$23 \cdot 7$ A regular tetrahedron $ABCD$ has its four vertices on the surface of a sphere with radius $R$, then the length of $AB$ is
(A) $\sqrt{2} R$.
(B) $\frac{2 \sqrt{2}}{3} R$.
(C) $\frac{2 \sqrt{6}}{3} R$.
(D) $\frac{3 \sqrt{6}}{4} R$.
(High School Mathematics Competition, Taiyuan City, Shanxi Province, China, 1990)
[Solution] Draw the median plane $D E C$ of the regular tetrahedron through $D, C$. If we set $A B=x$, it is easy to calculate that
$$
D E=C E=\frac{\sqrt{3}}{2} x,
$$

If $D H \perp E C, E F \perp D C$,
by $S_{\triangle D E C}=\frac{1}{2} E C \cdot D H=\frac{1}{2} D C \cdot E F$,

then $D H=\sqrt{\frac{2}{3}} x$, and in $\triangle D H C$ we can find $H C=\sqrt{\frac{1}{3}} x$.
Let the center of the sphere be $O$, then by $O D=O C=R$, we have
$$
O D^{2}=O C^{2}=O H^{2}+H C^{2} \text {, }
$$

If we set $O D=t \cdot D H$, then $O H=(1-t) D H$.
$$
\therefore t^{2}\left(\sqrt{\frac{2}{3}} x\right)^{2}=\left(\sqrt{\frac{1}{3}} x\right)^{2}+(1-t)^{2}\left(\sqrt{\frac{2}{3}} x\right)^{2} \text {. }
$$

Solving this, we get $t=\frac{3}{4}$. Thus $\frac{3}{4} \sqrt{\frac{2}{3}} x=R$,
$\therefore$
$$
x=\frac{2 \sqrt{6}}{3} R \text {. }
$$

Therefore, the answer is $(C)$.