20. In the ring bounded by concentric circles of radii $R$ and $r$ with center $O$, two points $A$ and $B$ are taken; the distance $AB$ is equal to 1. What is the smallest value that the angle $AOB$ can have?
20. It is clear that if $R<\frac{1}{2}$, then the required segment $A B$ does not exist; therefore, we can assume that $R \geqslant \frac{1}{2}$ (and $R \geqslant r$). If the segment $A B$ is moved parallel, say in the direction of the perpendicular dropped from $O$ to $A B$, moving $A B$ away from $O$, then $\angle A O B$ will decrease; therefore, we can assume that one of the endpoints of $A B$, say $A$, belongs to the larger circle $S$; in this case, $B$ belongs to

![](https://cdn.mathpix.com/cropped/2024_05_21_d4ffc236376258137647g-139.jpg?height=38&width=981&top_left_y=384&top_left_x=210)
where $B_{1}$ belongs to $S$, and $B_{2}$ belongs to the smaller circle $s$ or the segment $O A$ (make a drawing!).

Draw the tangent $O B_{0}$ from $O$ to the circle containing $\sigma$. If the point $B_{0}$ belongs to $\sigma$, then as the point $B$ moves along $\sigma$, the angle $A O B=\alpha$ first increases to the value $A O B_{0}$, and then begins to decrease; if $B_{0}$ does not belong to $\sigma$, then $\alpha$ changes monotonically as $B$ moves from $B_{1}$ to $B_{2}$. Thus, in all cases, $\min \alpha$ corresponds to $B$ coinciding with $B_{1}$ or $B_{2}$.

If $R-r \geqslant 1$, then $\min \alpha=0$, and it corresponds to the point $B_{2}$ on the segment $O A$. If $R-r<1$, then from the (isosceles!) $\triangle O A B_{1}$ and from $\triangle O A B_{2}$ we obtain

$$
\sin \frac{\angle A O B_{1}}{2}=\frac{1}{2 R} \quad\left(\mathrm{a} \cos \angle A O B_{1}=1-\frac{1}{2 R^{2}}\right)
$$

and

$$
\cos \angle A O B_{2}=\frac{R^{2}+r^{2}-1}{2 R r}
$$

But

$$
\begin{aligned}
& \frac{R^{2}+r^{2}-1}{2 R r}-\left(1-\frac{1}{2 R^{2}}\right)=\frac{R^{3}+R r^{2}-R-2 R^{2} r+r}{2 R^{2} r}= \\
& =\frac{(R-r)\left(R^{2}-R r-1\right)}{2 R^{2} r}=\frac{R-r}{2 R r}\left(R-\frac{1}{R}-r\right)
\end{aligned}
$$

from which it follows that

$$
\text { if } r\left\{\leqslant R - \frac { 1 } { R } , \text { then } \operatorname { c o s } \angle A O B _ { 2 } \left\{\leqslant \cos \angle A O B_{1},\right.\right.
$$

and consequently,

$$
\angle A O B_{2}\left\{\gtrless \angle A O B_{1}\right.
$$

Finally, we obtain: if $r \geqslant R-\frac{1}{R}$, then $\min \alpha=\angle A O B_{1}=$ $=2 \arcsin \frac{1}{2 R} ;$ if $R-1 \leqslant r \leqslant R-\frac{1}{R}$, then $\min \alpha=\angle A O B_{2}=$ $=\arccos \frac{R^{2}+r^{2}-1}{2 R r} ;$ if $r \leqslant R-1$, then $\min \alpha=0$.

Note. Similarly, we can solve the problem of the maximum possible value of $\angle A O B=\alpha$; in this case, we can assume that the point $B$ belongs to the circle $s$, and $A$ belongs to the arc with center $B$ and radius 1, contained within the considered annulus. The analysis of the corresponding problem of possible values of the angle $\alpha$ requires distinguishing 14 (!) cases (see problem 44 of the book [4]).