HMMT November

Let $X$ be the foot of the perpendicular from $A$ to $\triangle B C D$. Since $A B=A C=A D$, it follows that $X$ is the circumcenter of $\triangle B C D$. Denote $X B=X C=X D=r$. By the Pythagorean Theorem on $\triangle A B X$, we have $A X=\sqrt{4-r^{2}}$. Now, from the extended law of sines on $\triangle B C D$, we have the well-known identity $$\frac{B C \cdot C D \cdot D B}{4 r}=[B C D]$$ where $[B C D]$ denotes the area of $\triangle B C D$. However, we have $$V=\frac{1}{3} A X \cdot[B C D]$$ where $V$ is the volume of $A B C D$, which yields the expression $$[B C D]=\frac{3}{\sqrt{4-r^{2}}}$$ Now, given that $B C \cdot C D \cdot D B=16$, we have $$\frac{4}{r}=\frac{3}{\sqrt{4-r^{2}}}$$ Solving, we get $r=\frac{8}{5}$. Now, let $O$ be the center of $\omega$. Since $O B=O C=O D$, it follows that the foot of the perpendicular from $O$ to $\triangle B C D$ must also be the circumcenter of $\triangle B C D$, which is $X$. Thus, $A, X, O$ are collinear. Let $R$ be the radius of $\omega$. Then we have $$\begin{aligned} R & =O A \\ & =O X+X A \\ & =\sqrt{R^{2}-r^{2}}+\sqrt{4-r^{2}} \\ & =\sqrt{R^{2}-\frac{64}{25}}+\frac{6}{5} \end{aligned}$$ Solving, we get $R=\frac{5}{3}$. (Note: solving for $R$ from $O A=O X-X A$ gives a negative value for $R$.)