## 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
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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

[N.B.:I used $E'$ instead of $F$]
$\frac {BF}{CF}=\frac {sin BCF}{sin CBF}=\frac {sin BEF}{sin CEF}$
now $\frac {BD}{CD}=\frac {BE.sin BEF}{CE.sin CEF}$
so $\frac {BF}{CF}=\frac {BD}{CD}.\frac {CE}{BE}=\frac {c.cos B}{b.cos C}.\frac {CE}{BE}$
so it suffices to prove that $\frac {CE}{BE}=\frac {cos C}{cos B}$
let $E'$ be the second point of intersection of $MH$ and $\Gamma$.
Lemma:Let $H.O,M$ be the orthocenter,circumcenter and the midpoint of $BC$ of $\triangle ABC$,then $HM,AO$ intersect on the circumcircle.
Proof:Let $E"$ be the intersection point.It is well-known that $AH \parallel OM$ and $AH=2OM$
so a homothety with center $E"$ and ratio $2$ sends $\triangle E"OM$ to $E"AH$.
Hence $E"H=2E"O$or,$AO=E"O$ meaning $E"$ is the diametrically opposite point WRT $A$.
from this lemma we get the second point of intersection of $MH$ and $\Gamma$ is $E'=E"$ indeed.
so,$\angle ABE'=\angle ACE'=90^{\circ}$.....(1)
now $\frac {BM}{CM}=\frac {BE.sin BEE'}{CE.sin CEE'}=\frac {BE.sin BCE'}{CE.sin CBE'}$
since $BM=CM$ we get $\frac {CE}{BE}=\frac {sin BCE'}{sin CBE'}=\frac {sin BAE'}{sin CAE'}$
$=\frac {cos AE'B}{cos AE'C}$[using (1)]$=\frac {cos C}{cos B}$
বড় ভালবাসি তোমায়,মা

Ananya Promi
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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.