Serbian Mathematical Olympiad

Solution:

Denote by $k_{1}$ and $k_{2}$ the circles $MNP$ and $BOC$, respectively. The circle $k_{1}$ is the Euler circle of $\triangle ABC$ and passes through the feet of the altitudes $D$, $E$ from $B$, $C$ and the midpoint $O_{1}$ of the segment $AH$, where $H$ is the orthocenter of $\triangle ABC$.

Let us show that the second intersection point $Z$ of the line $AY$ and the circle $k_{1}$ lies on the Euler circle $k_{3}$ of triangle $ADE$. We shall consider the case where

$Z$ is between $A$ and $Y$; the proof in the other case is analogous. Let $D_{1}$ and $E_{1}$ be, respectively, the midpoints of the segments $AD$ and $AE$. Since $AY \cdot AZ = AD \cdot AN = AD_{1} \cdot AC$, the points $Y$, $Z$, $C$, $D_{1}$ are concyclic, so $\text{AZD 1 = ACY}$.

Analogously $\text{AZE 1 = ABY}$, so $\text{D 1 ZE 1 = AZD 1 + AZE 1 = ACY + ABY = BYC - BAC = BAC}$. From this it follows that $Z$ lies on $k_{3}$.

Since $O_{1}$ is the circumcenter of $\triangle ADE$, the similarity transformation that maps $\triangle ABC$ to $\triangle ADE$ also maps $k_{1}$ to $k_{2}$ and $k_{2}$ to $k_{3}$, so the image of the point $X \in k_{1} \cap k_{2}$ is the point $Z \in k_{2} \cap k_{3}$. Therefore, $\text{BAX = DAZ = CAY}$.