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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.

$\hspace{4cm}$ enter image description here

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.

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2 Answers

up vote 2 down vote accepted

Let's denote distance from intersection point to the nearest side of square as : $x$ , then :


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 :


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(a) OP says it's a unit square. (b) You should point out the gray quadrilateral is a square, which is deduced from symmetry. –  anon Feb 19 '12 at 8:54
@anon,thanks..fixed –  pedja Feb 19 '12 at 9:01
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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.

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