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 $\triangle ABD$. It is known that $\triangle ABD$ 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 $\triangle DMN$,
$$\begin{aligned} MN^2 &= DM^2 + DN^2 - 2 \cdot DM \cdot DN \cdot \cos \theta \\ &= 1^2 + (\sqrt{3})^2 - 2 \cdot 1 \cdot \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 2, \end{aligned}$$
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 $R = \sqrt{3}$.