# Area enclosed by the curve $\lfloor |x''| \rfloor +\lfloor |y''| \rfloor = 2$

The area enclosed by the curve $$\bigg\lfloor \frac{|x-1|}{|y-1|}\bigg\rfloor +\bigg\lfloor \frac{|y-1|}{|x-1|}\bigg\rfloor = 2\;,$$ Where $-2 \leq x,y\leq 0$

$\bf{My\; Try::}$ Let $x-1=x'$ and $y-1 = y'\;,$ Then $$\bigg\lfloor \frac{|x'|}{|y'|}\bigg\rfloor+\bigg\lfloor \frac{|y'|}{|x'|}\bigg\rfloor = 2\Rightarrow \bigg\lfloor \left|\frac{x'}{y'}\right|\bigg\rfloor+\bigg\lfloor \left|\frac{y'}{x'}\right|\bigg\rfloor=2$$

So here $-3\leq x',y'\leq -1.$ Now Put $\displaystyle \frac{x'}{y'}=x''$ and $\displaystyle \frac{y'}{x'} = y''$ and Here $\displaystyle \frac{1}{3}\leq x'',y''\leq 3$

So we get $$\lfloor |x''| \rfloor +\lfloor |y''| \rfloor = 2$$

Is my Process is Right , If not then how can I calculate it, Help me

Thanks.

• You have made a mistake in typing – Archis Welankar Apr 29 '16 at 6:01
• Note that if we let $x^\prime=1$ and $y^\prime=-1$ then $x^{\prime\prime}=-1$ so the statement $\dfrac{1}{3}\le x^{\prime\prime}$ cannot be correct. – John Wayland Bales Apr 29 '16 at 6:25
• You may be able to use the observation that $\left\lfloor u\right\rfloor + \left\lfloor \frac{1}{u}\right\rfloor=2$ implies that either $\frac{1}{3}<u\le\frac{1}{2}$ or $2\le u<3$. – John Wayland Bales Apr 29 '16 at 6:38
• @juantheron : Does $-2\le x,y\le 0$ mean that $-2\le x\le 0$ and $-2\le y\le 0$? – mathlove Apr 29 '16 at 8:20
• Yes Mathlove....,, – juantheron Apr 29 '16 at 8:52

Is my Process is Right

I think it is right.

We can separate it into cases as the following : $$\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor+\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor=2$$

$$\iff \left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(0,2),(1,1),(2,0)$$ since both $|(x-1)/(y-1)|$ and $|(y-1)/(x-1)|$ are positive.

Case 1 : \begin{align}&\left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(0,2)\\&\iff 0\le\left|\frac{x-1}{y-1}\right|\lt 1\quad\text{and}\quad 2\le\left|\frac{y-1}{x-1}\right|\lt 3\\&\iff 0\le\left|\frac{x-1}{y-1}\right|\lt 1\quad\text{and}\quad \frac 13\lt\left|\frac{x-1}{y-1}\right|\le \frac 12\\&\iff \frac 13\lt\left|\frac{x-1}{y-1}\right|\le \frac 12\\&\iff \frac 13\lt\frac{x-1}{y-1}\le \frac 12\\&\iff \frac 13(y-1)\gt x-1\ge \frac 12(y-1)\\&\iff 3x-2\lt y\le 2x-1\end{align}

Case 2 : \begin{align}&\left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(1,1)\\&\iff 1\le\left|\frac{x-1}{y-1}\right|\lt 2\quad\text{and}\quad 1\le\left|\frac{y-1}{x-1}\right|\lt 2\\&\iff 1\le\left|\frac{x-1}{y-1}\right|\lt 2\quad\text{and}\quad \frac 12\lt\left|\frac{x-1}{y-1}\right|\le 1\\&\iff \left|\frac{x-1}{y-1}\right|=1\\&\iff \frac{x-1}{y-1}=1\\&\iff y=x\end{align}

Case 3 : By symmetry about $y=x$, \begin{align}&\left(\left\lfloor\left|\frac{x-1}{y-1}\right|\right\rfloor,\left\lfloor\left|\frac{y-1}{x-1}\right|\right\rfloor\right)=(2,0)\\&\iff 3y-2\lt x\le 2y-1\\&\iff \frac 12x+\frac 12\le y\lt \frac 13x+\frac 23\end{align}

Therefore, we want to find the area of the following two triangles in red.

$\qquad\qquad\qquad$

Hence, from $A(-2,0),B(-2,-1/2),C(-1,0)$, the answer is $$2\times [\triangle{ABC}]=2\times\frac 12\times (-1-(-2))\times (0-(-1/2))=\color{red}{\frac 12}.$$