User:Quantum-phantom
Revision as of 05:11, 15 April 2024 by Quantum-phantom (talk | contribs)
By the law of cosines,
![\[\cos\angle DAC=\frac{221-CD^2}{220}=\cos\angle DBC=\frac{169-CD^2}{120},\]](http://latex.artofproblemsolving.com/4/3/c/43c0c5000729e6db20b4267810123cb4243361f4.png)
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,
![\[\frac{IA}{ID}=\frac{IA}{IC}\cdot\frac{IC}{ID}=\frac{AB}{CD}\cdot\frac{AC}{BD}=\frac{55}{78},~\frac{DJ}{JA}=\frac{BD}{AC}=\frac{12}{11}.\]](http://latex.artofproblemsolving.com/0/8/3/083897553949df1fb809306c771b2be590e97ead.png)
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]
