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Problem: In triangle ABC with incenter I and circumcenter O. Let P be the reflection of A over OI and Q be the second intersection of line OI with the circumcircle of BIC. Prove that lines AI, BC, and PQ concur. 

 

So for this problem I got that from the incenter-excenter lemma, the center of the circumcircle of BIC must be the midpoint of arc BC or triangle ABC, also, I let AI intersect BC at E and it is not to show that P, E and Q and collinear, but I don't really know how to do that, can anyone give me hints or solutions? Thanks. Also I would prefer not advanced topics such as inversion, complex bash, or barcentrics, thanks. 

 
 Jul 5, 2021
 #2
avatar+114012 
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Well, I drew the pic, and it does, on my graph anyway  (I replaced P with A' )

 

https://www.geogebra.org/classic/jwf7kzkh

 

Proving it looks like an undertaking though   indecision  angry

 

 

 

 

 

 
 Jul 11, 2021
 #4
avatar+57 
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Do you have a full solution? I really need help.

 
hellothere  Jul 13, 2021
 #3
avatar+110 
+1

Are you on AoPS? If so I would recommend asking there. I'm OlympusHero and you can get answered fast to this harder stuff there, though there are some really talented people here (melody, cphill, etc... I'm looking at you :P)

 
 Jul 11, 2021

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