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 $BAD$ and $BCD$ respectively, with the first touching side $AB$ at point $L$, and the second touching side $BC$ at point $F$.

a) Find the radius of circle $\Omega_{2}$, if $AL=\sqrt{2}, CF=2\sqrt{2}$.

b) Suppose it is additionally known that point $O_{2}$ is the center of the circle circumscribed around triangle $BOC$. Find the angle $BDC$.
Answer. a) $r=2$, b) $\angle B D C=\operatorname{arctg} \frac{\sqrt{3}-1}{\sqrt{2}}$.

Solution. a) Segments $A O_{1}$ and $C O_{2}$ are the bisectors of angles $B A D$ and $B C D$ (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 $B A D$ and $B C D$ is $180^{\circ}$, so the sum of their halves - angles $L A O_{1}$ and $F C O_{2}$ - is $90^{\circ}$. Let $\angle O_{1} A L=\alpha$. Then $\angle F O_{2} C=90^{\circ}-\angle F C O_{2}=\alpha$. Expressing $\operatorname{tg} \alpha$ in two ways, we get: $\operatorname{tg} \alpha=\frac{O_{1} L}{A L}=\frac{C F}{O_{2} F}, \frac{r}{\sqrt{2}}=\frac{2 \sqrt{2}}{r}, r=2$.

“Phystech-2015”, 10th grade, solutions for ticket 3

b) $O_{2} B=O_{2} C$ as radii of the circle circumscribed around triangle $B O C$, so the height of this triangle $O_{2} F$ is also its median. Points $O, O_{2}$, and $F$ lie on the same line (on the perpendicular bisector of segment $B C$). We then find: $O_{2} C=\sqrt{O_{2} F^{2}+C F^{2}}=2 \sqrt{3}, O_{2} O=O_{2} C=2 \sqrt{3}$ (as radii of the same circle).

There are two cases: points $O_{2}$ 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: $\angle B D C=\frac{1}{2} \angle B O C=\angle C O F=\operatorname{arctg} \frac{C F}{O F}=\operatorname{arctg} \frac{2 \sqrt{2}}{2+2 \sqrt{3}}=\operatorname{arctg} \frac{\sqrt{3}-1}{\sqrt{2}}$.

Consider the second case. Point $O_{2}$ lies on the bisector of angle $B C D$, so

$$
\angle B C D=2 \angle O_{2} C F=2 \operatorname{arctg} \frac{O_{2} F}{C F}=2 \operatorname{arctg} \frac{1}{\sqrt{2}}=\operatorname{arctg}(2 \sqrt{2})
$$

This means that the inscribed angle $B C D$ subtends arc $B D$, which is equal to $2 \operatorname{arctg}(2 \sqrt{2})$. At the same time, arc $B C$ is equal to angle $B O C$, and $\angle B O C=2 \angle C O F=2 \operatorname{arctg} \frac{C F}{O F}=2 \operatorname{arctg} \frac{C F}{O_{2} O-r}=2 \operatorname{arctg} \frac{\sqrt{2}}{\sqrt{3}-1}=2 \operatorname{arctg} \frac{\sqrt{3}+1}{\sqrt{2}}, \quad$ which is impossible, as in this case arc $B D$ must be smaller than arc $B C$.