
Problem 8.1. Let $F$ be the set of points with coordinates $(x, y)$ such that ||$x|-| y||+|x|+|y|=2$.

(a) Draw $F$.

(b) Find the number of points in $F$ such that $2 y=|2 x-1|-3$.


Solution: (a) If $|x| \geq|y|$, then ||$x|-| y||+|x|+|y|=|x|-|y|+$ $|x|+|y|=2|x|=2$, thus $|x|=1$ and therefore $1 \geq|y|$, so $-1 \leq y \leq 1$. We conclude that the segments $-1 \leq y \leq 1$ on the lines $x=1$ and $x=-1$ belong to $F$.

If $|x| \leq|y|$, then ||$x|-| y||+|x|+|y|=-|x|+|y|+|x|+|y|=$ $2|y|=2$, thus $|y|=1$ and therefore $1 \geq|x|$, so $-1 \leq x \leq 1$. We conclude that the segments $-1 \leq x \leq 1$ on the lines $y=1$ and $y=-1$ also belong to $F$.

Thus we have determined that $F$ consists of the sides of a square with vertices $A(-1,-1), B(1,-1), C(1,1), D(-1,1)$.

![](https://cdn.mathpix.com/cropped/2024_06_03_e679746950d7d3832835g-213.jpg?height=366&width=458&top_left_y=126&top_left_x=779)

(b) We find the number of solutions of $2 y=|2 x-1|-3$ on each of the segments $A B, B C, C D, D A$.

The segment $C D$ consists of all points $(x, y)$ such that $-1 \leq x \leq$ 1, $y=1$. The equation $2=|2 x-1|-3$ has no solution $x$ when $-1 \leq x \leq 1$. Therefore $2 y=|2 x-1|-3$ has no solution on $C D$.

The segment $A B$ consists of all points $(x, y)$ such that $-1 \leq x \leq$ $1, y=-1$. The equation $-2=|2 x-1|-3$ has two solutions: $x=0$ and $x=1$. Therefore $2 y=|2 x-1|-3$ has two solutions on $A B$.

As above we get that $2 y=|2 x-1|-3$ has a unique solution on $A D:(x, y)=(-1,0)$ and a unique solution on $B C:(x, y)=$ $(1,-1)$. Note that the last one has already been obtained as a point on $A B$. Thus there are three solutions of $2 y=|2 x-1|-3$ in $F$ : $(x, y)=(-1,0),(0,-1),(1,-1)$.
