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I am trying to find the equations to calculate:

  1. The intersection surface area generated by the intersection of 3 circles (3 circles
    like a Venn Diagram).
  2. The 3 circle's radius could be be different one from others but always 0 < radius <= 1
  3. The circles centres positions are fix and they are separated by 1 unit each from other (the circle's centres are located in the vertexs of an equilateral triangle of side=1)

To be clearer... the intersection "of the 3 circles" area will result in a figure like an "irregular Reuleux triangle". That means a Reauleaux triangle where the internal triangle could be any (and not only an equilateral triangle) and the three radius could be different one from the others

Thanks a lot in advance

Georges L

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very interesting, but first of all is such an equation exists? –  Ram Sep 13 '12 at 13:44
    
I can see that trigonometry will work quite well. If $A_1, B_1$ and $C_1$ is your Reuleux triangle, and $ABC$ is your equilateral triangle all you really need are the angles $\angle B_1 A C_1$ and its other two counterparts. Those can be derived by trigonometry. –  ivan Sep 13 '12 at 14:03

2 Answers 2

Given only two similarly intersecting circles, can you find the area of their intersection?

Say $S_1, S_2, S_3$ are the circles, $T$ is the triangle and $A(s)$ is the function calculating the area of some set $s$. Then $A(\bigcap S_i) = A(T) - \sum \limits_{i <j} \Big(A(S_i \cap T) - A(S_i \cap S_j)/2 \Big)$.

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One should make use of the symmetries inherent in the problem. Therefore I propose to operate in the complex plane, and to assume that the three centers are the points $\omega^0=1$, $\omega$, and $\omega^2=\bar\omega$, where $\omega:=e^{2\pi i/3}$. The vertex $z_0$, where the circles centered at $\omega$ and $\omega^2$ intersect, can be written in the form $$z_0=s + t (\omega -\omega^2)\ ,$$ where $s$ and $t$ are real. Now one has to solve the system $$|z_0-\omega|^2=r_1^2\ ,\qquad |z_0-\omega^2|^2=r_2^2\ .$$ The unknown $t$ will be uniquely determined, and for $s$ one has to chose the larger of the two solutions. So we arrive at a formula of the form $$z_0=f(r_1,r_2)\omega^0+g(r_1,r_2)(\omega^1-\omega^2)\ .$$ Having this formula in hand one immediately deduces the formulas for $z_1$ and $z_2$ by cyclic permutation of the indices and exponents.

Now the intersection in question is a union of the triangle $\Delta:=\Delta(z_0,z_1,z_2)$ and three circle segments. The area of $\Delta$ is computed easily from the $z_k$, whereas the area of the circle segment $S_k$ centered at $\omega^k$ is the difference of a sector area and a triangle area: $$|S_k|= {1\over2}r_k^2\ \angle(z_{k+1},\omega^k,z_{k-1})-|\Delta(z_{k+1},\omega^k,z_{k-1})|\ .$$

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