6. A circle $\Omega$ is circumscribed around an acute isosceles triangle $BCD$ with base $CD$. The distance from the midpoint of the arc $BD$, not containing point $C$, to the side $BD$ is 3, and the distance from the midpoint of the arc $CD$, not containing point $B$, to the side $CD$ is 0.5. Find the radius of the circle $\Omega$ and the area of triangle $BCD$.
Answer: $R=4, S=\frac{15 \sqrt{15}}{4}$.

Solution. Let $Q$ and $T$ be the midpoints of arcs $CD$ and $BD$; $A$ and $H$ be the midpoints of segments $BD$ and $CD$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $OT \perp BD, OQ \perp CD; OA=OT-TA=R-3, OH=OQ-QH=R-\frac{1}{2}$.

From the Pythagorean theorem for triangles $OAD$ and $OHD$, we get that $AD^{2}=R^{2}-(R-3)^{2}=6R-9$, $HD^{2}=R^{2}-\left(R-\frac{1}{2}\right)^{2}=R-\frac{1}{4}$. We find the sides of triangle $BDH: BH=BO+OH=2R-\frac{1}{2}$, $BD=2AD=2\sqrt{6R-9}, DH=\sqrt{R-\frac{1}{4}}$.

By the Pythagorean theorem for triangle $BDH$, we get $\left(2R-\frac{1}{2}\right)^{2}+\left(R-\frac{1}{4}\right)=24R-36$, from which $4R^{2}-25R+36=0, R=4$ or $R=\frac{9}{4}$. Only $R=4$ fits (in the second case, $AO=R-3<0$).

Thus, $BH=\frac{15}{2}, DH=\frac{\sqrt{15}}{2}, S_{BCD}=DH \cdot BH=\frac{15 \sqrt{15}}{4}$.