We are given in the plane a line $\ell$ and three circles with centres $A,B,C$ such that they are all tangent to $\ell$ and pairwise externally tangent to each other. Prove that the triangle $ABC$ has an obtuse angle and find all possible values of this this angle.

[i]Mircea Becheanu[/i]
1. Let \( a, b, c \) be the radii of the circles with centers \( A, B, C \) respectively. Assume that \( c \) is the smallest radius. The circles are tangent to the line \( \ell \) and pairwise externally tangent to each other. 

2. A well-known relation for three such circles is given by:
   \[
   \sqrt{ab} = \sqrt{ac} + \sqrt{bc} \tag{*}
   \]

3. We need to show that triangle \( ABC \) has an obtuse angle and find the possible values of this angle. Let \( \angle ACB = \gamma \). Using the cosine law in \( \triangle ABC \), we have:
   \[
   \cos \gamma = \frac{c(a+b+c) - ab}{(a+c)(b+c)}
   \]

4. For \( \gamma \) to be obtuse, \( \cos \gamma < 0 \). This implies:
   \[
   c(a+b+c) - ab < 0
   \]

5. Using the relation \((*)\), we substitute and simplify:
   \[
   c(a+b+c) - ab < 0 \implies c(a+b+c) < ab
   \]

6. Rearranging the inequality, we get:
   \[
   c < \frac{ab}{a+b+c}
   \]

7. From the relation \((*)\), we know:
   \[
   \sqrt{ab} = \sqrt{ac} + \sqrt{bc}
   \]

8. Squaring both sides, we get:
   \[
   ab = ac + bc + 2\sqrt{ac \cdot bc}
   \]

9. Simplifying further, we get:
   \[
   ab = ac + bc + 2c\sqrt{ab}
   \]

10. Rearranging, we get:
    \[
    ab - ac - bc = 2c\sqrt{ab}
    \]

11. Dividing both sides by \( c \), we get:
    \[
    \frac{ab}{c} - a - b = 2\sqrt{ab}
    \]

12. Let \( x = \sqrt{\frac{c}{a}} \) and \( y = \sqrt{\frac{c}{b}} \). Then, we have:
    \[
    x + y = 1
    \]

13. We need to minimize:
    \[
    \Delta = (1 + x^2)(1 + y^2)
    \]

14. Substituting \( y = 1 - x \), we get:
    \[
    \Delta = (1 + x^2)(1 + (1-x)^2)
    \]

15. Simplifying, we get:
    \[
    \Delta = (1 + x^2)(1 + 1 - 2x + x^2) = (1 + x^2)(2 - 2x + x^2)
    \]

16. Expanding, we get:
    \[
    \Delta = 2 + 2x^2 - 2x + x^4 - 2x^3 + x^2
    \]

17. Simplifying further, we get:
    \[
    \Delta = x^4 - 2x^3 + 3x^2 - 2x + 2
    \]

18. To find the minimum value, we take the derivative and set it to zero:
    \[
    \frac{d\Delta}{dx} = 4x^3 - 6x^2 + 6x - 2 = 0
    \]

19. Solving this cubic equation, we find the critical points. The minimum value occurs when \( x = y = \frac{1}{2} \).

20. Substituting \( x = y = \frac{1}{2} \), we get:
    \[
    \Delta = (1 + \left(\frac{1}{2}\right)^2)(1 + \left(\frac{1}{2}\right)^2) = \left(1 + \frac{1}{4}\right)\left(1 + \frac{1}{4}\right) = \frac{5}{4} \cdot \frac{5}{4} = \frac{25}{16}
    \]

21. Therefore, the maximum value of \( \sin \frac{\gamma}{2} \) is:
    \[
    \sin \frac{\gamma}{2} = \sqrt{\frac{1}{\Delta}} = \sqrt{\frac{16}{25}} = \frac{4}{5}
    \]

22. Hence, the maximum value of \( \gamma \) is:
    \[
    \gamma = 2 \sin^{-1} \left(\frac{4}{5}\right) \approx 106.26^\circ
    \]

The final answer is \( \boxed{ 90^\circ < \gamma \leq 106.26^\circ } \)