Competition problem — Geometry — Maths Olympiad Prep

Library / /51 of 133

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

Therefore, we have
$\angle N Q C = \frac{1}{2}(\overparen{NC} + \overparen{DB}) = \frac{1}{2}(\overparen{NC} + \overparen{CD}) = \frac{1}{2} \overparen{ND} = \angle N M P.$
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
$AP \cdot AQ = AM \cdot AN = AB \cdot AC.$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.