1st solution: The quadrilateral ABA'B' is cyclic with perpendicular diagonals, hence, its anticenter (concurency point of its maltitudes) is identical with its diagonal intersection S. Let P, Q, P', Q' be the midpoints of the sides AB, BA', A'B', B'A. The centroid G of the cyclic quadrilateral ABA'B' (the intersection of lines PP', QQ' connecting the midpoints of the opposite sides) is the midpoint between its circumcenter O and anticenter S. Since PQP'Q' is a parallelogram, its diagonals PP', QQ' cut each other in half at G. Thus the diagonals OS, PP' of the quadrilateral OPSP' also cut each other in half and this quadrilateral is also a parallelogram. Using the parallelogram rule, $PP'^2 + OS^2 = 2(OP^2 + SP^2)$ $GP^2 + GS^2 = \frac{OP^2 + SP^2}{2}$ The segment $SP$ is the median of the right angle triangle $\triangle ABS$, hence, $AP = BP = SP$. Since P is the midpoint of the chord AB of the circle (O), $OP \perp AB$ and $OP^2 + SP^2 = OP^2 + AP^2 = OA^2 = R^2$ where R is radius of the circle (O). Consequently, the distance GP is given by $GP^2 = \frac{r^2}{2} - GS^2$ and similarly, $GP'^2 = GQ^2 = GQ'^2 = \frac{r^2}{2} - GS^2$ Thus the midpoints P, P', Q', Q' all lie on the circle (P) with center G and radius independently on the direction of the perpendicular lines $AA' \perp BB$ meeting at S. The distance MM' between the vertices M, M' of the rectangles SABM, SA'B'M' is twice the distance PP' between their diagonal intersections P, P' and since the midpoint of PP' is G, the midpoint of MM' is on the line GS twice the distance GS from the common vertex S of the 2 rectangles, i.e., the midpoint of MM' is O, the center of the circle O. As a result, the locus of M, M' (and similarly,the locus of N, N') is a circle comcentric with the circle (O) and with radius $r = 2 GP = 2 \sqrt{\frac{R^2}{2} - GS^2} = \sqrt{2R^2 - OS^2}$ 2nd solution (similar to All the positions compatible with our problem): $(\vec {OS} + \vec{SA})^2 = R^2,\ (\vec {OS} + \vec{SB})^2 = R^2,\ \vec {SA} \cdot \vec{SB} = 0$ $OM^2 = (\vec {OS} + \vec{SA} + \vec{SB})^2 = (\vec {OS} + \vec{SA})^2 + 2(\vec {OS} + \vec{SA}) \cdot \vec{SB} + SB^2 =$ $= R^2 - OS^2 + OS^2 + 2\vec {OS} \cdot \vec{SB} + SB^2 = R^2 - OS^2 + (\vec {OS} + \vec{SB})^2 =$ $= 2R^2 - OS^2$ which is constant and independent on the position of the point A on the circle (O).

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1995Problem4 Vo Duc Dien