Example 8. The edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1. Find the distance between a point on the incircle of the base $A B C D$ and the vertices $A, C$, and $B_{1}$.
The figure shows two circles, each lying on one of two concentric spheres centered at the symmetry center of a cube, namely, the sphere that is tangent to all edges of the the cube (with radius $\frac{\sqrt{2}}{2}$, referred to as the small sphere) and the circumscribed sphere of the cube (with radius $\frac{\sqrt{3}}{2}$, referred to as the large sphere). The minimum distance between points on these two spheres is the difference in their radii, $d = \frac{\sqrt{3}-\sqrt{2}}{2}$. If there is a point on each circle that lies on the same ray emanating from the center $O$, then $d$ is the minimum value.

Considering the homothety with center $O$ and ratio $\sqrt{\frac{3}{2}}$, the small sphere is transformed into the large sphere, and the image of the circle on the small sphere is a circle on the large sphere.

Notice that the points $E$ and $F$, where the incircle $\odot O_{1}$ of $ABCD$ intersects the line segment $BD$, have images under this homothety that lie on opposite sides of the plane $AB_{1}C$ (since $\angle O_{1} O F = 45^{\circ} > \angle B B_{1} O_{1}$, the ray $OF$ does not intersect the plane $AB_{1}C$). Therefore, the image of $\odot O_{1}$ (a circle) will intersect the circle passing through the 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 circles, which is $d = \frac{\sqrt{3}-\sqrt{2}}{2}$.