Olimpiade Italiana di Matematica

First Solution: Consider the first hexagon, and call $O$ its center, $MN$ one of its sides, and $MNPQ$ the rectangle constructed on $MN$. The radius of the circle passing through the outer vertices is $OQ$. Since the hexagon is regular, $OMN$ is an equilateral triangle, and therefore $MN = OM = a$, and the angle $OMN$ measures $60^{\circ}$. Hence the angle $OMQ$ measures $60^{\circ} + 90^{\circ} = 150^{\circ}$. The triangle $OMQ$ therefore has two sides of lengths $OM = a$ and $MQ = b$, and the included angle of $150^{\circ}$.
Consider now the second hexagon, with center $O'$, side $M'N'$ and rectangle $M'N'P'Q'$, analogously to the first case. The radius of the second circle is therefore $O'Q'$. The same reasoning shows that the triangle $O'M'Q'$ has two sides of lengths $O'M' = b$ and $M'Q' = a$, and the included angle $O'M'Q'$ measures $150^{\circ}$. By the first congruence criterion for triangles, $OMQ$ and $O'M'Q'$ are congruent, and therefore $O'Q' = OQ$.

Second Solution: Let $O$ be the center of the circle, $MNPQ$ one of the six rectangles, where $MN$ is a side of the regular hexagon and $PQ$ is a chord of the circle. Let $R$ be the midpoint of $MN$ and $S$ the midpoint of $PQ$. The radius of the circle is evidently equal to $OP$. By the Pythagorean theorem,
$$r^{2} = OS^{2} + SP^{2} = (OR + b)^{2} + \left(\frac{a}{2}\right)^{2}.$$
It is easy to verify that $OR = \frac{\sqrt{3}}{2} a$ (the height of an equilateral triangle with side $a$), hence
$$r^{2} = \left(\frac{\sqrt{3}}{2} a + b\right)^{2} + \left(\frac{a}{2}\right)^{2} = a^{2} + \sqrt{3}ab + b^{2}$$
Since this expression remains the same if $a$ and $b$ are swapped, we obtain the claim.