I have the following equation:$$\bigg\lfloor \sqrt {c^2(x^2+y^2)+2c(x+y)+2} +\frac{1}{2} \bigg\rfloor = c$$For a given positive integer $c$, I am trying to define algebraically, in terms of $x$ & $y,$ the region in 2D space for which the above equation and the following conditions hold:
- $c\in \mathbb N$
- $0\leq x, y \leq 1$
I.E, which constraints/inequalities $x$ & $y$ must fulfill (in terms of $c$)
This is what I have done so far:
Getting rid of the floor yields $2$ quadratic inequalities:
$$(1)\quad \sqrt {c^2(x^2+y^2)+2c(x+y)+2} +\frac{1}{2}\geq c$$
$$(2)\quad \sqrt {c^2(x^2+y^2)+2c(x+y)+2} +\frac{1}{2}\lt c + 1$$
It is possible to substitute $u = x^2+y^2, v=x+y$ (and adjust constraints accordingly), and after some algerbra arrive at the $2$ linear inequalities (assuming I haven't made any silly algebraic mistakes):
$$v\geq \frac {4c^2-4c-4c^2u-7}{8c}=-\frac c2u+\frac c2-\frac 7{8c}-\frac 1 2$$
$$v\lt \frac {4c^2+4c-4c^2u-7}{8c}=-\frac c2u+\frac c2-\frac 7{8c}+\frac 1 2$$
So I have $2$ lines with equal negative slope $=-\frac c2,$ and $y$-interception differing by $1$ unit. Fulfilling just these $2$ inequalities yields a region of interest bounded between the lines. Adding the constraints $0\leq u,v \leq 2,$ which stem from $0\leq x,y \leq 1$ results in a yet more bounded region. Sketching this allows me to define the desired region (in 2D space of $u$ & $v$) graphically:
The picture was created in powerpoint so it is not 100% precise but the idea is clear. It is important to note:
- The desired region is colored in RED
- The square has dimensions $2\times 2$
- The axis represent $u$ & $v,$ rather than the originally desired $x$ & $y$
- The line with larger $y$-interception itself is NOT part of the region (if my logic is correct)
- The line with smaller $y$-interception itself IS part of the region (again, if my logic is correct)
- Different values of $c$ result in different polygons of the desired region, which stem from alternating boundaries of $u$ and $v$ – either the polygon is bounded by $0$ & $2,$ or by the inequalities – depending on which is maximal/minimal. For example, for some values of $c,$ a trapezoid would symbolize the region of interest.
I was hoping I could keep working my way from here to algebraic constraints regarding $x$ & $y$ as is originally intended. But this leads to a system of many inequalities and a big algebraic mess. Perhaps this is possible to do, but I am not sure how to do this. And perhaps a different approach to the original equation yields a more appealing, simpler, solution.
I also tried feeding this into Mathematica, but I think I did it wrong, and it produced constraints for $y$ which depend on $x,$ not what I am looking for. I want to find constraints on $x$ and $y$ only in terms of $c.$ I can assume WLOG $x\geq y$ because the original equation is symmetric, but again I am not sure where this leads me.
Any guidance and algebraic help / the solution of $x$ and $y$ is appreciated. Would also like to hear if anybody spots any algebraic mistakes and logic faults!
Best Answer
As it is pointed out in OP, the inequalities $(1),(2)$ can be written in one as $$ c \leq \sqrt {c^2(x^2+y^2)+2c(x+y)+2} +\frac{1}{2} < c+1$$ or $$ c-\frac{1}{2} \leq \sqrt {c^2(x^2+y^2)+2c(x+y)+2} < c+\frac{1}{2}$$ Since $c$ is a positive integer, the last double inequality is equivalent to $$ \left(c-\frac{1}{2}\right)^2 \leq {(cx+1)^2+(cy+1)^2} < \left(c+\frac{1}{2}\right)^2$$ and further to $$ \left(1-\frac{1}{2c}\right)^2 \leq {\left(x+\frac{1}{c}\right)^2+\left(y+\frac{1}{c}\right)^2} < \left(1+\frac{1}{2c}\right)^2$$ Therefore, the domain is intersection of the SQUARE defined by $0\leq x,y\leq 1$ and the RING centered at a point $\left(-\frac{1}{c},-\frac{1}{c}\right),$ with inner radius $r=1-\frac{1}{2c}$ and outer radius $R=1+\frac{1}{2c}.$
The convenient points on the small bounding circle of the ring belong to the domain, but those on the larger do not.