Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it

16. G1 (ROM) ${ }^{\mathrm{IMO} 1}$ Let $A B C$ be an acute-angled triangle with $A B \neq A C$ . The circle with diameter $B C$ intersects the sides $A B$ and $A C$ at $M$ and $N$ , respectively. Denote by $O$ the midpoint of $B C$ . The bisectors of the angles $B A C$ and $M O N$ intersect at $R$ . Prove that the circumcircles of the triangles $B M R$ and $C N R$ have a common point lying on the line segment $B C$ .

Solution

16. Note that $\triangle A N M \sim \triangle A B C$ and consequently $A M \neq A N$ . Since $O M = O N$ , it follows that $O R$ is a perpendicular bisector of $M N$ . Thus, $R$ is the common point of the median of $M N$ and the bisector of $\angle M A N$ . Then it follows from a well-known fact that $R$ lies on the circumcircle of $\triangle A M N$ . Let $K$ be the intersection of $A R$ and $B C$ . We then have $\angle M R A = \angle M N A = \angle A B K$ and $\angle N R A = \angle N M A = \angle A C K$ , from which we conclude that $R M B K$ and $R N C K$ are cyclic. Thus $K$ is the desired intersection of the circumcircles of $\triangle B M R$ and $\triangle C N R$ and it indeed lies on $B C$ .

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