The area problem! We have to find area of the quadrilateral formed by joining the point of intersection of the four quarter circles that are drawn from each vertex in a unit square.
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The challenge is only to use planer geometry (not even coordinate or calculus), I was wondering how could we do this?
PS: This is actually an extension of this problem.
 A: Let's denote distance from intersection point to the nearest side of square as : $x$ , then :
$x=1-\frac{\sqrt{3}}{2}$
Note that small quadrilateral is square by symmetry . 
Next , lets denote side of the small square as $~b~$ and diagonal of this square as $~d~$ .
So :
$d=1-2x =(\sqrt{3}-1)~$ , hence :
$b= \frac{d}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}~$ , hence :
$A=\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)^2$
A: The ends of one side of the quadrilateral ($A$ and $B$) are on the quarter circle whose center $M$ is a vertex of the square. Let us draw two lines joining $M$ to $A$ and $M$ to $B$. Thus we have formed a triangle $MAB$ whose angle $AMB$ is $30$ degrees. If the midpoint of $AB$ is denoted as $K$ and a line is drawn from $M$ to $K$, then the angle $KMB$ is $15$ degrees and the tangent of $15$ degree is equal to $KB / MB$. Using the value of $\tan 15$ enables us to calculate the length of $KB$; and doubling this length gives us the length of the $AB$ side of the quadrilateral (which, in fact, is actually a square. $AB$ squared is the area of the quadrilateral (SQUARE).
The area beteen $AB$ and the curve of the quarter circle is a $30$-degree segment of the circle.
The area of this circle's segment is equal to: 
$$
\pi\cdot (MA)^2\cdot \frac{15}{360}-\frac{1}{2}\cdot (AB)^2\cdot \sin 15.
$$
When we add the area of the quadrilateral (the square) to $4$ times the area of the calculated circle's segment, the addition sum is the answer we are seeking.      
