Dear friends, let P be a point on the circumcircle of triangle ABC with incenter I. denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB. I, Pa, Pb,...
Obviously the incenters Pa, Pb, etc. are on the bisectors of angles of the quadrangle ABCP and it seems that the condition is that the four bisectors pass...
Dear Bernard ... Suppose that P lies on the arc BC of the circumcircle opposite to A. The lines PaPb and PaPc go respectively through the excenters Ib and Ic. ...
... I prefer to replace "cannot exist" by "does not lie in real space"; I encountered many geometrical problems where e.g. a locus is imaginary except one...
Dear Bernard and Jean-Pierre, [BG] Let P be a point on the circumcircle of triangle ABC with incenter I denote by Pa, Pb, Pc the incenters of triangles PBC,...
Dear Jean-Pierre and Paul, ... I arrived to the same conclusions with the use of rotated circles about I, angle +/-90°. I have found that the triangle with...
In ETC I came across X(1285) which has the following description. X(1285) = LEMOINE HOMOTHETIC CENTER Trilinears f(a,b,c) : f(b,c,a) : f(c,a,b), where...
Dear Steve and Clark, [SS]: In ETC I came across X(1285) which has the following description. X(1285) = LEMOINE HOMOTHETIC CENTER Trilinears: f(a,b,c) :...
Dear Darij I tried to invent a transformation like isogonal and isotomic transformations. I thought that if Isogonal keep invariant incenter and isotomic keep...
Dear All (especially Steve and Wilson), In message #11703, I made some incorrect assumptions (which I think Steve tried to tell me in subsequent messages.)...
about my last post, I got another result. Given a triangle ABC and given a point P in its plane, let be O the circumcenter of ABC and Pa, Pb, Pc the simetrics...
Dear José Carlos, you wrote, ... the circumcenter of ABC and Pa, Pb, Pc the simetrics of P respect the perpendicular bisector of sides BC, AC, AB respectly....
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2005volume5/FG200524index.html The editors Forum...
ForumGeom
ForumGeom@...
Dec 6, 2005 8:45 pm
11795
Dear friends, does anybody know of a resource in which geometry is treated from constructive standpoint? I mean a presentation of geometry through...
Dear Vladimir, I obtained a copy of Clark Kimberling's "Enumerative Triangle Geometry" a few years ago and I have seen on his website additional references...
Dear José Carlos I tried to invent a transformation like isogonal and isotomic transformations. I thought that if Isogonal keep invariant incenter and ...
Alexey.A.Zaslavsky
zasl@...
Dec 7, 2005 1:18 pm
11798
Dear All, We have the following fact : If F is a Fermat Point of triangle ABC, then F is also a Fermat Point of its Cevian, and of its Anticevian Triangle. A...
Dear All, but especially Paul, In Bernard's CTC, K278 is described as the locus of P for which the Euler lines of P and its Anticevian triangle are parallel....
Dear Wilson, ... Indeed ! ... a circumsextic, isogonal conjugate of a non-circum cubic through X6, 15, 16, 399. the polar conic at K is the Stammler hyperbola,...
Dear Jeff, thanks a lot. The reference you gave may be helpful. But actually I'm thinking of something different. The project I'm currently engaged in is of...
Dear Vladimir, Paul Yiu in his work Elegant Geometric Constructions (See Hyacinthos links) has a list of references. Perhaps this list can help you. e.g. [13]...
If you have didactical goals maybe this site could be interesting: http://www.polarprof.net/geometriagon It's a site where the students are faced with...
Dear Wilson, ... this is direct consequence of the fact that the angular coordinates of a point wrt ABC, its cevian and anticevian triangles are the same. for...
Dear Bernard, ... X6, ... Thank you! I had thought that the locus would have high degree, so didn't compute it. I have now checked it and observe that the...
Dear All Sorry, I pressed the send key prematurely on message #11806. I meant to add that the fact that the sextic is the TA-transform of K278 is implicit in...
The line Fn--Fs that joins the two Fermat points is parallel to G--T (T = Tarry point) for each triangle, so these lines in the three triangles are either the...
... T ... T ... Sorry - perhaps my original post was ambiguous. If F is a Fermat point of ABC, then it is also a Fermat Point of the cevian triangle for F, and...
Each of these triangles has a point U = midpoint of GH. The Fermat line goes through this point, so either the U points of the three triangles involved are...
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