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In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.

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    $\begingroup$ Notice anything special about $\triangle ABC$? $\endgroup$ – Blue Jan 27 at 1:58
  • $\begingroup$ Hmmn! It's equilateral! $\endgroup$ – Abdulhameed Jan 27 at 2:02
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    $\begingroup$ Then just use one of the previous solutions, since you know the angle. $\endgroup$ – Andrei Jan 27 at 2:03
  • $\begingroup$ Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues? $\endgroup$ – Abdulhameed Jan 27 at 2:06
  • $\begingroup$ @Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ... $\endgroup$ – Blue Jan 27 at 2:08
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If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these (American!) football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:

$\pi=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)

OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $\sqrt{3}$, and this area includes $4$ whole footballs, $4$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:

$\sqrt{3}=6A+4B$

Now you can easily solve for $A$

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  • $\begingroup$ @Bram28 really nice picture, +1 $\endgroup$ – Zubin Mukerjee Jan 27 at 3:06
  • $\begingroup$ This is so beautiful! Its well appreciated $\endgroup$ – Abdulhameed Jan 27 at 3:10
  • $\begingroup$ @Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2\left(\frac{1}{6}\pi^2 - \frac{\sqrt{3}}{4}\right)$$ $\endgroup$ – Zubin Mukerjee Jan 27 at 3:16
  • $\begingroup$ @Abdulhameed You're welcome! :) $\endgroup$ – Bram28 Jan 27 at 3:33
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    $\begingroup$ "football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world. $\endgroup$ – ilkkachu Jan 27 at 16:03
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Notice that the area of the equilateral triangle with edge $R$ plus $\frac12$ the area of the shaded region is $\frac16$ the area of the circle. The height of the triangle is $\frac{R\sqrt3}{2}$ and the area is $\frac12\cdot R\cdot\frac{R\sqrt3}{2}=\frac{R^2\sqrt3}{4}$. The area of the shaded region is: $$2\left(\frac{\pi R^2}{6}-\frac{R^2\sqrt3}{4}\right)=\frac{2\pi R^2}{6}-\frac{3R^2\sqrt3}{6}=\frac{R^2(2\pi-3\sqrt3)}{6}$$ Also notice that the height of the triangle plus $\frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is: $$2\left(R-\frac{R\sqrt3}{2}\right)=2R-R\sqrt3=R(2-\sqrt3)$$

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$AC=BC=AB=AD=BD=R$.Now $AB$ and $CD$ are perpendicular bisectors of each other, say at $O$.Then $DO^2=AD^2- AO^2= R^2- ({R \over 2})^2$ i.e. $DO={ \sqrt{3}R \over 2}$.Then $height=2(R- { \sqrt{3}R \over 2})=(2- \sqrt{3})R$. Now if the shaded area be $∆$, as $\angle ADB=60°$ , ${∆ \over 2}={πR^2 \over 6}- { \sqrt{3}R^2 \over 4}$(it's area of $∆ABD$)so, required shaded region area $∆={2π-3√3 \over 6}R$.

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