Points closest to the edge of a square Let $S$ be the square formed by the four vertices $(1,1),(1,-1),(-1,1), $and $(-1,-1).$ Let the region $R$ be the set of points inside $S$ which are closer to the centre than to any of the four sides. Find the area of region $R.$
I figure that the area is the area between 4 parabolas, which I can't seem to find.
Some help needed here! (Full proof will be appreciated but hints don't hurt :))
 A: For the point to be closer to the centre $C(0,0)$ of the square than any side,$\sqrt{x^2+y^2}<\min\{|x-1|,x+1,|y-1|,y+1\}$. 
In the portion of the square in the $1^{st}$ quadrant, 
$|x-1|=1-x<x+1, |y-1|=1-y<y+1\\\therefore\sqrt{x^2+y^2}<\min\{1-x,1-y\}$
When $x\ge y, \sqrt{x^2+y^2}<1-x$ and when $x<y, \sqrt{x^2+y^2}<1-y$
Simplify these to obtain $y^2<1-2x, x\ge y$ and $x^2<1-2y, y>x$ as the desired regions in the first quadrant. This area looks something like this:

$y^2=1-2x, x^2=1-2y$ intersect at $(\sqrt2-1,\sqrt2-1)$ in the $1^{st}$ quadrant. The area of this region is given by $A=\int_{0}^{\sqrt2-1} \frac{1-x^2}2dx+\int_{0}^{\sqrt2-1} \frac{1-y^2}2-(\sqrt2-1)dy$
By symmetry, the total required region looks like this:

The area of this region is just $4A$.
A: Due to symmetry, it is enough to calculate the case $x\in[0,1]$ and $y\in[0,x]$ and multiply the answer by $8$. So let's calculate the parabola first: on the parabola you are at the same distance from $(0,0)$ as from the side $x=1$. Then $$x^2+y^2=(1-x)^2$$You can rewrite this as $$y^2=1-2x$$
Now calculate the intersection with the $x$ axis and with $x=y$. For the $x$ axis you get $y=0$ and $x=1/2$. For the $x=y$ you get $x=\pm\sqrt 2 -1$, and only the positive solution is of interest in this case. Note that it is less then $1/2$. So $$A_{1/8}=\int_0^{\sqrt 2 -1} xdx+\int_{\sqrt 2 -1}^{1/2}\sqrt{1-2x} dx$$ 
