Dear friends! Is the following property of Gergonne's point well-known? ABC is given triangle. A1 is the point on ray BA, so that AA1=BC, and A2 is the point...
alex
alex_geom@...
Feb 1, 2003 11:23 am
6462
Dear Alex, My first remark is that A1, A2, B1, B2, C1, C2 lie on one circle, called the Conway circle of triangle ABC. The center of this circle is the...
I apologize for a typo in Hyacinthos message #6445 (Grebe triangle): I wrote A' ( 2D + b² + c² : -ab : ca ), but this should be A' ( 2D + b² + c² : -ab :...
Dear Triangle Geometers and Historians, One of the first books on Triangle Geometry I have seen is Emil Donath: Die merkwürdigen Punkte und Linien des ebenen ...
Dear Darij, and all Hyacinthists [DG] ... exciting ... but I hope ... If P* is the isogonal conjugate of P then it is known that the perpendicular from A to...
Dear Hyacinthists, it is known (Brocard theorem) that if ABC is a triangle and P is an arbitrary point then the perpendiculars at P to AP, BP, CP meet the...
Dear Nikolaos Dergiades, Many thanks! Although I have not read both messages, let me make some ... Some time ago, I have found this Brocard theorem in Paul ...
Dear Floor van Lamoen, Paul Yiu and other Hyacinthians, Let me put together some old remarks: 1. (Floor van Lamoen in message #4547) Let A'B'C' be the...
Perhaps it was already discussed at EMHL, but I can't find it in my archives, so I ask it now. Let XYZ be the pedal triangle of a point P with respect to a...
The following paper has been published in Forum Geometricorum; it can be viewed at http://forumgeom.fau.edu/FG2003volume3/FG200303index.html The Editors, Forum...
ForumGeom
ForumGeom@...
Feb 3, 2003 2:13 pm
6472
Dear Hyacinthists, P is any point in the plane, P* = isogonal conjugate of P, P' = inverse of P in the circumcircle, Pb,Pc = feet of the cevian lines of P,...
Dear Jean-Pierre Ehrmann, ... Interesting... I don't have a proof, but may I ask you: How did you get it? Is it related to the Kosnita-point-inverted-in-the- ...
Dear Alex, Let A1 and C1 be on AC, B' be the vertix of the intouch triangle of ABC on AC. Then B' is the midpoint of A1C1. Let triangle, formed by ...
Emelyanov
emelyanov@...
Feb 3, 2003 10:53 pm
6475
Dear friends! A few days ago I've noted the following property of the line passing through incenter I and orthocenter H: ABC is given triangle. A1, B1, C1 is...
alex
alex_geom@...
Feb 4, 2003 2:47 pm
6476
Dear Alex, Your problem, in slightly changed wording, is: [Alexei Myakishev] A few days ago I've noted the following property of the line passing through...
P.S. And, of course, the proof is trivial, because A1,B1,C1 is homotetic to ABC with center I, so the point of perpendicular intersection is ortocenter of ABC,...
alex
alex_geom@...
Feb 4, 2003 5:12 pm
6478
Dear Darij! Sorry for repeating Your massage in my previous, because I've recieved it after sending my. Yours, Alex ... From: <darij_grinberg@...> To:...
alex
alex_geom@...
Feb 4, 2003 5:37 pm
6479
Dear Frans Gremmen, Paul Yiu, Eckard Specht, and other Hyacinthians, I am currently working on a list of formulas in Triangle Geometry (mostly trigonometric...
Dear Darij and other Hyacinthos! ... We can try to make some generalization of that easy property: Let's try to find the points A1,B1, C1 on interior...
alex
alex_geom@...
Feb 4, 2003 7:54 pm
6482
Dear Darij and others Hyacinthos. Little addition to the previous massage: the equation x(p-a)(b-c)+y(p-b)(c-a)+z(p-c)(a-b) is equation of the line, passing...
alex
alex_geom@...
Feb 4, 2003 9:32 pm
6483
... Correct conjecture. It was a simple coordinate calculation. -- Barry Wolk __________________________________________________ Do you Yahoo!? Yahoo! Mail...
Dear Barry Wolk, and other Hyacinthians, ... Thank you. I am trying to redo the proof online: Let P(x:y:z) in trilinears, and P[x1|y1|z1] in actual (exact)...
Dear Darij ... Suppose that P is trlinear x:y:z; up to the factor area(ABC)/2/(ap+bq+cr)^2, we have area(BXP):area(CYP):area(AZP):area(CXP):area(AYP):area(BZP)...
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