Saudi Arabian IMO Booklet

Note that the length of the segment of the common tangent to $o_B$ and $o_C$ joining the tangency points is equal to $$AB-AC=LB-LC=KC-KB,$$ SAUDI ARABIAN IMO Booklet 2022

which means that points $A$, $L$ lie on the one, and point $K$ on the other leg of some hyperbola $\eta$ with foci $B$, $C$. Denote by $S$ the midpoint of the segment $BC$ (the center of symmetry of $\eta$) and denote by $A^{\prime }$, $K^{\prime }$, $M^{\prime }$ the central reflections in $S$ of $A$, $K$, $M$, respectively. Then $A^{\prime }$, $K^{\prime }\in \eta$ and $A^{\prime }$, $K^{\prime }$, $C$ are collinear. From the optical property of a hyperbola follows that $A^{\prime }{M}^{\prime }$ is tangent to $\eta$ (as it is the bisector of $B{A}^{\prime }C$). Therefore Pascal's theorem applied to the degenerate hexagon $A{A}^{\prime }{A}^{\prime }{K}^{\prime }KL$ inscribed in $\eta$ gives the collinearity of points $$A{A}^{\prime }\cap K{K}^{\prime }=S,A^{\prime }{A}^{\prime }\cap KL,A^{\prime }{K}^{\prime }\cap LA=C.$$ Therefore the three lines $A^{\prime }{A}^{\prime }$ (tangent to $\eta$ in $A^{\prime }$), $KL$, $BC$ are concurrent, which means that $M^{\prime }=A^{\prime }{A}^{\prime }\cap BC=KL\cap BC=P$ and in consequence $CM=BP$. $\square$