Tangential Circles For a natural number $n$, the circles $C_{3n-2}$, $C_{3n-1}$, and $C_{3n}$ have the same radius, tangential to each other externally, and tangential to the circle $C_{3n+1}$ intenally. If the radius of $C_{1}=1$ unit, what is the radius of $C_{100}$?
And in general;
For a natural number $n$, the circles $C_{3n-2}$, $C_{3n-1}$, and $C_{3n}$ have the same radius, tangential to each other externally, and tangential to the circle $C_{3n+1}$ intenally. If the radius of $C_{1}=r_{1}$ unit, what is the radius of $C_{m}$, where $m$ is a natural number?
I have tried Descartes' theorem, but I got confused.
 A: We first determine $C_4$'s radius. The centres of $C_{1,2,3}$ form an equilateral triangle with side length 2, hence circumradius $\frac2{\sqrt3}$. The radius of $C_4$ is one more radius of $C_1$ from this, or $\frac2{\sqrt3}+1$.
Iterating this construction 33 times from $C_1$ will get us to $C_{100}$, and the radius multiplies by $\frac2{\sqrt3}+1$ each time, since the construction operates on the same configuration of three equal-sized mutually tangent circles. $C_{100}$'s radius is thus
$$\left(\frac2{\sqrt3}+1\right)^{33}$$
and in the more general case where $C_1$'s radius is $r_1$, $C_m$'s radius is
$$r_1\left(\frac2{\sqrt3}+1\right)^{\lfloor(m-1)/3\rfloor}$$
A: I thought you all might like to see the image.


We can think of the circle centered at C as $C_1$ and the circle centered at O as $C_4$.  To find the radius of $C_4$, look at the triangle $OAC$.  Notice that 
$$\sqrt3/2 = \cos(30^\circ)=AC/OC.$$
We know that $AC$ has length 1, so solving for $OC$ gives $OC=2/\sqrt3$.  To get $OB$, the radius of $C_4$, we just add $OC$ and the radius of $C_1$, $CB$, to get
$$
OB = OC+CB= 2/\sqrt3+1.
$$
To get the radius of the remaining 98 circles, see Parcly's answer!
