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Baltic Way competition problems

Geometry Difficulty 5.8 AIME, harder Prove it Baltic Way

The points $M$ and $N$ are chosen on the bisector $AL$ of a triangle $ABC$ such that $\angle ABM = \angle ACN = 23^\circ$ . $X$ is a point inside the triangle such that $BX = CX$ and $\angle BXC = 2\angle BML$ . Find $\angle MXN$ .

Solution

Answer: $\angle MXN = 2\angle ABM = 46^\circ$ .

Let $\angle BAC = 2\alpha$ . The triangles $ABM$ and $ACN$ are similar, therefore $\angle CNL = \angle BML = \alpha + 23^\circ$ . Let $K$ be the midpoint of the arc $BC$ of the circumcircle of the triangle $ABC$ . Then $K$ belongs to the line $AL$ and $\angle KBC = \alpha$ . Both $X$ and $K$ belong to the perpendicular bisector of the segment $BC$ , hence $\angle BXK = \frac{1}{2}\angle BXC = \angle BML$ , so the quadrilateral $BMXK$ is inscribed. Then

$\angle XMN = \angle XBK = \angle XBC + \angle KBC = (90^\circ - \angle BML) + \alpha = 90^\circ - (\angle BML - \alpha) = 67^\circ.$

Analogously we have $\angle CXK = \frac{1}{2}\angle BXC = \angle CNL$ , therefore the quadrilateral $CXNK$ is inscribed also and $\angle XNM = \angle XCK = 67^\circ$ . Thus, the triangle $MXN$ is equilateral and

$\angle MXN = 180^\circ - 2 \cdot 67^\circ = 46^\circ.$

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