Four Circle Problem

[bob012345]

TL;DR

Determine the fraction of the darker area to the total area of the figure.

This is a little exercise for fun. Determine the area of the darker petals to the total area. The four circles are identical. Have fun!

 

[.Scott]

Each half a petal is bounded by a chord and a 90-degree slice of the circle. So the area would (pi (r^2)/4) - (r^2)/2 = (r^2)(pi/4 - 1/2). Since there are 8 of those, the total petal area is 8(r^2)(pi/4 - 1/2)=(2pi-4)(r^2).
The total area is 4 circles minus the overlap: 4pi(r^2)-(2pi-4)(r^2) = (2pi+4)(r^2).
Dark petals to total area is: (2pi-4)(r^2)/(2pi+4)(r^2) = (2pi-4)/(2pi+4) = (pi-2)/(pi+2)

 

[bob012345]

Spoiler

I got $(\pi-2)/(\pi+2)$

 

[Gavran]

TL;DR: Determine the fraction of the darker area to the total area of the figure.

This is a little exercise for fun. Determine the area of the darker petals to the total area. The four circles are identical. Have fun!View attachment 371094

Spoiler

$\begin{align} f=&A/(A+B)\nonumber\\ =&(r^2\pi/4-r^2/2)/(r^2\pi/4+r^2/2)\nonumber\\ =&(\pi/4-1/2)/(\pi/4+1/2)\nonumber\\ =&(\pi-2)/(\pi+2)\nonumber\\ \end{align}$

 

[bob012345]

If anyone is interested, here is the case of five identical circles with a common point as in this diagram. Notice that there are now two levels of petals, the larger and the smaller. In this case determine the fraction of the figure with any petals to the whole figure.

Spoiler

I got $\large \frac{2\pi- \gamma}{3\pi+\gamma}$ with $\gamma=5/4\left(\sqrt{10+2\sqrt{5}}-\sqrt{10-2\sqrt{5}}\right )$