4. In acute triangle $A B C, M$ and $N$ are the midpoints of sides $A B$ and $B C$, respectively. The tangents to the circumcircle of triangle $B M N$ at $M$ and $N$ meet at $P$. Suppose that $A P$ is parallel to $B C$, $A P=9$ and $P N=15$. Find $A C$.
Answer: $20 \sqrt{2}$
Solution: Extend rays $P M$ and $C B$ to meet at $Q$. Since $A P \| Q C$ and $M$ is the midpoint of $A B$, triangles $A M P$ and $B M Q$ are congruent. This gives $Q M=M P=P N=15$ and $Q B=A P=9$. Observe that the circumcircle of triangle $B M N$ is tangent to $Q M$, so by power of a point, we compute $Q M^{2}=Q B \cdot Q N$, which yields $Q N=25$. Applying Stewart's theorem on triangle $Q N P$, we have
$$
15 \cdot 15^{2}+15 \cdot 25^{2}=30 \cdot 15 \cdot 15+30 M N^{2} \text {, }
$$
which implies that $M N^{2}=\frac{1}{2}\left(25^{2}-15^{2}\right)=200$. Hence, as $M N$ is the midline of triangle $A B C$, we obtain $A C^{2}=4 M N^{2}=800$ and $A C=20 \sqrt{2}$.