Dear Nik and Antreas, Happy New Year to you and all friends of Hyacinthos. ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2 are equilateral triangles...
Dear Paul, we have the same result as for A* . . . but I am impressed with the factorization you had in the equation of the cubic. Happy New Year. Best regards...
Dear Nikos, I also did the calculations but when I was about to post them, I felt gladly overwhelmed by the impressive answer by Paul. Happy New Year to all...
Dear Nik and Francisco Javier, ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2 are equilateral triangles and A* is the mid point of A1, A2 in the...
Dear Nikos, Paul Happy New Year!!! 1. We can take A'B'C' as the pedal triangle of P and ask for the locus. 2. We can take A*,B*,C* as triangle centers...
Dear Paul, A little generalization: If A'B'C' is the cevian triangle of a point P and A"BC, B"CA, C"AB are similar isosceles triangles outside of ABC with base...
Dear Antreas, if A'B'C' is the pedal triangle and not the cevian then the locus is a cubic but I think complicated. The same holds if we take circumenters. I...
Dear Nikos, Paul and friends, ... If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are...
Dear Nik, Bernard, and Francisco Javier, Thank you for these generalizations. FJ, I did not realize that the locus contains G! Best regards Sincerely Paul ...
Dear all, If A'B'C' is the pedal triangle of P (instead of cevian triangle of P), and A", B", C" as before, If A* is the homothetic of A" under h(A', k), B*...
X(5390) = EULER-MORLEY-ZHAO POINT Barycentrics (unknown) Let DEF be the classical Morley triangle. The Euler lines of the three triangles AEF, BFD, CDE ...
20. Euler-Morley-Zhao Point Reward: $50.00 Let A'B'C' be the Morley equilateral triangle of an arbitrary triangle ABC. Zhao Yong of Anhui, China, discovered...
I am wondering if there is ANY known proof that they are concurrent. Drawing evidence is not a proof...! aph ... [Non-text portions of this message have been...
Let ABC be a triangle and A'B'C' its internal Morley triangle. Zhao conjecture: The Euler Lines of AB'C', BC'A', CA'B' are concurrent. aph conjecture: The...
Dear Antreas, ... Indeed !... After some very ugly and painful computations, I did prove that the 3 Euler lines are concurrent. The SEARCH number of X(5390) is...
Dear Bernard Thank you! Naturally one may ask how about the case of the external Morley triangle ie the eq. triangle A'B'C' we get by the external trisectors...
Dear Nikos, Thanks. Now, we know some equilateral triangles which are homothetic to Morley internal equilateral triangle A'B'C'. One of these triangles is...
Dear all, First of all a Happy New Year to all Hyacinthians. Generalizating conjecture: Let t be a number. Consider the t-sector of an angle alpha the line...
Dear Antreas, Not only the Euler lines of triangles A'B"C", B'C"A", C'A"B" are concurrent but also the Euler lines of A"B'C', B"C'A', C"A'B' I am almost sure...
Dear Nikos, Thanks! I had thought about that (of A"B'C', B"C'A', C"A'B' ) and made a picture, but it seems that my picture was not good enough..... they were...
Dear All Is it correct to assume that all these conjectured Euler line concurrences are colinear and lie on the Euler line of the original triangle ABC? ...
Dear friends, I managed to calculate the coordinates of X(5390). Indeed the outcome is horrendous. You need several pages to print it. Just like Bernard I...
Let A'B'C', A"B"C" be two homothetic Equilateral triangles. Conjecture: The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent. (and of A"B'C', B"C'A',...
Antreas, I was able to confirm this computationally. Of course, not the same as a proof. Randy Hutson ... [Non-text portions of this message have been...
Dear Randy I am not sure what you mean by confirming computationally. If you used a coordinate system or other general algebraic method it is a proof! Not nice...
