Junior Balkan MO

Figure (Asymptote source)

size(15cm,0); draw((0,0)--(0,2)--(4,2)--(4,-3)--(0,0)); draw((-1,2)--(9,2)); draw((0,0)--(2,2)); draw((2,2)--(1,1)); draw((0,0)--(4,2)); draw((0,2)--(1,1)); draw(circle((0,1),1)); draw(circle((4,-3),5)); dot((0,0)); dot((0,2)); dot((2,2)); dot((4,2)); dot((4,-3)); dot((1,1)); dot((0,1)); label("A",(0,0),NW); label("B",(1,1),SE); label("M",(0,2),N); label("N",(4,2),N); label("$O_1$",(0,1),NW); label("$O_2$",(4,-3),NE); label("$k_1$",(-0.7,1.63),NW); label("$k_2$",(7.6,0.46),NE); label("$t$",(7.5,2),N); label("P",(2,2),N); draw(rightanglemark((0,0),(0,2),(2,2))); draw(rightanglemark((0,2),(1,1),(2,2))); draw(rightanglemark((0,2),(4,2),(4,0)));

Let $O_1$ and $O_2$ be the centers of circles $k_1$ and $k_2$ respectively. Also let $P$ be the intersection of $\overrightarrow{AB}$ and line $t$ .
Note that $\overline{O_1M}$ is perpendicular to $\overline{MN}$ since $M$ is a tangent of $k_1$ . In order for $\overline{AM}$ to be perpendicular to $\overline{MN}$ , $A$ must be the point diametrically opposite $M$ . Note that $\angle MBA$ is a right angle since it inscribes a diameter. By AA similarity, $\triangle ABM\sim\triangle AMP$ . This gives that $\angle BMA \cong \angle MPA$ .
By Power of a Point on point $P$ with respect to circle $k_1$ , we have that $PM^2=PB\cdot PA$ . Using Power of a Point on point $P$ with respect to circle $k_2$ gives that $PN^2=PB\cdot PA$ . Therefore $PM^2=PN^2$ and $PM=PN$ . Since $MN=2AM$ , $MA=MP$ . We now see that $\triangle APM$ is a $45-45-90$ triangle. Since it is similar to $\triangle MPA$ , $\angle PMB \cong \boxed {\angle NMB \cong 45^{\circ} \cong \frac{\pi}{4}}$ .
Solution by Someonenumber011 :)
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