1. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with $AB$ as the hypotenuse, $SA=SB=SC=2, AB=2$. Suppose points $S, A, B, C$ are all on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$ (1997, National High School Mathematics Competition)
(Tip: From the basic conclusion 5, we know that the projection of $S$ on the plane $ABC$ is $M$, which is the circumcenter of $\triangle ABC$, i.e., $M$ is the midpoint of the hypotenuse of $\triangle ABC$ and the radius of the circumcircle is $MC=1$. Also, $MS=\sqrt{3}$. From $OA=OB=OC$, we know that the projection of $O$ on the plane $ABC$ is also $M$. Furthermore, from $OS=OC$ and $MS > MC$, we know that $O$ lies on the line segment $MS$, as shown in Figure 9. Let $OC=OS=x$. Then $OM=\sqrt{3}-x$. We can set up an equation to solve for $x=\frac{2\sqrt{3}}{3}$.)