24-13 9 congruent spheres are placed inside a unit cube, given that the center of one of the spheres is located at the center of the cube, and the other 8 spheres are tangent to this sphere and to three faces of the cube, then the radius of each sphere is
(A) $1-\frac{\sqrt{3}}{2}$.
(B) $\frac{2 \sqrt{3}-3}{2}$.
(C) $\frac{\sqrt{2}}{6}$.
(D) $\frac{1}{4}$.
(E) $\frac{\sqrt{3}(2-\sqrt{2})}{4}$.
(41st American High School Mathematics Examination, 1990)
[Solution] As shown in the figure, construct the diagonal section $A A_{1} C_{1} C$ of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, which intersects to form five mutually tangent circles.
In the rectangle $A A_{1} C_{1} C$,
$$
A A_{1}=1, A C=\sqrt{1^{2}+1^{2}}=\sqrt{2} \text {, }
$$

Then $A C_{1}=\sqrt{(\sqrt{2})^{2}+1^{2}}=\sqrt{3}$.
Let the radius of the sphere be $r$. According to the figure, we have
$$
A O_{1}=C_{1} O_{2}=\sqrt{3} r \text {, }
$$

Thus, $4 r+2 \sqrt{3} r=\sqrt{3}$,
Solving for $r$ gives $r=\frac{2 \sqrt{3}(2-\sqrt{3})}{4^{2}-(2 \sqrt{3})^{2}}$
$$
=\frac{2 \sqrt{3}-3}{2} .
$$

Therefore, the answer is $(B)$.