Harvard-MIT Mathematics Tournament

Solution:

Answer: 8. Construct point $E'$ on the circumcircle of $A B C D E F$ such that $D E' = E F = 12$ and $E' F = D E = 9$; then $\overline{B E'}$ is a diameter. Let $B E' = d$. Then $C E' = \sqrt{B E'^2 - B C^2} = \sqrt{d^2 - 4}$ and $B D = \sqrt{B E'^2 - D E'^2} = \sqrt{d^2 - 144}$. Applying Ptolemy's theorem to $B C D E'$ now yields
$$9 \cdot d + 2 \cdot 12 = \sqrt{(d^2 - 4)(d^2 - 144)}$$
Squaring and rearranging, we find $0 = d^4 - 229 d^2 - 432 d = d(d-16)(d^2 + 16 d + 27)$. Since $d$ is a positive real number, $d = 16$, and the circumradius is $8$.