Harvard-MIT Mathematics Tournament

Solution:

Let $C'$ be the point on minor arc $BCD$ such that $BC' = 12$ and $C'D = 5$, and write $AC' = BD = C'E = x$, $AD = y$, and $BD = z$. Ptolemy applied to quadrilaterals $ABC'D$, $BC'DE$, and $ABDE$ gives
$$\begin{aligned} & x^2 = 12y + 5^2 \\ & x^2 = 5z + 12^2 \\ & yz = 14x + 5 \cdot 12 \end{aligned}$$
Then
$$\left(x^2 - 5^2\right)\left(x^2 - 12^2\right) = 5 \cdot 12 yz = 5 \cdot 12 \cdot 14x + 5^2 \cdot 12^2$$
from which $x^3 - 169x - 5 \cdot 12 \cdot 14 = 0$. Noting that $x > 13$, the rational root theorem leads quickly to the root $x = 15$. Then triangle $BCD$ has area $\sqrt{16 \cdot 1 \cdot 4 \cdot 11} = 8\sqrt{11}$ and circumradius $R = \frac{5 \cdot 12 \cdot 15}{4 \cdot 8 \sqrt{11}} = \frac{225 \sqrt{11}}{88}$.