Circle of centroids

Let $c_0 = \dfrac{1}{3} (A + B + C + D)$ , then the centroids are given by,

$c_1 = c_0 - \dfrac{1}{3} D$

$c_2 = c_0 - \dfrac{1}{3} A$

$c_3 = c_0 - \dfrac{1}{3} B$

$c_4 = c_0 - \dfrac{1}{3} C$

From which, we can write,

$(c_i - c_0) = -\dfrac{1}{3} . v_i$ for $i = 1,2,3,4$ , where $v_i = D, B, A, C$

Thus,

$(c_i - c_0)^T (c_i - c_0) = \dfrac{1}{9} . {v_i}^T v_i = \dfrac{1}{9}$

Because ${v_i}^T v_i = 1$ , since $A, B, C,$ and $D$ lie on the unit circle. Thus the $c_i$ 's lie on a circle with center $c_0$ and radius $\dfrac{1}{3}$ , making the answer $\boxed{0.3333}$ .