XVI-th Junior Balkan Mathematical Olympiad

Let $P$ be the symmetric of $A$ with respect to $M$ (figure 1). Then $\overline{AM}=\overline{MP}$ and $t\perp AP$, hence the triangle $APN$ is isosceles with $AP$ as its base, so $\angle NAP=\angle NPA$.

We have $\angle BAP=\angle BAM=\angle BMN$ and $\angle BAN=\angle BNM$.

Thus we have $$180^{\circ }-\angle NBM=\angle BNM+\angle BMN=\angle BAN+\angle BAP=\angle NAP=\angle NPA,$$ so the quadrangle $MBNP$ is cyclic (since the points $B$ and $P$ lie on different sides of $MN$). Hence $\angle APB=\angle MPB=\angle MNB$ and the triangles $APB$ and $MNB$ are congruent ($\overline{MN}=2\overline{AM}=\overline{AM}+\overline{MP}=\overline{AP}$). From that we get $\overline{AB}=\overline{MB}$, i.e. the triangle $AMB$ is isosceles, and since $t$ is tangent to $k_1$ and perpendicular to $AM$, the centre of $k_1$ is on $AM$, hence $AMB$ is a right-angled triangle. From the last two statements we infer $\angle AMB=45^{\circ }$, and so $\angle NMB=90^{\circ }-\angle AMB=45^{\circ }$.



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Alternative solution.

Let $C$ be the common point of $MN$, $AB$ (Figure 2). Then ${\overline{CN}}^2=\overline{CB}\cdot \overline{CA}$ and ${\overline{CM}}^2=\overline{CB}\cdot \overline{CA}$. So $\overline{CM}=\overline{CN}$. But $\overline{MN}=2\overline{AM}$, so $\overline{CM}=\overline{CN}=\overline{AM}$, thus the right triangle $ACM$ is isosceles, hence $\angle NMB=\angle CMB=\angle BCM=45^{\circ }$.