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Bulgaria competition problems

Geometry Difficulty 6.0 National Olympiad Prove it Bulgaria

Isosceles $\triangle ABC$ ( $AC = BC$ ) is inscribed in a circle $k$ . A point $M$ lies on the side $BC$ . A point $N$ from the ray $AM$ ( $M$ lies between $A$ and $N$ ) is such that $AN = AC$ . The circumcircle of $\triangle MCN$ intersects $k$ at $C$ and $P$ , where $P$ is from the arc $BC$ , not containing $A$ . The lines $AB$ and $CP$ intersect at $Q$ . Prove that $\angle QMB = \angle QMN$ .

Solution

Denote $\angle MAC = \alpha$ and $\angle ACB = \beta$ . Let $J$ be the incenter of $\triangle AMC$ . It follows that $\angle AJM = 90^\circ + \beta/2$ and $\angle ABM = 90^\circ - \beta/2$ , which implies that the quadrilateral $ABMJ$ is inscribed in a circle $k_1$ . Analogously we have that the quadrilateral $MNCJ$ is inscribed in a circle $k_2$ which in fact is the the circumcircle of $\triangle MCN$ . Therefore $P$ is a point of $k_2$ . Since $QA \cdot QB = QC \cdot QP$ , we conclude that $Q$ has one and the same degree with respect to $k_1$ and $k_2$ . Hence $Q$ lies on $JM$ , which is the angular bisector of $\angle BNC$ . This implies the desired equality $\angle QMB = \angle QMN$ .

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