11.4. Consider triangle $ABC$ and $BC=a$. Let $M$ be a point on the ray $[CA)$ such that $\angle MBC = \angle BAC$. Find the smallest possible distance between the centers $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$ of the circumcircles of triangles $ABC$ and $ABM$.
## Solution.

There are 2 possible cases:

1. $\angle A B C > \angle B A C$. Then $M \in [A C]$.
2. $\angle A B C \leq \angle B A C$. Then $M$ is on the extension of line $(A C)$ beyond point $A$ and $M \notin [A C]$.

Consider case 1. $\triangle A B C$ is similar to $\triangle B M C$ (since $\angle C$ is common, and $\angle M B C = \angle B A C$ according to the condition). Therefore,

$$
\frac{A B}{M B} = \frac{B C}{M C} = \frac{A C}{B C}
$$

Then

$$
B C^{2} = A C \cdot M C
$$

so $B C$ is tangent to the circle with center $O_{2}$, circumscribed around triangle $\triangle A B M$.

Let $D$ be the midpoint of $B C$. Then the center $O_{1}$ of the circle circumscribed around triangle $\triangle A B C$ lies on the perpendicular bisector (perpendicular line through the midpoint of the segment)

$$
D O_{1} ; D O_{1} \perp B C
$$

Since $B C$ is tangent at point $B$ to the circle with center $O_{2}$, then $O_{2} B \perp B C$.

Then the segment $[B D]$ of length $\frac{a}{2}$ is the orthogonal projection of the segment $\left[\mathrm{O}_{2} \mathrm{O}_{1}\right]$ on the line $(B C)$. Therefore,

$$
O_{2} O_{1} \geq \frac{a}{2}
$$

Equality holds when $\angle B = 90^{\circ}$ and $O_{1}$ is the midpoint of $A C$, $O_{2}$ is the midpoint of $A B$, and $B M \perp A C$.

It follows that the minimum value of the distance between the centers $O_{1}$ and $O_{2}$ is equal to $\frac{a}{2}$.

Case 2 is investigated similarly and gives the same result.