Area Between Three Circles of Differing Radii 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?

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?)
 A: 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
$$\sqrt{r_1r_2r_3(r_1+r_2+r_3)}-\frac{1}{2}\left(\begin{align}
r_1^2&\arccos\left(1-\frac{2r_2r_3}{(r_1+r_2)(r_1+r_3)}\right)
\\&+r_2^2\arccos\left(1-\frac{2r_1r_3}{(r_2+r_1)(r_2+r_3)}\right)
\\&+r_3^2\arccos\left(1-\frac{2r_1r_2}{(r_3+r_1)(r_3+r_2)}\right)
\end{align}\right).$$
A: 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. 
A: At first join the centers of the circles to get a big triangle with known sides. First calculate the sides of triangle-
a=R1+R2
b=R2+R3
c=R3+R1 [Here R1,R2,R3 are radii of the above three triangles]

Now calculate half-perimeter to calculate the area of large triangle :
s=(a+b+c)/2

Area of big triangle is :
S = √(s*(s-a)*(s-b)*(s-c))

Now we have to find area of three circular sectors. For that we have to find three angles of the large triangle first using this formula :
X=acos((b*b+c*c-a*a)/(2*b*c)) [Here acos means cos inverse]
Y=acos((a*a+c*c-b*b)/(2*c*a))
Z=acos((a*a+b*b-c*c)/(2*a*b))

Now we will calculate the area of the three circular sectors as follows:
 S1= 0.5*((Y*r1*r1)+(Z*r2*r2)+(X*r3*r3))

Now we have area of large triangle and area of three sectors in our hand. 
By subtracting area of three sectors from area of large triangle we will finally got the area of shaded region.
Area of shaded region= Area of large triangle- Area of three sectors

A: $$\frac{(r_1+r_2)\sqrt{(r_1+r_3)^2-r_1}}{2} - \frac{\pi x r_1^2}{\frac{\angle 1}{360}} + \frac{\pi x r_2^2}{\frac{\angle 2}{360}} + \frac{\pi x r_3^2}{\frac{\angle 3}{360}}$$
where $r_1$, $r_2$, and $r_3$ are the radii of the circles; $\angle 1$, $\angle 2$, and $\angle 3$ are the angles of the triangle created by the centres of the circles. This equation takes the areas of the sectors of the circles and subtracts them from the triangle in which they are contained.
Hope this helped
A: In the special case of all 3 circles of the same radius $r$, the area is given by the equation $k r^2$  where $k=0.161254481$.
