Example 12

How many circles with radius $\frac{r}{2}$ are needed to cover $\odot(O, r)$?
First, we prove that six circles with a radius of $\frac{r}{2}$ cannot cover $\odot(O, r)$.
Because each circle with a radius of $\frac{r}{2}$ can cover at most $\frac{1}{6}$ of the circumference of $\odot(O, r)$. If six such small circles can cover $\odot(O, r)$, then each small circle must exactly cover $\frac{1}{6}$ of the circumference of $\odot(O, r)$. However, these circles cannot cover the center $O$ of $\odot(O, r)$. This can be used to show that if a circle $\odot\left(O_{1}, \frac{r}{2}\right)$ covers the point $O$, then this circle can contain at most one point on the circumference of $\odot(O, r)$, and will not cover $\frac{1}{6}$ of the circumference of $\odot(O, r)$. This is because the distance $O O_{1} \leqslant \frac{r}{2}=r-\frac{r}{2}$, so $\odot\left(O_{1}, \frac{r}{2}\right)$ is covered by $\odot(O, r)$, and has no common points or only one common point (the tangent point) with the circumference of $\odot(O, r)$.
Second, we prove that seven circles with a radius of $\frac{r}{2}$ can cover $\odot(O, r)$. For this, construct the inscribed regular hexagon $A B C D E F$ of $\odot(O, r)$, and draw circles with each side of this hexagon as the diameter. These six circles, together with $\odot\left(O, \frac{r}{2}\right)$, cover $\odot(O, r)$
(Figure 1.18).
In fact, take the midpoints $A_{1}, B_{1}, C_{1}, D_{1}, E_{1}, F_{1}$ of $O A, O B, O C, O D, O E, O F$. Since $\triangle O A B$ is an equilateral triangle, $\angle B B_{1} A=90^{\circ}$, and $B_{1}$ belongs to the circle with $A B$ as the diameter. Similarly, $A_{1}$ also belongs to this circle, and the circle with $A B$ as the diameter covers the quadrilateral $A A_{1} B_{1} B$. From this, the conclusion is not difficult to derive. The detailed derivation is left to the reader.