... orthocenter. ... Beta. ... Dear Jack, playing with C.a.R. around this interesting figure, it seems that one can notice two facts (I have no proof, only...
Dear Francois! You are right, I proved the theorem by this way. Sincerely Alexey Dear Alexey That reminds me something...
Alexey.A.Zaslavsky
zasl@...
Nov 1, 2008 7:07 am
16941
Dear Alexey To sum up your sketch, we start with a triangle ABC and a point P. <Gamma> is the ABC -inconic with perspector P. The tripolar line L of P wrt ABC...
Dear Hyacinthists, my question is again very poor: what means R. C. J. for the English geometer Nixon and what is his date of birth and die? Where can I get...
Dear François, that is an interesting observation. The proof for the configuration is really not that hard, but your observation might be hard to prove. I ...
Dear Eisso I would be happy to have your proof of this beautiful theorem. Friendly Francois ... [Non-text portions of this message have been removed]...
... Dear Alexey, Last year I have handled this subject and several other similar questions in my Gallery http://www.math.uoc.gr/~pamfilos/eGallery/Gallery.html...
Dear friends Looking at a quadrangle of 4 points A1, A2, A3, A4, I name: T1 the pedal triangle of A1 wrt A2A3A4. T2 the pedal triangle of A2 wrt A1A3A4. T3 the...
Dear François, some (perhaps even all) of your observations are discussed in: Schroeder, Eberhard “Zwei 8-Kreise-Saetze fuer Vierecke.” Mitteilungen der...
Dear Francois! Let B1B2B3 be the pedal triangle of A4 wrt A1A2A3. Then for example angle (oriented) B1B3B2=A1A3A2+A1A4A2. So all four pedal triangles have...
Alexey.A.Zaslavsky
zasl@...
Nov 13, 2008 1:04 pm
16954
Dear Paris Thank you very much for your help. In fact, I have the book but the idea to look at it never occur to me! You are right, this theorem (and many...
Dear Eisso Thank you very much. As I said to Paris, many other theorems were proved in this chapter XIV. So nothing is new in triangle geometry as we all know...
Dear Alexey Thank you very much! That's just the Johnson angle chasing. I like your characterization of the common center of similarity as a center of the...
Dear Alexey In case A1A2A3A4 is orthocentric, all vertices share the same pedal triangle. So the center of similitude is everywhere, oops! Friendly Francois. ...
Dear Francois! Yes, this is a singular case Sincerely Alexey Dear Alexey In case A1A2A3A4 is orthocentric, all vertices...
Alexey.A.Zaslavsky
zasl@...
Nov 14, 2008 7:54 am
16959
Dear friends Starting from two triangles ABC and A'B'C', what is the shortest way to construct the point P if it exists such that the pedal triangle PaPbPc of...
Dear Hyacinthists, let ABC be a triangle, 0 the circumcircle of ABC, I the incenter of ABC, A*, B*, C* the point of contact of the A, B, C-mixtilinear circles ...
Dear Jean-Louis, this is a nice property, and I don't recall seeing it before. The conjecture is true, the second common point is the inverse of I in the ...
Francois, The point you are looking for is the isogonal conjugate w.r.t. ABC of what you call the pivot of the triangles circumscribing ABC and directly ...
Dear Eisso Thank you very much. Of course, I knew your construction but I thought that there was a more direct one. It is very important that there is no such...
Dear Francois, I think that your point P is the point such that <BPC = <A + <A' <CPA = <B + <B' <APB = <C + <C' where <A is the angle BAC of triangle ABC. Best...
Dear Hyacinthists, let ABC be un triangle, 0the circumcircle of ABC 1a, 1b, 1c the A, B, C-mixtilinear incircles of ABC, Ab, Ac the points of contact of 1a wrt...
Dear Francisco and Hyacintists, Thank for your precision. I think that the two triangle in question are orthologic. Sincerely Jean-Louis ... perspective...
Yes, the triangle UVW bounded by Da, Db, Dc and ABC are orthologic, being the orthology centers the circumcenter and the point {a^2 (a^4 + 2 a^3 b - 4 a^2 b^2...
