Saudi Arabia Mathematical Competitions

Consider the case when point $P$ is between $C$ and $B$. Quadrilateral $MCPN$ is cyclic and from Ptolemy's relation it follows $$PM\cdot CN=PC\cdot MN+CM\cdot PN\text{\hspace{1em}(1)}$$



We have $CM=CN=R$, $MN=R\sqrt{2}$, and replacing in (1) we obtain $PM=PC\sqrt{2}+PN$, hence $$\frac{PM-PN}{PC}=\sqrt{2}.$$

If $P$ is between $C$ and $A$, then in similar way we get $$\frac{PN-PM}{PC}=\sqrt{2}.$$

Combining these two cases we obtain $$\frac{|PM-PN|}{PC}=\sqrt{2}.$$