Dear Antreas, Interesting question generates interesting result: The rectangular circumhyperbola passing through A, B, C, H and P* has center at concurrent...
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 9:31 am
13842
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...
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 9:53 am
13843
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...
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...
Dear Antreas, Francisco, Please forget my previous message about ''Three Concurrent Pedal Circles''. We don't need these very complicated ways. I have already...
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 2:09 pm
13846
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...
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 2:57 pm
13847
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...
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 3:37 pm
13848
Dear Tuan, 1. Exeter points of {HaEbEc, HbEcEa, HcEaEb} are also perspective to the Euler triangle at the intersection of lines ...
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...
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 5:26 pm
13850
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 - ...
... 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,...
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...
Quang Tuan Bui
bqtuan1962@...
Aug 1, 2006 7:06 pm
13853
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...
... 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...
... 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...
... 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...
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...
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...
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...
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...
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,...
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...
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...
Quang Tuan Bui
bqtuan1962@...
Aug 3, 2006 12:15 pm
13865
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...
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....
[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...
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# ...
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...
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,...
