6. An isosceles obtuse triangle $P Q T$ with base $P T$ is inscribed in a circle $\Omega$. Chords $A B$ and $C D$, parallel to the line $P T$, intersect the side $Q T$ at points $K$ and $L$ respectively, and $Q K=K L=L T$. Find the radius of the circle $\Omega$ and the area of triangle $P Q T$, if $A B=2 \sqrt{66}, C D=2 \sqrt{114}$.
Answer: $R=12.5, S=108$.

Solution. The line $Q O$ is perpendicular to the chords $A B, C D, P T$ and bisects each of them. Let points $M$, $E$, and $H$ be the midpoints of $A B, C D$, and $P T$. Denote the radius of the circle $\Omega$ as $R; Q M=M E=E H=x$.

Since the triangle is obtuse, the center of its circumscribed circle lies outside the triangle. Then $O M=O Q-M Q=R-x, O E=O Q-Q E=R-2 x$ and by the Pythagorean theorem for triangles $O M A$ and $O E C$ we get $R^{2}=66+(R-x)^{2}, R^{2}=114+(R-2 x)^{2}$, from which after expanding the brackets and combining like terms, we have

$$
\left\{\begin{array}{l}
x^{2}-2 R x+66=0 \\
2 x^{2}-2 R x+57=0
\end{array}\right.
$$

Subtracting the second equation from the first, we find that $-x^{2}+9=0, x=3$. Then $R=\frac{x^{2}+66}{2 x}=\frac{25}{2}$ $O H=O Q-Q H=R-3 x=\frac{7}{2}, T H^{2}=O T^{2}-O H^{2}=\left(\frac{25}{2}\right)^{2}-\left(\frac{7}{2}\right)^{2}=144$.

Therefore, $S_{P Q T}=Q H \cdot T H=9 \cdot 12=108$.