BdMO National Secondary 2015#6

[samiul_samin]

Trapezoid $ABCD$ has sides $AB=92,BC=50,CD=19,AD=70$ $AB$ is parallel to $CD$ A circle with center $P$ on $AB$ is drawn tangent to $BC$ and $AD$.Given that $AP=\dfrac mn$ (Where $m,n$ are relatively prime).What is $m+n$?

[samiul_samin]

Hint

Extend $AD$ and $BC$.

Answer

$\fbox {164}$

Problem Source

$AIME$ $1992$ $P9$

Solution source

Here.

Just a duplicate problem!So sad.

[Thephysimatician]

Let X be the intersection of AD and BC. Then XA/XB=XD/XC =(XA-XD)/(XB-XC)=AD/BC=70/50=7/5. Then P lies on the bisector of <AXB. Therefore P is the intersection of side AB and the bisector of its opposite angle. So AP/PB=XA/XB=7/5. So AP=7AB/(7+5)=7×92/12=161/3.

So the answer is 161+3=164

[Mehrab4226]

Let X be the intersection of AD and BC. Then XA/XB=XD/XC =(XA-XD)/(XB-XC)=AD/BC=70/50=7/5. Then P lies on the bisector of <AXB. Therefore P is the intersection of side AB and the bisector of its opposite angle. So AP/PB=XA/XB=7/5. So AP=7AB/(7+5)=7×92/12=161/3.

So the answer is 161+3=164

You should learn LaTeX. It is very simple and absolutely not Rocket science.

[Thephysimatician]

Thank you for the suggestion. I will do that.