Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 :


share|cite|improve this answer
(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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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