Hard problem from elementary geometry - how do I discard this option for x? I came across this problem in a FB group.
Seems like a math competition problem.
See the drawing below.

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
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.

Since the area of the yellow circle is less than the area of the big quarter circle we have
$$\pi r^2 \lt \frac{1}{4} \pi R^2 $$ i.e.
$$r / R < \frac{1}{2}$$
So $x = r/R$ cannot be $3/5$
$\Rightarrow x = r/R = 3/29$
