Ok got it. This seems correct. I'll have to figure out why [my approach](https://en.wikipedia.org/wiki/Descartes%27_theorem) is giving a different result. I likely must have applied it incorrectly. Please don't spend more time on this, I'm assuming the error must be on my side at this point.

For $$O_3$$

![circles2](https://i.ibb.co/RSRGwnM/circles2.png)

(the top-left corner of the triangle is supposed to be the center of $$O$$)

The hypotenuse is 1 + $$r$$, where $$r$$ denotes the radius of $$O_3$$.

The horizontal leg is simply 1 (this is true for all $$O_n$$).

The vertical leg is $$1$$ minus the sum of all **diameters** $$O_1$$ and $$O_2$$ minus $$r$$ of $$O_3$$.

So, Pythagoras gives $$r$$ of $$O_3$$.

Scroogey

 (as in the right-hand side of the formula below)? And same comment for the subsequent 

I used a different approach to solve this problem (I'll post it as a solution with the next iteration), but that one gives me 

science

[Daily puzzle] Diameter of circle

south_korea_ln

Aren't you using the same symbol $$r$$ to denote the different radii of $$O_1$$ (as in the drawing) and $$O_2$$ (as in the right-hand side of the formula below)? And same comment for the subsequent $$O_3$$ equation?

> For $$O_2$$
> 
> $$(1 - \frac{1}{2} - r)^2 + 1 = 1 + 2*r + r^2$$
> 
> For $$O_3$$
> 
> $$(1 - \frac{1}{2} - \frac{1}{6} - r)^2 + 1 = 1 + 2*r + r^2$$

I used a different approach to solve this problem (I'll post it as a solution with the next iteration), but that one gives me $$1/18$$ for $$O_3$$. 