# Variation of Russian Olympiad

**Geometry**Level 4

Let \(M\) be the midpoint of the side \(AC\) of \( \triangle ABC\). Let \(P\in AM\) and \(Q\in CM\) be such that \(PQ=\frac{AC}{2}\). Let circle \((ABQ)\) intersect with \(BC\) at \(X\not= B\) and circle \((BCP)\) intersect with \(BA\) at \(Y\not= B\). If \(\angle BYM = 45^{\circ}\), find \(\angle BXM\) in degrees.