4. In triangle $ABC$ with angle $A$ equal to $60^{\circ}$, the angle bisector $AD$ is drawn. The radius of the circumcircle of triangle $ADC$ with the center at point $D$ is $2 \sqrt{3} / 3$. Find the length of segment $BM$, where $M$ is the intersection point of segments $AD$ and $BO$, if $AB=1$.
Solution: 1) $D C=2 \sqrt{3} / 3$

$\triangle O D C$

is equilateral,

since

$2 \angle D A C=\angle D O C=60^{\circ}$.

2) Let $A D=l, B D=x, A C=z$. By the properties of the angle bisector, we have $\frac{A B}{A C}=\frac{B D}{D C} \Rightarrow \frac{1}{z}=\frac{x \sqrt{3}}{2}$. Since $S_{\triangle A B D}+S_{\triangle A D C}=S_{\triangle A B C}, \quad \frac{1 \cdot l}{2}+\frac{l \cdot z}{2}=1 \cdot z \sin 60^{\circ} \Rightarrow l=\frac{z \sqrt{3}}{z+1}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-09.jpg?height=642&width=571&top_left_y=450&top_left_x=1254)

Using the angle bisector theorem $l^{2}=A B \cdot A C-B D \cdot D C$, we get $l^{2}=z-\frac{2 \sqrt{3} x}{3}$.

We have the system of equations $\left\{\begin{array}{c}x=\frac{2}{z \sqrt{3}}, \\ l=\frac{z \sqrt{3}}{z+1}, \\ l^{2}=z-\frac{2 \sqrt{3} x}{3},\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{2}{z \sqrt{3}}, \\ l=\frac{z \sqrt{3}}{z+1}, \\ \left(\frac{z \sqrt{3}}{z+1}\right)^{2}=z-\frac{4}{3 z},\end{array} \Rightarrow 9 z^{3}=\left(3 z^{2}-4\right)(z+1)^{2}\right.\right.$,

$3 z^{4}-3 z^{3}-z^{2}-8 z-4=0,(z-2)\left(3 z^{3}+3 z^{2}+5 z+2\right)=0, z=2$, no other solutions $(z \geq 0)$.

Then $x=\sqrt{3} / 3, l=2 \sqrt{3} / 3$. Triangle $A O D$ is equilateral, $\angle O A B=\angle A B D=90^{\circ}$.

We have $\frac{B M}{M O}=\frac{1}{2}, B M=\frac{1}{3} B O=\frac{1}{3} \sqrt{1+\frac{4}{3}}=\frac{\sqrt{21}}{9}$.

Answer: $\frac{\sqrt{21}}{9}$.