Dear friends, I think this one is right but I haven't thought about it at all. So I think I 'll call this one my conjecture : Let ABC be a triangle and I_a,...
Dear Hyacithists ... all. ... axises of (W_a) ... of ABC are concurent. What is its barycentric coordinate ? ... Your radical center is the complement of...
Dear Darij I don't quote your long and very interesting message; your results suggest me the following : A'B'C' is a triangle inscribed in ABC; ga,gb,gc are...
Dear Jean-Pierre Ehrmann, ... Very nice! Before I give a synthetic proof, I rewrite your result using my notations: Let g be a line in the plane of a triangle...
Let ABC be a triangle, and P a point. The parallel though A to BC intersects the circles (PAB),(PCA) at Ab, Ac again [the other intersection is A] A' := BAb /\...
Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. A' := the orth. projection of A on the line PPa A" := the reflection of A' in P. Similarly...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2004volume4/FG200418index.html The editors, Forum...
ForumGeom
ForumGeom@...
Oct 1, 2004 9:44 pm
10571
Dear Antreas, ... P. ... Reckon a Cubic, cyclic[(SB^2 SC + SC^2 SA + SA SB SC) y z^2 - (SA SB^2 + SB SC^2 + SA SB SC) y^2 z] + (SB^2 SC + SC^2 SA + SA^2 SB -...
Dear Peter ... Variation: A" := the reflection of P in A' (instead of A' in P) Similarly B",C" (ie A", B", C" are the reflections of P in the sidelines of the...
Dear Darij and Floor ... May be, it is interesting to notice that this remains true with three concurrent lines - not necessarily parallel - ga,gb,gc. Let Pa,...
Let ABC be a triangle. A' := (Reflection of BC in BA) /\ (Reflection of BC in CA). B' := (Reflection of CA in CB) /\ (Reflection of CA in AB). C' :=...
Dear JPE, DG and FvL ... The "dual" (in some sense) theorem (by Lemoine, somewhere in FG-M): Keeping DG's wording: Let g be an arbitrary line in the plane of...
Dear Antreas, ... This was known to me. By the way, the circles A'BC, B'CA, C'AB concur at O. Speaking in newer terms, the triangle A'B'C' is the Schaal /...
Let ABC be a triangle. A' := (Reflection of BC in BA) /\ (Reflection of BC in CA). B' := (Reflection of CA in CB) /\ (Reflection of CA in AB). C' :=...
Dear Antreas, ... This is what was called "Bliss theorem" before you found out that it is due to Lemoine. I remember you posting this theorem at Hyacinthos ...
... I owe to Conway not only the existence of Hyacinthos ! He was the man who made me to re-love geometry; (I had left her since my high-school days.) I hope...
Dear Darij and Antreas ... Note that the result remains true if, instead of the midpoints of the sides, you take for MaMbMc a triangle inscribed in ABC and ...
Generalizations: Let ABC be a triangle, A'B'C' the pedal (or cevian) triangle of P =(x:y:z), and L a fixed line, tripolar of Q = (u:v:w). La: = the parallel to...
Let ABC be a triangle, and A'B'C', A"B"C" its orthic, medial triangles, resp. ha,hb,hc := the lines through A',B',C' parallel to the Euler line of ABC ma,mb,mc...
In Hyacinthos message #10485, Antreas presented a beautiful conjecture about the Feuerbach point. Here I give a synthetic proof of this conjecture. With ...
Dear Jean-Pierre, ... Indeed, that's clear (it's all about the angles). [...] ... Yes, of course, it doesn't remain true. Maybe this could give a good locus......
Dear Darij ... Yes, but I like this with-no-circles construction of the circlecevians of both O,H: Denote: (mn) := the reflection of the sideline m in the...
Dear Khoa Lu, Jean-Pierre and Paul, ... I don't have a synthetic proof, but I have some remarks. After Theorem F2 of my Hyacinthos message #10584 (with...
Dear friends: Let ABC be triangle. The tangency points of its excircles (Ia),(Ib), (Ic) with (AB,AC,BC),(AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K), (H,I,L)...
Dear Juan Carlos, you wrote ... (Ib), ... The first barycentric coordinate of P is (b+c)(a^2(3a^2-2b^2-2c^2)-(bb-cc)^2)/(b+c-a) This point is not in the ETC...
Dear Darij ... Thank you for your nice synthetic proof ! What remains now is to find the locus of P such that the Euler lines of the triangles AAbAc, BBcBa and...
Dear Juan Carlos and Eric, Very nice results! In Hyacinthos message #10588, Juan Carlos Salazar ... Indeed, if Ia, Ib, Ic are the excenters of triangle ABC,...
Dear Eric, Darij and Paco García: Thank you very much for your remarks. Futher, if we consider:IL cut KF at U,HL cut JE at V,DJ cut KG at W. Also UVW and ABC...
