6. Around an acute isosceles triangle $N P Q$ with base $N Q$, a circle $\Omega$ is circumscribed. Point $F$ is the midpoint of the arc $P N$ not containing point $Q$. It is known that the distances from point $F$ to the lines $P N$ and $Q N$ are 5 and $\frac{20}{3}$, respectively. Find the radius of the circle $\Omega$ and the area of triangle $N P Q$.
Answer: $R=6, S=\frac{35 \sqrt{35}}{9}$.

Solution. Let $O$ be the center of the circle, $G$ be the intersection point of segments $O F$ and $N P$ (then $G$ is the midpoint of $N P$ and $O G \perp N P$); $P H$ be the height of the triangle ($O \in P H$), $F D$ be the perpendicular dropped from point $F$ to line $N Q$, $O J$ be the perpendicular dropped from point $O$ to line $F D$. Let the radius of the circle be $R$, and the angle $N P H$ be $\gamma$.

Then $O G=O P \sin \gamma=R \sin \gamma, F G=O F-O G=R-R \sin \gamma, F J=O F \sin \gamma=R \sin \gamma$. Since triangle $O P N$ is isosceles ($O N=O P=R$), $\angle O N P=\angle O P N=\gamma$, and by the external angle theorem of a triangle, $\angle N O H=2 \gamma$. Therefore, $O H=O N \cos 2 \gamma=R \cos 2 \gamma, \quad D J=O H=R \cos 2 \gamma$ (JDHO is a rectangle). Then $F D=F J+J D=R(\sin \gamma+\cos 2 \gamma)=R\left(1+\sin \gamma-2 \sin ^{2} \gamma\right)=R(1-\sin \gamma)(1+2 \sin \gamma)$.

By the condition $R(1-\sin \gamma)=5, R(1-\sin \gamma)(1+2 \sin \gamma)=\frac{20}{3}$. Dividing the second equation by the first, we get $1+2 \sin \gamma=\frac{4}{3}, \sin \gamma=\frac{1}{6}$; then $R=\frac{5}{1-\sin \gamma}=6$.

"Phystech-2016", mathematics, solution to ticket 5

We find the area $S$ of triangle $N P Q$:

$S=N H \cdot P H=(N O+O H) \cdot P H=(R+R \cos 2 \gamma) R \sin 2 \gamma=R^{2} 2 \cos ^{2} \gamma \sin 2 \gamma=4 R^{2} \sin \gamma \cos ^{3} \gamma=4 \cdot 36 \cdot \frac{1}{6} \cdot \frac{35 \sqrt{35}}{216}=\frac{35 \sqrt{35}}{9}$.