In message #1006 by Edward Brisse I read: "Construct the common pole P to incenter and circumscribed circle to ABC" I don't know what this is and how to...
Dear Giovanni I suppose Edward means by pole of a pair of circles, a circle of null radius in the bundle of circles they define. So two non secant circles have...
... A nightmare in y^24 :-) The trouble is that for a given center of concurrence, three lines through it intersect the sides of ABC in 9 points, thus giving...
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the Hyacinthos group. File :...
Hyacinthos@yahoogroup...
Feb 2, 2009 11:50 am
17190
I recap: Draw three diameters UX,VY,WZ of a conic K with midpoint M such that their endpoints lie on the sides AB,BC,CA of a triangle. (See uploaded file.) ...
See file. By default the blue quadrangle UVWX has one green "complete triangle" ABC (at least that's how they are called in German?!), and counting free...
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the Hyacinthos group. File : /radius.bmp ...
Hyacinthos@yahoogroup...
Feb 3, 2009 7:31 pm
17193
... Uploaded file shows this is not correct. What am I doing wrong? Sincerely, Jeff...
... Nothing. First, note the "an". There are *lots* of solutions, the one I gave is the *simplest* (and doesn't have to appear with the simplest configuration...
See uploaded file 2UMKR. Let the c.q. consist of the straight lines ABE,ADF,BCF,DCE. Assume now that ABCD and BEFD lie on circles with centers P and Q,...
Hauke, here is how to construct a cyclic quadrilateral ABCD from given EFG under the assumption that EBFD is cyclic as well: You are right to assume that for...
Dear Hyacinthists, a first article on the mixtilinear incircles has been put on my website http://perso.orange.fr/jl.ayme   vol. 4 Sincerely Jean-Louis ...
Dear Hyacinthians, I was playing with an obvious generalization of the Taylor circle: instead of a triangle ABC and its orthocenter H, consider ABC and any ...
Dear Eisso Looking at P as a parameter, you can find the subset U of the plane such that if P is in U, there exists at least an (euclidian) metric structure ...
It's obvious how to do the analogous constructions of, say, the barycenter, circumcenter, incenter and orthocenter for a spherical triangle. But do the lines...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2009volume9/FG200902index.html The editors Forum...
ForumGeom
ForumGeom@...
Feb 9, 2009 4:16 pm
17203
Jeff, I never looked at Hauke's drawing. As it turns out, it seems I labeled the three diagonal points differently from Hauke. His G is my E and vice versa...
Hauke, The general rule is that all theorems in plane geometry are true in spherical geometry provided that the "parallel line postulate" is not evoked and...
Dear Eisso, Sorry, I got focused on your comment "for a cyclic quadrilateral the angle bisectors for the angles at the diagonal points E, F, and G have to be...
Let ABC be a triangle P = (x:y:z) a point and L a line through P. The Reflections of AP, BP, CP in L intersect BC, CA, AB at A',B',C', resp. The points...
Dear Hyacinthists, let ABC be a triangle, I the incenter of ABC, UVW the cevian triangle of I wrt ABC, O the center of the circumcircle of ABC, X the meetpoint...
Dear Jean-Louis You say that X is the meeting point of the parallel to AI with BC but this line, parallel to AI, is on which point? (Cette droite parallèle à...
Dear François, sorry... "X is the intersection of BC with parallel to AI through O". Francisco Javier Garcia Capitan has verified my conjecture and the point...
Let X be the intersection of AB and the Neuberg cubic of ABC. Let Y be the intersection of BC and the Neuberg cubic of ABC. Let Z be the intersection of CA and...
For the moment, I see that L(P) and L(P*) are tangent to the in-conic with foci P and P* but that's all I can do! Friendly Francois [Non-text portions of this...
Dear Antreas and François, ... The intersection Q of L(P) and L(P*) is a 15th degree point with respect to the coordinates p:q:r of P ! Very ugly ! When P =...
Dear Jean-Louis Just the point I thought! It's a long way to Manapany! As for me, I can only look at the river, just one mile south from the Tower. Friendly ...
... point of concurs is not in ETC. ... solved. ... Dear Jean-Louis, It's always thrilling when three lines (constructed in the same way) coincide. I checked...
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