[ Tangent Circles ]  [ Homothety helps solve the problem $\quad$]

In triangle $A B C$, it is known that $\angle A=120^{\circ}$, sides $A C=1$ and $B C=\sqrt{7}$. On the extension of side $C A$, a point $M$ is taken such that $B M$ is the altitude of triangle $A B C$. Find the radius of the circle passing through points $A$ and $M$ and tangent at point $M$ to the circle passing through points $M, B$ and $C$.
Prove that the circles touch each other internally. Let $K$ be the point of intersection of the first circle with the segment $B M$, distinct from $M$. Triangles $M A K$ and $M C B$ are similar.

## Solution

Since the diameter $B C$ of the circle circumscribed around triangle $M B C$ is seen from point $A$ at an obtuse angle, point $A$ lies inside this circle. Therefore, the given circles touch each other internally.

Let $Q$ and $O$ be the centers of the smaller and larger circles, respectively, and $K$ be the point of intersection of the smaller circle with the segment $B$, distinct from $M$. Since

$$
\angle M A K = \angle A M Q = \angle C M O = \angle O C M = \angle B C M
$$

triangles $M A K$ and $M C B$ are similar.

Let $A B = x$ and apply the cosine rule to triangle $A B C$:

$$
B C^{2} = A B^{2} + A C^{2} - 2 A B \cdot A C \cos 120^{\circ},
$$

or

$$
7 = 1 + x + x^{2} \text{, or } x^{2} + x - 6 = 0.
$$

From this, we find that $x = 2$, and since $\angle M A B = 60^{\circ}$, then

$$
A M = \frac{1}{2} A B = \frac{x}{2} = 1
$$

Therefore,

$$
A K = B C \cdot \frac{A M}{M C} = \frac{\sqrt{7}}{2}, O A = \frac{1}{2} A K = \frac{\sqrt{7}}{4}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-24.jpg?height=420&width=509&top_left_y=1156&top_left_x=777)

## Answer