This problem is the same level as last year's P2 or a bit harder, I feel.
No hand diagram because I didn't use any diagram~ (I head solved it)
Problem: Let ABC be a triangle with AC>AB , and denote its circumcircle by \Omega and incentre by I. Let its incircle meet sides BC,CA,AB at D,E,F respectively. Let X and Y be two points on minor arcs \widehat{DF} and \widehat{DE} of the incircle, respectively, such that \angle BXD = \angle DYC. Let line XY meet line BC at K. Let T be the point on \Omega such that KT is tangent to \Omega and T is on the same side of line BC as A. Prove that lines TD and AI meet on \Omega.
We begin with the following claim!
Points B,X,Y,C-are concyclic
Because CD-is tangent to the incircle, we get that \angle CYD=\angle BXD and \angle CDY=\angle DXY. So \angle BXD+\angle DXY+YCB=180 \implies \angle BXY+\angle YCB=180.
Also note that K-B-C is radical axis of the incircle and BCXY.
By power of point, we get that KT^2=KX\cdot KY=KB\cdot KC=KD^2\implies KD=KT.So \angle BTD = \angle KTD - \angle KTB = \angle KDT - \angle TCD = \angle DTC where \angle KTB=\angle TCD by tangency.So TD is angle bisector of \angle BTC. So done!
I did see that T is the sharkydevil point but I didn't find it useful. This property was pretty nice though! I can see someone trying to harmonic or use angle chase in it! Really nice problem.Guessing BCXY cyclic was the first thing that came into my mind. Then radicals axis and angles started coming out. Pretty generic problem to be honest. Like there is nothing new in this problem.
Overall, this was a nice problem! Not sure if it was EGMOish level but it is cute!
~Sunaina 14/04/24
THATS MY ROOMMATE IM SO PROUD OF HER
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