I came across this problem in a FB group.  

Seems like a math competition problem.

See the drawing below.

[![pic1](https://i.stack.imgur.com/XiuYn.jpg)](https://i.stack.imgur.com/XiuYn.jpg)

First we draw the big quarter circle,
then the blue semi-circle, then the blue full circle.

Finally we draw the small yellow circle.

In this order we do the drawing.

We need to prove $r/R = 3/29$.

OK, I denote $x=r/R$.

The easy part is to compute the radius of the blue full circle. It turns out to be $3R/8$. Then it took me a lot of time and pain (about 8 hours in total maybe) but I was finally able to come up with an equation for $x=r/R$.

Here it is

[WA Equation](https://www.wolframalpha.com/input?i=%28%285%2F4%29%20*%20%283%20*%20%285-4*x%29%20%2F%20%2825*%20%281%2B4*x%29%29%20-%20%2820*x-3%29%20%2F%20%2812*x%2B3%29%29%29%20%20%5E%202%20%20%2B%20%20%28%20%285-4*x%29%20%2F%20%285%20*%20%281%2B4*x%29%29%20%29%20%5E%202%20%3D%201%20)

I can explain how I got this equation (I use twice the cosine rule, other things, etc. etc.), but this is not really my issue.

Now I have one final issue to overcome. The algebraic equation has 2 roots for x.

So how do I prove $x$ cannot be $3/5$?

From the drawing it's kind of obvious $r/R$ cannot be $3/5$ but how do I discard this option in a more rigorous way? I know that "kind of obvious" doesn't count in maths.