Dear Chystyan and Charles Worall and all, We have triangle ABC. We draw the bisectors BB', CC' which intersect themselves in point I. We also draw the bisector...
Dear friends, Below is a different approach to the quadrilateral problem (a.k.a. Christopher Bradley conjecture). In cicrcumscribed quadrilateral ABCD...
Dear Floor [FvL] ... [JPE] ... [FvL] ... so that ApBpCp must be a perspector, i.e. Darboux. ... parallel (perpendicular to PP*). This is a particular case of a...
Consider a triangle ABC and a point P. Let phi be an arbitrary angle, actually a directed angle modulo 180 degrees. Now, let A', B', C' be some points on the...
Dear Darij, ... your locus is always a cubic but the general equation is a bit ugly : with cot(phi) = T, I find : Darboux + k1 T (a nK(K,K) ) + k2 T^2 Thomson...
Dear Darij ... Your locus is the circumcubic with asymptots the three lines La, Lb, Lc going respectively through the midpoints of BC, CA, AB and such as (La,...
I was just thinking about writing a similar vein. I was thinking that our age of geometry might very well end up being one of the golden ones in triangle...
Dear Darij [DG] ... [JPE] ... Lb, ... Note that, when P moves on L(phi), the perspector of ABC and A'B'C' moves on a cubic member of the pencil generated by...
Let ABC be a triangle, P a point, PaPbPc the pedal triangle of P, and A'BC, B'CA, C'AB three triangles erected on the triangle sides (out/inwardly ABC). Ab =...
Dear Bernard, ... Thank you very much for the comments. I have been trying to calculate the equation, but I didn't succeed. Your result is also very nice since...
Dear Steve, ... Thanks for the feedback and for all the kind words. Yes, I could expect that our opinions would be different - all of us are, more or less,...
Dear Rafi, ... where x = AP, y = BP, c = CP, d = DP. (I mention this just for completeness.) ... This should be 2S1 = xy * sin(2A), where S1 is the area of...
Dear Nikolaos, ... Thanks for this brilliant generalization of the Napoleon theorem and of the contest problem! ... [...] This proof assumes that triangle ABC...
I think that there were discussions of the loci below but for particular values of the parameter t. (Sorry I don't have time to search the archive, which...
Dear Jean-Pierre ... It follows from your nice remark that the triangle formed by the asymptotes is similar at O to ABC and its circumcircle is centered at O ...
Dear Antreas ... In this case, your locus is the member of the Euler pencil going through the 6 vertices of the Kiepert triangles with basis angles w and -w...
It's just a thought but.... Is it possible to construct any triangle if ANY three linear elements are given ? Standard conditions apply to construction. We all...
Dear Antreas [APH] ... sides ... [JPE] ... w ... pivot ... More precisely, if U is the pivot of the cubic locus of P and V is the pivot of the cubic locus of...
Here's a puzzle for you - I haven't solved it. Can you constuct a triangle given its isodynamic and Fermat points (Kimberling's 13, 14, 15, and 16)? You can...
Lawrence Evans
75342.3052@...
Mar 3, 2004 9:22 am
9452
Dear Antreas, I am sending a copy to the McCay list, because: ... The locus for La, Lb, Lc to be concurrent is the union of sides, Linf (triply), circumcircle...
Dear Floor and Antreas ... resp. ... Linf ... sides, Linf ... t. ... bound a ... May be I've missed something, but the lines AQ, BQ, CQ are perpendicular to...
Dear Jean-Pierre and Antreas, ... [APH] ... [FvL] ... [JPE] ... I see. I have taken A*, B* and C* to be such that PA* / PA" = PB* / PB" = PC* / PC" = t in...
Dear Jean-Pierre and Antreas, ... But of course it doesn't make to much sense to introduce parameter t here. We could have restricted to A"B"C" for (1) and...
Dear Peter, ... No. For instance, the problem to construct a triangle ABC with compass and ruler if you have given the sidelength a, the altitude from C, and...
Dear Nikolaos, ... And I think this can be still generalized further: < C'AB = < C'BA = w; < A'BC = < B'AC = u; < A'CB = < B'CA = 90 - u - w. Then, triangle...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2004volume4/FG200404index.html The Editors, Forum...
A fascinating problem just posed on http://www.mathlinks.ro/viewtopic.php?p=13838#13838 (Notations changed.) If O is the circumcenter of a triangle ABC, and D,...
... Well, for any P, if D', E', F' are the reflections of P in the lines BC, CA, AB, then the lines D'A', E'B', F'C' concur. This follows from Jean-Pierre's...
Dear Darij ... I think that there was a discussion on the general problem: Let P, Q=P* be two isogonal conjugate points, and PaPbPc, QaQbQc their pedal...
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