Example 12 The edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1. Find the minimum distance between a point on the circumference of the incircle of the base $A B C D$ and a point on the circumference of the circle passing through the vertices $A_{1} 、 C$ and $B_{1}$.
(19th All-Soviet Union Olympiad)
Consider Figure 21-5, examining two circumferences on the concentric spherical surfaces with the symmetry center of the cube as the center, i.e., the spherical surface tangent to all edges of the cube (radius $\frac{\sqrt{2}}{2}$) and the circumscribed spherical surface of the cube (radius $\frac{\sqrt{3}}{2}$). The minimum distance between points on these two spherical surfaces is the difference in their radii, $d=\frac{1}{2}(\sqrt{3}-\sqrt{2})$. If

there is a point on each of the two circumferences that lies on the same ray emanating from the center $O$, then $d$ is the minimum value.
Consider the transformation under a homothety with $O$ as the center and $\sqrt{\frac{3}{2}}$ as the ratio, which transforms the smaller spherical surface into the larger spherical surface, and the image of the circumference on the smaller spherical surface is the circumference on the larger spherical surface.

Notice that the points $E$ and $F$ where the incircle $\odot O_{1}$ of $ABCD$ intersects the line segment $BD$ are on opposite sides of the plane $AB_{1}C$ under this homothety (since $\angle O_{1} O F=45^{\circ}>\angle B B_{1} O_{1}$, the ray $OF$ does not intersect the plane $AB_{1}C$), thus, the image of $\odot O_{1}$ (the circumference) will intersect the circumference passing through vertices $A, C$, and $B_{1}$. Let one of the intersection points be $N$, and the preimage of $N$ be $M$, then the distance between $M$ and $N$ is the minimum distance between points on the two circumferences, which is $d=\frac{1}{2}(\sqrt{3}-\sqrt{2})$.