Dear Tuan thanks you very much for your verification and also for your nice email wrote for some months ago about "A little result". I continue my rechearch in...
Dear friends Sorry, I mixed up the labels of my points because of too much burgundy, hence I restore the true text: Let <s> be a direct similitude of center...
Dear friends I give you an strange theorem that I have found in a french book of exercices published more than 10 years ago. As it links euclidian and circular...
Dear All My Friends, Consider one point P with first barycentric coordinates: (b^2 - c^2)^2*(-a^6*(b^2 + c^2) + (b^2 - c^2)^2*(b^4 + 3*b^2*c^2 + c^4) +...
Dear Tuan, Your point is indeed on the Steiner Deltoid. Copied below are some Deltoid bits that I sent Clark a while ago. (Some bits may be irrelavent to...
Dear Peter, Thank you very much for your verification and plentiful results, information about Steiner deltoid. It is interesting that we have one common idea:...
Dear friends To show you how this theorem is useful, I will give you some of its applications but first I write it again more symmetrically: Let (A, B, C) and...
Dear Friends, There exists well-known JAVA export capabilities with our mainstream geometry software, but is anyone aware of a low-cost dynamic geometry ...
Dear Hyacinthians, I did some computations on the Lemoine quadrilateral as defined by Alexey Zaslavsky. I find that this class of quadrilaterals is broader ...
Given triangle ABC, one point U with barycentrics (u : v : w), one point P with barycentrics (p : q : r) and point X with barycentrics (x : y : z). A line from...
Dear Alexei ... there exists the point L such that the distances from L to the sidelines of ABCD are proportional to respective sides. Does anybody know which...
Dear friends In the affine choo-choo theory, I had to think how a metric structure can give a new look at it and that's why I was led up to this famous "3...
Dear friends, I've uploaded in Hyacinthos files a figure (Lemoine quadrilateral) showing the locus of D such as ABCD is a Lemoine quadrilateral : - the...
Dear Alexei and other Hyacinthists in fact, it is not difficult to construct when A,B,C,L are given the points D such as ABCD is a Lemoine quadrilateral with L...
Dear Jean-Pierre! Thanks for your remarks. I considered the case of oriented distances from L to the sidelines, i.e if L_1, L_2, L_3, L_4 are the projections...
Alexey.A.Zaslavsky
zasl@...
Nov 6, 2007 12:13 pm
15769
Dear Alexei ... distances from L to the sidelines, i.e if L_1, L_2, L_3, L_4 are the projections of L then the vector sum of LL_i is zero. Your approach is the...
Dear Jean-Pierre! If the quadrilateral is harmonic then its diagonals are the symmedians of correspondent triangles. So the common point of the diagonals is...
Alexey.A.Zaslavsky
zasl@...
Nov 7, 2007 1:34 pm
15774
Dear Alexei ... symmedians of correspondent triangles. So the common point of the diagonals is the Lemoine's point. OK, I was stupid and I see my confusion. ...
Dear friends ... Of course this macro works only with points in the euclidian plane. Now I explain you how to get the image M' = f(M) of any point M by a...
Good grief! Like a small child just learning to describe the world around him, I see tonight I was asking essentially for the center zx of a direct similitude....
Dear all, now in the section 'Charlas con JBRM' (*) of my site you can see: If we construct enclosing rectangles BCCaBa, CAAbCb, ABBcAc to the triangle ABC,...
Dear Garcia, ... If A'B'C' is the orthic triangle and A1B1C1 is the medial triangle of ABC then the line AbBa is the symmetric line of A'B' relative to C1 the...
Dear Nikolaos, thank you for your nice generalization, I have included it at http://garciacapitan.auna.com/problemas/jb/20071108/ Also, since the sum of...
Dear Nikos, I have sent to Hyacinthos a response to your message... ... but I haven't seen it published, Thank you for your nice generalization. I have put it...
... Yes, a rectangle triangle with legd 2 and 1. Trace the heihgt over the hipotenuse, and divide the greatest resultant triangle in four congruent ones. The...
Dear Ignacio Larrosa Cañestro, Thank you very much for your solution. It is interesting that in Vietnamese Yahoo Answer (I posted already some days) they are...
... By fact, in the same way it is possible divide a triangle in n^2 + 1 congruent triangles and similar to the initial. The legs of the triangle must be n and...
Dear Bernard and dear friends It is known that the shapes of repeated triangles wrt a fixed point P recur with period 3 and the Mac Cay cubic is the locus for...
Dear Francois ... P recur ... the third ... isometric ... ratio k > ... If I don't mistake,if P is inside ABC, the third pedal triangle cannot be isometric...
