Dear friends let P be a point in the plane of triangle ABC let I be the incenter of the cevian triangle of P what are the regions of the plane which cannot...
Dear all, Sorry, my mistake, there is no typo in the description of X(1276) and X(1277) in ETC. For the concyclic properties of X(1), X(484), X(1276), X(1277),...
Let abc be a triangle (ps SODDY is a quinlirateral, hence the rearragnement isn't valid) Let P and Q be random points inside the triangle. Using, for example,...
Bernard, What a wonderful problem. Let the seven regions of the Plane of ABC be numbered as 0 for the central region 1,2,3 for the regions found by crossing...
Chia-Lin Hwang wrote on December 25, 2005 ... Did you come to this result by observation of by proof? Why for Q you mentioned ETC points only? It holds for...
... and C- ... pencil of ... circles ... Soddy ... Here is "my" :-) way to define the Fletcher point: In the framework of my "power line" approach, if the...
... Just for fun, replace the Soddy circles radii with a,b,c instead of s-a,s-b,s-c. The "power line" and radical center stays the same (neither negating all...
Dear Nikolaos, Thank you for the response. I got this result (conjecture more precisely) by a lot of computation. and only check it for those points in ETC. So...
Dear Chia-Lin If the point P on the Brocard axis is such that the simple ratio k = (OKP) is constant (independent from the side lengths of the triangle) then...
Im's sorry there was some confusions between Lemoine and Brocard. It should be ( I hope that it is correct now): First I wish that 2006 will be a very fruitful...
Dear Nikos and all, Forget the "index version" of Xcm[P|Q] (based on index of ETC) and redefine Xcm[P|Q] as the barycentrics of the point P of ABC wrt...
Dear Chia-Lin Hwang and other Hyacinthists ... and ... Here, you have supposed that u and v are constants. But, for most triangle centers lying on the Brocard...
As I already mentioned, Darij proved that #389 is the RC of the three circles centered in A,B,C and having radius A-HA,B-HB,C-HC where the Hi are the...
Dear Jean-Pierre and Nikolaos, Thanks a lot. The result should be The circumcevian triangle of Q lying on the Lemoine axis or the symedian point has the same...
Dear Jean-Pierre ... Let two triangles ABC and A'B'C' be inscribed in the same circle and let the Lemoine axis of ABC meets the line AA' at Q. If we fix the...
Dear friends, If ABC is a triangle, P ( x : y : z ) is a point in barycentrics and P* is the isogonal conjugate of P then the inconic (The conic tangent to the...
Dear Nikolaos [JPE] ... [ND] ... This is not true : Suppose that A'B'C' has the same circumcircle and the same orthocenter than ABC; then A'B'C' circumscribes...
Dear Nikolaos ... The center of the conic is the midpoint of PP* and the perspector is the isotomic conjugate of the anticomplement of the center. Friendly....
Dear Nikolaos ... I don't understand: In case P = H the orthocenter of ABC and A'B'C', it's true that both triangles are circumscribed to an inconic of foci H...
Hello, a problem: Let ABC be a triangle given w_c ,w_b bisector lengths and length of |AB| side. Is it possible construct that triangle? Any proof? I met that...
Dear Jean-Pierre and Francois [JPE] ... [JPE] ... Of course you are right. Thank you very much. [FR] ... Here I was confused. Thanks for disproving a false...
Dear Riccardo ... The answer is no. a and b are given by an equation of degree 8 and it seems, azccording to Maple, that the Galois group of the equation is ...
Dear Ricardo Once again, Happy New year! I began to do some calculations about your problem but they are very tedious so I have left them as lazy as I am. Here...
Dear all, By coincidence I wrote yesterday to Paul (no answer yet) regarding this problem. In the following you are going to understand why. I meant the same...
Dear all, A very happy new year to all on Hyacinthos. May it be fruitful! Consider the midcircle of the in- and circumcircles, i.e. the circle that defines the...
Dear Hyacinthians, a friend of mine Nikos Kyriazis, found interesting results in the following configuration: On the sides of a triangle ABC we erect...
Dear Hyacinthians, To forget Euclidian Geometry for a few moment, yeah, here a little result of affine geometry I found thinking about 3 chouf-choufs running...
Sorry, I correct the little typo in the last sentence: Given a triangle ABC in the plane and abc an inscribed triangle. Till here that sounds OK. We suppose we...
Dear François, I am not sure the following solution is more elementary than yours, but let me try. Let O be a point outside the plane ABC. Then the oriented...
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