Team Selection Test for IMO 2007

Draw the diameter $DL$ perpendicular to the secant $AB$. Let $\alpha =\angle LDA=\angle LDB$. Assume that the point $P$ lies on the arc $ALB$ and between the points $A$ and $L$, and let $\beta =\angle LDP$. Also let $K$ be the point of intersection of the line $LD$ and the perpendicular to the line $DM$ through $M$. Let $R$ be the radius of $\Gamma$. We have $PD=2R\cos \beta$. Moreover, since $\angle PAB=\angle PDB=\angle PDL+\angle LDB=\beta +\alpha$, and $\angle PBA=\angle PDA=\angle LDA-\angle LDP=\alpha -\beta$, we have $PB=2R\sin (\alpha +\beta )$ and $AP=2R\sin (\alpha -\beta )$. Then $$\begin{array}{rl}MD& =MP+PD=AP+PB+PD\\ & =2R(\sin (\alpha -\beta )+\sin (\alpha +\beta )+\cos \beta )\\ & =2R\cos \beta (1+2\sin \alpha ),\end{array}$$ and $KD=MD/\cos \beta =2R(1+2\sin \alpha )$. This means that the point $K$ does not depend on the point $P$, and as the point $P$ varies on the arc $ALB$, the point $M$ traces the arc of the circle with diameter $DK$ lying inside $\angle ADB$.



Similarly, if $K^{\prime }$ is the point lying on the line $LD$ and on the same side of the line $AB$ as $D$, and satisfying the condition $K^{\prime }L=2R(1+2\sin (90^{\circ }-\alpha ))$; then as the point $P$ varies on the arc $ADB$, the point $M$ traces the arc of the circle with diameter $K^{\prime }L$ lying inside $\angle ALB$.

Geometric locus is the union of these two arcs.