User:Quantum-phantom

By the law of cosines, $\cos\angle DAC=\frac{221-CD^2}{220}=\cos\angle DBC=\frac{169-CD^2}{120},$ so $CD=\sqrt{\tfrac{533}{5}}$ . Similarly, $AB=\tfrac{2 \sqrt{5} \sqrt{533}}{13}$ . Let $AD\cap BC=I$ , $AB\cap CD=J$ , $OE\cap IJ=F$ , then $OF\perp IJ$ by Brocard's theorem. Since $ON\perp DC$ , $OM\perp AB$ , then $\begin{aligned} \frac{M P}{N P} & = \frac{M O \sin ∠ M O E}{N O \sin ∠ N O E} = \frac{\sin ∠ M O E}{\sin ∠ N O E} = \frac{\sin ∠ I J A}{\sin (π - ∠ D J I)} = \frac{\sin ∠ I J A}{\sin ∠ D J I} \\ = \frac{J A \cdot J I \sin ∠ I J A}{I J \cdot D J \sin ∠ D J I} \cdot \frac{D J}{J A} = \frac{[I J A]}{[D J I]} \cdot \frac{D J}{J A} = \frac{I A}{I D} \cdot \frac{D J}{J A} . \end{aligned}$ By the law of sines, $\frac{IA}{ID}=\frac{IA}{IC}\cdot\frac{IC}{ID}=\frac{AB}{CD}\cdot\frac{AC}{BD}=\frac{55}{78},~\frac{DJ}{JA}=\frac{BD}{AC}=\frac{12}{11}.$ So the answer is $\frac{M P}{N P} = \frac{12}{11} \cdot \frac{55}{78} = \frac{10}{13}$ . [img]https://img.picgo.net/2024/04/07/IMG_3135a148c4fcc55ace27.jpeg[/img]

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User:Quantum-phantom