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The area of regin traced by the point in the cartesian plane which satisfy the equation $\sin^2(\pi x)+\sin^2(\pi y)>1$ , where $x,y\in[-1,1]$

My Try:

We can write

$\sin^2(\pi x)+\sin^2(\pi y)>1\Longrightarrow 2\sin^2(\pi x)+2\sin^2(\pi y)>2$

$\displaystyle 1-\cos(2\pi x)+1-\cos(2\pi y)>2$

$\displaystyle \cos(2\pi y)<-\cos(2\pi x)=\cos(\pi-2\pi x)$

$\cos(2\pi y)<\cos(\pi-2\pi x)\Longrightarrow 2\pi y<2n\pi\pm (\pi-2\pi x)$

$\displaystyle \Longrightarrow 2y\pm 2x<(2n\pm 1)\cdots \cdots (1)$

Here $x,y\in[-1,1]\Longrightarrow 2x,2y\in[-2,2]$

$\displaystyle 2y\pm 2x\in[-4,4]$

$\displaystyle 2y\pm 2x<-3,-1,1,3$

Added:

enter image description here

How do i find the bounded regin enclosed by that graphs

Help required, Thanks

jacky
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  • You're almost done. If you draw your lines, and the square $[-1,1]^2$, and shade appropriately, you'll see the region is four squares. You can find the area of four squares! – anon Mar 31 '23 at 03:11
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    This is a very similar question to the 25th problem on the 2020 AMC 12B. – Bumblebee Mar 31 '23 at 03:33
  • @runway44 I have added graph and from graph we have seen that one square with vertices $(\pm \frac{1}{2},0)$ and $(0,\pm \frac{1}{2})$ . So we get area $1$. Is I am right – jacky Mar 31 '23 at 03:34
  • https://math.stackexchange.com/questions/3825756/finding-the-area-under-the-inequality-sin2-pi-x-sin2-pi-y-le-1-for-x?rq=1 – jacky May 14 '23 at 15:25

1 Answers1

3

In a comment, you said

I have added graph and from graph we have seen that one square with vertices $(±\frac 12,0)$ and $(0,±\frac 12)$.

I think this is not correct since $(x,y)=(0,0)$ (which is inside the square) does not satisfy $\sin^2(\pi x)+\sin^2(\pi y)>1$.

After getting $$\sin^2(\pi x)+\sin^2(\pi y)>1\iff \cos(2\pi y)<\cos(\pi-2\pi x)$$ one can do as follows : $$\begin{align}&\cos(\pi-2\pi x)-\cos(2\pi y)\gt 0 \\\\&\iff -2\sin\bigg(\frac{\pi-2\pi x+2\pi y}{2}\bigg)\sin\bigg(\frac{\pi-2\pi x-2\pi y}{2}\bigg)\gt 0 \\\\&\iff \sin\bigg(\frac{\pi-2\pi x+2\pi y}{2}\bigg)\sin\bigg(\frac{\pi-2\pi x-2\pi y}{2}\bigg)\lt 0 \\\\&\iff \sin\bigg(\frac{\pi-2\pi x-2\pi y}{2}\bigg)\lt 0\lt\sin\bigg(\frac{\pi-2\pi x+2\pi y}{2}\bigg) \\&\qquad\qquad \text{or}\ \sin\bigg(\frac{\pi-2\pi x+2\pi y}{2}\bigg)\lt 0\lt\sin\bigg(\frac{\pi-2\pi x-2\pi y}{2}\bigg)\end{align}$$

$$\small\iff \bigg(2k_1\pi\lt \frac{\pi-2\pi x+2\pi y}{2}\lt (2k_1+1)\pi\ \text{and}\ (2k_2+1)\pi\lt \frac{\pi-2\pi x-2\pi y}{2}\lt (2k_2+2)\pi\bigg)$$ $$\small\text{or}\ \bigg((2k_3+1)\pi\lt \frac{\pi-2\pi x+2\pi y}{2}\lt (2k_3+2)\pi\ \text{and}\ 2k_4\pi\lt \frac{\pi-2\pi x-2\pi y}{2}\lt (2k_4+1)\pi\bigg)$$

$$\small\iff \bigg(x+2k_1-\frac 12\lt y\lt x+2k_1+\frac 12\ \text{and}\ -x-2k_2-\frac 32\lt y\lt -x-2k_2-\frac 12\bigg)$$ $$\small\text{or}\ \bigg(x+2k_3+\frac 12\lt y\lt x+2k_3+\frac 32\ \text{and}\ -x-2k_4-\frac 12\lt y\lt -x-2k_4+\frac 12\bigg)$$ where $k_1,k_2,k_3,k_4\in\mathbb Z$.

enter image description here

With $x,y\in[-1,1]$, we see that the region is four squares.

  • a square whose vertices are $(1/2,0),(1,1/2),(1/2,1),(0,1/2)$

  • a square whose vertices are $(0,1/2),(-1/2,1),(-1,1/2),(-1/2,0)$

  • a square whose vertices are $(-1/2,0),(-1,-1/2),(-1/2,-1),(0,-1/2)$

  • a square whose vertices are $(0,-1/2),(1/2,0),(1,-1/2),(1/2,-1)$

Therefore, the area of the region is $4\times (\frac{\sqrt 2}{2})^2=\color{red}2$.

mathlove
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