EGMO 2024 P2

 

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