1. Observe that the circle's centers must lie on the internal bisector $AM$ of $\angle BAC$, since their distance from the triangle side 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 $$\overline{AP_4}<\overline{AM}<\overline{AP_5}.$$