by Tomás Feder
A parallelogram can be circumscribed in a circle (with all four sides of the parallelogram touching the circle), up to rotation and scaling, if and only if the parallelogram is a rhombus. Conversely, any rhombus can be circumscribed in any convex figure (with all four sides of the rhombus touching the convex figure), up to rotation and scaling. Proof: Let ab be the diameter of the convex figure, and let A and B be the perpendiculars to ab at a and b respectively. If the rhombus has an angle with d degrees, rotate the touching parallels A and B clockwise d degrees to obtain the touching parallels C and D respectively, and rotate them counterclockwise d degrees to obtain the touching parallels C' and D' respectively. Now A,B are farther apart than C,D and C',D'. Thus rotating two sets of parallel lines E,F and E',F' that are d degrees apart, from C',D' and A,B, to A,B and C,D, we must at some intermediate angle have the distance E,F equal to the distance E',F', so EE'FF' is the desired rhombus.
A parallelogram can be inscribed in a circle (with all four vertices of the parallelogram touching the circle), up to rotation and scaling, if and only if the parallelogram is a rectangle. We would like to show that any rectangle can be inscribed in any convex figure (with all four corners of the rectangle touching the convex figure), up to rotation and scaling. This is easy to prove if the convex figure has either an axis of symmetry (just align two sides of the rectangle in the direction of this axis, and bring these parallels closer until the lengths coincide with the proportions of the rectangle) or a center of symmetry (just make this center be the center of the rectangle we are looking for, make two diagonals at the correct angle go through this center, and rotate until the length of the diagonals coincide). Another easy case is if the rectangle is a square. Proof: We show that a square can be inscribed in any convex figure. Just choose an arbitrary direction with two parallel lines A,B touching the convex figure at the boundary, and start bringing A,B closer while maintaining the length of A,B across the convex figure the same, until the resulting parallelogram ACBD has the lenghts of C,D the same as the lengths of A,B, that is we have a rhombus. Thus we obtain a rhombus in any direction. If the rhombus ACBD has the angle AC acute, then the rhombus A'C'B'D'=CBDA has the angle A'C'=CB obstuse. Turning clockwise from the directions A,B to the direction A',B' we must thus find a rhombus A''C''B''D'' where the angle A''C'' is straight, thus the rhombus A''C''B''D'' is a square.
It is conjectured more generally that any closed curve in the plane (convex or not) contains four vertices forming a square. It is easy to show that for any closed curve in the plane (convex of not) and any triangle, there exists three vertices in the curve that form a triangle similar to the given triangle.
Reference: T. Feder, On quadrilaterals and convex figures in the plane.