6. Around an acute isosceles triangle $F K T$ with base $K T$, a circle $\Omega$ is described. Point $M$ is the midpoint of the arc $F T$ that does not contain point $K$. It is known that the distances from point $M$ to the lines $K T$ and $F T$ are $\frac{9}{5}$ and 1, respectively. Find the radius of the circle $\Omega$ and the area of triangle $F K T$.
Answer: $R=\frac{5}{3}, S=\frac{56 \sqrt{7}}{25 \sqrt{3}}$.

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

Then $O G=O F \sin \gamma=R \sin \gamma, G M=O M-O G=R-R \sin \gamma, J M=O M \sin \gamma=R \sin \gamma$. Since triangle $OFT$ is isosceles ($O F=O T=R$), $\angle O T F=\angle O F T=\gamma$, and by the exterior angle theorem of a triangle, $\angle T O H=2 \gamma$. Therefore, $O H=O T \cos 2 \gamma=R \cos 2 \gamma, J Q=O H=R \cos 2 \gamma$ (since $J Q H O$ is a rectangle). Then $M Q=M J+J Q=R(\sin \gamma+\cos 2 \gamma)=R\left(1+\sin \gamma-2 \sin ^{2} \gamma\right)=R(1-\sin \gamma)(1+2 \sin \gamma)$.

By the problem's condition, $R(1-\sin \gamma)=1, R(1-\sin \gamma)(1+2 \sin \gamma)=\frac{9}{5}$. Dividing the second equation by the first, we get $1+2 \sin \gamma=\frac{9}{5}, \sin \gamma=\frac{2}{5}$; then $R=\frac{1}{1-\sin \gamma}=\frac{5}{3}$.

We find the area $S$ of triangle $F K T$:

“Phystech-2016”, mathematics, solution to ticket 6 $S=F H \cdot H T=(F O+O H) \cdot H T=(R+R \cos 2 \gamma) R \sin 2 \gamma=R^{2} 2 \cos ^{2} \gamma \sin 2 \gamma=4 R^{2} \sin \gamma \cos ^{3} \gamma=4 \cdot \frac{25}{9} \cdot \frac{2}{5} \cdot \frac{21 \sqrt{21}}{125}=\frac{56 \sqrt{7}}{25 \sqrt{3}}$.