Mutually Tangent Circles

The radius of $C_1$ is half the difference of the diameter of $C_A$ and the diameter of $C_B$ , which is $r_1 = \frac{1}{2}(2 \cdot 3 - 2 \cdot 2) = 1$ .

Then by Descartes’ Theorem , the radius $r_n$ of $C_n$ is defined by $\frac{1}{r_n} = \frac{1}{r_B} + \frac{1}{r_{n-1}} - \frac{1}{r_A} + 2\sqrt{\frac{1}{r_Br_{n-1}} - \frac{1}{r_Ar_B} - \frac{1}{r_Ar_{n-1}}}$ , which can inductively shown to be $r_n = \frac{r_Ar_B(r_A-r_B)}{(r_A-r_B)^2(n-1)^2+r_Ar_B}$ .

For $r_A = 3$ , $r_B = 2$ , and $n = 145$ , $r_n = \frac{3 \cdot 2 \cdot (3 - 2)}{(3 - 2)(145-1)^2+3 \cdot 2} = \frac{1}{3457}$ , so $\boxed{m = 3457}$ .

Brute force, I calculated the centers and radii. Therefore, my solution is closer to that of Felipe Lorenzzon.

I'm not sure if my method was the easier one. Initially, I plotted $C_A$ , $C_B$ and $C_1$ as functions on the plane. Then I used the distance formula to calculate the distance between the centers of $C_B$ and $C_2$ and set that equal to the radius of $C_B$ plus the radius of $C_2$ (because they are tangent). I found the distance between the center of $C_A$ and $C_2$ and set that equal to the radius of $C_A$ minus the radius of $C_2$ (since the latter is internally tangent to the former). I Then found the distance between $C_2$ and $C_1$ and set that equal to the sum of their radii (again, because they are tangent to each other). Finally, I solved for the radius using basic algebra and found $\dfrac{6}{7}$ .

After that, I repeated the same process, but, instead of finding the distance between $C_2$ and $C_1$ , I did it with a general $C_{n+1}$ and $C_n$ . By isolating $C_{n+1}$ in terms of $C_n$ , I found the formula $C_{n+1}= \dfrac{6C_n^2+36C_n-12C_n \sqrt{6C_n-6C_n^2}}{25C_n^2-12C_n+36}$

Using an online recursion calculator, I plugged in $C_1=1$ and found $C_{145} \approx 0,0002892681215$ . Since I knew the solution was on the form $\dfrac{1}{m}$ , I tried a bunch of integers until the number 3457 matched my result.

Maybe you could find the explicit formula for $C_n$ , however, I wasn't able to do so. Nice question, by the way.