National BDMO Secondary P8
 Kazi_Zareer
 Posts: 86
 Joined: Thu Aug 20, 2015 7:11 pm
 Location: Malibagh,Dhaka1217
National BDMO Secondary P8
$\triangle ABC$ is inscribed in circle $\omega$ with $AB = 5$, $BC = 7$, $AC = 3$. The bisector of $\angle A$ meets side $BC$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $DE$. Circles $\omega$ and $\gamma$ meet at $E$ and at a second point $F$. Then $AF^{2} = \frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
We cannot solve our problems with the same thinking we used when we create them.
 Kazi_Zareer
 Posts: 86
 Joined: Thu Aug 20, 2015 7:11 pm
 Location: Malibagh,Dhaka1217
Re: National BDMO Secondary P8
We cannot solve our problems with the same thinking we used when we create them.
Re: National BDMO Secondary P8
Please help to solve:
In $\triangle ABC$, $AD$ is an angle bisector. So, $\frac{BD}{CD}=\frac{AB}{AC} \Rightarrow \frac{BD}{CD}=\frac{5}{3} \Rightarrow \frac{BC}{CD}=\frac{8}{3} \Rightarrow CD=\frac{3 \times 7}{8} \Rightarrow CD=\frac{21}{8}$

 Posts: 8
 Joined: Thu Feb 01, 2018 11:56 am
Re: National BDMO Secondary P8
[After Tasnood]
In the cyclic $BCFE$ quadrileteral,we get $/angleBFC=120$.$/angleDFE=90$ so,$/angleCFD=30$.Now $FD$ bisects $/angleBFC$
then
https://artofproblemsolving.com/communi ... 80p2644107
[it is a well known geometry problem from AIME 2012.So,sad that it is th 8th problem in 2012 national problem.
In the cyclic $BCFE$ quadrileteral,we get $/angleBFC=120$.$/angleDFE=90$ so,$/angleCFD=30$.Now $FD$ bisects $/angleBFC$
then
https://artofproblemsolving.com/communi ... 80p2644107
[it is a well known geometry problem from AIME 2012.So,sad that it is th 8th problem in 2012 national problem.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: National BDMO Secondary P8
Actually 2016 National Olympiad is a showcase of well known problems.Some other problems were also very well known.But,this problem was really different and tough one(for whom who didn't see it before the exam).prottoydas wrote: ↑Tue Mar 13, 2018 9:13 pm[After Tasnood]
In the cyclic $BCFE$ quadrileteral,we get $/angleBFC=120$.$/angleDFE=90$ so,$/angleCFD=30$.Now $FD$ bisects $/angleBFC$
then
https://artofproblemsolving.com/communi ... 80p2644107
[it is a well known geometry problem from AIME 2012.So,sad that it is th 8th problem in 2012 national problem.