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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.0 AIME, harder Prove it Saudi Arabia

On a semicircle of diameter $A B$ and center $C$ consider variable points $M$ and $N$ such that $M C \perp N C$ . The circumcircle of triangle $M N C$ intersects $A B$ for the second time at $P$ . Prove that $\frac{|P M-P N|}{P C}$ is a constant and find its value.

Solution

Consider the case when point $P$ is between $C$ and $B$ . Quadrilateral $M C P N$ is cyclic and from Ptolemy's relation it follows
$P M \cdot C N = P C \cdot M N + C M \cdot P N \tag{1}$

We have $C M = C N = R$ , $M N = R \sqrt{2}$ , and replacing in (1) we obtain $P M = P C \sqrt{2} + P N$ , hence
$\frac{P M - P N}{P C} = \sqrt{2} .$

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

Combining these two cases we obtain
$\frac{|P M - P N|}{P C} = \sqrt{2} .$

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