42nd Balkan Mathematical Olympiad

Let $D$ be the point such that $ABDC$ is a parallelogram and $O_A$ be the center of ${\omega }_2$. Let $AM$ meet ${\omega }_1$ at $E\ne A$, and $AO$ meet ${\omega }_1$ at $X\ne A$. Furthermore, let $L$ be the reflection of $K$ across $M$, and $P^{\prime }$ be the reflection of $P$ across $M$. $LBKC$ is a parallelogram, so $\angle BLC=\angle BKC=180^{\circ }-\angle BHC=\angle BAC$, which means $L$ lies on ${\omega }_1$. We similarly obtain that $D$ lies on ${\omega }_2$ and $P^{\prime }$ lies on ${\omega }_1$. $ALDK$ is also a parallelogram, so $\angle ALK+\angle LKH=\angle LKD+\angle PKH=\angle PKD+\angle PKH=\angle HKD=\angle HCD=90^{\circ }$, i.e. $HK\perp AL$. Since $HK$ passes through $O$, it is the perpendicular bisector of $AL$. From the power of point $M$ we deduce $MA\cdot ME=MB\cdot MC=MP\cdot MK$, which means $APEK$ is cyclic. It is well-known that $AH{O}_{A}O$ is a parallelogram, so $H{O}_A=OA=OL$ and $O{O}_A=AH=LH$. Therefore, $LHO{O}_A$ is an isosceles trapezoid, so $$L{O}_A\parallel OH\Rightarrow L{O}_A\perp AL\Rightarrow \angle AL{O}_A=90^{\circ }=\angle ALX,$$ which means that $O_A$ lies on $LX$. We also have $\angle O_{A}LE=\angle XLE=\angle XAE=\angle XAD=\angle ED{O}_A$, where the last equality holds because $AX\parallel D{O}_A$. Hence $DE{O}_{A}L$ is cyclic. Next we have $\angle BLP=\angle BL{P}^{\prime }=\angle BC{P}^{\prime }=\angle PBC=\angle PBM$, so $BM$ is tangent to $(BLP)$. From here we obtain $ME\cdot MD=ME\cdot MA=MB\cdot MC=M{B}^2=MP\cdot ML$, so $DEPL$ is also cyclic. This implies that points $L,P,O_A,E,D$ all lie on a circle, and so $\angle O_{A}PE=\angle O_{A}LE=\angle XLE=\angle XAE=\angle OAE$. Now let the tangent line to ${\omega }_1$ at $A$ and the tangent line to ${\omega }_2$ meet at $T$. Then we have $$\angle TAE=\angle TAO+\angle OAE=90^{\circ }+\angle O_{A}PE=\angle O_{A}PT+\angle O_{A}PE=\angle EPT,$$ and so $TAPE$ is cyclic. Therefore, points $A,P,E,K,T$ are concyclic, meaning that $T$ lies on $(APK)$, as needed.