle and P a point. Denote (Iaa), (Iab), (Iac) = the three excircles of PBC (corresponding to P,B,C, resp.) (Ibb), (Ibc), (Iba) = the three...
20585
yiuatfauedu
Dec 25, 2011 1:54 am
Dear Antreas, [APH}: Let ABC be a triangle and A'B'C' the cevian triangle of I. Denote: L1 = Radical axis of (the reflection of the excircle (Ic) in IB) and...
20586
Bernard Gibert
bernardgibert
Dec 25, 2011 2:18 pm
Dear Antreas and Paul, ... This is K311 = pK(X2, X320), ... and more generally, the point of concurrence lies on K510, a central pK with center X(214), pole...
20587
Sung Hyun Lim
bbblow
Dec 25, 2011 4:41 pm
Dear everyone, Merry Christmas! In Korea it's already over but I think it's still 25th in Europe and USA now. Could someone help me with following...
20588
Bernard Gibert
bernardgibert
Dec 25, 2011 4:57 pm
Dear Sung Hyun Lim, ... this should be : a(aP1 x cP2) where a, c, x denote anticomplement, complement, barycentric product. Best regards Bernard [Non-text...
20589
Sung Hyun Lim
bbblow
Dec 25, 2011 5:04 pm
Dear Bernard, You are right! Thanks a lot. This will help me in my research now greatly. Regards, Sung Hyun [Non-text portions of this message have been...
20590
Antreas Hatzipolakis
xpolakis
Dec 25, 2011 11:06 pm
Let ABC be a triangle. Denote: (Ica), (Iba) = the inversions of AI in the excircles (Ic), (Ib) resp. (Iab), (Icb) = the inversions of BI in the excircles (Ia),...
20591
Bernard Gibert
bernardgibert
Dec 26, 2011 10:58 am
Dear Antreas, ... a rather ugly point with first barycentric coordinate a (a - 3 b - 3 c) (a - b - c) (a^2 - b^2 + 6 b c - c^2) (a^2 + 2 a b + b^2 + 2 a c - 2...
20593
Antreas
xpolakis
Dec 26, 2011 9:19 pm
Let ABC be a triangle, P a point and (Oa),(Ob),(Oc) the circumcircles of PBC,PCA,PAB, resp. Denote: (Oab) = the inversion of (Ob) in (Oa) (Oac) = the inversion...
20594
Antreas
xpolakis
Dec 26, 2011 9:22 pm
... Not true (Thanks to Francisco) APH...
20595
Antreas Hatzipolakis
xpolakis
Dec 26, 2011 11:36 pm
Let ABC be a triangle, A'B'C' the orthic triangle and A"B"C" the Euler triangle (ie A",B",C" are the midpoints of AH,BH,CH, resp.) The circle (H,HA")...
20596
yiuatfauedu
Dec 27, 2011 4:58 am
Dear Antreas, [APH]: Let ABC be a triangle, A'B'C' the orthic triangle and A"B"C" the Euler triangle (ie A",B",C" are the midpoints of AH,BH,CH, resp.) The...
20597
Alexey Zaslavsky
zasl@...
Dec 27, 2011 6:46 am
Dear colleagues! Correspondence round of VIII Sharygin olympiad is available on http://www.geometry.ru/olimp/sharygin/2012/zaochn-e.pdf Sincerely...
20598
Antreas Hatzipolakis
xpolakis
Dec 27, 2011 8:06 am
The Problem #11 reads: Given triangle ABC and point P. Points A′, B′, C′ are the projections of P to BC, CA, AB. A line passing through P and parallel to...
20599
Antreas Hatzipolakis
xpolakis
Dec 27, 2011 12:32 pm
The Problem #15 reads: Given triangle ABC. Consider lines l with the next property: the reflections of l in the sidelines of the triangle concur. Prove that...
20600
Antreas Hatzipolakis
xpolakis
Dec 27, 2011 10:05 pm
Sorry I do not have time to check whether the problems I post are known or have been discussed here before. Here is a locus: Let ABC be a triangle, P a point...
20601
Nikolaos Dergiades
ndergiades
Dec 28, 2011 7:31 am
Dear Antreas, ... I think that the locus is the circumcircle and the quintic CYCLIC SUM {x^3*S_A*[(cy)^2-(bz)^2]}=0 and special cases are P = X(n) n = 1, 2, 4,...
20602
Antreas
xpolakis
Dec 28, 2011 7:54 am
Another one: http://anthrakitis.blogspot.com/2011/12/circumcenters-triangle-perspecti\ ve-with.html ...
20603
Bernard Gibert
bernardgibert
Dec 28, 2011 8:25 am
Dear Nikolaos, ... This is the Euler-Morley quintic Q003 http://bernard.gibert.pagesperso-orange.fr/curves/q003.html but X1557 is not on the curve. Best wishes...
20604
Nikolaos Dergiades
ndergiades
Dec 28, 2011 8:37 am
Dear Bernard, I found my mistake a little ago visiting your site before you write. Thank you. Yes 1557 is not on the curve. Best regards Nikos Dergiades...
20605
Antreas Hatzipolakis
xpolakis
Dec 28, 2011 9:22 pm
[APH] ... P. ... [...] ... [Francisco]: For P=O, the parallelogic center on NPC is X128. The other center is : ......... See it here: ...
20606
Chris Van Tienhoven
chris.vantie...
Dec 28, 2011 10:15 pm
Dear friends, The second presentation on "Perspective Fields" is now available on my site://www.chrisvantienhoven.nl/ This part includes the algebraic...
20607
Antreas Hatzipolakis
xpolakis
Dec 29, 2011 10:36 pm
Let ABC be a triangle, P a point, A1B1C1 the circumcevian triangle of P. The perpendiculars from A1,B1,C1 to the sidelines of ABC, BC,CA,AB resp. intersect...
20608
Nikolaos Dergiades
ndergiades
Dec 30, 2011 1:01 am
Dear Antreas, the locus is Line at infinity + Circumcircle + McCay cubic. Best regards Nikos...
20609
Nikolaos Dergiades
ndergiades
Dec 30, 2011 8:31 am
Dear Antreas, I forgot to say that if ABC is right angled A = pi then there is no locus. For every point P the circles are concurrent at the intersection of...
20610
Nikolaos Dergiades
ndergiades
Dec 30, 2011 8:33 am
Dear Antreas, I repeat my message because of a typo I forgot to say that if ABC is right angled A = pi/2 then there is no locus. For every point P the circles...
20611
Antreas Hatzipolakis
xpolakis
Dec 30, 2011 10:28 am
[APH] ... [ND] ... [ND] ... Dear Nikos Thanks. KALH XRONIA! Antreas PS: There are two "natural" variations. Ab := the orthogonal projection of A2 on BB1 Ac :=...
20612
Nikolaos Dergiades
ndergiades
Dec 30, 2011 2:11 pm
Dear Antreas, [APH] ... For this second case if A*B*C* is the tangential triangle of ABC then the locus is the line at infinity and the circles A*(B), B*(C),...
20613
Quang Tuan Bui
bqtuan1962
Dec 31, 2011 6:12 am
Dear All My Friends, ABCD is inscribed convex quadrilateral. Two diagonals AC, BD intersect each other at E. A' is symmetry of E in A B' is symmetry of E in...
20614
shokoshu2
Dec 31, 2011 3:53 pm
This wouldn't have happen to one of us, right? :-) http://www.goominet.com/unspeakable-vault/vault/58/ (Not deade yet, I'm still watching this group - but I'm...
