6. A circle $\Omega$ is circumscribed around an acute isosceles triangle $A D E$ with base $A D$. The distance from the midpoint of the arc $D E$, not containing point $A$, to the side $D E$ is 5, and the distance from the midpoint of the arc $A D$, not containing point $E$, to the side $A D$ is $\frac{1}{3}$. Find the radius of the circle $\Omega$ and the area of triangle $A D E$.
Answer: $R=6, S=\frac{35 \sqrt{35}}{9}$.

Solution. Let $M$ and $T$ be the midpoints of arcs $A D$ and $D E$; $B$ and $H$ be the midpoints of segments $D E$ and $A D$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $O T \perp D E, O M \perp A D ; O B=O T-T B=R-5, O H=O M-M H=R-\frac{1}{3}$.

From the Pythagorean theorem for triangles $O B D$ and $O H D$, we get that $B D^{2}=R^{2}-(R-5)^{2}=10 R-25$, $H D^{2}=R^{2}-\left(R-\frac{1}{3}\right)^{2}=\frac{2 R}{3}-\frac{1}{9}$. $\quad$ We find the sides of triangle $\quad D E H: \quad E H=E O+O H=2 R-\frac{1}{3}$, $D E=2 B D=2 \sqrt{10 R-25}, D H=\sqrt{\frac{2 R}{3}-\frac{1}{9}}$.

By the Pythagorean theorem for triangle $P N H$, we get $\left(2 R-\frac{1}{3}\right)^{2}+\left(\frac{2 R}{3}-\frac{1}{9}\right)=40 R-100$, from which $6 R^{2}-61 R+150=0, R=6$ or $R=\frac{25}{6}$. Only $R=6$ fits (in the second case, $B O=R-5<0$).

Thus, $E H=\frac{35}{3}, D H=\frac{\sqrt{35}}{3}, S_{A D E}=D H \cdot E H=\frac{35 \sqrt{35}}{9}$.

"Phystech-2016", mathematics, solution to ticket 19