Vijetnam 2007

Let $A^{\prime }$ be the image of the point $A$ by the reflection through the point $O$. We prove that $N$, $M$, $A^{\prime }$ are collinear and therefore the line $MN$ passes the fixed point $A^{\prime }$. First, $DE$ is the radical axis of the circle $(O)$ and the circle (${\gamma }_1$) with the diameter $PD$.

Since $\angle PN{A}^{\prime }=90^{\circ }$ the line $N{A}^{\prime }$ is the radical axis of the circle $(O)$ and the circle (${\gamma }_2$) with the diameter $P{A}^{\prime }$.

The line $D{A}^{\prime }$ meets the line $BC$ at the point $F$; since $\angle PF{A}^{\prime }=90^{\circ }$, $\angle AD{A}^{\prime }=90^{\circ }$ $\Rightarrow \angle PF{A}^{\prime }=90^{\circ }$, therefore $BC$ is the axis of the circle (${\gamma }_1$) and (${\gamma }_2$), thus the radical axis $DE$, $BC$ and $N{A}^{\prime }$ are concurrent at the radical center $M$, thus the points $M$, $N$ and $A^{\prime }$ are collinear.