Dear Antreas, Very nice idea! I try only some special case and we should do together to get some general results. I am trying with first case: Q is fixed, and...
Quang Tuan Bui
bqtuan1962@...
Jul 1, 2006 8:10 am
13486
Dear Antreas, I have got a very nice first result: - Q is fixed and is incenter I - Line is OH The locus (or at least all points on this line) is line...
Quang Tuan Bui
bqtuan1962@...
Jul 1, 2006 8:33 am
13487
Dear Tuan [BQT] ... That is, Let ABC be a triangle, and P a point. The perpendicular to IA through P intersects AB,AC at Ab,Ac, resp. Similarly Bc,Ba, and...
Dear Antreas and All My Friends, Here are some results for the case: Q=incenter I and fixed, line is OH, P is any point on the line connected Gergonne point...
Quang Tuan Bui
bqtuan1962@...
Jul 1, 2006 1:09 pm
13490
On 1-07-06, Quang Tuan Bui <bqtuan1962@...> wrote: Dear Tuan ... Hmmm... but what is the locus of Pe as P moves on the Mit-Gerg line? APH...
Dear Antreas, The locus of Pe as P moves on the Mit-Gerg line looks as one line, but I am not sure. I am now trying construction it. If I can construction it,...
Quang Tuan Bui
bqtuan1962@...
Jul 1, 2006 7:12 pm
13492
Dear Antreas, I can confirm: it is one line! Best regards, Bui Quang Tuan Quang Tuan Bui <bqtuan1962@...> wrote: Dear Antreas, The locus of Pe as P moves...
Quang Tuan Bui
bqtuan1962@...
Jul 1, 2006 7:58 pm
13493
Dear Tuan A variation of our locus is for Q = P That is Let ABC be a triangle, and P a point. The perpendicular to PA at P intersects AB,AC at Ab,Ac, resp. ...
Dear Friends I thank Ricardo for all these nice summer problems. As I am going back to Britanny next Thursday far away from Hyacinthos, I give you too my...
Sunday, July 02, 2006 1:29 PM [GMT+1=CET], ... In the interior, on it, and in the exterior of de Steiner inner-ellipse. On the sides of ABC, the locus is a par...
Bravo, Ignacio: That's right! The inner Steiner ellipse is the separator! In, {Gamma} is an ellipse. On, {Gamma} is a parabola. Out,{Gamma} is an hyperbola. ...
... Dear Tuan Very Good! Can somebody find out the equation of the line, or any known point(s) on it? Antreas PS Reminding what the line is Let ABC be a...
Let ABC be a triangle, A'B'C' a fixed equilateral triangle, and P a point. Denote La,Lb,Lc three lines passing through P and parallels to the sidelines of...
Dear All My Friends, There are three points on circumcircle each bound with reference triangle one bicentric quadrangle. Please construct these points! Best...
Quang Tuan Bui
bqtuan1962@...
Jul 3, 2006 6:21 am
13504
... triangle one bicentric quadrangle. Please construct these points! ... Dear Bui Quang Tuan (I think that I made a mistake with your name in a previous...
Dear Tuan I don't look at these solutions : this is a 1962 IMO problem. You can see it at http://www.kalva.demon.co.uk/imo/imo62.html It solution is at ...
Dear Francois, Please continue play with this. It is interesting. Dear Francisco Javier, Thank you for your reference. I have visited and see my solution is...
Quang Tuan Bui
bqtuan1962@...
Jul 3, 2006 7:46 am
13508
Dear Bui Quang Tran ... triangle one bicentric quadrangle. Please construct these points! When M moves on the arc BC of the circumcircle opposite to A, the ...
Dear Tuan and all my friends. There is a property of cyclic quadrilaterals which, I think, may be help us. If a quadrilateral is bicentric, then the line...
Anh Tuan I draw again the figure with Cabri. It's very interesting. I found that the triangle of the centers of the 3 incircles is in perpective with ABC. But...
Dear Francois, Thank you very much. If it is true so I am very happy. I will check it. Best regards, Bui Quang Tuan Francois Rideau <francois.rideau@...>...
Quang Tuan Bui
bqtuan1962@...
Jul 3, 2006 11:36 am
13513
Dear Jean-Pierre and All My Friends, Thank you for your simple construction. My construction more complicated as following: Let's I is incenter, Ia is touch...
Quang Tuan Bui
bqtuan1962@...
Jul 3, 2006 11:38 am
13515
Dear Jean-Pierre. I have just learned, from your very simple construction, another interesting property of cyclic quadrilaterals. Thank you and best regards. ...
Dear Tuan, Jean-Pierre and all. Based on Jean-Pierre's construction we have also other three points P'1, P'2, P'3, on arcs BC, AC, AB respectively, as the 4th ...
Let ABC be a triangle and HaHbHc its orthic triangle. Let A'B'C' be the triangle formed by the parallels to sidelines of HaHbHc through Ha,Hb,Hc (ie A' =...
Dear Tuan, ... The perpendicular bisector of BC meets the circumcircle of ABC at the points A' , A". If AB=AC then the required point D is A'. If AC > AB then...
