IMO MOCK5 (iii)

[Nadim Ul Abrar]

Let $O$ and $H$ be the circumcenter and orthocenter of acute $△ABC$. The bisector of $\angle BAC$ meets
the circumcircle $Γ$ of $△ABC$ at $D$. Let $E$ be the mirror image of $D$ with respect to line $BC$. Let $F$
be on $Γ$ such that $DF$ is a diameter. Let lines $AE$ and $FH$ meet at $G$. Let $M$ be the midpoint of
side $BC$. Prove that $GM⊥AF$.

[Nadim Ul Abrar]

Solution

MOCK 5 (iii).JPG (37.21KiB)Viewed 5476 times

[najif]

Solution:
Since AD is the bisector of the angle BAC; D is the midpoint of the arc BDC.
then D,M,O,E,F are collinear. Since H is the orthocentre, the reflection in line BC takes H' to H.
Now suppose l is a line through O that is parallel to BC. then reflection in line l would take DH' to FA.
And from previous facts reflection in BC takes DH' to EH.
Since l and BC r parallel, FA and EH must be parallel. and obviously FA=EH.
then AFEH is a parallelogram implying G is the midpoint of EH.
In triangle AED, G and M are the midpoints of the sides EA and ED. that is GM is parallel to AD.
Since FD is a diameter FA is perpendicular to AD.
As a result, we get GM is perpendicular to FA.

[Tahmid Hasan]

Nadim vai, I don't see your solution, only a figure. Anyways here's mine
$EF=FM-EM=FM-DM=FC.\cos \frac{1}{2}A-BD.\sin \frac{1}{2}A$
$=2R(\cos^2 \frac{1}{2}A-\sin^2 \frac{1}{2}A)=2R \cos A=AH$
Again $AH \parallel EF$[Both perpendicular to $BC$]
So $AHEF$ is a parallelogram $\Rightarrow G$ is the midpoint of $AE$.
In $\triangle ADE;G,M$ are midpoints of $AE,DE$ respectively.
So $GM \parallel AD$. But $AD \bot AF \Rightarrow GM \bot AF$.

[*Mahi*]

Let reflection of $A$ on $O$ be $A'$.
We know $A'$ is also the reflection of $H$ on $M$.
So, $HEA'D$ is a parallelogram, and thus $HE || A'D || AF$.
Again, $EF || AH$, and thus $FAHE$ is a parallelogram.
So, $G$ is the midpoint of $AE$. And as $M$ is the midpoint of $DE$, $GM || AD$.
And as $AD \perp AF, GM \perp AF$

[SANZEED]

I used trigonometry in the following way.
From Sine-law, $a=2R\sin A\Rightarrow a\sin \frac{A}{2}=2R\sin A\sin \frac{A}{2}=2R\cdot 2\sin \frac{A}{2}\cos \frac{A}{2}\sin \frac{A}{2}=2R\cdot 2\sin^{2}\frac{A}{2}\cos \frac{A}{2}$. Thus $a\tan \frac{A}{2}=2R\cdot 2\sin^{2}\frac{A}{2}$. Using the fact that $\sin B=\pm \sqrt(\frac{1-\cos B}{2})$, we have that $1-\cos A=2\sin^{2}A$,and so,
$a\tan \frac{A}{2}=2R(1-\cos A)\Rightarrow 2R\cos A=2R-a\tan \frac{A}{2}$. But we know that ,$AH=2R\cos A,FE=FD-2MD=2R-a\tan \frac{A}{2}$,so $AH=FD$. Now $AH\parallel FD$ since both of them are perpendicular to $BC$. This implies that $AFEH$ is a parallelogram.Thus $AG=GE$. Again $DM=ME$ by reflection. So, $GM\parallel AD$ and $GM$ is perpendicular to $AF$ since $DF$ is a diameter and $AD$ is perpendicular to $AF$.

[Phlembac Adib Hasan]

My solution is same as Najif, so there is no need to post it.

[Nadim Ul Abrar]

My solution is same as Najif, so there is no need to post it.

Same here