## Task B-1.1.

Let $a>1$. The points $(x, y)$ for which the coordinates satisfy $|x|+y=a, x^{2}+y=a|x|$ determine a figure in the coordinate plane with an area of 120 square units. Calculate the real number $a$.
## Solution.

From the first equation of the system, we have $y=-|x|+a$, which, when substituted into the second equation, gives

$$
x^{2}-|x|+a=a|x| \text {. }
$$

This leads to a series of equivalent equations

$$
\begin{aligned}
& x^{2}-|x|+a-a|x|=0 \\
\Longleftrightarrow & |x|^{2}-|x|+a-a|x|=0 \\
\Longleftrightarrow & |x|(|x|-1)-a(|x|-1)=0 \\
\Longleftrightarrow & (|x|-1)(|x|-a)=0
\end{aligned}
$$

We have two possibilities: $|x|=1$ or $|x|=a$. Both solutions are possible since $a>0$.

In the first case, $x=1$ or $x=-1$, and in the second case, $x=a$ or $x=-a$.

For $x=1, y=-1+a$, and for $x=-1, y=-1+a$.

For $x=a, y=0$ and for $x=-a, y=0$.

Then the vertices of the given figure are the following points:

$$
(-a, 0), \quad(a, 0), \quad(1, a-1), \quad(-1, a-1)
$$

These points in the coordinate system determine an isosceles trapezoid with the y-axis as its axis of symmetry, so its area is

$$
P=\frac{2 a+2}{2} \cdot(a-1)=(a+1)(a-1)=a^{2}-1
$$

From $120=a^{2}-1$, we get $a^{2}=121 \Rightarrow a=11(a>1)$.

Note: The problem can also be solved by distinguishing the cases $x \leqslant 0$ and $x>0$. In the first case, we get the equation $(x+1)(x+a)=0$, and in the second case, $(x-1)(x-a)=0$, whose solutions for $x$ are the numbers $-1,-a, 1, a$. We proceed as in the previous solution.

## Problem B-1.2

Solve the inequality

$\frac{x-8}{2012}+\frac{x-7}{2013}+\frac{x-6}{2014}+\frac{x-5}{2015}+\frac{x-4}{2016} \leqslant \frac{x-2012}{8}+\frac{x-2013}{7}+\frac{x-2014}{6}+\frac{x-2015}{5}+\frac{x-2016}{4}$.

## Solution.

$\frac{x-8}{2012}+\frac{x-7}{2013}+\frac{x-6}{2014}+\frac{x-5}{2015}+\frac{x-4}{2016} \leqslant \frac{x-2012}{8}+\frac{x-2013}{7}+\frac{x-2014}{6}+\frac{x-2015}{5}+\frac{x-2016}{4}$.

Add -1 five times to both sides

$$
\begin{aligned}
& \frac{x-8}{2012}-1+\frac{x-7}{2013}-1+\frac{x-6}{2014}-1+\frac{x-5}{2015}-1+\frac{x-4}{2016}-1 \leqslant \\
& \leqslant \frac{x-2012}{8}-1+\frac{x-2013}{7}-1+\frac{x-2014}{6}-1+\frac{x-2015}{5}-1+\frac{x-2016}{4}-1 \\
& \frac{x-2020}{2012}+\frac{x-2020}{2013}+\frac{x-2020}{2014}+\frac{x-2020}{2015}+\frac{x-2020}{2016} \leqslant \\
& \leqslant \frac{x-2020}{8}+\frac{x-2020}{7}+\frac{x-2020}{6}+\frac{x-2020}{5}+\frac{x-2020}{4} \\
& \quad(x-2020)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}-\frac{1}{5}-\frac{1}{4}\right) \leqslant 0
\end{aligned}
$$

The expression in the second parenthesis is negative because each of the first 5 fractions is smaller than each of the other 5 fractions, making their difference negative. Therefore, the entire expression is less than or equal to zero only if $x-2020 \geqslant 0, x \geqslant 2020, x \in[2020,+\infty\rangle$.