5. For which $a$ does the set of solutions of the inequality $x^{2}+(|y|-a)^{2} \leq a^{2}$ contain all pairs of numbers $(x ; y)$ for which $(x-4)^{2}+(y-2)^{2} \leq 1$?

Problem 5 Answer: $a \in\left[\frac{19}{2} ;+\infty\right)$
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_74adfed0f5eb8a4ea54fg-5.jpg?height=625&width=488&top_left_y=1592&top_left_x=704)

Case 1. $a \leq 0$. The first inequality is satisfied by a single point (0, 0), so the situation where the solutions of the second inequality (points of the circle of radius 1) are contained in the solutions of the first inequality is impossible.

Case 2. $a>0$. The solutions of the first inequality are the points of circles $K_{1}$ and $K_{2}$, where $K_{1}$ is a circle of radius $a$ centered at point $O_{1}(0 ; a)$, and $K_{2}$ is a circle of radius $a$ centered at point $O_{2}(0 ;-a)$. The solutions of the second inequality are the points of circle $K_{3}$ of radius 1 centered at point $O(4 ; 2)$. The conditions of the problem will be satisfied if circle $K_{3}$ is inside circle $K_{1}$. This situation occurs when $O_{1} O \leq a-1$. We obtain the inequality:

$$
\sqrt{4^{2}+(a-2)^{2}} \leq a-1
$$

This is equivalent to the system

$$
\left\{\begin{array}{l}
a-1 \geq 0 \\
4^{2}+(a-2)^{2} \leq(a-1)^{2}
\end{array}\right.
$$

Solving this system, we find $a \geq \frac{19}{2}$.

Problem 6 Answer: $s=7.5$

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_74adfed0f5eb8a4ea54fg-6.jpg?height=531&width=588&top_left_y=614&top_left_x=657)

Let the chords $B C$ and $A B$ subtend central angles $\alpha$ and $\beta$ respectively. Then, due to the perpendicularity of the diagonals, the chords $A D$ and $C D$ subtend central angles $\pi-\alpha$ and $\pi-\beta$. The sum of the areas of triangles $B C O$ and $A B O$ is:

$$
S_{B C O}+S_{A B O}=\frac{1}{2} R^{2}(\sin \alpha+\sin \beta)
$$

where $R$ is the radius of the circumscribed circle.

The sum of the areas of triangles $A D O$ and $C D O$ is:

$$
S_{A D O}+S_{C D O}=\frac{1}{2} R^{2}(\sin (\pi-\alpha)+\sin (\pi-\beta))=\frac{1}{2} R^{2}(\sin \alpha+\sin \beta)=S_{B C O}+S_{A B O}
$$

Thus, the area of quadrilateral $A B C O$ is half the area of quadrilateral $A B C D$, which is $\frac{1}{2} d_{1} \cdot d_{2}=\frac{1}{2} \cdot 5 \cdot 6=15$.

## Variant 2

Problem 1 Answer: 1) 6 pairs; 2) (3;5), (27;5), (243;5), (2187;5), (6561, 25),(1;625).

Problem 2 Answer: $S_{\text {min }}=\frac{\pi^{2}}{14}$

Problem 3 Answer: 24 circles (15+9)

Problem 4 Answer: $P(A)=\frac{\operatorname{arctg} 2-\operatorname{arctg}(1 / 2)}{\pi}=\frac{\operatorname{arctg} 2-\operatorname{arcctg} 2}{\pi}=\frac{2 \operatorname{arctg} 2}{\pi}-\frac{1}{2}$

Problem 5 Answer: $a \geq \frac{5}{2}$

Problem 6 Answer: $s=7$

## Variant 3

Problem 1 Answer: 1) 2 pairs; 2) (4;49), (1024;49)

Problem 2 Answer: $S_{\min }=\frac{\pi^{2}}{6}$

Problem 3 Answer: 30 circles. (18+12)

Problem 4 Answer: $P(A)=\frac{\operatorname{arctg}\left(\frac{4}{3}\right)-\operatorname{arctg}\left(\frac{3}{4}\right)}{\pi}=\frac{\operatorname{arctg}\left(\frac{4}{3}\right)-\operatorname{arctg}\left(\frac{4}{3}\right)}{\pi}=\frac{2 \operatorname{arctg}(4 / 3)}{\pi}-\frac{1}{2}$

Problem 5 Answer: $a<\frac{7}{3}$

Problem 6 Answer: $s=1.5$

## Variant 4

Problem 1 Answer: 1) 2 pairs; 2) (8;125), (64;125)

Problem 2 Answer: $S_{\min }=\frac{\pi^{2}}{10}$

Problem 3 Answer: 9 circles. (9+0)

Problem 4 Answer: $P(A)=\frac{\operatorname{arctg}\left(\frac{3}{2}\right)-\operatorname{arctg}\left(\frac{2}{3}\right)}{\pi}=\frac{\operatorname{arctg}\left(\frac{3}{2}\right)-\operatorname{arcctg}\left(\frac{3}{2}\right)}{\pi}=\frac{2 \operatorname{arctg}(3 / 2)}{\pi}-\frac{1}{2}$

Problem 5 Answer: $a=-\frac{8}{5}$

Problem 6 Answer: $s=12$