Dear Hyacinthos members, here is one olympiad problem. ABC is a given acute angled triangle. The point D is the foot of the perpendicular from A to the side...
Dear Martin, consider the direct similarity S (rotation followed by dilation) centered at F that takes B to D (its angle = 90 deg., and ratio k=FD/FB). Since...
Dear friends, I'm sorry, I meant "squared distance" so what I am looking for are two "easy" points P and Q whose squared distance is a^2 + 2 a b + b^2 + 2 a c...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2008volume8/FG200816index.html The editors Forum...
ForumGeom
ForumGeom@...
Jun 2, 2008 4:40 pm
16440
... centered at ... AD/ED=CB/DB, we ... the ... perpendicular ... Very nice and clear solution, much better than the official. Thank you very much Vladimir. ...
Dear Friends, Do the conjugacies or conjugate points in triangle geometry have anything to do with the conjugacy classes or conjugate elements in group theory?...
Dear Martin and Vladimir, There is no need for the triangle to be acute angled. Another proof is : If for vectors CP = DA then since DE/CP = BD/BC the points...
Dear Jean-Louis This is an interesting projective configuration. As you know I am addict with affine geometry, I prove your theorem by sending the vertices B...
Dear Jean-Louis Of course, sending points B and C to infinity is not needed to prove your theorem but I think it is a good exercise to prove it this way. So...
Dear colleagues! Is next fact known? Let O, H, N be the circumcenter, the orthocenter and the Nagel point of triangle ABC. Then the oriented angle NOC=2CHN. ...
Alexey.A.Zaslavsky
zasl@...
Jun 5, 2008 6:02 am
16447
Dear Hyacinthists, the problem: let ABC be a triangle, 0 the circumcircle of ABC, P,Q two points on 0, M the perpendicular bissector of PQ, A' the point of...
Dear Jean-Louis! We discussed this problem some times ago. There is the solution. Let X, Y be the common points of 0 and M. Considering the projection of 0 to...
Alexey.A.Zaslavsky
zasl@...
Jun 5, 2008 10:11 am
16449
Dear Jean-Louis I feel like you send us a similar configuration some months ago. If O is the ABC-circumcenter and D is the symmetric of A wrt O, your meeting ...
Dear Alexey and dear friends As I have an old book of Laisant written in 1893 compiling all problems in triangle geometry set down in this period, I was unable...
Dear Alexey, ... I don't know if your result is known but it can be written as: Let L be the perpendicular bisector of HN that meets the lines HA, HB, HC at...
Dear Francois If I understand your problem 1 correctly, it's a particular case of problem 42 from vol. 1 of Yaglom's "Geometric Transformations": given three...
Dear Vladimir Thank you for your swift reply. I have the Yaglom book , translated into english, but my exercise n°42 is not the same as you tell me. Are you...
In my edition of Yaglom (of 1955) #42 is the last but one problem in Part One (Isometries), Chapter Two (called Symmetries in Russian; in English, possibly,...
Dear Francois! I know only the solution using the complex numbers. If the circumcircle of ABC is the unit circle of complex plane, a, b, c, n are the complex...
Alexey.A.Zaslavsky
zasl@...
Jun 6, 2008 6:49 am
16456
Dear François, Alexey and Hyacinthists, yes, my question is a continuation of the problem I send you some time ago. If we find a solution, a nice result will...
Dear Francois and Vladimir, [FR] ... [VD] ... For vectors OA, OB, OC it is known that if p.OX + q.OY + r.OZ = 0, the points X, Y, Z are collinear and...
Dear Alexey and Wladimir Thank you for your nice remarks. At last I find Yaglom exercice with number 46 in my English edition, so Joe must be happier than Ivan...
Dear Nikos What is funny in your proof is that you follow an affine way, using the affine ratios x, y, z. That just a choo-choo calculation in a very special...
Dear Francois ... I think the simplest way to get the golden ratios is just to write out the Menelaos theorem: it immediately yields the "golden" equation for ...
Dear Wladimir Yes of course, the Menelaus theorem gives the golden ratio. That's the affine way to get it. Using intersections of the Yaglom circles with the...
Dear Vladimir, ... I think you are right. Francois' problem is not exactly the same with Yaglom's. And in my solution the equation of x is now different. Hence...
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