Dear Antreas, Interesting question generates interesting result: The rectangular circumhyperbola passing through A, B, C, H and P* has center at concurrent...
13842
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 9:53 am
Dear Antreas and Francisco, Please refer to my message ''Three Concurrent Pedal Circles'' #12300 and a lot of related interesting messages. We can apply some...
13843
Jeff Brooks
jbrooks_tulsa
Aug 1, 2006 12:47 pm
Dear Steve, Thank you for the reference. I periodically read your paper. Can you please help me to understand how the cyclocevian conjugate fits into this...
13844
Francois Rideau
francoisrideau
Aug 1, 2006 1:10 pm
Dear friends Here a construction of the 2 parabola on four points to compare with the euclidian Newton's construction you can find in the Dörrie book:100...
13845
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 2:09 pm
Dear Antreas, Francisco, Please forget my previous message about ''Three Concurrent Pedal Circles''. We don't need these very complicated ways. I have already...
13846
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 2:57 pm
Dear Kostas Vittas and All My Friends, I am thinking now: how we can construct similar configuration for isotomic conjugate? I think it is more difficult but...
13847
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 3:37 pm
Dear Antreas, Francisco, I am sorry for my wrong. There is not new triangle center here but interesting that - Three lines AYa, BYb, CYc are concurrent at...
13848
Moses, Peter J. C.
peter_mows
Aug 1, 2006 4:02 pm
Dear Tuan, 1. Exeter points of {HaEbEc, HbEcEa, HcEaEb} are also perspective to the Euler triangle at the intersection of lines ...
13849
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 5:26 pm
Dear Antreas, Francisco, Now we consider three lines: La connected midpoints of AaXa and BbCc Lb connected midpoints of BbXb and CcAa Lc connected midpoints of...
13850
garciacapitan
Aug 1, 2006 6:25 pm
Dear Tuan, the search number in ETC is 2.81278568... Its barycentric coordinates are: {2*a^12 - 6*a^10*b^2 + 6*a^8*b^4 - a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - ...
13851
Kostas Vittas
vittasko
Aug 1, 2006 6:41 pm
... Dear Antreas, Tuan and all. I would like to present a simple proof for the perspectivity of the triangles HaHbHc, NaNbNc, based on the below proposition,...
13852
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 7:06 pm
Dear Francisco, Antreas and All My Friends, Thank you for your calculation! I am very happy to see it. I generalize the problem as following: Given triangle...
13853
Steve Sigur
steve_sigur
Aug 1, 2006 8:30 pm
Jeff, I had to look up the meaning of cyclocevian conjugate of P. In Conway notation it is tdgmt P. The term "conjugacy" is fair for this concept. It is in...
13854
Kostas Vittas
vittasko
Aug 2, 2006 5:52 am
... ABC. ... Dear Antreas Tuan and all. A friend of mine Kostas Helatiotis, said me that there is another, well known, formulation of this proposition, as...
13855
Kostas Vittas
vittasko
Aug 2, 2006 6:54 am
... Dear Antreas, Tuan and all. By the well known V. Thebault's theorem, the Euler lines of the triangles AhaHc, BhcHa, ChaHb, in your configuration, are...
13856
Kostas Vittas
vittasko
Aug 2, 2006 9:04 am
... Dear Antreas, Tuan and all. I think that we can easy to prove this result by the fact that the Euler lines of two similar triangles, form congruent...
13857
garciacapitan
Aug 2, 2006 10:35 am
Dear Tuan, ... It seems that these three lines are always concurrent (???) Best regards, Francisco Javier García Capitán...
13858
Antreas P. Hatzipolakis
xpolakis
Aug 2, 2006 12:11 pm
Dear Kostas ... A more general theorem is The reflections of any line passing through H, in the sidelines of ABC, are concurrent on the circumcircle of ABC APH...
13859
Moses, Peter J. C.
peter_mows
Aug 2, 2006 12:46 pm
Dear Tuan & Francisco, The locus looks to be composed of The circumcircle .. lines are the Simson line of P. Line at Infinity. ABC sidelines. 3 circles...
13860
Kostas Vittas
vittasko
Aug 2, 2006 1:16 pm
Dear Antreas, Tuan and all. Now, is better for me to present the proof I have in mind, about this result. In the configuration which is mentioned in message...
13861
Kostas Vittas
vittasko
Aug 2, 2006 1:30 pm
Dear Antreas. Thank you for your reminding of this general theorem, which we have been learned in high school ( many years ago for me ). Best regards. Kostas...
13862
garciacapitan
Aug 2, 2006 2:29 pm
Dear Peter and Tuan, now I see that the results of Peter are correct, in fact, I saw them with Mathematica, but I was not able to read the imaginary conics,...
13863
Antreas P. Hatzipolakis
xpolakis
Aug 2, 2006 7:22 pm
Dear Tuan and Peter [Bui Quang Tuan] ... [PM] ... I am wondering what is the locus of the perspectors as P moves on the Darboux cubic. I guess it is some...
13864
Quang Tuan Bui
bqtuan1962@...
Aug 3, 2006 12:15 pm
Dear All My Friends, When P is on the locus so six points Aa, Bb, Cc, Xa, Xb, Xc are on one conic, say (Co). This conic is a circle when P is incenter and...
13865
Antreas P. Hatzipolakis
xpolakis
Aug 3, 2006 6:39 pm
Probably we have discussed the following before. Anyway... Let ABC be a triangle, P,P* two isogonal points, A'B'C', A"B"C" their pedal triangles. A* := the 2nd...
13866
Moses, Peter J. C.
peter_mows
Aug 3, 2006 7:33 pm
Dear Antreas, A# = {2 a^2 p, a^2 q - p SC, a^2 r - p SB} 1. Thompson cubic. {1,2,3,4,6,9,57,223,282,1073,1249} The perspector looks to live on the Lucas cubic....
13867
Antreas P. Hatzipolakis
xpolakis
Aug 3, 2006 8:05 pm
[APH] ... *********** Sorry, the A#s are not identical! In the first case, we have that A*P* = A'A#, and in the second, that A*A" = A'A# So we have two...
13870
Moses, Peter J. C.
peter_mows
Aug 4, 2006 11:03 am
Dear Antreas, The problem I solved was A# = reflection of P* in the midpoint of {P, A*}. ... also A# reflection of A' in P. So I have a parallelogram PP*A*A# ...
13873
Quim Castellsaguer
fqces
Aug 4, 2006 3:19 pm
Dear friends, I have noticed this fact: P, Q being two isogonal points, the cevians of P and the cevians of Q intersect at P, Q and six more points. The lines...
13875
garciacapitan
Aug 4, 2006 5:46 pm
Dear Quim, in the general case, if P={u,v,w} (barycentric coordinates), then X={a^2*b^2*u*w^2, b^2*c^2*u^2*v, a^2*c^2*v^2*w} Y={a^2*c^2*u*v^2, a^2*b^2*v*w^2,...
