Prove that the line joining the -circumcenter of a triangle to a vertex- is perpendicular to the line -joining the foot of altitudes of adjoining sides.- Some...
... I don't see exactly what you mean here.Can you be more explicit? ... As for the sides of the medial triangle of the diagonal triangle of ABCD are ... . ......
Note: Only two weeks left for early registrations! Who: The Synergetics Collaborative (SNEC) in collaboration with the Philomorphs and The Edna Lawrence...
dear colleague There is a very simple response the orthic triangle is homothetic to the tangential one The line joining the -circumcenter of a triangle to a ...
Dear Franηois, I hoped the drawing that I uploaded would clear that up. Anyway, what I mean is that if the line through the midpoints of EFG intersects AD in...
... altitudes of ... Let E, F be the feet of the altitudes through B, C upon AC, AB respectively, Let X be the midpoint of AH (H orthocenter) X is the center...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2007volume7/FG200721index.html The editors Forum...
ForumGeom
ForumGeom@...
Oct 1, 2007 6:49 pm
15582
Dear Alexey Here how I checked your property with Cabri (and GSP). I change your notations a little bit, calling G the ABC-centroid, O the ABC-circumcenter and...
Dear Alexey With Cabri, I found 2 other properties of the map f leading, I think, to a solution of your problem. I label I and J' the limit points of the...
Dear Alexey and Francois you wrote that if A'B'C' is the triangle with sides the medians of ABC that <OGK = <O'G'K' and that there exists a proof through some...
Dear Nikos You are our Savior! And now what I am going to do with my circular map, the best is still to go to bed! Friendly Francois ... [Non-text portions of...
Dear Alexey The existence of the direct circular map f defined in my previous post gives indeed a synthetic solution of yout theorem: 1° In the sequel I label...
Dear Francois! Thank you for your proof. I have another proof. Let the sidelines of the triangle A'B'C' are perpendicular to the medians of ABC and pass...
Alexey.A.Zaslavsky
zasl@...
Oct 2, 2007 12:12 pm
15589
Dear Alexey, dear Nikos and dear friends. Thank you for your very elementary proof. I will look at it and compare it with mine. In your proof, triangles ABC...
Dear Geometers, The GREAT book Advanced Euclidean Geometry – Roger A. Jonhson is available again. www.amazon.com. Dover publications ISBN: 0486462374. ...
Dear Alexey and Nikos and dear friends Here a short summary (without proof) of the theory of the metapoles: We start with 2 triangles ABC and A'B'C' and to...
Dear Francois, You wrote ... I used MathCad and complicated distance formulas. Another more simple method I see now is the following general. Let L: px+qy+rz =...
Dear friends By curiosity, I look at the configuration of a pair of metapolar four points (A, B, C, D) and (A', B', C', D') , (checking Alexey equality each...
Dear Nikos Thank you for your explanation. Of course, we can calculate the distance of any pair of points knowing their barycentrics but I am always scared by...
Dear Francois, ... Yes. Let T the triangle ABC, G the centroid, M the mid point of BC and G' the reflection of G in M. GG' = (2/3)ma, BG = (2/3)mb, BG' = CG =...
Dear friends, in the ETC X(175) is the center of the outer Soddy-circle and is called the isoperimetric point. This isoperimetric property is only valid if the...
Dear Eric, You are right about all your conclusions. (I did not test your integer value option). If x, x' are the radii of the inner and outer Soddy circles...
Dear friends, sorry for the error in my previous message If angle B = angle C then I found that AB=AC=5.BC I limited my analysis to triangles in which the...
Dear Nikos yes and this is why the triangle I have labeled A'B'C' , parallelogic with ABC with centers of parallelogy their centroids G and G' has its sides ...
In message #6452 (30/01/2003), I wrote "Related with that, the limit max(A,B,C) < 2 arcsin(4/5) is only valid if the triangle is isosceles. If the triangle is...
Dear Francois, ... We can notice that such a triangle T' is similar to the pedal triangle (or the circumcevian triangle) of the symedian point K. If A'B'C' is...
Dear Jean-Pierre As you notice, this is the opposite orientation equality which do the job and Alexey only ask for an equality with "ordinary" angles. That's...
Dear Jean-Pierre Thanks for your help. I had already made the drawing in this case. Note that the metapoles are the respective centroids G and G' of ABC and...
I correct a little typo ... Of course A'B'C' is the pedal triangle of K. So the centroid of A'B'C' is G' = K. ... [Non-text portions of this message have been...
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