Difference between revisions of "User:Quantum-phantom"
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| − | < | + | By the law of cosines, |
| − | + | <cmath>\cos\angle DAC=\frac{221-CD^2}{220}=\cos\angle DBC=\frac{169-CD^2}{120},</cmath> | |
| − | + | so <math>CD=\sqrt{\tfrac{533}{5}}</math>. Similarly, <math>AB=\tfrac{2 \sqrt{5} \sqrt{533}}{13}</math>. Let <math>AD\cap BC=I</math>, <math>AB\cap CD=J</math>, <math>OE\cap IJ=F</math>, then <math>OF\perp IJ</math> by Brocard's theorem. Since <math>ON\perp DC</math>, <math>OM\perp AB</math>, 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, |
| − | \ | + | <cmath>\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}.</cmath> |
| − | \ | + | 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] |
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Latest revision as of 06:11, 15 April 2024
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]
