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From the link in wikipedia

http://web.gnowledge.org/wiki/index.php/Area_Between_Three_Circles_of_Differing_Radii

OPEN QUESTION:

What is the equation, in three variables, relating the radii of three circles to the area between them, when each is tangent to the other two?

Three circles, different radii

But wiki says that it is an open question and I am interested what is main reason of it? or is the author simply trying to understand if anybody can solve it (while it's answer is known?)

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I'm thinking it's simply an "open question" in the sense of an exercise with no answer provided. (Perhaps the question-poser doesn't know it, or is intentionally withholding it.) The solution is pretty straightforward, given knowledge of basic trigonometry (specifically, Heron's formula and the Law of Cosines) ... and, of course, familiarity with the basic geometry of tangency of circles. –  Blue Jul 6 '11 at 9:41
    
@user: I inserted the image and text from your link directly into the problem. –  amWhy Jul 6 '11 at 13:34

2 Answers 2

up vote 4 down vote accepted

Open question? Really?

Join the centers of the circles to get a big triangle with known sides and, therefore, known angles. Now make a triangle of the three tangency points; from what you know of the big triangle, you can get the sides and so the area of the small one. The shaded area is the small triangle minus three circular bits. The circular bits are circular segments of known parameters minus triangles of known parameters.

It may not be much fun to follow this recipe to its conclusion, but it ought to work.

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Instead of "small triangle, minus three circular segments", you can calculate "large triangle, minus three circular sectors". There's no need to consider the small triangle at all. –  Blue Jul 6 '11 at 12:37
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That would indeed be much easier. The sound you hear is me kicking myself for not seeing it. –  Gerry Myerson Jul 6 '11 at 12:56

As a follow-up to Gerry's answer and Day Late Don's comment there, let's consider the triangle formed by the centers of the circles and subtract the sectors. Call the circles 1, 2, and 3, with radii $r_1$, $r_2$, and $r_3$, respectively, and let $\theta_1$, $\theta_2$, and $\theta_3$, be the interior angles of the triangle formed by the centers of the circles, at the centers of circles 1, 2, and 3, respectively.

Applying the Law of Cosines (and simplifying some) to find $\theta_1$, $\theta_2$, and $\theta_3$ gives: $$\theta_1=\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)$$ $$\theta_2=\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)$$ $$\theta_3=\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)$$ so the sector areas are $$\frac{1}{2}\theta_1r_1^2=\frac{1}{2}r_1^2\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)$$ $$\frac{1}{2}\theta_2r_2^2=\frac{1}{2}r_2^2\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)$$ $$\frac{1}{2}\theta_3r_3^2=\frac{1}{2}r_3^2\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)$$

Applying Hero's formula to find the area of the triangle (and simplifying) gives $$K_\triangle=\sqrt{r_1r_2r_3(r_1+r_2+r_3)}.$$

So the area of the target region is $$\sqrt{r_1r_2r_3(r_1+r_2+r_3)}-\frac{1}{2}\left(\begin{align} r_1^2&\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right) \\&+r_2^2\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right) \\&+r_3^2\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right) \end{align}\right).$$

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