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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Soviet Union

Problem:

$ABC$ is an acute-angled triangle with circumcenter $O$ . The circumcircle of $ABO$ intersects $AC$ and $BC$ at $M$ and $N$ . Show that the circumradii of $ABO$ and $MNC$ are the same.

Solution

Solution:

It is sufficient to show that $\angle MBN = \angle C$ . But $\angle MBN = \angle MBO + \angle OBN = \angle MAO + \angle OBN = \angle MCO + \angle OCN = \angle C$ .

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