Problem (Geometry):
Let $E U C L I D$ be a hexagon inscribed in a circle of radius 5 . Given that $E U=U C=L I=I D=6$, and $C L=D E$, compute $C L$.

Solution:
Let $C L=x$. Because the quadrilaterals $E U C L$ and $L I D E$ are congruent, $\overline{E L}$ is a diameter of the circle in which the hexagon is inscribed, so $E L=10$. Furthermore, because $\overline{E L}$ is a diameter of the circle, it follows that the inscribed $\angle E U L$ is a right angle, hence $U L=8$.





Using Ptolemy's Theorem for cyclic quadrilaterals and the fact that $\triangle E C L$ is also a right triangle,

$$
\begin{aligned}
& U C \cdot E L+E U \cdot C L=E C \cdot U L \\
\Longrightarrow & 6(10+x)=8 \sqrt{100-x^{2}} \\
\Longrightarrow & 36(10+x)^{2}=64(10+x)(10-x) \\
\Longrightarrow & 6 \sqrt{10+x}=8 \sqrt{10-x} \\
\Longrightarrow & 36(10+x)=64(10-x) \\
\Longrightarrow & 360+36 x=640-64 x \\
\Longrightarrow & 100 x=280 \\
\Longrightarrow & x=\frac{\mathbf{1 4}}{\mathbf{5}} .
\end{aligned}
$$

Final answer: $\frac{14}{5}$