6. Around an acute isosceles triangle $A B C$ with base $B C$, a circle $\Omega$ is circumscribed. Point $T$ is the midpoint of the arc $A C$ that does not contain point $B$. It is known that the distances from point $T$ to the lines $A C$ and $B C$ are 3 and 7, respectively. Find the radius of the circle $\Omega$ and the area of triangle $A B C$.
Answer: $R=9, S=40 \sqrt{5}$.

Solution. Let $O$ be the center of the circle, $G$ be the intersection point of segments $OT$ and $AC$ (then $G$ is the midpoint of $AC$ and $OG \perp AC$); $AH$ be the height of the triangle ($O \in AH$), $TQ$ be the perpendicular dropped from point $T$ to line $BC$, and $OJ$ be the perpendicular dropped from point $O$ to line $TQ$. Let the radius of the circle be $R$, and the angle $CAH$ be $\gamma$.

Then $OG = OA \sin \gamma = R \sin \gamma, GT = OT - OG = R - R \sin \gamma, JT = OT \sin \gamma = R \sin \gamma$. Since triangle $OAC$ is isosceles ($OA = OC = R$), $\angle OCA = \angle OAC = \gamma$, and by the external angle theorem of a triangle, $\angle COH = 2\gamma$. Therefore, $OH = OC \cos 2\gamma = R \cos 2\gamma, JQ = OH = R \cos 2\gamma$ ($JQHO$ is a rectangle). Then $TQ = TJ + JQ = R(\sin \gamma + \cos 2\gamma) = R(1 + \sin \gamma - 2 \sin^2 \gamma) = R(1 - \sin \gamma)(1 + 2 \sin \gamma)$.

By the problem's condition, $R(1 - \sin \gamma) = 3, R(1 - \sin \gamma)(1 + 2 \sin \gamma) = 7$. Dividing the second equation by the first, we get $1 + 2 \sin \gamma = \frac{7}{3}, \sin \gamma = \frac{2}{3}$; then $R = \frac{3}{1 - \sin \gamma} = 9$.

We find the area $S$ of triangle $ABC$:

$S = AH \cdot HC = (AO + OH) \cdot HC = (R + R \cos 2\gamma) R \sin 2\gamma = R^2 2 \cos^2 \gamma \sin 2\gamma = 4 R^2 \sin \gamma \cos^3 \gamma = 4 \cdot 81 \cdot \frac{2}{3} \cdot \frac{5 \sqrt{5}}{27} = 40 \sqrt{5}$.