32. [18] Cyclic pentagon $A B C D E$ has side lengths $A B=B C=5, C D=D E=12$, and $A E=14$. Determine the radius of its circumcircle.
$A C^{\prime}=B D=C^{\prime} E=x, A D=y$, and $B D=z$. Ptolemy applied to quadrilaterals $A B C^{\prime} D, B C^{\prime} D E$, and $A B D E$ gives
$$
\begin{array}{l}
x^{2}=12 y+5^{2} \\
x^{2}=5 z+12^{2} \\
y z=14 x+5 \cdot 12
\end{array}
$$

Then
$$
\left(x^{2}-5^{2}\right)\left(x^{2}-12^{2}\right)=5 \cdot 12 y z=5 \cdot 12 \cdot 14 x+5^{2} \cdot 12^{2} \text {, }
$$
from which $x^{3}-169 x-5 \cdot 12 \cdot 14=0$. Noting that $x>13$, the rational root theorem leads quickly to the root $x=15$. Then triangle $B C D$ has area $\sqrt{16 \cdot 1 \cdot 4 \cdot 11}=8 \sqrt{11}$ and circumradius $R=\frac{5 \cdot 12 \cdot 15}{4 \cdot 8 \sqrt{11}}=$ $\frac{225 \sqrt{11}}{88}$.