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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.4 AIME, harder Prove it Saudi Arabia

Let $A$ , $B$ , $C$ be colinear points in this order, $\mathcal{C}$ an arbitrary circle passing through $B$ and $C$ , and $l$ an arbitrary line different from $BC$ , passing through $A$ and intersecting $\mathcal{C}$ at $M$ and $N$ . The bisectors of the angles $\angle C M B$ and $\angle C N B$ intersect $BC$ at $P$ and $Q$ . Prove that $AP \cdot AQ = AB \cdot AC$ .

Solution

Let $D$ be the midpoint of the arc $\text{BC}$ of the circle $\mathcal{C}$ opposite to both $M$ , $N$ . Because $\text{CD = DB}$ , the bisectors of angles $\angle C M B$ and $\angle C N B$ , both, intersect the arc $\widehat{CB}$ at $D$ .

Therefore, we have
$\text{N Q C = 1 2 ( NC + DB ) = 1 2 ( NC + CD ) = 1 2 ND = N M P.}$
This proves that quadrilateral $M N Q P$ is cyclic. But quadrilateral $M N C B$ is also cyclic and lines $M N$ , $BC$ intersect at $A$ , hence by the power of the point $A$ with respect to these two circles we have
$AP \cdot AQ = AM \cdot AN = AB \cdot AC.$

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