Official solution: Denote by $O, O'$ two possible positions of the center of the largest sphere (among the five marked points) and denote by $A, B$, and $C$ the other three marked points. Consider the points O, O', A, and B. In the configuration of spheres where O is the center of the largest sphere, denote by $R, r', r_a$, and $r_b$ the radii of the spheres centered at O, O', A, and B, respectively. Then we have $OO = R - r', OA = R - r_a, OB = R - r_b, O'A = r' + r_a, O'B = r' + r_b$, and $AB = r_a + r_b$, which yields $OO' - AB = OA - O'B = OB - O'A$; denote this common difference by $d$. Similarly, from the configuration with $O'$ being the center of the largest sphere we obtain $d = OO' - AB = O'A- OB = O'B - OA = -d$. Thus $d = 0$, and therefore $OO' = AB, OA = O'B$, and $OB = O'A$. Applying similar arguments to the tuples $(O, O', A, C)$ and $(O, O', B, C)$ we learn $OO' = AB = AC = BC$ and $OA = O'B = OC = O'A = OB = O'C$. So, the triangle ABC is equilateral (let its side length be $2\sqrt{3}$), and the regular pyramids $OABC$ and $O'ABC$ are congruent. Thus the points O and O' are symmetrical to each other about (ABC). Moreover, we have $OO' = 2\sqrt{3}$, so the altitude of each pyramid has the length \sqrt{3}. Let H be the common foot of these altitudes, then $HO = HO' = \sqrt{3}$ and $HA = HB = HC = 2$, thus $OA = O'A = \sqrt{7}$. So the radii of the spheres centered at $A, B, C$ are equal to $\sqrt{3}$, while the radii of the other two spheres are equal to $\sqrt{7} - \sqrt{3}$ and $\sqrt{7} + \sqrt{3}$, hence the answer.