## APMO 2012/04

Discussion on Asian Pacific Mathematical Olympiad (APMO)
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### APMO 2012/04

Let $ABC$ be an acute triangle. Denote by $D$ the foot of the perpendicular line drawn from the point $A$ to the side $BC$, by $M$ the midpoint of $BC$, and by $H$ the orthocenter of $ABC$. Let $E$ be the point of intersection of the circumcircle $\Gamma$ of the triangle $ABC$ and the half line $MH$, and $F$ be the point of intersection (other than $E$) of the line $ED$ and the circle $\Gamma$ .
Prove that $\frac{BF}{CF} = \frac{AB}{AC}$ must hold.
[Here we denote by $XY$ the length of the line segment $XY$ ].
$\frac{1}{0}$

FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

### Re: APMO 2012/04

It totally uses the idea from ISL 2005 G5. But it's quite a nice one! I loved solving it.

Tahmid Hasan
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Joined: Thu Dec 09, 2010 5:34 pm

### Re: APMO 2012/04

বড় ভালবাসি তোমায়,মা

Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm
We'll have to show, $BF/CF=AB/AC$ which means $BF/CF=sin\angle{BCF}/sin\angle{CBF}=(BK*AC)/(AB*CK)$
We have to show now, $AB^2/AC^2=BK/CK$ or $AF$ is a symmedian.
We get $A,O,P$ collinear
Then $AEDM$ is cyclic
Again, $\angle{AMC}=180-\angle{AMB}=\angle{AEF}=\angle{ABF}$
Also $\angle{AFB}=\angle{ACM}$
Then $\angle{BAF}=\angle{MAC}$
So, $AF$ is a symmedian.