Baltic Way 2023 · Problem 15

Since $M P N$ is an isosceles triangle, we have $\angle P M A=\angle P M N=\angle M N P=\angle B N P$. By tangent and chord theorem, $\angle M C A=$ $\angle P M A=\angle B N P=\angle B D N$.

Since $\angle M C P=\angle M N P$, the quadrilateral $C M P N$ is cyclic. Analogously, from $\angle P D N=\angle P M N$, we get that $N D M P$ is cyclic. Since $C$ and $D$ both lie on the circumcircle of $N P M$, points $P, N, M, C$ and $D$ are concyclic.

From inscribed angles subtending arcs with the same length, we get that $\angle M D P=$ $\angle M C P=\angle M N P=\angle P D N=\angle P M N=\angle P C N$.

The power of $P$ with respect to $\omega_{1}$ gives us that $P M^{2}=P A \cdot P C$. The power of $P$ with respect to $\omega_{2}$ gives us that $P N^{2}=P B \cdot P D$. Since $P M=P N$, the powers of $P$ with respect to $\omega_{1}$ and $\omega_{2}$ are equal ( $P$ lies on the radical axis). Hence, $P A \cdot P C=P B \cdot P D$, which implies that $A B D C$ is cyclic. From inscribed angles subtending the $\operatorname{arc} A B$, we get that $\angle A C B=\angle A D B$.

Hence, $\angle B C N=\angle A C N-\angle A C B=\angle M D B-\angle A D B=\angle M D A$.

2nd Solution: Since $M P N$ is an isosceles triangle, we have $\angle P M A=\angle P M N=$ $\angle M N P=\angle B N P$. By tangent and chord theorem, $\angle M C A=\angle P M A=\angle B N P=$ $\angle B D N$.

Since $\angle M C P=\angle M N P$, the quadrilateral $M P N C$ is cyclic, which means that $P$ lies on the circumcircle of $M N C$. Since $M P N$ is isosceles, the perpendicular bisector of $M N$ passes through $P$. Since the intersection point of the angle bisector and the perpendicular bisector of the opposite side of the triangle lies on the circumcircle, it follows that $C P$ bisects angle $\angle M C N$. Hence, $\angle M C P=\angle P C N$. Analogously, since $\angle P D N=\angle P M N$, it follows that $N D M P$ is cyclic and the circumcircle of $M N D$, the perpendicular bisector of $M N$ and the angle bisector of $\angle M D N$ meet at $P$. Hence, $\angle M D P=\angle P D N=\angle M C P=\angle P C N$.

Now we continue as in the previous solution.