From the link in wikipedia



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?)

  • 1
    $\begingroup$ 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. $\endgroup$
    – Blue
    Jul 6 '11 at 9:41
  • $\begingroup$ @user: I inserted the image and text from your link directly into the problem. $\endgroup$
    – amWhy
    Jul 6 '11 at 13:34

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


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.

  • 4
    $\begingroup$ 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. $\endgroup$
    – Blue
    Jul 6 '11 at 12:37
  • 3
    $\begingroup$ That would indeed be much easier. The sound you hear is me kicking myself for not seeing it. $\endgroup$ Jul 6 '11 at 12:56

At first join the centers of the circles to get a big triangle with known sides. First calculate the sides of triangle-

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 :


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]

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

$$\frac{(r_1+r_2)\sqrt{(r_1+r_3)^2-r_1}}{2} - \frac{\pi x r_1^2}{\frac{\angle 1}{360}} + \frac{\pi x r_2^2}{\frac{\angle 2}{360}} + \frac{\pi x r_3^2}{\frac{\angle 3}{360}}$$ 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

  • $\begingroup$ Please see this tutorial on how to format mathematics on this site. $\endgroup$ Mar 9 '15 at 3:05

In the special case of all 3 circles of the same radius $r$, the area is given by the equation $k r^2$ where $k=0.161254481$.

  • 2
    $\begingroup$ The title specifically states "Differing Radii" so this doesn't answer the question. $\endgroup$
    – Dijkgraaf
    Dec 21 '14 at 22:28
  • $\begingroup$ Equation? What equation? $\endgroup$
    – HDE 226868
    Dec 21 '14 at 22:41

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