Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
1

Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.

Thus we have

$$
180^{\circ}-\angle N B M=\angle B N M+\angle B M N=\angle B A N+\angle B A P=\angle N A P=\angle N P A
$$

so the quadrangle $M B N P$ is cyclic (since the points $B$ and $P$ lie on different sides of $M N$ ). Hence $\angle A P B=\angle M P B=\angle M N B$ and the triangles $A P B$ and $M N B$ are congruent ( $M N=2 A M=A M+M P=A P$ ). From that we get $A B=M B$, i.e. the triangle $A M B$ is isosceles, and since $t$ is tangent to $k_{1}$ and perpendicular to $A M$, the centre of $k_{1}$ is on $A M$, hence $A M B$ is a right-angled triangle. From the last two statements we infer $\angle A M B=45^{\circ}$, and so $\angle N M B=90^{\circ}-\angle A M B=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_0ffcf6c39e8d83d98455g-2.jpg?height=423&width=809&top_left_y=1162&top_left_x=632)

Figure 1

## Solution 2

Let $C$ be the common point of $M N, A B$ (Figure 2). Then $C N^{2}=C B \cdot C A$ and $C M^{2}=C B \cdot C A$. So $C M=C N$. But $M N=2 A M$, so $C M=C N=A M$, thus the right triangle $A C M$ is isosceles, hence $\angle N M B=\angle C M B=\angle B C M=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_0ffcf6c39e8d83d98455g-2.jpg?height=308&width=902&top_left_y=1962&top_left_x=583)

Figure 2