Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.3 AIME Prove it JBMO

Problem:

Two circles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$ . A circle $k_{3}$ centered at $A$ meets $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$ , such that $N$ and $Q$ are on different sides of $MP$ and $AB > AM$ .
Prove that the angles $\angle MBQ$ and $\angle NBP$ are equal.

Solution

Solution:

As $AM = AP$ , we have
$\angle MBA = \frac{1}{2} \operatorname{arc} AM = \frac{1}{2} \operatorname{arc} AP = \angle ABP$
and likewise
$\angle QBA = \frac{1}{2} \operatorname{arc} AQ = \frac{1}{2} \operatorname{arc} AN = \angle ABN$
Summing these equalities yields $\angle MBQ = \angle NBP$ as needed.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.