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Three circles of different radii are tangent to each other externally. The distance between their centers are $9\ cm$, $8\ cm$, and $11\ cm$.

  1. Find the radius of each circle.

  2. Find the area in between the three circles.

The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$\begin{align}r_1+r_2&=9\\r_2+r_3&=8\\r_3+r_1&=11\end{align}$$ From this we can find $r_1,r_2,r_3$.

Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.

Is this correct?

Would I have to use the $\arccos(x)$ function to get the final answer or there is a simpler way?

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  • $\begingroup$ What did you tried so far ? $\endgroup$ – servabat Jan 24 '15 at 13:01
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    $\begingroup$ The first time I read the title I thought it said "Solid menstruation (cycles)". $\endgroup$ – Pp.. Jan 25 '15 at 14:16
  • $\begingroup$ $(r_1,r_2,r_3)=(6,3,5)$, by the way. $\endgroup$ – Akiva Weinberger Nov 3 '15 at 17:05
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Solving for the radii, we get $\{3,5,6\}$.

For the area we get $$ \begin{align} &\sqrt{14(14-8)(14-9)(14-11)}\\[9pt] &-\frac{(8+9-11)^2}8\arccos\left(\frac{8^2+9^2-11^2}{2\cdot8\cdot9}\right)\\ &-\frac{(11+8-9)^2}8\arccos\left(\frac{11^2+8^2-9^2}{2\cdot11\cdot8}\right)\\ &-\frac{(9+11-8)^2}8\arccos\left(\frac{9^2+11^2-8^2}{2\cdot9\cdot11}\right) \end{align} $$ which is $3.05537320587455$.

enter image description here

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Mathematica solving for 3 variables.

Clear[a, b, c];
Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]

(* {{a -> 6, b -> 3, c -> 5}} *)
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here is one way to do this:

(a) construct a triangle $ABC$ with sides $8, 9$ and $11.$

(b) find the incenter $I$ of $ABC$

(c) the common value $AI = BI = CI$ is the radius you want.

the same can be done algebraically by finding

(a) two angles using the cosine rule

(b) use herons formula to find the area of $ABC$

(c) in-radius = $\dfrac{area}{semi perimeter}$

(d) $AI = \dfrac{r}{\tan A/2}$

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