69th Belarusian Mathematical Olympiad

Let $N$ be the midpoint of $AM$ and $D$ be the reflection of $A$ in $M$. We will prove that the quadrilateral $BPND$ is cyclic by showing the equality $$AP\cdot AB=AN\cdot AD.$$ Since $AM$ is tangent to the circumcircle of the triangle $BPM$, $$AP\cdot AB=A{M}^2=2AN\cdot 0.5AD=AN\cdot AD.$$ Hence $AP\cdot AB=AN\cdot AD$, the quadrilateral $BPND$ is cyclic and therefore $\angle APN=\angle BDA$. The lines $BD$ and $AC$ are parallel (since $ACDB$ is the parallelogram), whence $\angle BDA=\angle CAD$. This leads us to the equality $\angle APN=\angle CAN$, which means that the circumcircle of the triangle $APN$ is tangent to the line $AC$.

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Alternative solution.

Since the circumcircle of the triangle $BPM$ is tangent to the line $AM$, the triangles $AMP$ and $ABM$ are similar and, in particular, $\angle AMP=\angle ABC$. From the similarity follows the equalities $\frac{PM}{MN}=\frac{2PM}{AM}=\frac{2BM}{BA}=\frac{CB}{BA}$, hence the triangles $PMN$ are similar to $CBA$ and $\angle PNM=\angle CAB$. Wherein $\angle PNM=\angle PAN+\angle APN$ as an external angle of the triangle $APN$, and $\angle CAB=\angle PAN+\angle CAN$, whence as in the first solution $\angle APN=\angle CAN$.