2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You have made a mistake in typing $\endgroup$ – Archis Welankar Apr 29 '16 at 6:01
  • $\begingroup$ 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. $\endgroup$ – John Wayland Bales Apr 29 '16 at 6:25
  • $\begingroup$ 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 $2\le u<3$. $\endgroup$ – John Wayland Bales Apr 29 '16 at 6:38
  • $\begingroup$ @juantheron : Does $-2\le x,y\le 0$ mean that $-2\le x\le 0$ and $-2\le y\le 0$? $\endgroup$ – mathlove Apr 29 '16 at 8:20
  • $\begingroup$ Yes Mathlove....,, $\endgroup$ – juantheron Apr 29 '16 at 8:52
2
$\begingroup$

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.

$\qquad\qquad\qquad$enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.