# BDMO 2016 National Secondary Problem 8.

$\triangle{ABC}$ is inscribed in circle $\omega$ with $AB = 5, BC = 7\; \text{and}\; AC = 3$ the bisector or $\angle A$ meets $BC$ at $D$ and circle $\omega$ at another point $E$. Let $\gamma$ be the circle with diameter $DE$. Circle $\omega$ and $\gamma$ meet at point $E$ and another point $F$. Given $AF^2 = \frac{m}{n}$. Where $m,n$ are co-prime positive integers. Find $m+n$.

I dont even able to draw the picture. Here is my work -

I can't understand how the 2 circles will meet at a second point. Even if I understand that I am not able to solve this problem. Any hint will be helpful. This problem was worth of 30 points.
Source: BDMO 2016 National Secondary Problem 8

• Circle γ is NOT tangent to ω.
– Mick
Jan 17, 2017 at 17:09
• $E$ is the midpoint of the $BC$-arc in the circumcircle of $ABC$. From that, you may compute every length. There probably is a slick solution, but this kind of problems is easy to tackle through geometric brute-force. Jan 17, 2017 at 18:11

We may notice that $E$ is the midpoint of the $BC$-arc in $\omega$.
Additionally, $\gamma$ goes through $M$, the midpoint of $BC$. If you prove that the second intersection between $AF$ and $\gamma$, depicted above as $G$, is just the symmetric of $M$ with respect to $AE$ (hint: $MGFE$ is a cyclic quadrilateral, hence $AF$ and $ME$ are antiparallel), you may exploit the fact that $$AG\cdot AF = AD\cdot AE$$ together with the fact that both $AD\cdot AE$ and $AG$ are simple to compute from the side lengths of $ABC$, since $AD\cdot DE=BD\cdot DC$ and $AG=AM$. If I did not mess up with the involved computations, Stewart's theorem (for computing the squared length of medians and bisectors) leads to $AF^2=\color{red}{\frac{900}{19}}$.

Interesting facts:

1. $AF$ and the tangents to $\omega$ at $B$ and $C$ concur, since $AF$ is a symmedian in $ABC$;

2. If $N$ is the antipode of $E$ in $\omega$, $D\in FN$. Additionally, $NA$ and $EF$ intersect at $T\in BC$
and $MAF$ is the orthic triangle of $TNE$.

• Before this edit I was thinking how $AD \cdot AE = AM$ :D Thanks :) Jan 17, 2017 at 19:27
• @RezwanArefin: sorry for the confusion. I checked my computations again and they should be fine. By Stewart's theorem (giving the squared length of medians and bisectors), the above argument leads to: $$AF^2 = \left(\frac{3\cdot 7}{8}\cdot\frac{5\cdot 7}{8}+\frac{3\cdot 5}{8^2}(8^2-7^2)\right)^2/\frac{2\cdot 3^2+2\cdot 5^2-7^2}{4} = \frac{900}{19}.$$ Jan 17, 2017 at 19:34
• By 'Both $AD \cdot AE$ and $AG = AM$'.. Did you mean $AD \cdot AE$ = AM ? Jan 17, 2017 at 19:36
• @RezwanArefin: I mean that both $AD\cdot AE$ and $AG$ are simple to compute, since $AD\cdot DE=BD\cdot DC$ and $AG=AM$. Jan 17, 2017 at 19:37
• Got it now :D Thanks :) Jan 17, 2017 at 19:38