Dear Antreas ... In both cases, your assertions are true and the locus of the common point of the three NP-circles is the line joining H to X(143) = the ...
While working on Feuerbach theory, I have found the following conjecture: Let the excircles of triangle ABC touch the corresponding sides BC, CA, AB at Aa, Bb,...
Let ABC be a triangle, A'B'C' its orthic triangle, P a point and Pa, Pb, Pc the orthogonal projections of P on AA',BB',CC', resp. Denote Nab = The Nine Point...
Dear Darij ... Of course, it is true. Note that your point is barycentric x=(s-a)/(a.SA)^2 or equivalently x=cot(A/2) sec^2(A) The point is the common point of...
Dear Antreas ... The common points of the NP-circles of PaAB and PaAc are A' and the midpoint of APa. Hence, the line NabNac is the perpendicular bisector of...
Dear Hyacinthists, apart the NP-circle, does there exist other circles having an nonempty intersection with the incircle and the three excircles? Jean-Pierre...
Dear Jean-Pierre, ... Thanks for the verification! ... Could you please explain this to me? I fear this is not correct in the sense I understand it. Happy...
Dear friends, Let A1,B1,C1 be the points of intersections of internal bisectors and sides of triangle ABC and K1(So,ro) be the circumcircle of triangle A1B1C1....
Dear Darij [DG] ... [JPE] ... [DG] ... If Ia = A-excenter, the circumcircle of the extouch triangle AaBbCc meets again the A-excircle at the reflection X of...
Dear Hyacinthos, Inscriptable quadrilateral is divided into 4 non-overlapping triangles by it's diagonals. Prove that the 4 incenters are on a circle. A simple...
Dear Rafi ... This is not true - draw a figure, for instance - May be, you were meaning a circumscribed quadrilateral (the four sides are tangent to a circle)?...
Dear Milorad Stevanovic, ... Your center So is X(1158) in Kimberling's ETC, and your barycentrics should be equivalent to the trilinears given in the ETC. The...
Dear Jean-Pierre, ... [...] ... Ah, thanks! Obviously, my coordinates were incorrect; now, we obtain the trilinear coordinates ( cot(A/2) sec^2 A csc A : ... )...
Dear Rafi, ... ^^^^^^^^^^^^ Meaning: A quadrilateral with an incircle. [This notion can have different meanings. You call "inscriptable" what Jean-Pierre...
Dear Darij and Milorad ... I think that there is a confusion; Milorad has computed the circumcenter of the cevian triangle of the incenter (this point is not...
Dear Milorad and Jean-Pierre, ... Yes, actually. Thanks for the correction. I was thinking about my triangle AaBbCc from message #8901. Sincerely, Darij...
... Hi Jean-Pierre, When I say 'inscriptable' I mean the circle is inside of the quadrilateral (sides are tangents). 'Inscribed' means the quadrilateral is...
Dear Rafi ... I'm sorry; I didn't know that inscriptable quadrilateral = circumscribed quadrilateral = tangential quadrilateral. I thought that inscriptable...
... I have never written to Hyacinthos, but I have been fascinated reading it since this summer. However, I happen to have written a paper on circumscribable...
Let ABV be a triangle and A'B'C' its orthic triangle. Denote Aa = (perpendicular to AB at B) /\ (perpendicular to AC at C) Ab = AA' /\ (perpendicular to AB at...
Dear Jean-Pierre, ... = ( csc^2(A/2) sec^2 A : ... ) ... = ( sec^2(A/2) cos^2 A : ... ) ... = ( cos^2(A/2) sec^2 A : ... ). These are even simpler! No wonder...
... I have tried for a while to prove that the incenters of the four triangles are on a circle, but I can't seem to get it. But I have found some interesting...
... triangles ... Dear Charles, The fact that the 2 cirles from your theorem are tangent follows immediately from the property of inscriptable quadrilateral: ...
... Dear Rafi, I still need to finish unpacking your last message, but in the meantime I will provide this proof, since I like it so much. Again, the theorem...
Dear Charles Worrall, I have read both of your mails with much interest and was struck by the results, especially by the one locating the center of the circle....
Dear Rafi, ... Nice to hear that this has a name... ... This is very interesting: you state that the external homothetic center of the incircles of triangles...
... Darij, This is intriguing. I'd be very interested to know if there is a general principle here about whether an angle can possibly be 1/3 of another or ...
Dear Darij, ... The theorem of 3 circles states that their homothety centers ( external ones, and the circles have to be outside of each other) are collinear....
