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Junior Balkan Mathematical Olympiad

Geometry Difficulty 4.3 AIME Prove it JBMO

Problem:

Two circles k1k_{1} and k2k_{2} intersect at points AA and BB. A circle k3k_{3} centered at AA meets k1k_{1} at MM and PP and k2k_{2} at NN and QQ, such that NN and QQ are on different sides of MPMP and AB>AMAB > AM.
Prove that the angles ∠MBQ\angle MBQ and ∠NBP\angle NBP are equal.

Solution

Solution:

As AM=APAM = AP, we have
∠MBA=12arc⁡AM=12arc⁡AP=∠ABP \angle MBA = \frac{1}{2} \operatorname{arc} AM = \frac{1}{2} \operatorname{arc} AP = \angle ABP
and likewise
∠QBA=12arc⁡AQ=12arc⁡AN=∠ABN \angle QBA = \frac{1}{2} \operatorname{arc} AQ = \frac{1}{2} \operatorname{arc} AN = \angle ABN
Summing these equalities yields ∠MBQ=∠NBP\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.