HMMT February

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