Dear Francois, ... Yes you are right if the areas are signed. The circumcevian triangle A1B1C1 of P is similar to the pedal(P) and the circumcevian triangle...
Dear Nikos Yes, I forgot this question of sign but may be you did the same and your locus must be: pqrxyz(x+y+z)^3 = (+)(-)(pyz+qzx+rxy)^3 So it has 2...
Dear Francois, ... Yes. You are right. The locus must be: pqrxyz(x+y+z)^3 = (+)(-)(pyz+qzx+rxy)^3 What do you mean to write explicitly the factorization when p...
... I just want to know if the other component is also reducible or not? I have no Maple nor Mathematica on my computer and I am lazy! Friendly Francois ...
Dear Nikos and Francois, Thank you very much for nice results! I would like to inform you some my small efforts: With any point X we denote CT(X) as triangle...
Dear friends! May be somebody has any information about analogies of Fermat-Torricheli points (or Appolonius points) in the tetrahedron? Best regards, Alexei...
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alex_geom@...
Feb 2, 2007 3:02 pm
14815
Dear friends As Tuan enjoys to solve area problems, I give you this riddle I found when I was thinking about l [Non-text portions of this message have been...
Sorry, I send my post too soon! So I said that when I was thinking about last Tuan problem on area, I create this riddle that I hope you will find interesting ...
Dear Alexei, ... How much analogies would they be if any? The first Fermat point exist always by the construction of equilateral triangles outside ABC. If ABCD...
Dear Nicos! Of course, but what about equil pedal tetrahedron? and angles 120? Yours, Alexei ... From: "Nikolaos Dergiades" <ndergiades@...> To:...
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alex_geom@...
Feb 2, 2007 9:47 pm
14820
Dear Francois, Thank you very much for your hints. I think they are complete proof already and not only normal proof. It is special proof: proof by picture. By...
[ND] ... {HR] ... abc is the orthic triangle and H1,H2,H3 are the orthocenters of Abc, Bca, Cab. K = X(389) is the center of homothety of the triangles abc,...
Dear All My Friends, Given triangle ABC and any point X moving on sideline BC. C1 = orthogonal projection of X on AB (projection f1) B1 = orthogonal...
Anh Tuan oi It's typically a chowchow problem that is to say composition of some affine maps. For example you must prove that the map A1 --> A2 is a...
Dear Hyacintheans,I have found this result Given equilateral triangle ABC, P is an arbitrary point on its incircle (C), P* is isogonal conjugate of P then...
I can give you some hint: P* is the contact point of the Simson line of the point Q image of P by the homothety of center O and ratio -4/3 wrt the triangle...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2007volume7/FG200705index.html The editors Forum...
ForumGeom
ForumGeom@...
Feb 5, 2007 11:01 pm
14832
Dear All My Friends, Given triangle ABC with PaPbPc is pedal triangle of P and QaQbQc is pedal triangle of Q. Please prove that: Area(APaQa)*a^2 +...
Dear Quang Tuan Bui ... construct point X? Your point is the isotomic conjugate of the point of the circumcircle whose Simson line is parallel to PQ Friendly....
Dear Hauke and Nikos, I would like to add some spices in your delicious altitude salad. Let three altitudes of reference triangle ABC are: AHa, BHb, CHc. We...
Dear Hyacinthists, I liked this way of constructing the golden ratio. Segment MN in Figure 1 is quadratic mean = Q = \sqrt{(a^2+b^2)/2}. Consider A = (a+b)/2...
Dear Jean-Pierre, Thank you very much for interesting construction. By my calculations, if barycentrics of P = (p : q : r) Q = (u : v : w) then the first...
Dear Geometers, I found these books. The modern geometry of the triangle. By William Gallatly. http://www.amazon.com/modern-geometry-triangle-William- ...
Dear Quang Tuan Bui, In your problem, if X is orthogonal projection of A on BC then the perpendicular bisectors of A1A2, B1B2, C1C2 are concurrent in a point...
Dear friends, a lot of books can be found at The University of Michigan Historical Mathematics http://www.hti.umich.edu/u/umhistmath/ Also is interesting this...
Dear Francois and Petrisor, Thank you very much for your valuable advices and interesting remarks. All my results bellow are for orthogonal projection case: -...
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