6. An isosceles triangle $ABC$ with base $BC$ is inscribed in a circle $\Omega$. Chords $LM$ and $PQ$, parallel to the line $BC$, intersect side $AB$ at points $D$ and $T$ respectively, and $AD = DT = TB$. Find the radius of the circle $\Omega$ and the area of triangle $ABC$, if $LM = \frac{10}{\sqrt{3}}, PQ = \frac{2 \sqrt{26}}{\sqrt{3}}$, and the center $O$ of the circle $\Omega$ is located between the lines $LM$ and $PQ$.
Answer: $R=\frac{37}{12}, S=6$.

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

Then $O H=A H-O A=2 x-R, O N=O A-A N=R-x$ and by the Pythagorean theorem for triangles $O H P$ and $O L N$ we get $R^{2}=\frac{26}{3}+(2 x-R)^{2}, R^{2}=\frac{25}{3}+(R-x)^{2}$, from which after expanding the brackets and combining like terms, we have

$$
\left\{\begin{array}{l}
2 x^{2}-2 R x+\frac{13}{3}=0 \\
x^{2}-2 R x+\frac{25}{3}=0
\end{array}\right.
$$

Subtracting the second equation from the first, we find that $x^{2}-4=0, \quad x=2$. Then $\quad R=\frac{6 x^{2}+13}{6 x}=\frac{37}{12}$; $O E=A E-A O=3 x-R=\frac{35}{12}, B E^{2}=O B^{2}-O E^{2}=\left(\frac{37}{12}\right)^{2}-\left(\frac{35}{12}\right)^{2}=1$. Therefore, $S_{A B C}=A E \cdot B E=6 \cdot 1=6$.