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 $A{A}_1{B}_{1}B$, $ABCD$, and $A{A}_1{D}_{1}D$; while the other lies on the three surfaces not including $A$, that is $C{C}_1{D}_{1}D$, $A_1{B}_1{C}_1{D}_1$ and $B{B}_1{C}_{1}C$.



On surface $A{A}_1{B}_{1}B$, the intersection curve is arc $\widehat{EF}$ which lies on a circle with $A$ as the center. Since $AE=\frac{2\sqrt{3}}{3}$, $A{A}_1=1$, so $\angle A_{1}AE=\frac{\pi }{6}$. 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 $B{B}_1{C}_{1}C$, 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 $$3\times \frac{\sqrt{3}}{9}\pi +3\times \frac{\sqrt{3}}{6}\pi =\frac{5\sqrt{3}\pi }{6}.$$