Determine the possible values for the radius of $C_1$

Question

Let $ABC$ be an equilateral triangle with side 3. A circle $C_1$ is tangent to $AB$ and $AC$. A circle $C_2$, with a radius smaller than the radius of $C_1$, is tangent to $AB$ and $AC$ as well as externally tangent to $C_1$. Successively, for $n$ positive integer, the circle $C_{n+1}$, with a radius smaller than the radius of $C_n$, is tangent to $AB$ and $AC$ and is externally tangent to $C_n$. Determine the possible values for the radius of $C_1$ such that 4 circles from this sequence, but not 5, are contained on the interior of the triangle $ABC$

Solution:

I do not understand this solution so much because it is in Portuguese, so I kindly hope for a new solution here

What I understood: $ x $ is the radius of $ C_1 $, $ y $ radius of $ C_2 $, $ M $ midpoint $ \overline BC $

$ \sin (30 °) = \frac {y-x} {x + y} = \frac {1} {2}$

Answer 1

I'd say the demonstration goes as follows.

1. Observe that the circle's centers must lie on the internal bisector $AM$ of $\angle BAC$, since their distance from the triangle sides is the same.
2. Connect the centers of $\mathcal C_1$ and $\mathcal C_2$, and draw, from those points also segments perpendicular and parallel to $AB$. Thus generate a $90^\circ-30^\circ-60^\circ$ triangle. The hypothenuse of this triangle is $y+x$, and the smaller side $y-x$, thus the ratio $\frac{y-x}{y+x}$ is $\frac12$, which gives $y=3x$.
3. Iterating the same procedure yields, for $n=2,3,\dots,$ \begin{eqnarray}\overline{AP_n} &=& \overline{AO_1}+\overline{O_1O_2}+\overline{O_2O_3}+\cdots +\overline{O_{n-1}O_n}+\overline{O_nP_n}=\\ &=&2x + (x + 3x) + (3x+9x) + \cdots + \left(3^{n-2}x+3^{n-1}x\right)+3^{n-1}x,\end{eqnarray} where $O_k$ is the center of $\mathcal C_k$ and $O_nP_n$ is the radius of the rightmost circle.
4. You want $x$ so that $$\overline{AP_4}<\overline{AM}<\overline{AP_5}.$$