TITLE: In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.
QUESTION [14 upvotes]: In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.

I first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $\frac{1}{4}$, but the correct answer was $\frac{2}{9}$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.

REPLY [14 votes]: @SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.

Let the quarter-, semi-, and full-circles have radius $a$, $b$, $c$, respectively.

From the right triangle, we have 
$$a^2+(a-b)^2=(a+b)^2 \quad\to\quad a=4b \tag{1}$$
Considering the side of the square a circle of curvature $0$, that special case of the Kissing Circles theorem implies
$$\frac1{c} = \frac{1}{a}+\frac{1}{b}\pm 2\sqrt{\frac{1}{a}\cdot\frac{1}{b}} = \frac{5}{4b}\pm 2\sqrt{\frac{1}{4b^2}} = \frac{5\pm 4}{4b}\quad\to\quad c = \frac49 b \;\text{or}\; 4b\;\text{(extraneous}) \tag{2}$$
Then, with $a=2$, we have $b=1/2$, so that $c=2/9$. $\square$