User:Quantum-phantom

By the law of cosines, so . Similarly, . Let , , , then by Brocard's theorem. Since , , then \begin{align*} \frac{MP}{NP}&=\frac{MO\sin\angle MOE}{NO\sin\angle NOE}=\frac{\sin\angle MOE}{\sin\angle NOE}=\frac{\sin\angle IJA}{\sin(\pi-\angle DJI)}=\frac{\sin\angle IJA}{\sin\angle DJI}\\ &=\frac{JA\cdot JI\sin\angle IJA}{IJ\cdot DJ\sin\angle DJI}\cdot\frac{DJ}{JA}=\frac{[IJA]}{[DJI]}\cdot\frac{DJ}{JA}=\frac{IA}{ID}\cdot\frac{DJ}{JA}. \end{align*} By the law of sines, So the answer is \(\tfrac{MP}{NP}=\frac{12}{11}\cdot\frac{55}{78}=\frac{10}{13}\). [img]https://img.picgo.net/2024/04/07/IMG_3135a148c4fcc55ace27.jpeg[/img]