16. (1997 National High School Competition Question) Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, $SA=SB=SC=2, AB=2$. 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
16. Fill in $\frac{\sqrt{3}}{3}$. Reason: Let the midpoint of the hypotenuse $AB$ of the isosceles right $\triangle ABC$ be $D$, then $D$ is the circumcenter of $\triangle ABC$. Given $SA = SB = SC = 2$, it follows that the projection of $S$ on the base $ABC$ is $D$. Therefore, the center of the sphere $O$ lies on $SD$, and $OA = OB = OS$, so $O$ is the center of $\triangle SAB$. Since $\triangle SAB$ is an equilateral triangle, $SD = \sqrt{2^2 - 1^2} = \sqrt{3}$, thus the distance from $O$ to the plane $ABC$, $OD = \frac{1}{3} SD = \frac{\sqrt{3}}{3}$.