6. In triangle $A B C$ with angle $A$ equal to $60^{\circ}$, the angle bisector $A D$ is drawn. The radius of the circumcircle of triangle $A D C$ with center at point $O$ is $\sqrt{3}$. Find the length of the segment $O M$, where $M$ is the intersection point of segments $A D$ and $B O$, if $A B=1.5$.
Solution: 1) $D C=\sqrt{3}$

$\triangle O D C \quad-\quad$ 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 $\frac{A B}{A C}=\frac{B D}{D C} \Rightarrow \frac{3}{2 z}=\frac{x}{\sqrt{3}}$. Since $S_{\triangle A B D}+S_{\triangle A D C}=S_{\triangle A B C}, \quad \frac{1.5 \cdot l}{2}+\frac{l \cdot z}{2}=1.5 \cdot z \sin 60^{\circ} \Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_06_dbf1970871b608540faag-15.jpg?height=651&width=574&top_left_y=1945&top_left_x=1255)
$l=\frac{z \cdot 3 \sqrt{3}}{2 z+3}$. Using the statement for the angle bisector $l^{2}=A B \cdot A C-B D \cdot D C, \quad$ we get $l^{2}=1.5 z-\sqrt{3} x$.

We have the system of equations $\left\{\begin{array}{c}x=\frac{3 \sqrt{3}}{2 z}, \\ l=\frac{z \cdot 3 \sqrt{3}}{2 z+3}, \\ l^{2}=1.5 z-\sqrt{3} x,\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{3 \sqrt{3}}{2 z}, \\ l=\frac{z \cdot 3 \sqrt{3}}{2 z+3}, \\ \left(\frac{z \cdot 3 \sqrt{3}}{2 z+3}\right)^{2}=1.5 z-\frac{9}{2 z},\end{array} \Rightarrow 54 z^{3}=\left(3 z^{2}-9\right)(2 z+3)^{2}\right.\right.$, $4 z^{4}-6 z^{3}-3 z^{2}-36 z-27=0,(z-3)\left(4 z^{3}+6 z^{2}+15 z+9\right)=0, z=3$, there are no other solutions $(z \geq 0)$. Then $x=\sqrt{3} / 2, l=\sqrt{3}$. Triangle $A O D$ is equilateral, $\angle O A B=\angle A B D=90^{\circ}$.

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

Answer: $\frac{\sqrt{21}}{3}$.