C1. Find the areas of the three polygons described by the following conditions (a), (b) and (c), respectively.
(a) the system of inequalities $|x| \leq 1$ and $|y| \leq 1$
(b) the inequality $|x|+|y| \leq 10$
(c) the inequality $|x|+|y|+|x+y| \leq 2020$
Solution for $\mathrm{C1}$ (a):
Here we have all points $(x, y)$ such that $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$.
This is a square with vertices $(1,1),(1,-1),(-1,1),(-1,-1)$. The area is 4 .
Solution for $\mathbf{C 1}$ (b) Consider the first quadrant $x \geq 0, y \geq 0$. Then $|x|=x,|y|=y$ and we have $x+y \leq 10$. So, we have all points $(x, y)$ inside the triangle with vertices $(0,0),(10,0),(0,10)$.
Consider the second quadrant $x \leq 0, y \geq 0$. Then $|x|=-x,|y|=y$ and we have $-x+y \leq 10$. So, we have all points $(x, y)$ inside the triangle with vertices $(0,0),(-10,0),(0,10)$.
Consider the third quadrant $x \leq 0, y \leq 0$. Then $|x|=-x,|y|=-y$ and we have $-x-y \leq 10$, equivalently, $x+y \geq-10$. So, we have all points $(x, y)$ inside the triangle with vertices $(0,0),(-10,0),(0,-10)$.
Consider the fourth quadrant $x \leq 0, y \leq 0$. Then $|x|=x,|y|=-y$ and we have $x-y \leq 10$. So, we have all points $(x, y)$ inside the triangle with vertices $(0,0),(10,0),(0,-10)$.

The area of each of the four right triangles is $(10 \times 10) / 2=50$. They form the square with vertices $( \pm 10,0),(0, \pm 10)$. The total area of the square is 200 .

C1(b) Alternative solution:
Consider the first quadrant $x \geq 0, y \geq 0$. Then $|x|=x,|y|=y$ and we have $x+y \leq 10$. So, we have all points $(x, y)$ inside the triangle with vertices $(0,0),(10,0),(0,10)$. By symmetry, the entire figure is the rotated square centered at the origin. Its side length is $\|(10,0)-(0,10)\|=\sqrt{\left(10^{2}+10^{2}\right)}=\sqrt{200}$. Thus, the square has area $\sqrt{200}^{2}=200$.

Solution for $\mathbf{C 1}(\mathrm{c})$ :
Consider first the region bounded by $|x|+|x+y|+|y|=2$.
We work out the cases:
1) If $x \geq 0, y \geq 0$ then $x+x+y+y \leq 2$ so $x+y \leq 1$. This gives a triangle with vertices $(0,0),(0,1),(1,0)$. The area is 0.5 .
2) If $x \leq 0, y \geq 0$ and $x \geq-y$ then $-x+x+y+y \leq 2$ so $y \leq 1$. This gives a triangle with vertices $(0,0),(0,1),(-1,1)$. The area is 0.5 .
3) If $x \leq 0, y \geq 0$ and $x \leq-y$ then $-x-x-y+y \leq 2$ so $x \geq 1$. This gives a triangle with vertices $(0,0),(-1,0),(-1,1)$. The area is 0.5 .
4) If $x \leq 0, y \leq 0$ then $-x-x-y-y \leq 2$ so $x+y \geq-1$ This gives a triangle with vertices $(0,0),(0,-1),(-1,0)$. The area is 0.5 .
5) If $x \geq 0, y \leq 0$ and $x \geq-y$ then $x+x+y-y \leq 2$ so $x \leq 1$. This gives a triangle with vertices $(0,0),(1,0),(1,-1)$. The area is 0.5 .
6) If $x \geq 0, y \leq 0$ and $x \leq-y$ then $x-x-y-y \leq 2$ so $y \geq-1$. This gives a triangle with vertices $(0,0),(0,-1),(1,-1)$. The area is 0.5 .
Consequently, the six triangles form a hexagon with vertices $( \pm 1,0),(0, \pm 1)$ and $(-1,1),(1,-1)$, which has area 3. Scaling by a linear factor of 1010 gives an area of $3(1010)^{2}=3060300$.

Alternatively, without re-scaling, area of each triangle will be $1010^{2} / 2=505 \times 1010=510050$. The six triangles give $510050 \times 6=3060300$.

Alternative solution C1(c):
This area is fully contained within the square $\left\{\begin{array}{l}|x| \leq 1010 \\ |y| \leq 1010\end{array}\right.$.
What parts of the square $(|x| \leq 1010 \&|y| \leq 1010)$ are in $|x|+|y|+|x+y| \leq 2020$ ?
The complement is $|x| \leq 1010,|y| \leq 1010$ and $|x+y|>1010$. That is, we subtract 2 triangles of area $\frac{1}{2}(1010)^{2}$ (a total of $1010^{2}$ ) from a big square of area $2020^{2}$. This is, $1010^{2}(4-1)=1020100 \times 3=3060300$.