(J.-L.Ayme, France) Points $ P$, $ Q$ lie on the circumcircle $ \omega$ of triangle $ ABC$. The perpendicular bisector $ l$ to $ PQ$ intersects $ BC$, $ CA$, $ AB$ in points $ A'$, $ B'$, $ C'$. Let $ A"$, $ B"$, $ C"$ be the second common points of $ l$ with the circles $ A'PQ$, $ B'PQ$, $ C'PQ$. Prove that $ AA"$, $ BB"$, $ CC"$ concur.
Problem
Source: Sharygin contest. The final raund. 2008. Grade 9. Second day. Problem 8
Tags: geometry, circumcircle
darij grinberg
11.09.2008 19:21
Why is it necessary for $ P$ and $ Q$ to lie on $ \omega$ ? dg
jayme
12.09.2008 07:10
Dear Darij and Mathlinkers, I found this situation by solving another problem. Perhaps you could see any developpment with your remark. Sincerely Jean-Louis
vulalach
15.09.2010 07:01
I've found a solution for this problem by using Inversion. But i think there is another solution.
jayme
15.09.2010 08:38
Dear Mathlinkers, this problem is in relation with the Seimiya's line. Sincerely Jean-Louis
hatchguy
15.09.2010 12:58
jayme wrote: Dear Mathlinkers, this problem is in relation with the Seimiya's line. Sincerely Jean-Louis Can you give the solution Jayme? The problem is very interesting by itself and the point where the lines concurr makes it even more.
jayme
15.09.2010 18:55
Dear Mathlinkers, for a synthetic proof of my result you can see http://perso.orange.fr/jl.ayme vol. 3 La droite de Seimiya p. 5. Sincerely Jean-Louis
armpist
16.09.2010 02:40
Dear MLs In the proposed form this problem is a very particular case of Blaikie theorem ......................................... Why is it necessary for P and Q to lie on \omega ? dg ...................... Again, Darij the Great is precise in his , so appealing to me , vagueness. M.T.
hatchguy
16.09.2010 05:30
jayme wrote: Dear Mathlinkers, for a synthetic proof of my result you can see http://perso.orange.fr/jl.ayme vol. 3 La droite de Seimiya p. 5. Sincerely Jean-Louis I will wait until someone else posts a solution since I don;t understand french.
jayme
16.09.2010 13:50
Dear Armpist (M.) and Mathlinkers, I am always ready for new and nove approach of a problem. Dear Armpist, how do you imagine the proof by using the Blaikie theorem and perhaps can you recall this result? Thanks in advance. Sincerely Jean-Louis
armpist
16.09.2010 16:23
Dear J-L This is the link to the problem proposed by DG (Darij the Great) some time ago: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=38178 M.T.
jayme
16.09.2010 16:42
Dear Armpist (M.) and Mathlinkers, thank you for the links I have lost. But how do you use this resultat for the problem in question? Sincerely Jean-Louis
jred
11.12.2017 17:02
We may crack this problem by Desargues's involution theorem, although it was quite possible that the projective solution would not be recognized in the contest.