16th Turkish Mathematical Olympiad

Let $O$ be the center and $r$ be the radius of the circle $\Gamma$. If $PQ\cap RS=\{A\}$ and $PS\cap QR=\{B\}$ where the points $P$, $Q$, $R$, $S$ lie on the circle $\Gamma$; then by Miquel's theorem for the triangle $ABQ$, the circumcircles of the triangles $APS$ and $BRS$ intersect at a point $K$ lying on the side $AB$.



Considering the powers of the points $A$ and $B$ with respect to these circles we obtain $A{O}^2-r^2=AS\cdot AR=AK\cdot AB=A{K}^2+AK\cdot KB$ and $B{O}^2-r^2=BS\cdot BP=BK\cdot BA=B{K}^2+BK\cdot KA$. In particular, $A{O}^2-A{K}^2=B{O}^2-B{K}^2$, and $OK$ is perpendicular to $AB$. Then we also have $AK\cdot KB=O{K}^2-r^2$. Hence, for any pair $\{A,B\}$ satisfying the conditions of the problem, the circle with diameter $AB$ passes through two points on the line $OK$ which are at a distance $\sqrt{O{K}^2-r^2}$ away from the line $\ell$. Since $O$, $K$ and $r$ are independent of $A$ and $B$, these two points are constant and form the intersection set.