A chess board 8x8 is placed on top of an identical board and rotated through 45 degrees about its center?

Question

A chess board 8x8 is placed on top of an identical board and rotated through 45 degrees about its center.

What is the area S which is black in both boards?

Answer 1

Both chessboards intersect in a regular octagon, whose area, as far as I could get, is 64(2√2 - 2) and 25% of its area is black-black as physicsmastersdegree notes above,
i.e. 32(√2 - 1). The proof involves central symmetry. Both initial squares and the resulting octagon are central symmetric (we take into account the coloring along with the shape). Take a black-black point - its central symmetric is also black-black, the same is true if we take instead white-white, white-black and black-white. Initially there are 50% white-white and 50% black-black. Rotate at 90 degrees - obviously now there are 50% black-white and 50% white-black points. Symmetry considerations (intermediate value etc.) imply that in the middle position (45 degrees) all 4 color combinations will be presented with 25% share: now the symmetry of the shape only (regular octagon - union of smaller polygons) is described by the dihedral group D_4. Take any of the polygons, say P, of this partition, let its coloring is x'-y'. The 8 symmetrical images of P (P itself and its images by 7 other symmetries of D_4) in turn have colorings x'-y", x"-y", x"-y' and then again x'-y', x'-y", x"-y", x"-y', each color combination encountered twice.

I personally prefer "Black & White", You understand what I mean.

Answer 2

If the area of overlapping black is G before the rotation, then G = 50% of the board. a 45 degree rotation will reduce by 1/2, so S= 25%. But I can't proove this.