Area enclosed by the curve $\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$

Question

What is the area enclosed by the curve $$\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$$ $$x\ge y, \forall x, y \ge 0$$

$\lfloor x\rfloor$ stands for the Greatest Integer Function.

I think that the curve will start from $\left(\cfrac{5}{2},\cfrac{5}{2}\right)$ and then be a straight line parallel to the $y$ axis till $\left(\cfrac{7}{2},\cfrac{5}{2}\right)$ (not including this last point). But then I am finding it difficult to manipulate it further.

Answer 1

The given equation is: $$ \lfloor x+y \rfloor + \lfloor x-y \rfloor = 5 $$ Let $t_1 = x+y \text{ and } t_2 = x-y$. We can begin by observing that LHS is the sum of two non-negative integers. So the ordered pair $(t_1, t_2)$ can possibly be $(5,0),\;(4,1),\;(3,2)$ and so on. Consider $(t_1, t_2)=(5,0)$. For this pair, we can write the inequalities: $$ 5\le x+y < 6 \\ 0 \le x-y < 1$$ The intersection of these inequalities gives a part of the total graph. Similarly we may write all inequalities for all pairs, and draw the graph.

Now it is easy to observe the pattern and use the inequalities to find the complete graph. With non-negativity restrictions on $x$ and $y$, the graph of the solution is the three diamond (tilted square) regions whose centres lie on $x=3$.

Thus the area of the graph is $3/2 \text{ sq. units}$.

^{Image credit: Desmos Graphing Calculator}

Answer 2

Blue tiles: $$ \lfloor{x + y}\rfloor + \lfloor{x - y}\rfloor = 5 $$ Gray triangle $$ y > 0, \quad x \le y $$ Find the area of the blue tiles within the gray triangle.

The black dots are the vertices of the first square:

$$ \left( \begin{array}{c} 3 \\ 0 \end{array} \right), \quad % \left( \begin{array}{c} 3.5 \\ 0.5 \end{array} \right), \quad % \left( \begin{array}{c} 3 \\ 1 \end{array} \right), \quad % \left( \begin{array}{c} 2.5 \\ 0.5 \end{array} \right) % $$ The side length for each square is $s =\frac{1}{\sqrt{2}}$. Therefore each square has area $A=s^{2} = \frac{1}{2}.$ There are three squares within the second region. Therefore the total area is $$A_{total} = 3 A = \frac{3}{2}$$