Define on the given sheet a coordinate system with axes parallel to the grid lines, and with the origin in one of the nodes. Let the “parity” of the node be the parity of the sum of its coordinates. Assume one of the compass parts as "first", and another – as the "second" and see how the node parity changes while the second part moves. Denote the vector connecting the first end with the second after the i-th step as vi = (xi, yi), i = 1, 2, …, then (xi)² + (yi)² = r² (r is the radius of the compass, integer).
Consider three possible cases:
a) If r² is odd, then in all of the pairs (xi, yi) one of the numbers is even and another is odd, thus, xi + yi is odd and on the i-th step the parity of the moving part is changing on xi + yi − x(i-1) − y(i-1) = 2ki, i.e. it remains the same. The parity of the nonmoving part, obviously, also remains the same. Since in the beginning the parities of the parts were different (because r is
odd) the parts of the compass can not swap their places.
b) If r² = 4c + 2, then all of the numbers xi, yi are odd and the coordinates of the parts change on the i-th step on xi − x(i-1) = 2ki or yi − y(i-1) = 2li keeping the parity. And again, since in the beginning the parities of the parts were different (because x0, y0 are odd), the parts of the compass can not swap their places.
c) If r² = 4c, then all of the numbers xi, yi are even. Then consider a new grid with the length of a cell two times bigger than before. Then the compass moves are made in the nodes of a new grid and one of the cases a), b) or c) will be performed. Besides, the square of the length in a new grid is 4 times smaller than before, and the case c) cannot be performed up to infinity.
Everything is proved. Answer. No