IMO MOCK5 (V)

[Nadim Ul Abrar]

In acute $△ABC$, $AB>AC$. Let $M$ be the midpoint of $BC$. The exterior angle bisector of $\angle BAC$
meets ray $BC$ at $P$. Points $K$ and $F$ lie on line $PA$ such that $MF⊥BC$ and $MK⊥PA$. Prove that
$BC^2=4PF.AK$

[Nadim Ul Abrar]

My solution :

MOCK 5 (v).JPG (40.03KiB)Viewed 2762 times

Let the circumcircle of the triangle $ABC$ intersect the line $MF$ at the
point $F'$ (where $F',F$ lie on the same side of $BC$) and $BC \cap AF'=P'$

Now as $M$ is the midpoint of $BC$ and $MF'? BC$ ,
$\angle BF'M=\angle CF'M=\frac{1}{2} \angle BF'C=\frac{A}{2}$

So $\angle MCF'=\angle BAF'=exterior \angle BAP'=90-\frac{A}{2}$ .
That imply $AP'$ bisects exterior $\angle BAC$ or $P'=P$ . $F'=F$ .

So $ACBF$ is cyclic .

$BC$ is tangent to the circumcircle of triangle $FKM$ as $\angle FKM=\angle FMC=90^o$

Using power of point , $PM^2=PK.PF$

$PK.PF=PF(AK+PA)=PF.AK+AP.AF=PF.AK+PB.CP$
$PM^2=(MC+CP)^2=MC^2+2MC.CP+CP^2=MC^2+CP(2MC+CP)$
$=MC^2+PB.CP$

So $4MC^2=4(\frac{BC}{2})^2=4PF.AK$