MHB Find the exact shaded area of the region in 4 overlapping circles

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The discussion focuses on calculating the exact shaded area of a region formed by four overlapping circles and understanding the geometry involved. It is established that each part of the intersection corresponds to one-quarter of the circumference of the circles. The area of the shaded region is derived as A = 2(π - 2), while the area of the non-shaded region in each circle is calculated to be 2r², leading to a total non-shaded area of 8r² for all four circles. The final result intriguingly does not include π, highlighting the unique nature of the problem. The conversation emphasizes the need for clearer explanations of the geometric principles at play.
Zekes
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So, say you got 4 circles intersecting this way:

View attachment 9061

Now, I am looking for two things:


  1. A proof that each part of the circle which is in an intersection is 1/4 the size of the whole circle's circumference

  • The exact area of the non-shaded region.

Now, in my search to finding the answer to this, I stumbled upon this Circle-Circle Intersection -- from Wolfram MathWorld. The only problem? I have no idea what this article is trying to say, and how it can help me. I did find the equation to get the area of the shaded region ( it's $$A=2(\pi-2)$$ ) which I can use in Part 2 but I still don't understand how the solution got to there, and how to do Part 1. Please help me in learning what is trying to be said here in simpler terms! Thanks!
 

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Hi Zekes.

Let the radius of each circle be $r$; let P, Q, R, S be the centres of the cirlces and O, A, B, C, D the points of intersections marked as follows:

View attachment 9063

Now PQRS is a square with diagonal $2r$ and so the length of each side is $r\sqrt2$, i.e. $|\mathrm{PQ}|=r\sqrt2$. So the “thickness” of each light-blue lens-shaped region of overlap between circles is $(2-\sqrt2)r$. if T is the point of intersection of the line segments AO and PQ, then

$$|\mathrm{PT}|\ =\ r-\frac{2-\sqrt2}2r\ =\ \frac r{\sqrt2}.$$

Hence $\angle\mathrm{APT}\ =\ \arccos\frac1{\sqrt2}\ =\ 45^\circ$; i.e. $\angle\mathrm{APO}=90^\circ$. That is to say, each circular arc drawn from O is one-quarter the circumference of each circle.

Now:

  • area of quadrant APO = $\dfrac{\pi r^2}4$
    ;
  • are of triangle APO = $\dfrac12r^2$
    ;
  • therefore area of each light-blue shaded region = $2\times\left(\dfrac{\pi r^2}4-\dfrac12r^2\right)=\dfrac{(\pi-2)r^2}2$
    ;
  • therefore area of non-shaded region in each circle = $\pi r^2-2\times\dfrac{(\pi-2)r^2}2=2r^2$
    ;
  • therefore total non-shaded area = $4\times2r^2=8r^2$.

Interesting to note that the final answer does not contain $\pi$.
 

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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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