The following puzzle was asked in company interview round.I have no idea ,how to do it?

What is the diameter of the largest circle that can be drawn on a 
chessboard so that its entire circumference
gets covered by the black squares and no part of the circumference 
falls on any white space.
The side of the chess square is 2 cm.

How to solve this Puzzle?

Thank You!

  • 1
    $\begingroup$ I think it's $2\sqrt{10}$, this fits in to a 5x5 square grid with a black square in the middle, going through the corners of all the black squares. If it were any bigger, there would be no circle that would fit the corners of the black squares, that is, the corners of the black squares would not be equidistant from an origin. $\endgroup$ Jul 29 '12 at 0:18

It's one that passes through $8$ black squares arranged in a diamond shape, passing through the corners of each. Any larger and it couldn't curve in one direction while passing only through corners, which is has to do to never hit a white square.

Any circular arc can be repeated on the board simply by reflecting it across each axis and diagonal. In other words, a circle can be considered fully defined by an $8^{th}$ of its circumference. Of course it could be defined by any section, but that's irrelevant since we want to use the board's eightfold symmetry. Our definition (on an $8^{th}$ circumference arc) involves centering the circle in a central black square and specifying the radius to the single vertex the arc must pass through. We can clearly specify such an arc, and due to the symmetry of the board we know that the circle will also pass through all the other vertices we need it to.

Find the actual diameter by looking at opposite vertices. This involves going $2\ cm$ down and $6\ cm$ across (or equivalent), giving a diameter of $2\sqrt{10}$.


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