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Let $c_1$, $c_2$ and $c_3$ three external tangent circles. Their centers are $O_1$, etc. The three circles are tangent to a single line $r$. The problem is to prove that $O_1O_2O_3$ is a triangle with an obtuse angle.
Tries: I have tried Euclidean geometry, analytical geometry and trigonometry and even inversion. No success so far.
Ok thanks for the clarification. Two observations to start. First, the three circles must be on the same side of the line if they are all tangent to it. Therefore, we will choose the common tangent line to be the $Ox$ axis and the three circles to lie in the upper half plane. Second, note that one of the radii of the circles has to be smaller than the other two to allow the other two circles to be tangent. By dilating and choosing the coordinate axes nicely, we can assume the smaller circle has radius $1$ and is centered at the point $(0,1)$.
Let the other radii be $a$ and $b$ and recall that $a>1$ and $b>1$. Let the centers of these two other circles be $(x,a)$ and $(y,b)$ respectively (since they are both tangent to the horizontal axis, clearly the heights off the horizontal axis will be $a$ and $b$ respectively). Moreover, $x$ and $y$ cannot have the same sign, so wlog let's take $x<0<y$.
Now let's figure out the value of $x$. By the assumption of the $a$ circle and the small circle being tangent, we must have that the distance between the centers of these two circles is exactly $a+1$. Therefore $x^2+(a-1)^2=(a+1)^2$ and so $x=-2\sqrt{a}$. Similarly we get $y=2\sqrt{b}$. So we reduced to the case of the circles having centers $(-2\sqrt{a},a), (0,1), (2\sqrt{b},b)$ and radii $a,1,b$ respectively. Let us denote the centers by $A,O,B$ respectively.
Next, the 'outer' circles are also tangent, meaning that the distance between $(-2\sqrt{a},a)$ and $(2\sqrt{b},b)$ is exactly $a+b$. In other words, $4(\sqrt{a}+\sqrt{b})^2+(b-a)^2=(a+b)^2$. After simplification, this implies $ab=a+b+2\sqrt{ab}$.
Now we show that the angle $\angle AOB$ is greater than $\frac{\pi}{2}$. To do this it suffices to show $AB^2>AO^2+BO^2$, i.e., that $(a+b)^2>(a+1)^2+(b+1)^2$. Simplifying, this condition boils down to $ab>a+b+1$. But we know that $ab=a+b+2\sqrt{ab}$ and both $a>1$ and $b>1$ by construction. Therefore $2\sqrt{ab}>2$.
