6. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $ABC$ and $ADC$ respectively, with the first touching side $BC$ at point $K$, and the second touching side $AD$ at point $T$.

a) Find the radius of circle $\Omega_{1}$, if $BK=3\sqrt{3}, DT=\sqrt{3}$.

b) Suppose it is additionally known that point $O_{1}$ is the center of the circle circumscribed around triangle $BOC$. Find the angle $BDC$.
Answer. a) $r=3$, b) $\angle B D C=30^{\circ}$.

Solution. a) Segments $B O_{1}$ and $D O_{2}$ are the angle bisectors of angles $A B C$ and $A D C$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a circle, the sum of its opposite angles $A B C$ and $A D C$ is $180^{\circ}$, so the sum of their halves - angles $K B O_{1}$ and $T D O_{2}$ - is $90^{\circ}$. Let $\angle O_{1} B K=\alpha$. Then $\angle T O_{2} D=90^{\circ}-\angle T D O_{2}=\alpha$. Expressing $\operatorname{tg} \alpha$ in two ways, we get: $\operatorname{tg} \alpha=\frac{O_{1} K}{B K}=\frac{D T}{O_{2} T}, \frac{r}{3 \sqrt{3}}=\frac{\sqrt{3}}{r}, r=3$.

b) $O_{1} B=O_{1} C$ as radii of the circle circumscribed around triangle $B O C$, so the height $O_{1} K$ of this triangle is also its median. Points $O, O_{1}$, and $K$ lie on the same line (on the perpendicular bisector of segment $B C$). We then find: $O_{1} B=\sqrt{O_{1} K^{2}+K B^{2}}=6, O_{1} O=O_{1} B=6$ (as radii of the same circle).

There are two cases: points $O_{1}$ and $O$ can lie either on the same side of line $B C$ or on opposite sides of it.

In the first case, we get: $\angle B D C=\frac{1}{2} \angle B O C=\angle B O K=\operatorname{arctg} \frac{B K}{K O}=\operatorname{arctg} \frac{3 \sqrt{3}}{9}=30^{\circ}$.

Consider the second case. Point $O_{1}$ lies on the bisector of angle $A B C$, so

$$
\angle A B C=2 \angle O_{1} B K=2 \operatorname{arctg} \frac{O_{1} K}{B K}=2 \operatorname{arctg} \frac{1}{\sqrt{3}}=60^{\circ}
$$

This means that the inscribed angle $A B C$ subtends an arc $A C$ equal to $120^{\circ}$. At the same time, the arc $B C$ is equal to the angle $B O C$, and $\angle B O C=2 \angle B O K=2 \operatorname{arctg} \frac{B K}{O K}=2 \operatorname{arctg} \frac{B K}{O_{1} O-r}=2 \operatorname{arctg} \sqrt{3}=120^{\circ}$, which is impossible, as in this case the arc $A C$ must be smaller than the arc $B C$.