Dear Floor! ... I'm sorry, I considered another concurrence. The lines joining the points A', B', C' with the projections of A, B, C to d concur in point Q....
Alexey.A.Zaslavsky
zasl@...
Nov 1, 2004 12:44 pm
10850
Dear Floor and Milorad ... Two lines L, L' have the Droz-Farny property if and only if they touch the same inscribed parabola (in which case the line joining ...
Dear Jean-Pierre and Milorad, ... [JPE] ... Nice! In accordance with H for the perpendicular pairs of lines, for any P the pairs of P-perpendicular through P...
Dear Floor and Milorad ... through P ... these ... P the ... each P ... Of course, we have another characterization : A pair of lines L, L' intersecting at P...
If P<> H, there is an unique point Q on the circle through ABC so that HP is the Steiner-line of Q wrt ABC. Then the Droz-Farny line of P is the perpendicular...
Dear Francois, Floor and Milorad ... Then ... Yes. Consider the points U1, U2 on the line HP such as PU1 = PU2 = PQ; the perpendicular bisector L1 - or L2 - of...
My dear Jean-Pierre You know what? I am happy. It's the first time I get an answer. I was looking at Hyacinthos Forum since a few week without understanding...
Dear Francois, ... what do you exactly mean by "flow" ? ... I suppose you call "isotomic lines" 2 lines meeting each sideline of ABC at two points points...
My dear Bernard I apologyze for "flow". It's not the best word. I want to suggest: flow of a vector field but what I mean is to search the plane curves so that...
Dear all, In fact DF is a projective theorem: Let ABC, I1, I2 and P be inscribed in a conic. Let A1 be the point where Inf1P meets BC, A2 where Inf2P meets BC....
Dear all, [Fvl] ... Yes, there is. Note that the sides of triangles ABC and I1I2P are tangent to a conic K, as the triangles are inscribed in a conic (in the...
My dear Antreas For the problem n°1, the six points are on two lines. But this is an affine therem, there is no need for lines 1,2,3 and 1', 2', 3' to be...
Dear friends, Let L1 and L2 be two perpendicular lines through H. Let M1, M2 and M3 be the midpoints of the intercepts on L1, L2 and BC of the pair of lines AB...
Dear Floor May be I don't understand the terms of your problem or what do you call circumcevian triangle (see Kimberling?) but in drawing the figure, I have...
Dear friends, if ABC is a triangle and A'B'C' is the cevian triangle of a point P then if the Euler lines of the triangles A'AB, A'AC, A'PB are concurrent at Q...
Dear Friend: The MacBeath inconic of a triangle is the wonderful inconic Its foci the circumcenter O and the orthocenter H, giving the center as the...
Dear Ricardo, I don't know about MacBeath, but I know that this conic was described recently in this forum by Paul Yiu: The Droz-Farny lines of perpendicular...
If I understand this correctly, MacBeath was a student at Cambridge who posted a problem regarding this inconic in one of the "journals" of the math clubs at...
A nice way to generate this conic is to connect H to a variable point P on the circumcircle. The perpendicular bisectors of HP envelope the conic. Steve...
... And If I understand correctly your comment, then the journal should be the "Eureka" of the Archimedeans club. A search on the back issues of the journal ...
... If I understand things correctly Cambridge is divided into colleges. The "Eureka" journal was Trinity college. But other colleges had math clubs with...
Dear Hyacinthians (especially Clark Kimberling) In the early 90's I wrote a Visual Basic program to make the basic geometric constructions When the ETC...
Dear Hyacinthians (especially Eric Danneels) Thanks, Eric, and others, for finding errors in ETC. The corrections listed below should be made within a few...
Dear friends, let M be a point. denote by : -- Aa, Ab, Ac the areas of the triangles MBC, MCA, MAB, -- Da, Db, Dc the squares of the distances MA, MB, MC. I...
Dear Bernard as it appears that it was impossible to read my mail, I hope that now it will be better. ... proportional ... inellipse. ... Yes. They are the...
Dear Jean-Pierre ... It's OK now. Thank you. Antreas, can you please delete JP's previous posting. ... of course, it is O and not K. this hyperbola contains a...
Dear friends, I have found(with proof) three easy constructible points on the MacBeath ellipse. If we take perpendicular bisector of line AH,then intersection...
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