Let the centers of four circles with the radius $R=a$ be on 4 vertexs a square with edge size $a$. How calculate the shaded area in this picture?

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Let $(APD)$ be the area of the figure $APD$.

And let $x,y,z$ be $(KEPM),(PAD),(MPD)$ respectively.

First, we have $$(\text{square}\ ABCD)=a^2=x+4y+4z.\tag1$$

Second, we have $$(\text{sector}\ BDA)=\frac{\pi a^2}{4}=x+2y+3z.\tag2$$

Third, note that $KA=KD=a$ and that $(\triangle KAD)=\frac{\sqrt 3}{4}a^2$ since $\triangle KAD$ is a equilateral triangle.

So, since we have $$\begin{align}(K(E)AD(M))&=(\text{sector}\ AKD)+(\text{sector}\ DKA)-(\triangle KAD)\\&=\frac{\pi}{6}a^2+\frac{\pi}{6}a^2-\frac{\sqrt 3}{4}a^2\\&=\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2,\end{align}$$ we have $$\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2=x+y+2z.\tag3$$

Solving $(1),(2),(3)$ gives us $$(KEPM)=x=\left(1+\frac{\pi}{3}-\sqrt 3\right)a^2.$$


Here is a geometric solution.

Let's find the angle $EDK$. Since $KA=KD=a=AD$ hence the triangle $AKD$ is equiliteral and the angle $KDA$ is $\pi/3$. Therefore angle $KDC$ is $\pi/6$. In the same way the angle $EDA$ is also $\pi/6$ and we get that the angle $EDK$ is $\pi/6$.

Now the shaded area $$(EPMK) = ({\rm square\,} EPMK) + 4\times ({\rm segment\,}EKE).$$

The edge of the square $EPMK$ (from the triangle $EDK$) is $$EK=2a\sin\frac{\pi}{12}=a\frac{\sqrt{3}-1}{\sqrt{2}},$$ hence the area $$({\rm square\,} EPMK) =(2-\sqrt{3})a^2.$$

And the area $$({\rm segment\,}EKE) = \frac{1}{2}\left(\frac{\pi}{6}-\sin\frac{\pi}{6}\right)a^2=\frac{1}{4}\left(\frac{\pi}{3}-1 \right)a^2.$$

So $$(EPMK) = (2-\sqrt{3})a^2 + \left(\frac{\pi}{3}-1 \right)a^2=\left(\frac{\pi}{3}+1-\sqrt{3}\right)a^2.$$


Well this one is very easy to solve. We put a coordinate system directly in the middle of your square. The upper arc from the top left to bottom right is a quater circle and can be considered the graph of a function $g$. Now, since it is on a circle that must have its middle point at $(-a/2,-a/2)$, $g$ satisfies $$ (g(x)+a/2)^2 + (x + a/2)^2 = a^2,$$ because $a$ is its radius. Now we solve this equation for $g(x)$, keeping in mind, that it is the upper half of the circe and get $$ g(x) = \sqrt{a^2 - (x+a/2)^2} - a/2. $$ Due to the symmetry along the line $x=y$, it must be $g(g(0)) = 0$. Due to further symmetries, we get the area as $$ A = 4\int_0^{g(0)} g(x)\,dx. $$ We can easily see, by standard integration techniques, that $$ G(x) = \frac{1}{8} \left(4 a^2 \tan ^{-1}\left(\frac{a+2 x}{\sqrt{(a-2 x) (3 a+2 x)}}\right)-4 a x+(a+2 x) \sqrt{(a-2 x) (3 a+2 x)}\right) $$ is an antiderivative. Now we use the fundamental theorem of calculus and finally get $$ A = 4(G(g(0)) - G(0)) = 4 \frac1{12} a^2 (3 - 3\sqrt 3 + \pi) = a^2(1 - \sqrt 3 + \pi/3).$$

  • $\begingroup$ Sorry for sounding so cocky. I wrote this for someone who asked the same question, but was very rude in formulating it. It was closed before I could post the answer but I didn't want it to go to waste. :) $\endgroup$ – Stefan Hante Nov 8 '17 at 13:38

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