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
$$\begin{aligned} MD &= MP + PD = AP + PB + PD \\ &= 2R(\sin(\alpha - \beta) + \sin(\alpha + \beta) + \cos \beta) \\ &= 2R \cos \beta (1 + 2 \sin \alpha), \end{aligned}$$
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 $K'$ is the point lying on the line $LD$ and on the same side of the line $AB$ as $D$, and satisfying the condition $K'L = 2R(1 + 2\sin(90^\circ - \alpha))$; then as the point $P$ varies on the arc $ADB$, the point $M$ traces the arc of the circle with diameter $K'L$ lying inside $\angle ALB$.

Geometric locus is the union of these two arcs.