## National BDMO Secondary P8

Kazi_Zareer
Posts: 86
Joined: Thu Aug 20, 2015 7:11 pm
Location: Malibagh,Dhaka-1217

### 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,Dhaka-1217

### Re: National BDMO Secondary P8

I solved by using angle bisector theorem, stewart's theorem, power of the point, law of cosines etc. I think I just messed up but got an answer $AF^2 = 900/19$. So, $m + n = 900 + 19 = 919$

Anyone?
We cannot solve our problems with the same thinking we used when we create them.