China Mathematical Competition

As shown in the figure, the surface of the ball intersects all of the six surfaces of the cube. The intersection curves are divided into two kinds: One kind lies on the three surfaces including vertex $A$ respectively, that is $AA_1B_1B$, $ABCD$, and $AA_1D_1D$; while the other lies on the three surfaces not including $A$, that is $CC_1D_1D$, $A_1B_1C_1D_1$ and $BB_1C_1C$.

On surface $AA_1B_1B$, the intersection curve is arc $\widehat{EF}$ which lies on a circle with $A$ as the center. Since $AE = \frac{2\sqrt{3}}{3}$, $AA_1 = 1$, so $\angle A_1AE = \frac{\pi}{6}$. In the same way $\angle BAF = \frac{\pi}{6}$. Therefore $\angle EAF = \frac{\pi}{6}$. That means the length of arc $\widehat{EF}$ is $\frac{2\sqrt{3}}{3} \cdot \frac{\pi}{6} = \frac{\sqrt{3}\pi}{9}$. There are three arcs of this category.

On surface $BB_1C_1C$, the intersection curve is arc $\widehat{FG}$ which lies on a circle centred at $B$. The radius equals $\frac{\sqrt{3}}{3}$ and $\angle FBG = \frac{\pi}{2}$. So the length of $\widehat{FG}$ is $\frac{\sqrt{3}}{3} \cdot \frac{\pi}{2} = \frac{\sqrt{3}\pi}{6}$. There are also three arcs of this category.

In summary, the total length of all intersection curves is
$$3 \times \frac{\sqrt{3}}{9}\pi + 3 \times \frac{\sqrt{3}}{6}\pi = \frac{5\sqrt{3}\pi}{6}.$$