XIX Silk Road Mathematical Competition

Suppose that $AC\le BC$. Let $D$ be a point on the side $AB$ such that $DM\parallel BC$. Then $$\frac{DB}{DA}=\frac{CM}{AM}=\frac{BL}{CL},$$ i. e. $DL\parallel AC$. On the tangent line to $\omega$ at point $C$ let's choose a point $T$, such that $KT\parallel BC$. Then $\angle TKA=\angle CBA=\angle TCA$. Therefore, $AKCT$ is cyclic. Let segments $KT$ and $AC$ intersect at point $S_1$. Then $$\frac{S_{1}P}{S_{1}C}=\frac{KP}{KL}=\frac{KA}{KD}=\frac{S_{1}A}{S_{1}M}⟹S_{1}P\cdot S_{1}M=S_{1}C\cdot S_{1}A=S_{1}K\cdot S_{1}T.$$ Thus, $TMKP$ is inscribed into the circumcircle of triangle $KMP$. It is known that the common chords of three pairs of circles, centers of which are not collinear, are concurrent. It means that the common chord of $\omega$ and the circumcircle of $KMP$ passes through point $S_1$. Therefore, point $S_1$ coincides with $S$, and $S_{1}K\parallel BC$ follows from the definition of $T$.

Solution 2:

Let's introduce some notation: $$b=AC,m=AM,s=AS,p=AP.$$ Since $S$ lies on the radical axis of the circumcircles of $ABC$ and $PKM$, then $$SM\cdot SP=SA\cdot SC⟹(m-s)(s+p)=s(b-s)⟹s=\frac{pm}{b+p-m}.$$ So, $$\frac{AS}{SC}=\frac{s}{b-s}=\frac{pm}{(b-m)(b+p)}.$$ According to Menelaus' theorem for triangle $ABC$ and transversal line $PKL$: $$\frac{AK}{KB}\cdot \frac{BL}{LC}\cdot \frac{CP}{PA}=1⟹\frac{AK}{KB}=\frac{CL}{LB}\cdot \frac{AP}{PC}=\frac{AM}{MC}\cdot \frac{AP}{PC}=\frac{m}{b-m}\cdot \frac{p}{b+p}=\frac{AS}{SC}⟹SK\parallel BC,$$ Q. E. D.