Dear Cosmin, I like your argument too, although I would like to observe that your Lemma 1 (or at least the part that you need for your final statement) is ...
Let ABC be a triangle and P = (x:y:z) a point. Denote: A* :=(Perpendicular from B to CP) /\ (Perpendicular from C in BP) B* :=(Perpendicular from C to AP) /\...
[APH] ... Francisco told me it is not true in general. However, it seems it is true for P = O For P = H, it seems that La,Lb,Lc are concurrent. Antreas...
[APH] ... Francisco verified that it is true for P = O,H,G,N. (is it true for all points on the Euler line ?) He has also computed the coordinates of the...
Let ABC be a triangle, A'B'C' its medial triangle and P a point. Let La, Lb, Lc be the reflections of PA', PB', PC' in the bisectors AI, BI, CI, resp. Which is...
Dear Antreas, ... The locus is the linf + the conic with center X(1112) that passes through the vertices of the cevian triangles of X(4) Orthocenter and X(648)...
[APH] ... [ND] ... Dear Nikos It is nice! Thanks. Probably there are also nice results for pedal triangles of other points. For the pedal triangle of I, for...
Dear Hyacinthists, an article intitled "La promesse - le Tour -Le prestige ou le Géométrie magique" has been put on my site http://perso.orange.fr/jl.ayme ...
Let ABC be a triangle. Denote: L := The Euler Line of ABC H1, H2, H3 := The Reflections of AH, BH, CH in L, resp. La := The Euler Line of the triangle bounded...
Let ABC be a triangle and A'B'C' the cevian triangle of H (orthic triabgle). Denote: (Na), (Nb), (Nc) := The Reflections of (N) = the NPC of ABC, in AA', BB',...
Dear Antreas, [APH] ... If A', B', C' are arbitrary not collinear points with barycentrics A'=(0:q1:r1), B'=(p2:0:r2), C'=(p3:q3:0) then the locus is the linf...
[APH] ... Let (Q) be the circumcircle of NaNbNc and Sa, Sb, Sc the radical axes of [(Q),(Na)], [(Q),(Nb)], [(Q),(Nc)], resp. The Triangles ABC, Triangle...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2009volume9/FG200914index.html The editors Forum...
ForumGeom
ForumGeom@...
Jul 6, 2009 1:20 pm
17971
[APH] ... The circumcenter of NaNbNc is H. Since Sa,Sb,Sc are perpendiculars to HNa, HNb, Hnc resp., the problem is equivalent to: The parallels from A,B,C to...
Let ABC be a triangle and A'B'C' the cevian triangle of I. If the Euler Lines of ABC, BB'C, CC'B are concurrent, then is ABC necessarily isosceles? Antreas...
Dear Antreas, [APH] ... No. For example construct the triangle ABC with angles A=22.5 B=90 C= 67.5 The triangle B'BC is isosceles (BB' = BC) and the triangles...
Dear Nikos [APH] ... [ND] ... So, if these Euler Lines are concurrent the triangle is either right angled or isosceles or ??? Note that we can replace Euler...
Dear Nikos, that was a really nice example. I get that the condition -a^4 b^3 + 2 a^2 b^5 - b^7 + a^5 b c - a^4 b^2 c - a^3 b^3 c + a^2 b^4 c - a^4 b c^2 - a^3...
Let ABC be a triangle and A'B'C' the cevian triangle of the incenter I. Let La, Lb, Lc be the perpendiculars to AA',BB',CC' at A', B', C' resp. Are the Euler...
Dear Antreas, [APH]: Let ABC be a triangle and A'B'C' the cevian triangle of the incenter I. Let La, Lb, Lc be the perpendiculars to AA',BB',CC' at A', B', C'...
Dear Paul Very nice! Thanks! ... I am wondering how could we generalize it. If the perpendiculars are drawn at A,B,C (instead of A',B',C'), then the La,Lb,Lc...
Antreas and Paul: This is true. If we call AA*:A*A' = BB*:B*B' = CC*:C*C' = t:1, then the point P of concurrence of the Euler lines of triangles bounded by the...
[APH] ... Generalization: Let A'B'C'be the cevian triangle of P. Which is the locus of P such that the Euler lines of (La, BB',CC'), (Lb, CC', AA'), (Lc, AA',...
Dear Antreas and Alexey, [AZ] ... Here is a proof: If IA = a, IB = b, IC = c, ID = d angle x = (A+C)/2 and r is the radious of incircle ABCD then a =...
Having problems with message search?
Fill out this form to ensure your group is one of the first to be migrated to the new message search system.
