In the angle with vertex $A$, equal to $60^{\circ}$, a circle with center $O$ is inscribed. A tangent to this circle intersects the sides of the angle at points $B$ and $C$. Segment $B C$ intersects segment $A O$ at point $M$. Find the radius of the circle inscribed in triangle $A B C$, if $A M: M O=2: 3$ and $B C=7$.

#
Let $R$ be the radius of the given circle, $P$ and $Q$ be its points of tangency with the lines $B C$ and $A B$ respectively, $A H$ be the altitude of triangle $A B C$, $r$ be the radius of the inscribed circle of triangle $A B C$, and $p$ be the semiperimeter of triangle $A B C$. Then

$$
p=A Q=O Q \operatorname{ctg} \angle O A Q=R \operatorname{ctg} 30^{\circ}=R \sqrt{3}
$$

From the similarity of the right triangles $A H M$ and $O P M$, it follows that $\frac{A H}{O P}=\frac{A M}{M I O}=\frac{2}{3}$, so $A H=\frac{2}{3} O P=\frac{2}{3} R$. Then

$$
S_{\triangle A B C}=\frac{1}{2} B C \cdot A H=\frac{1}{2} \cdot 7 \cdot \frac{2}{3} R=\frac{7}{3} R
$$

On the other hand,

$$
S_{\triangle A B C}=p r=R \sqrt{3} \cdot r=R r \sqrt{3}
$$

From the equality $\frac{7}{3} R=R r \sqrt{3}$, we find that $r=\frac{7}{3 \sqrt{3}}$.

## Answer

$\frac{7}{3 \sqrt{3}}$.