Team Selection Test for IMO 2010

Let $A^{\prime }$ be the midpoint of the side $BC$, and let $A^{\prime \prime }$ be the second point of intersection of the line $A{A}^{\prime }$ and the circumcircle $\Gamma$ of the triangle $BDC$. Since both $D,E,A,F$ and $D,C,A^{\prime \prime },B$ are concyclic, $\angle B{A}^{\prime \prime }C=\angle BAC$. Hence $A^{\prime }$ is the midpoint of $A{A}^{\prime \prime }$. In particular, the circle $\Gamma$ and $J$, the second point of intersection of $\Gamma$ and $A{A}^{\prime \prime }$, do not depend on the point $D$. We will show that $J$ lies on ${\Gamma }_D$. If $D$ lies on the same side of $A{A}^{\prime }$ as $B$, then we have $\angle JCB=\angle JDE$ by concyclicity of $B,D,J,C$. Considering the power of $A^{\prime }$ with respect to $\Gamma$ we obtain $A^{\prime }J\cdot A^{\prime }{A}^{\prime \prime }=A^{\prime }B\cdot A^{\prime }C$, hence $A^{\prime }J\cdot A^{\prime }A=A^{\prime }{C}^2$ implying that the line $A^{\prime }C$ is tangent to the circumcircle of the triangle $AJC$. In particular, we have $\angle JCB=\angle JAC$. Therefore $\angle JDE=\angle JAE$, and $J$ lies on ${\Gamma }_D$. A similar reasoning works if $D$ lies on the same side of $A{A}^{\prime }$ as $C$.