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 $\angle NMB = 90^\circ - \angle AMB = 45^\circ$.

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 $\angle NMB = \angle CMB = \angle BCM = 45^\circ$.