5 Olympic rings

Let's put another unit circle on top of the existing circles as shown below:

$\angle{EOP}=\alpha \Rightarrow |OE|=\frac{1}{\sin{\alpha}}=|OC_{1}|=|OA_{1}|=|OA_{2}|.$

$\angle{C_{1}OE}=2\alpha \Rightarrow \angle{OEA_{3}}=\angle{A_{3}OE}=90^\circ-\alpha \Rightarrow \angle{OA_{3}E}=2\alpha$ .

$A_{1}A_{3}C_{1}B$ is parallelepiped, so $\angle{A_{3}BC_{1}}$ should be $2\alpha$ .

Solution 1: $\Rightarrow \angle{A_{1}BC_{1}}=4\alpha, \angle{A_{1}OC_{1}}=180^\circ-6\alpha$

Cosine rule for finding trianlge sides: $|A_{1}C_{1}|=x$

$|A_{1}C_{1}|^2=|A_{1}B|^2+|C_{1}B|^2-2\cdot|A_{1}B|\cdot|C_{1}B|\cdot\cos{4\alpha} \Rightarrow x^2=2^2+2^2-2\cdot2\cdot2\cdot\cos{4\alpha} \Rightarrow x^2=8\cdot(1-\cos{4\alpha})$

$|A_{1}C_{1}|^2=|OA_{1}|^2+|OC_{1}|^2-2\cdot|OA_{1}|\cdot|OC_{1}|\cdot\cos{(180^\circ-6\alpha)} \Rightarrow x^2=\left(\frac{1}{\sin{\alpha}}\right)^2+\left(\frac{1}{\sin{\alpha}}\right)^2-2\cdot\frac{1}{\sin{\alpha}}\cdot\frac{1}{\sin{\alpha}}\cdot\cos{(180^\circ-6\alpha)}=\frac{2}{\sin^2{\alpha}}\cdot(1+\cos{6\alpha})$

$\Rightarrow x^2=8\cdot(1-\cos{4\alpha})=\frac{2}{\sin^2{\alpha}}\cdot(1+\cos{6\alpha})=\frac{4}{1-\cos{2\alpha}}\cdot(1+\cos{6\alpha}) \Rightarrow 2\cdot(1-\cos{4\alpha})\cdot(1-\cos{2\alpha})=1+\cos{6\alpha}$

$\cos{4\alpha}=2\cos^2{2\alpha}-1; \cos{6\alpha}=4\cos^3{2\alpha}-3\cos{2\alpha}$

$(\cos{2\alpha}=a): \Rightarrow 2(2-2a^2)(1-a)=1+4a^3-3a \Rightarrow 4a^2+a-3=0 \Rightarrow (a+1)(4a-3)=0 \Rightarrow a=\frac{3}{4}.$

$cos{2\alpha}=\frac{3}{4}=1-\sin^2{\alpha} \Rightarrow \sin{\alpha}=\frac{1}{\sqrt{8}} \Rightarrow |OE|=\frac{1}{\sin{\alpha}}=\sqrt{8}. \Rightarrow R=|OK|=|OE|+|EK| \Rightarrow$ $R=\sqrt{8}+1 \approx 3.828$

Solution 2: I later realized the easiest way to find the value of $\sin{\alpha}$ .

$\triangle{OA_{3}E}\sim\triangle{EOC_{1}} \Rightarrow \frac{|A_{3}E|}{|OC_{1}|}=\frac{|OE|}{|EC_{1}|} \Rightarrow \frac{\frac{1}{\sin{\alpha}}}{4}=\frac{2}{\frac{1}{\sin{\alpha}}} \Rightarrow \frac{1}{\sin{\alpha}}=\sqrt{8}=|OE|=|OC_{1}| \Rightarrow R=|OE|+|EK|=\sqrt{8}+1\approx 3.828$

Maybe I solved it correctly through luck. I just put another circle on top of the existing circles and thought they formed a square. I calculatet the diagonal which is $\sqrt{8}$ and than added the +1 as radius of smaller circle. I later realized it wasnt a square, but surprisingly the answer was correct