China Mathematical Competition

Let the center of the sphere circumscribing $ABCD$ be $O$. Then $O$ is on the vertical line of plane $ABD$ through point $N$ the circumcenter of $△ABD$. It is known that $△ABD$ is regular, so $N$ is the center of it. Let $P$ and $M$ be the midpoints of $AB$ and $CD$, respectively. Then $N$ is on $DP$ with $ON\perp DP$ and $OM\perp CD$.



Let $\theta$ denote the angle between $CD$ and plane $ABD$. From $$\angle CDA=\angle CDB=\angle ADB=60^{\circ },$$ we find $\cos \theta =\frac{1}{\sqrt{3}}$, $\sin \theta =\frac{\sqrt{2}}{\sqrt{3}}$.

Since $DM=\frac{1}{2}CD=1$, $DN=\frac{2}{3}\cdot DP=\frac{\sqrt{3}}{2}\cdot 3=\sqrt{3}$, by cosine theorem we have, in $△DMN$, $$\begin{array}{rl}M{N}^2& =D{M}^2+D{N}^2-2\cdot DM\cdot DN\cdot \cos \theta \\ & =1^2+(\sqrt{3}{)}^2-2\cdot 1\cdot \sqrt{3}\cdot \frac{1}{\sqrt{3}}=2,\end{array}$$ that is $MN=\sqrt{2}$. The radius of the sphere circumscribing $ABCD$ is then $$OD=\frac{MN}{\sin \theta }=\frac{\sqrt{2}}{\sqrt{2}}=\sqrt{3}.$$ The answer is $R=\sqrt{3}$.