How to find the locus of $d(x,y)=max\{|x|,|y|\}$ when $d(x,y)=a$?

Question

If the distance of any point $(x,y)$ from the origin is defined as $d(x,y)=max\{|x|,|y|\}$ where $|.|$ represents the absolute value function, and $d(x,y)=a$ whre $a$ is a non-zero constant, then the locus of the resulting curve is:

(A) Circle

(B) Straight Line

(C) Square

(D) Triangle

I know the meaning of $max\{b,c\}$ which gives the maximum of $b$ or $c$ as its output. How to deal with the case $max\{|x|,|y|\}$? Since $d(x,y)=a$, I concluded the locus must be a circle, but the answer is a square. How can this be possible?

Answer 1

This problem is not as difficult as it looks!

Let us consider the graph of $|y|=|x|$ which is equivalent to $y=\pm x$ :

It can be seen that the lines $y=x$ and $y=-x$ divides the two-dimensional cartesian plane into four different regions. Any point on these lines are either of the form $(p,p)$ or $(p,-p)$ where $p$ is any real number. The points inside the four regions satisfy any of the two inequalities : $|y|<|x|$ or $|y|>|x|$

The following graph shows the regions $|y|<|x|$ and $|y|>|x|$ in blue and red respectively:

If we focus our attention on points in the four regions demarcated by $|y|=|x|$, two cases arise:

Case-I: When $|y|<|x|$ (Blue coloured region)

Here, $d(x,y)=max\{|x|,|y|\}=|x|=a$

So, $x=\pm a$

Case-II: When $|y|>|x|$ (Red coloured region)

Here, $d(x,y)=max\{|x|,|y|\}=|y|=a$

So, $y=\pm a$

From the above two cases, we get the following graph (assuming $a=4$, of course it can be any non-zero constant):

Click Here to open the graph in Desmos.

Clearly, the locus is a square.

Answer 2

Given $d(x,y)=\max\{|x|,|y|\}=a$. Let us assume $a>0$. Then

Case 1:If $(|x| \geq |y|)$ Then $|x|=a$ and $|y| \leq a$. This means we have $x=\pm a$ and $-a \leq y \leq a$. Geometrically these are (vertical) line segments at $x=a$ and $x=-a$ such that $y$ ranges in $[-a,a]$.

Case 2:If $(|x| \leq |y|)$ Then $|y|=a$ and $|x| \leq a$. This means we have $y=\pm a$ and $-a \leq x \leq a$. Geometrically these are (horizontal) line segments at $y=a$ and $y=-a$ such that $x$ ranges in $[-a,a]$.

Combining the two cases, we get a square with side length $2a$ and sides as described above.

Answer 3

Take a trip in the first quadrant:

0) $y=x$.

1) All points $(x,y)$ with $0 \le y \le x$ are on or below the line $y=x$ and on or above the positive $x-$axis.

2) Start at $(a,0)$ and move up along $x= a$ until you reach $(a,a)$: $d_1(x,y)=a$;

3) All points $(x,y)$ with $x< y \le a$ are above the line $y=x$ and on or below $y=a$.

4)Start from $(a,a)$ moving along $y=a$ towards the $y-$ axis until you reach $(0,a):$ $d_2(x,y)=a$.

$d(x,y)= a$ moving along sides of a square centred at $(0,0)$ , side length $2a$, in the first quadrant.

The other quadrants are dealt with similarly.