Circles in a Square – Part 10

Assume the radius of the two congruent semi-circle is $r$, the radius of the full circle is $s$. First, we need to find the value of $r$, as as the semi-circles are drawn first in the set up. Let’s draw line $\overline{EF}$, connecting the centers of the semi-circles, as shown below:

Obviously, $\triangle{BEF}$ is a right triangle. We have $$\overline{EF}^2=\overline{BE}^2+\overline{BF}^2$$ Since $\overline{AE}=\overline{CF}=r$, $\overline{BE}=\overline{BF}=\overline{AB}-\overline{AE}=1-r$, we have: $$(2r)^2=(1-r)^2+(1-r)^2$$ The solution of the above equation, we have $r=\pm\sqrt{2}-1$. Ignoring the negative $r$ value, we have the radius of the semi-circles as $$r=\sqrt{2}-1$$ Next, we consider the full circle center at point $G$. Connect the center of one semi-circle and the center point of the full circle, we have $\overline{EG}$ intersecting the semi and full circles at point $I$. We have $$\overline{EG}=\overline{EI}+\overline{GI}=\sqrt{2}-1+s$$ Draw line $\overline{KL}$ passing thru $G$ and parallel to $\overline{CD}$, and line $\overline{EM}$ perpendicular to $\overline{CD}$ and intersecting with line $\overline{KL}$ at point $N$. As $\triangle{EGN}$ is a right triangle, we have $$\overline{EG}^2=\overline{EN}^2+\overline{GN}^2$$ Since $$\overline{EN}=\overline{EM}-\overline{NM}=1-s$$ $$\overline{GN}=\overline{KN}-\overline{GK}=\overline{AE}-s=\sqrt{2}-1-s$$ Therefore $$(\sqrt{2}−1+s)^2=(1−s)^2+(\sqrt{2}−1−s)^2$$ Solve the above equation, we have $$s=-1+2\sqrt{2}\pm2\sqrt{2-\sqrt{2}}$$ Ignoring the $s$ value greater than 1, we have the radius of the full circle as $$s=-1+2\sqrt{2}-\sqrt{2-\sqrt{2}} \approx 0.2976933952$$