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The problem is credited to Vladimir Dubrosky.
Unfortunately the magazine with his solution is now defunct.
Several more or less elegant proofs are possible. I like Bractals' but here are two more.
The first is a sort of carpet proof in which I subtract equal areas from both sides until I am left with regions I can work with.
Here we let ABCD, the larger square, have side length r, and A'B'C'D', the smaller square have side length s, rotated by theta about its centre. We enclose A'B'C'D', in a square aligned to the sides of ABCD and remove this from the equation. It should be clear that the four trianglular areas removed from N, S, E and W are equal in area.
Next I construct and remove the four corner blocks with diagonals AA', BB' etc. Again it is easy to see that each block takes equally from the N+S sum and the E+W sum.
This leaves four rectangular blocks in N, S, E and W. Now it only remains to show that these (N', S', E' W') satisfy N'+S' = E' +W'.
Each has base s Cos(theta), and heights a, b, c and d corresponding to N', S', E' W' . But a + b = c +d = r - s Cos(theta) .
Hence N'+S' = E' +W'.
Alternatively, we can prove the invariance of the relation under affine transformations. We can consider a square, M, aligned with the larger and with the same centre.
Clearly N=S=E=W and the equality is satisfied. It also clear that this is true for all scalings of the smaller square.
Now if we rotate the smaller square about its centre, then again the four regions are congruent and equal in area so the equality is satisfied and invariant under rotations about the centre of M.
Now we move M in an east-west direction. It is easy to see that the N and S regions are congruent under reflection and equal to their original areas, hence N+S is unchanged and so, by subtraction, E+W must be unchanged. Similarly we move the square in an north-south direction. It is easy to see that the E and W regions are congruent under reflection and equal to their original areas, hence N+S is unchanged and so, by subtraction, E+W must be unchanged. Hence the equality is invariant under translations E-W or N-S.
Since all sizes, positions and orientations of smaller squares can be achieved by a composition of scaling, rotation and translation of M, all of which preserve the equality, the equality holds for all other smaller squares.
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