# Complex surface: Finding the perimeter of the union of multiple disc intersections

Consider a system of $N$ circles($N>40$) Each of radius $5$ and the center of these circles along a circle at origin of radius $3$ [as given in code].

Code for above description :-

Graphics[Table[{Hue[t/40],
Circle[{3 Sin[2 Pi (t - 1)/40], 3 Cos[2 Pi (t - 1)/40]}, 5]}, {t,
40}]]


I want to find the perimeter of the resultant figure formed by regions covered by the union of the "intersection of any two arbitrary circles".

Then, the perimeter of the resultant figure formed by regions covered by the union of the"intersection of any three arbitrary circles".

And so on till we intersect of all the circles.

• Then the end result is the intersection of $N$ circles, right? So there is no further need to form a union in the end. Then the final situation is actually easier to solve than the intermediate steps, especially if you're allowed to pick, e.g., three arbitrary circles out of the $N$ given circles. In the end, one has a shape with an $N$ fold rotation symmetry, but that can't be exploited for the intermediate steps. It would help a great deal if you showed what attempts at a solution you have already tried, and maybe to sketch the desired end product (maybe I misunderstood it). – Jens Sep 6 '13 at 23:08

If I understand correctly, you want to find lengths of these lines (defined by color)

n = 10;
R = 5;
r = 3;

Graphics[Table[{Hue[Max[k, -1-k]/n],
Circle[{r Cos[2 Pi t/n], r Sin[2 Pi t/n]},
R, {Pi (2 t + k)/n + ArcSin[r/R Sin[k Pi/n]],
Pi (2 t + k + 1)/n + ArcSin[r/R Sin[(k + 1) Pi/n]]}]
}, {t, n}, {k, 1 - n, n}]] From geometrical considerations (basically, the law of sines) we have lengths of arcs

$l_k = R \left[\dfrac{\pi}{n}+\arcsin\left(\dfrac{r}{R}\sin\left(\dfrac{k+1}{n}\pi\right)\right)-\arcsin\left(\dfrac{r}{R}\sin\left(\dfrac{k}{n}\pi\right)\right)\right],\quad k=0,\ldots,n.$

$k=0$ and $k=n$ correspond to outermost and innermost lines respectively. The code above is a proof. Finally, the perimeters are

$P_k = 2n l_k = 2 n R \left[\dfrac{\pi}{n}+\arcsin\left(\dfrac{r}{R}\sin\left(\dfrac{k+1}{n}\pi\right)\right)-\arcsin\left(\dfrac{r}{R}\sin\left(\dfrac{k}{n}\pi\right)\right)\right].$

• It's quite possible that the question is badly formulated and that your answer is what was intended. So +1. – Jens Sep 6 '13 at 23:58
• @Jens I agree, I understand this question as a question about only Orange line perimeter. – Kuba Sep 7 '13 at 8:07