Dear Paul and Bernard ... centroids of ... anticevian ... aP is ... known, ... intersections of ... They are the common points of the circumconic with center...
Dear Paul and Bernard ... the ... I just realize that the line above is homothetic of the trilinear polar of q = aP/G in (G,4); this is, of course, an easier ...
... Triangles EAD and EDC are similar. Consider the similarity that takes the first into the second. It is the composition of 90 degrees rotation about E and...
Dear Jean-Pierre and Paul, ... I have found the same thing. the 3 points also lie on three (easy to draw) conics : one of them passes through B, C, Ga (vertex...
* was Steiner circumconic and X(1962) Dear Jean-Pierre and Bernard, [PY] Given a point P, can you find a point Q so that the centroids of the cevian triangle...
Dear Hyacinthians, let A'B'C' be the orthic triangle of ABC Let A* be the intersection of the angular bisectors of AC'C and AB'B Define B* and C* similarly ...
Dear Hyacinthians, Let ABC be a triangle and (K) is an arbitrary circle such that there exist 3 circles (K_a), (K_b), (K_c) extouch it at M, N, P; and...
Dear Hyacinthians, I believe that the following result is known: A plane going through the midpoints of two opposite edges of a tetrahedron cuts the...
Dear Maurice, ... Let the midpoints, M and N, be taken on edges AB and CD, and let the other two vertices of the section be K (on AC) and L (on BD). Consider a...
Dear Vladimir, Thanks very much for this nice little proof (1); very simple... And as a bonus I learnt the "Cavalieri's Principle". I have an other question...
Dear Maurice, the converse "3D Pythagoras" seems to fail. One possible approach: For each face of the tetrahedron, consider its outward normal vector equal in...
Dear Vladimir, Thanks again for this nice counterexample. Strangely I have never seen the interesting Minkowski condition in one of my books about polyhedra; I...
Dear Eric, ... Unless I missed something, this formula is valid if you assume that : 1. the angular bisectors are the internal bisectors at C', B' in the ...
Dear friends, the 6 cevians of the two Steiner circum-ellipse foci have the same length. Hence these foci lie on the Ehrmann strophoid K025. any reference for...
My dear Bernard In an ellipse (E), there are infinitely many triangles ABC inscribed in (E) so that (E) is their Steiner circum-ellipse. Then the common...
... An other "simple" but less elementary proof (Georges Lion): Consider the symmetry with axis the line going through the midpoints of two opposite edges and...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2005volume5/FG200503index.html The editors Forum...
ForumGeom
ForumGeom@...
Feb 10, 2005 6:42 pm
11063
Dear Hyacinthians, Cyril Parry's daughter, Vivien Harrison, in England, has just sent me the message copied below. I hope that someone will post a list of ...
From Clark Kimberling we heard the sad news that Cyril F. Parry, a significant geometer, passed away. [CK] ... Here is a listing, after a quick Internet search...
Also, Cyril F. Parry contributed many beautiful problems and solutions to the Problem Corner of The Mathematical Gazette. His ingenious problems have always...
Dear friends, if P is a point and p, q, r are the distances of this point from the vertices of triangle ABC then I think that the minimum sum p^n + q^n + r^n...
Dear Nikolaos ... the ... is found ... Of course, it is true : the gradient of AP^n + BP^n + CP^n is n.AP^(n-2).Vector(AP)+cyclic. Note that, for n = =4, your...
Dear colleagues! Sorry for long absence in Hiacinthos forum, there were some technical problems with my e-mail. We (with Arsenij Akopyan) found interesting...
Alexey.A.Zaslavsky
zasl@...
Feb 18, 2005 6:32 am
11069
Dear Jean-Pierre thank you very much. For n=4 I misunderstood with the case n = -2 from message 11004. Best regards Nikolaos Dergiades ... ...
Dear Alexey, hence the isogonal conjugates wrt ABC of the four centers (incenter-excenters) of the cevian triangle A'B'C' of the point P are lying on the line...
Dear Alexey and Nikolaos ... equilateral ... ABC. ... hyperbola ... that the ... In fact, the two following results are old : they were known, for instance by...
Dear Hyacinthians, I've been asked the following question: do the three cevians each of which divides a given triangle into two triangles with equal inradii...
Hello Vladimir, ... two ... concurrence, but ... I have done a similar experiment and find that although close, I do not think they are perspective. Call A'...
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