Find the radius of the circle!

The shift method follows from the previous problem of mine. The proof is left for the readers to work out.

Let $r$ denote the radius of the circle. Since all $3$ inscribed circles have the same radius, we can shift equilateral triangle $XYZ$ as shown below.

Where the side length of $\Delta ABC$ is $12$ .

Since in the original diagram the small triangle $\Delta XYZ$ is positioned at the center of the large one, the altitude of $BYZC$ is $3r$ (from point $O$ to the bottom). Then, since the length of $\Delta XYZ$ is $2r$ , then $|\overline{XO}| = \sqrt{3}r$ , so the altitude of $\Delta ABC$ is $\sqrt{3}r + 3r$ . Therefore, the radius of the circle is $\begin{array}{rl} \text{radius of circle} &= \text{altitude of }\Delta ABC\\ r\left(3 + \sqrt{3}\right) &= 6\sqrt{3}\\ r &= \dfrac{6\sqrt{3}}{3 + \sqrt{3}} \end{array}$