Find the radius of the circle

I got a solution of sqrt(112/3), and brilliant says "correct" when I gave 112 as my answer. However, I maintain that this is wrong, because the problem stipulates that a must be square free; But 112 is divisible by 4, and is therefore not square free. Therefore, I claim there is no answer to this problem which satisfies the requested conditions. Ed Gray

One way of solving this problem is rearranging the six triangles in a more symmetrical way. Denote the center of the circle by $O$ , with $XY = 4$ and $YZ = 8$ adjacent sides of the rearranged hexagon. Notice that quadrilateral $XOZR$ is exactly one-third of the entire hexagon. Therefore $\angle XOZ=\dfrac{360}{3}=120$ . By the symmetry of the rearranged hexagon, all of its interior angles are equal, so $\angle XYZ = \dfrac{720}{6} = 120$ .

Let $XZ=a$ , by law of cosines in $\triangle XZO$ , we have

$a^2 = r^2+r^2-2r^2 \cos~120=2r^2-2r^2\left(\dfrac{-1}{2}\right)=3r^2$

by law of cosines in $\triangle XZY$ , we have

$a^2=4^2+8^2-2(4)(8) \cos~120=80-64\left(\dfrac{-1}{2}\right)=112$

$a^2=a^2$

$3r^2=112$

$r=\sqrt{\dfrac{112}{3}}$

Finally, the desired answer is $\color{#D61F06}\boxed{112}$ .