IMO-2011-G4

[Tahmid Hasan]

Let $ABC$ be an acute triangle with circumcircle $\Omega$.Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$.Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$.Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X \neq A$.Prove that the points $D,G$ and $X$ are collinear.

[Tahmid Hasan]

Let $\odot AB_0C_0=\alpha,\odot XB_0C_0=\omega,M=AD \cap B_0C_0$.
Let $AD' \bot B_0C_0,D' \in B_0C_0$,$O$ be the centre of $\Omega$.
Let $K$ be the intersection of the tangents to $\Omega$ from $A,X$.
$AK$ is the radical axis of $\Omega,\alpha$;$KX$ is the radical axis of $\Omega,\omega$ and $B_0,C_0$ is the radical axis of $\alpha,\omega$.
So $K \in B_0C_0$.
$\angle OAK+\angle OXK=90^{\circ}+90^{\circ}=180^{\circ} \rightarrow AKXO$ are cyclic.
A homothety of ratio $-2$ and centre $G$ sends $O$ to $H$,the orthocentre of $ABC$ and $\triangle A_0B_0C_0$ to $\triangle ABC$.
So $O$ is the orthocentre of $\triangle A_0B_0C_0$.
$\angle OD'K=\angle OAK=90^{\circ} \rightarrow AKXOD'$ is cyclic.
$AB_0C_0 \cong A_0B_0C_0 \rightarrow AM=A_0D'$
$AM \bot B_0C_0,A_0D' \bot B_0C_0 \rightarrow AM \parallel A_0D'$
which implies $AMA_0D'$ is a parallelogram.
Note that a homothety of ratio $-2$ and centre $G$ sends $D'$ to $D$.
Hence $D,G,D'$ are collinear.
So it suffices to prove that $X,D,D'$ are collinear.
$\angle KD'X=\angle KAX=\angle KXA=\angle KD'A=\angle AD'M$
$=\angle A_0MD'=\angle MA_0D=\angle MD'D$[The last arguement comes from the fact that $MDA_0D'$ is a rectangle.]
which implies $X,D,D'$ are collinear.