Anh Tuan oi Now, I criticize your formula: We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U)) I find it too much complex for X(6) is not needed. You deduct it...
Dear Francois thank you very much for explaining what is equicenter E and areal point Q. Another simple construction for Q comes from the following property: ...
Dear all, In what follows, I will use |(x, y)| for the norm of a vector from (0, 0) to (x, y). Consider 3 points with integral coordinates in the plane: (0,...
Dear Francois and Nikos, Thank you very much for your interesting and general analyses from general view point. I learn a lot from your messages. In fact I...
Dear friends, the point Q is on the line at infinity if the point (p-p' : q-q' : r-r') is on the Steiner circumellipse. Hence the denominator K can be written...
Dear Barukh, I think your general problem is complicated a partial solution is the following: p = aabb - ccdd q = 2abcd u = aacc - bbdd v = 2abcd x = -p y = q ...
Dear Nikos and Tuan Thanks for your interesting remarks. Of course in choo-choo theory, these lines La, Lb, Lc play a central role and I have called them equal...
Dear Tuan, ... Yes the same method I also used. If A is the intersection of XY, X'Y' and M, M' are the mid points of XY', X'Y then the locus such that the...
Dear All My Friends, Given triangle ABC with centroid G, median triangle MaMbMc and two point Q, X. XaXbXc is Cevian triangle of X. The locus of X such that: ...
Dear Nikos and Tuan ... So if abc and a'b'c' are 2 inscribed triangles in ABC and if we use ordinary area ( i.e not signed), the Tuan-Nikos six lines cut...
Dear Nikos I give you my own construction of the equal area axis based on vectors. So you could compare it with yours. Given 4 points A, B, A', B' such that...
Dear Francois, [Tuan] ... [Francois] ... Yes. The line La+ is parallel to the segment of mid points of bc', b'c and gives equal areas The line La- is parallel...
Dear Francois, I've sent you a message and now I saw your message that you sent it before mine. As you see we use almost the same symbolism. You conclude the...
Dear Nikos Yes, in fact your proof also gives the general case for on any line there exists always in general 2 isogonal conjugate points. friendly Francois ...
Dear Nikos Cheer up, I will look at your beautiful proof but notice that the areal point of any 2 pedal triangles is always on the circumcircle. I have several...
Dear Nikos As I cannot sleep owing to my age, I feel like to begin some summer serial about choo-choos. As this theory is rather complex, today I will only...
Dear Francois and Nikos, If we choose two inscribed triangles as Cevian triangles of two Brocard points then Q can be one triangle center with barycentrics: Q...
Dear Francois, and Tuan. Thanks Francois, for your proof that can be written another way as: Let A'B'C', A"B"C" be two inscribed triangles in ABC, vector BC =...
Dear Nikos I think Cabri have the same ability, though as I have said it I am not interested directly in triangle centers. Suppose we have a construction...
Dear Antreas Thanks for your useful remarks. I am not a specialist of search values like Anh Tuan , so I have to think about it before answering. Good news,...
Dear Antreas Sorry, I send my post too soon! I said from the 2 points equicenter and areal center, the first is the more important for we all know importance...
Dear Francois, You answered to Antreas, I think by mistake. I am not Antreas. An interesting case is when abc, a'b'c' are similar. If the triangle abc is...
Dear Nikos and (always understood of course !) Dear Antreas ... "constant areal center". If we look at 2 inscribed directly similar triangles abc and a'b'c'...
Greetings, I see that in Paul Yiu, Introduction to the geometry of the triangle, it is mentioned that the perspector of triangle formed by the points of...
Dear Nikos Look at 3 choo-choos t --> a(t); t --> b(t) ; t --> c(t) with areal center S and equicenter E. Then for time u fixed, the function t --> Area(S,...
Dear Nikos Of course, your proof is better and one can effectively define directly on the drawing the barycentrics (x:y:z) of the areal center. In fact the...
Dear Nikos It was not an answer to me, but a comment to me as list-owner. [ND] ... [FR] ... ... and was [ the post] incomplete APH -- [Non-text portions of...