# Problem 4

Find the area of the figure defined by the inequality

$$
|x|+|y|+|x-y| \leq \mathbf{2}
$$
Answer: The area of the figure is 3 sq.units.

## Solution.

It is easy to see that if the point $(\boldsymbol{x}, \boldsymbol{y})$ satisfies the original inequality $|\boldsymbol{x}|+|\boldsymbol{y}|+|\boldsymbol{x} \boldsymbol{y}| \leq 2$, then the point $(-x,-\boldsymbol{y})$ will also satisfy this inequality, since $|-\boldsymbol{x}|+|-y|+|-\boldsymbol{x}+\boldsymbol{y}| \leq \mathbf{2}$ and $|\boldsymbol{x}|+|\boldsymbol{y}|+|-(x-y)| \leq \mathbf{2}$, or $|\boldsymbol{x}|+|\boldsymbol{y}|+|\boldsymbol{x}-\boldsymbol{y}| \leq \mathbf{2}$. Geometrically, this will mean the symmetry of the figure with respect to the origin.

Due to this, we will construct the figure in the 1st and 2nd quadrants of the coordinate plane and perform a central symmetry transformation.

We will open the modulus for points in the 1st quadrant. In this case, there are two possibilities: $\left\{\begin{array}{c}x \geq 0 \\ y \geq 0 \\ x-y \geq 0 \\ x+y+x-y \leq 2\end{array}\right.$ and $\left\{\begin{array}{c}x \geq 0 \\ y \geq 0 \\ x-y \leq 0 \\ x+y-x+y \leq 2\end{array}\right.$. In the first case, we get $x \leq 1$ and $y \leq x$.

Therefore, we have a triangle lying below the line $y=x$.

In the second case, $y \leq 1$ and $y \geq x$. Therefore, we have a triangle lying above the line $\mathrm{y}=\mathrm{x}$. Combining the figures, we get a square with sides lying on the coordinate axes and equal to 1. For points in the second quadrant, there is only one situation:

![](https://cdn.mathpix.com/cropped/2024_05_06_bea8bd31fc71aedf7f6dg-3.jpg?height=265&width=345&top_left_y=1809&top_left_x=296)

That is, $x-y \leq 1, y \geq x-1$. We get a triangle formed by the coordinate axes and the line $y=x-1$. As a result, we get the figure: The area of this figure is 3 (sq.units) as the area of two squares with side 1 and two right isosceles triangles with legs 1.