BdMO National Secondary 2015#6
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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 $m+n$?
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Re: BdMO National Secondary 2015#6
Hint
Answer
Problem Source
Solution source
Just a duplicate problem!So sad.
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Re: BdMO National Secondary 2015#6
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
So the answer is 161+3=164
- Mehrab4226
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Re: BdMO National Secondary 2015#6
You should learn LaTeX. It is very simple and absolutely not Rocket science.Thephysimatician wrote: ↑Sat Dec 19, 2020 9:33 pmLet 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
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
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Re: BdMO National Secondary 2015#6
Thank you for the suggestion. I will do that.